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V.M. Lipunov

In the Worldof Binary Stars

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MirPublishersMoscow

Translated from the Russianby Alexander A. Kandaurov

First published 1989Revised from the t 986 Russian edition

Printed in the Union of Soviet Socialist Republics

ISBN 5-03-000528-5

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© English translation, Mir Publishers, f989

Contents

Preface 7

CHAPTER t. WHAT IS A BINARY STAR? 10

Kepler's Laws (12). Motion in a PotentialWell (17). About the Centre of Mass (25).Who's Chief in a Binary? (28) Let UsDetermine the Orbit of a Binary (30). TestParticles in a Binary (33). Stars' Shape (40).Who Was Right? (43) Formation of Bi-naries (45).

CHAPTER 2. ALGOL PARADOX 53

Whose Light Is "Louder"? (54) Light Curve(56). What Can an Alkol-Type Light CurveTell Us? (61) Doppler Effect (64). RadialVelocity Curve (71). Why Are Stars SoDifferent? (77) Finally, a Paradox (84).Changmg Roles (86).

CHAPTER 3. LIVING APART TO-GETHER M

In 40 Minutes Only (89). Getting Hot byCooling (93). As Little As 30 Million Years(96). But for Quantum Mechanics (99).Nuclear Evolution (101). Fleeing the MainSequence (106). White Dwarfs (110). NeutronStars (116). Black Holes (123). Ancestors andDescendants (t27).

CHAPTER 4. FIRST EXCHANGE 130

John Goodrike Again (130). The Spectrum of6 Lyrae (135). Gravity Wind (137). A Peculiarbevice (143). A Strait-Jacket for a Star(147). The Second Puzzle of p Lyrae (149).After the Exchange (i51). The Story WillBe Continued (156). How to Count Stars (157).

6 Contents

CHAPTER 5. MASSIVE X-RAY BINARYSTARS 161

"M with a Dot" (161). "Brighter Than aThousand Suns" (167). Uhuru (174). In theStellar Wind (18t). How an X-ray PulsarWorks (185). Why Neutron Stars? (190)"Discology" (192). Cygnus X-1 and OtherBlack Holes (198). Back to the Scenario(202).

CHAPTER 6. DWARF BINARY STARS 206

Pussy, Pussy, Hercules X-1 (207). Novae andRecurrent Novae (215). U GeminorumStars (224). Polars (231). Those SupernovaeThat Are So Much Alike (238). X-rayBursters (24t). But for General Relativity(247).

CI-IAPTEH 7. COSMIC MIHACLE 250

Moving Four Ways At Once (253). To MeetAgain (256). Why Two? (260) A KinematicModel (263). Something Familiar (270).Supercritical Accretion (273). The UN ight-mare" of the Second Exchange (281).

CI-IAPTEH 8. \VANDEREHS 285

Runaways (286). "Single" Wolf-Hayet Stars(291). The Second Explosion (298). BinaryRadio Pulsars (302). Options (309).

GLARING PEAKS 313

Summing Up (315). Evolution Squared (319).

Preface

Many of the stars in our Galaxy are binary.I daren't say most for neither the exactnumber nor the relative fraction of binariesare yet known. That they account for atleast ten percent of all the stars is so faras much as we know. Some estimates,though, come close to 90%. This, however,is not the only or even the major reasonwhy binary stars draw researchers' attention. In fact, there are two reasons fortheir interest.

The first one is purely practical. A singlestar is very difficult to study. The scientists'task is much easier when it has a companion. In a way, the binary arrangementis more "approachable" and lets astronomersattempt to understand it.

The second reason is that a binary starhas a fascinating life. I t is much more funto explore a binary. To see a star coupledwith another is like finding another dimension. The life of a couple is more variedthan that of a loner just like a plane isricher than a line or a space is richer thana plane.

8 Preface

Although mankind has been exploringbinary stars for several centuries, it hasonly been in the last decades that we havecome to understand the laws governingtheir existence. An especially great amounthas been done in the last 10-15 years thanksto the new science of X-ray astronomy. Inbinary star science, two rivals, theory andfact, take part in a never-ending race inwhich they push each other forward tounpredictable and stupendous discoveries.

By today, our understanding of star lifehas changed dramatically. There has alsobeen a change of attitude, and we now livein a permissive world. Could you imaginea serious astronomer of the 19508 describingone star as "swallowing" its neighbour?At present, this topic is a major issue investigated by observers and theorists. Couldyou imagine that in our Galaxy there areobjects that owe their lives entirely to theexistence of gravitational waves? Natureutilizes as a matter of routine what physicists have been trying to record for decades.

The diversity of the physical processesoccurring in binary stars is greatly interestingand they are often comprehensibleonly to a researcher who has knowledgein nearly every branch of today's physics.An attempt to go into all the new and interesting developments in binary star science

Preface 9

would be both unnecessary and futile.I shall not try to do the impossible.

In this book, I sought to present the fundamental ideas which have paved the wayto major research developments. When I wasshort of words, I took pen and ink and drewsomething.

The exploration of binary stars continuesand new horizons are coming within reach.I am sure that in years to come my readerswill witness fascinating developments oreven make important discoveries in thisfield themselves.

I dedicate this book to the memory ofVera Lipunova, my mother. She was alsomy first teacher as she was for many others.

I also want to express my profound gratitude to N. A. Lipunova for her help indiscussing the text and preparing the manuscript for the press. I am also grateful toT. A. Birulya for her help with the illustrations.

Vladimir Lipunov

Chapter 1

What Is a Binary Star?

The first answer that comes to mind is thata binary system is a pair of stars that arebeld together by their gravity. This statement, however, doesn't define binaries because any two stars, like any two masses,are attracted to each other by gravity. Thatmeans that every star comes within thisdefinition and thus it is utterly useless. Tounderstand what a true binary is, let usfirst simplify the situation.

Imagine a universe with nothing but twostars. I t may be represented by a blank sheetof paper with two points, 1 and 2, eachshowing a star with masses M 1 and M 2respectively (Fig. 1). Let us find the centre

Ch. 1. ",ThAt. Is a Binary Star? it

of mass of the tV{O stars (we label itO).To bring it closer to reality, we must fora moment come back to the normal universe, or to Earth. We connect the twomasses, Afl and M 2 , with a weightless rod

Ht r t - 112 "2

'r1 "z.-----.------------411( 0 "'2

Fig. 1. The distance between the centre of massand the stars is inversely proportional to theirmasses. 0 is the centre of mass of the binary

and suspend it wi th a string so that the rodis in a horizontal position. The point onthe rod to which the string is tied is thecentre of mass 0 of the binary star system.

So ,ve only have two stars and we havetheir centre of mass. Now we can makea good definition. The two stars comprisea binary system if they travel wi thina certain limited space. Purely kinematic,this definition is just as good as one basedon energy, namely, a star system is binaryif its total energy is negative. Accordingto Newton's laws of motion and the law ofuniversal gravitation, the requirements oflimited space and negative energy areinterchangeable. To prove this, you don'thave to know exactly how the bodies attracteach other. What is important is to realizethat, as the distance between the bodies

t2 In the World of Binary Stars

increases, the gravitational force decreasesrather rapidly. Let us prove it by contradiction. We assume that the stars' totalenergy, which is composed of kineticand potential energies, is negative and thatthe stars are infinitely far apart. Characteristically, their total energy remains constant while the energy of the gravitationalattraction, i.e. the potential energy, fallsto zero. Now the stars' total energy is entirely made up of their kinetic energy, whichis at least not negative. Here we have a contradiction. Hence bodies which have a negative total energy cannot move infinitelyfar apart.

This general defini tion is not, however,particularly lucid. First, we are surroundedby a multitude of stars, and so perhaps thedefinition is only theoretical. Second, weshould have a better understanding of howstars move in a binary system.

The first objection is not valid becausethe Universe, and even our Galaxy, hasenough space for two stars to find a spotto enjoy their privacy. As for the lawswhich govern the motions of two companions, they are discussed below.

Kepler's Laws

After graduating Irom the University ofTiibingen in 1593 the famous German astronomer Johannes Kepler was made professor

Ch. 1. What Is a Binary Star? 13

of mathematics and moral philosophy ata high school. This means he made his firststeps in science 17 years before the telescopewas invented. He was destined to becomea great theorist and an outstanding experimentor. Six years before Galileo pointed histelescope at the sky, Kepler wrote a bookabout the application of optics to astronomy.Today's school telescopes, in which boththe objective and the eyepiece magnify,were invented by Kepler in 1611.

Yet his most astonishing discoveriescame from the tip of his pen. Kepler wasnot what we now call a theorist but rathera master of interpretation. I t is commonlybelieved that interpretation implies neitherobservation nor theorizing and an interpreter only arranges facts and traces regularities. This, however, is not always true.Kepler was a devout believer in the fundamental harmony of the world. Though attimes this faith took him too far,- he placedprimary reliance on observation.

In 1601, Kepler "inherited" a pricelesstreasure, the great astronomer Tycho Brahe'sarchives. Kepler spent nine years analyzingthe observation of Mars. Scholars at thetime did not write scientific articles forthere were no periodicals where they couldbe published; instead they wrote books.Kepler published his analysis of the motionof Mars based on Tycho's observations inhis books Astronomia Nova and Commentaries

t4 In the World 01 Binary Stars

on the Motions of Mars. In these books,Kepler stated for the first time that theorbit of Mars is an ellipse with the Sunat one focus. Recall that an ellipse is aclosed plane curve consisting of all thepoints for which the sum of the distances

H ow to draw an ellipse

between a point on the curve and two fixedpoints (the foci) is the same. I t is easy todraw an ellipse. Take t\VO needles and pinthem to a sheet of paper. The pinholes willbe the foci of the ellipse. Take a piece of

Ch. t, What Is a Binary Star? i5

string and tie its ends to the needles. Thenpull the string tight with a pencil and movetile pencil slowly. Since the string remainsconstant in length the shape being drawnwill be an ellipse.

Kepler also noticed that Mars movesslower when it is farther from the SUD andfaster when it is close. Kepler never stopped

~-__ t4,

Fig. 2. In equal time intervals (t2 - t1 = t" - t3 ) ,

a planet's radius-vector sweeps equal areas (Kepler's law)

searching for beauty and harmony. Itturned out that the areas swept by theplanet's radius-vector in equal time intervals are equal (Fig. 2).

Ten years later Kepler wrote anotherbook, Harmony of the World, in which hedemonstrated that the other planets orbitingthe Sun behaved in the same manner asMars. In other words, each planet's orbitis an ellipse with the Sun at one focus(Kepler's first law) and the radius-vectorof the planet sweeps equal areas in equal

16 In the World of Binary Stars

time intervals (Kepler's second law). Thereader may wonder what all this has to dowith beauty and harmony and what is thebig difference between an oval and anellipse. Before I answer try to draw anellipse. Though it is a matter of taste, youwill probably admit that it is much morebeautiful because it is a shadow of a circle,which is the most symmetric plane figure.The projection of a circle is an ellipse, notan oval. Probably that's what makes it sobeautiful.

In Kepler's laws, there is another kindof harmony which has a profound physicalmeaning. I t is surprising how planets,which are so far apart and which seem tolive on their own, know that they shouldmove along elliptical orbits. When youdraw an ellipse on a sheet of paper everything is clear for there is a string holdingthe pencil. By changing the length of thestring you can draw many figures, but theywill all be ellipses. I t follows that theremust be some kind of string in the Universetoo. Kepler saw much beauty and harmonyin these "strings", which set the worldsin motion and which cannot be seen bya human eye. Another astonishing regularityin planets' motions he discovered proveshe was. right. Kepler's third law is thatthe squares of the periods of revolutionof any two planets are proportional to thecubes of the semimajor axes of their orbits.


The semimajor axis is half the distance between the two most distant points of anellipse.

What are these invisible strings? Itbecame more clear at the beginning of the20th century with Albert Einstein's general relativity. As far back as in the 17thcentury, however, Isaac Newton also yieldeda much better vision of the world by replacing Kepler's three laws with the lawof universal gravitation.

Motion in a Potential Well

An analysis of Kepler's laws brought Newton to the conclusion that the strength ofthe invisible strings holding all bodiestogether is inversely proportional to thesquare of the distance between them. Hecalled this force universal gravitation. Thelaw of universal gravitation is applicableto all matter irrespective of colour, taste,odour, or chemical composition. The universality of gravitation explains why allplanets hold to similar orbits.

Kepler, Newton, and every other inhabitant of the Earth were very lucky. Themass of the Sun exceeds hundreds of timesthe total mass of all the planets. Thismeans that the effect of all the planets onthe Sun is minor and may, therefore, beneglected. I t is for this reason that themotions of the planets about the Sun are

2-0667

is In the World of Binary Stars

simple. Even so it took the human racemore than a millennium to solve the riddleof epicycles and deferents. Somebody jokedthat if mankind had appeared on a planetin a binary, in which the planet's motionswould be beyond formal description, thelaw of universal gravitation would neverhave been discovered. So binaries are notsuitable as the birth places of sophisticatedcivilizations.

The problem is easier in our Solar System.The planets are kept in their orbits by theSun. The attraction between the planets isinsignificant. I t is easier to tackle planetarymotions in terms of energy rather thanforce, since the mathematics taught atschool is not enough to solve the equationsof motion. According to Newton's law,bodies' are surrounded by a gravitationalfield. The field may be described by a potential U. Let us look at the physicalmeaning of the potential.

Suppose there is a body with a mass M(Fig. 3). To measure its gravitational fieldwe need a test particle, i.e. a body whosemass is so small that its effect on the bodyM may be neglected. Usually we assumethe mass of a test particle to be equal tounity whatever measurement units beingused. Now we take a test particle and shiftit from point 1 to infinity. To do this, wehave to apply a certain force and do somework on overcoming the body's gravitation.

Ch. t. What Is a Binary Star? 19

The work done and taken with an oppositesign is termed a potential of the gravitationalfield of body M. By contrast, if a particle

Fig. 3. The work to be done to move a test mass Min a gravitational field is equal to the differencebetween the potentials, V., - U t , of the initialpoint and the final point and is independen t of theshape of a trajectory the test mass Iollows

falls from infini ty, the work done by thegravitational field is a potential with anopposite sign, i.e, - U.

The gravitational force varies only withthe distance and is always directed alongthe straight line toward body M. Thismeans that the potential is dependent onlyon the distance between the body and theparticle. The work to be done to move thetest mass from point 2 to point 1 is equalto the difference between potentials VI -2*

·20 In the World of Binary Stars

u2 • When moving freely the test particle's total energy, i.e. the sum of thekinetic. energy (K) and potential energy(U), remains constant, or K + U = const,The potential of point mass M is

GMU= --r-' (1)

where G is the gravitational constant andr is the distance between the body and the

U

r r

Fig. 4. The potential well around the Sun. In thecoordinates U (potential) and r (distance), theorbits of the planets are ovals

particle M. It is easier to give the gravitational constant as 1/G = 1.5 X 1010 kg.821m3•

Figure 4 shows qualitatively the potential of the Sun as a hyperbola. Rotatingthe hyperbola about the U axis yields apotential well. The idea of potential is

Ch. t. What Is a Binary Star? 2t

useful because a body "placed" at some pointr will "slide down" to where the potentialis the smallest.

The planets of the Solar System are liketest particles. Kepler and Newton treatedthe planets in the same way we treated testparticles when they studied the Sun's gravity. (Thus we are inhabitants of a testparticle called the Earth.)

How do the planets move in a potentialwell? In terms of the coordinates U (r),they move along circles (if the orbits arecircular) or ovals (not ellipses!). You mayask why the planets don't "slide" to thewell's bottom. The reason is because theypossess angular momentum caused by theirorbital motions. The planets move withoutfriction and therefore their angular momentum is everlasting. This, however, is notthe only type of motion possible in a potential well. We mentioned above that ifwe place a test particle on the side of thewell, it will slide to the centre, i.e. fallinto the Sun. If, however, we push itstrongly, it may flyaway to infini ty; wehave given it a velocity in excess of thesolar escape velocity.. The reason that theonly celestial motions we observe today areellipses is that everything else that wasdoomed to leave the Solar System or fallinto the SUD did so a long time ago.

It can be shown that all the possibletrajectories inside a potential well may be


obtained using a cone and a plane. Bycutting a cone with a plane, we obtain

Circle Ellipse Parabola

~==::~_:=:==~Hyperbola

Fig. 5. Various conic sections

three types of curves (Fig. 5): (a) a circleor an ellipse; (b) a parabola; and (c) a hyperbola.

There are four curves but three types.·There is an explanation for this. A circle

• A straight line is also possible. This degenera tecase occurs if a test particle does not possess angularmomentum. The total energy in this case may haveany sign.


and an ellipse correspond to closed motionin which the sum of the kinetic and potential energies is negative. For a parabola,the sum is zero: K + U = O. Finally,a hyperbola corresponds to positive totalenergy. If, say, the Solar System is visited

e = '/t-b2;a 2

Fig. 6. Illustration of an ellipse's eccentricity

by an interstellar spacecraft, it will followa hyperbola until the crew switch on theengines to satisfy the inequality K + U <0, which is indispensable for contact.

Now let us return to the planets. Animportant parameter of a planet's ellipticalorbit is its eccentricity, which is a measureof its oblateness (Fig. 6). The more oblatean ellipse, the higher its eccentricity is,and the closer it is to unity. A circle is infact an ellipse with zero eccentricity. TheEarth's eccentricity is e = 0.017, and it isclosest to the Sun in winter in the Northernhemisphere and farthest from it in summer.This difference is not, however, big enough


to produce a noticeable effect on the changeof seasons, which is mainly caused by therelative position of the Earth's ax is wi threspect to the Sun.

Kepler's first law is a special result ofthe law of universal gravitation. If gravitywere not subject to the inverse-square law,the orbits of test particles would not beellipses and might not even be closed,which means that having left one point8 particle would not follow the same patha second time. Periodic motion is a phenomenon brought about by forces whichare inversely proportional to the square ofthe distance.*

Kepler's second law results from theconservation of the test particle's angularmomentum in the gravitational field of apoint mass. Indeed, gravity is always directed along the line connecting the testparticle and the central body. The angularmomentum is determined by the velocitycomponent perpendicular to the radiusvector; this momentum is not changed bygravity. The same can be said about anycentral force field.

Kepler's third law may be formulatedin the following way:a3 Gpt=W M0 '

• This is also typical of a body moving in a fieldof force proportional to distance, e.g. the rotationof a weight tied to an elastic string.

Ch. t. What Is a Binary Star? 25

where P is the planet's period of revolutionabout the Sun, M0 = 2 X 1030 kg is themass of the Sun, and a is the semimajoraxis of the ellipse. For circular orbits, thispattern may be determined by every schoolpupil provided it is not forgotten that thesemimajor axis of a circle is its radius.

You may wonder why I have spent somuch time discussing motion about theSun instead of binary stars since-in a binarythe masses of the stars may be comparableand the above conclusions may not berelevant. The surprising thing, however,is that everything we have discussed isuseful for the binaries too.

About the Centre of Mass

A binary's components can hardly be calledtest particles and the motion of one staris determined by the position of its companion. Both revolve about each other. Weknow (don't we?) that in a closed systemthe centre of mass must either stay whereit is or move uniformly. The effect on thesystem of other stars is insignificant. Bothstars move in the same manner as the planets in the Solar System, i.e. along ellipsesat 'one focus of which lies the binary's centreof mass (Fig. 7). These ellipses are termedthe stars' absolute orbits. The name refleets the fact that these orbits are construct-


ed in the system of a binary's centre ofmass.

The elliptical shape of the orbits of binarystars has been proved by direct observations

Fig. 7. The stars' absolute orbits in a binary

of our nearest binaries (Fig. 8). Astronomers, though, find it easier to calculatea relative orbit.

Here the general practice is to assumethat one of the two stars is stationary andto measure the position of the second onewith respect to its stationary companion.The apparent orbit is the projection of thetrue relative orbit on the celestial sphere.It is also an ellipse. (If you think aboutit, you will see that if a projection of acircle is an ellipse, then a projection of anellipse is an ellipse too.) The foci of theapparent and true ellipses do not coincide.For this reason the stationary star in Fig. 8stays beyond the axis of the apparent orbit.

Ch. 1. What Is a Binary Star?

270_0 --~~~---

, , , , , , I , , , ,

. :~e. •

1870

27

Fig. 8. The apparent orbit of a star in the binarysystem of ex Centauri. This is our closest binary,therefore its orbit's shape can be plotted fromdirect observations. The bright star is at point 0,the position of its companion is shown by dots.(The observations were conducted in 1830-1940.The scale is seconds of arc.)


In a star's motion along an ellipse, Kepler's second law remains valid, but histhird law is slightly changed. In the formula involving the orbit's period and size,star masses must both be present, viz.

4 3 Gpr= 4n 2 (Mf + }lt12 ) . (2)

If M 2 is much smaller than M 1 (M 2 ~Ml)'

then M 2 may be neglected and we get Kepler's third law for planets.

The point on the orbit where the stars areclosest together is called the periastronand the point at which they are farthestapart is called the apastron.* These twopoints are connected by the major axis orthe line of apsides.**

Who's Chief in 8 Binary?

Which of the two stars in a binary is thedominant one when considering orbitalmotion? A naive student of astronomy maydecide it is the more massive partner, supporting the conclusion by taking the limitas the mass of one of the stars tends to zero.

• Periastron and apastron stem from the Greekwords peri meaning near and apo meaning awayfrom; and astron meaning star.•• The line of apsides (absis means an arch inLatin) is the line connecting the point of closestapproach and the point of farthest retreat.

Cb. 1. What Is a Binary Star? 29

In reali ty, this is wrong and the situationis not that simple.

An orbital motion in the centre-of-masssystem is described by three mechanicalquantities. First, each of the stars has amomentum, Mlv1 and M 2v2• Second, eachhas a kinetic energy, M1v:/2 -and M 2v:/2.Third, you should not forget about theangular momentum. For circular motions,the latter may easily be calculated: M1v1rland M 2v2r2' where r l and r 2 are the distances between the binary's centre of massand the stars. Each parameter is important.

Now let us compare the stars in terms ofthese parameters (Table 1). Let M1 > M 2 •

Table 1

More LessParameter massive massive Which is more

(absolute value) companion companion important?

MomentumKinetic energy

Angular momentum

equalthe smallerstarthe smallerstar

In the centre-of-mass system the binary'stotal momentum must be zero. It followsthat the absolute values of the stars'momenta are equal (but having oppositesigns). Now let us compare the energies.Since the momenta are equal. the ratio


between the kinetic energy of the moremassive star and that of the smaller oneis clearly equal to the ratio of their orbitalvelocities. I t follows from the equality ofthe momenta that vJv, = M 21M1 , i.e. thekinetic energy of the less massive star ishigher!

The same thing is true of the angularmomentum. By dividing the angular momentum of the more massive star by theangular momentum of its less massive companion, we get rI /T2 , that is M 2/M1 , whichis less than one. Now you can see that ina binary the smaller star is chief.

Let Us Determine the Orbitof 8 Binary

I t is clear that in order to describe an orbityou need quantities independent of time.Here we shall rely on the conservation laws,which determine quantities that are constant over time.

First, a binary's energy is conservedbecause it is a closed system. Second, thetotal momentum of the two stars is conserved. Third, the angular momentum associated with the orbital motion is constant.Note that we can afford to neglect the size,temperature, luminosi ty, etc. of the stars,as the orbital motion of the stars is independent of them.


If the orbits are not circular, there isanother invariant, namely, the directionof the line of apsides.

Let us count the quantities we need todetermine an orbit. We do this in a reference frame whose origin is at the binary'scentre of mass. The angular momentum ofa binary is a vector perpendicular to theorbital plane of the stars. It means thatby specifying the angular momentum weautomatically specify the orientation ofthe orbital plane. A vector is specified bythree scalar quantities and so the orbitalplane is determined by the same number ofquantities. Yet we are still far from a complete specification of an orbit. An ellipsemay rotate in this plane by changing thedirection of the major axis (the line of apsides). Consequently, we need two morequantities which determine the directionof the line of apsides, All in all, we havefive parameters. Having specified them, wecan fix the orbit in space, but so far we havesaid nothing about its size or oblateness.That adds two more parameters. Thus a complete specifics tion of an orbi t requiresseven parameters. We could use the binary'stotal energy (one parameter), the totalangular momentum (three parameters), avector in the direction of the line of apsides(three parameters though only two of themtndependent), and the eccentricity. Thistonlbination .. however, is not the best one


because the energy and the angular momentum cannot be directly observed.

In practice, the following parametersare used: the masses of the stars, M1 andM2; the semimajor axis a or the period ofthe binary, P (these are related through

Fig. 9. A star's orbital parameters in a binary

the masses by Kepler's third law); theeccentricity e; the angle i of the orbitalplane to the plane of the sky (the one perpendicular to the line of sight); the anglebetween the line of intersection betweenthe orbital plane and the plane of the sky(the nodal line) and the line of apsides, 00;and finally the position angle Q (Fig. 9).In this figure, N is the direction to theNorth celestial pole and n is the vector

ci, 1. \Vhat Is a Binary Star? 33

along the normal to the orbital plane. Theseseven parameters are conserved in Newtonian mechanics and these are the ones determined by researchers when studying binaries. In order to describe stars' motionsalong an orbit, we must also specify theposition of the stars at a particular moment.

Test Particles in a Binary

We now look to see what the orbits of theplanets in the Solar System would be likeif the Sun were accompanied by anotherstar with comparable-mass, One potentialwell is now supplementedby another. Clearly, the system is far more complicated andcalculating these orbits would have beena riddle for Kepler himself.

I t is easier to study the motions of testparticles in the gravitational field of abinary star in a reference frame whichrigidly rotates with the stars. The originof the frame coincides with the centre ofmass, the x axis is directed along the lineconnecting the stars, the y axis lies in theorbital plane, and the z axis is perpendicular to the plane (Fig. 10). Thus the rotation may be 'ignored. However, as 'soonas we get rid of rotation of the stars, wehave to deal with the centrifugal forcewhich appears in any noninertial referenceframe. But this is a minor headache. In8:·system with the stars moving in circular

3-0667


orbits, the angular velocity of the star doesnot vary with time. The forces attractinga test particle will depend only on theparticle's position and will be independentof time. This allows us to introduce a po-tential. .

In celestial mechanics, this is called arestricted three-body problem. I t is re-

z

Fig. 10. The motion of a test particle around oneof the stars

stricted because the third body, i.e, thetest particle, does not affect the motion ofits two companions. In the general case,the trajectory of a particle may only becalculated numerically and its shape willresemble that in Fig. to. By changing theinitial coordinates and velocity of theparticle, we can obtain a great many trajectories of this kind. I say "many" and I meanit. The best computers now available mayreduce the time needed to calculate orbits,but get us into trouble because we would

eh. 1. What Is a Binary Star? 35

drown in the flood of paper produced by theprinter.

Luckily, we don't need computers andthe motions of a test particle may be studiedqualitatively. The trajectory of an indi-


vidual particle is not important becauseits complexity makes the exercise useless.What we want is to see what happens tothe particle. Such an analysis was firstaccomplished by the American astronomerand mathematician George Hill at theend of the last century.

Let a test particle be innuenced by threeforces: the gravity of the two stars M 1and M 2' and a centrifugal force. All theseforces are potential and may be describedby a single effective potential U. Now letus see what happens to a particle movingin the binary's orbital plane. All the forcesact in the same plane, so the test particlemust stay within the plane too.

The behaviour of the potential U isshown qualitatively in Fig. 11. Here theplot shows three peaks separated by wellswhere the stars are located. You will noticethree special points. If we carefully puta test particle .atop any of the peaks, theywill be in equilibrium. Unstable, but itis an equilibrium! These points were firstdiscovered by the great French scientistLouis Lagrange in 1772.

Surfaces over which the gravitationalfield is constant are called equipotentialsurfaces. We can obtain them by cuttingthe potential by planes parallel to the xand y axes (Fig. 11). Their projections onthe xy plane are shown in Fig. 11. Unfortunately, it is not easy to follow the be-

Ch. 1. What is a Binary Star? 37

haviour of the potential in the xy plane.We must keep in mind that the potentialgrows when moving away from the x axisand along the y axis. I t follows that pointsL 1 , £2' and L 3 are, probably, passes rather

U

y""

Fig. i1. The behaviour of an effective potentialin a binary

than peaks (Fig. 12). An attentive readerwill notice that if the potential grows whenmoving away from points L1 , L 2 , and L:4and is small again at infinity, there mustbe peaks somewhere. And right he is. Thesepeaks correspond to two more Lagrangianpoints, otherwise known as libration points.They will be discussed below,


As it moves, the test particle's totalenergy, i. e. a sum of i ts kinetic energy Kand potential energy U, remains constant:

K+l!-=E const.

Suppose we make a thought experiment. Weshall throw a particle from the stars. Itis obvious that if we give the particle a

Fig. 12. The behaviour of a poten Hal near theinner Lagrangian point (a saddle)

push, it. will first go up the well and thenwill fall down it again. When it is farthestIrom t.he star, the particle halts and itskinetic energy becomes zero: K O. Thismeans that at this moment the particle'slotal energy E is equal t.o the potential atthe turning point. It is clear that the particlecan only move when {I ~ E. By passinga section through the poten tial along theline U E, we find the area within whicha particle wi t h energy A" CHn move. 1nthree-dimonsinnal space \\'C ~et equi potcn-

Ch. t. What is a Binary Star? 39

tial surfaces instead of lines. Hence, ifwe know a particle's energy, we can specifythe area within which it can move. Figure 13 shows a view of an equipotentialsurface in the binary's orbital plane. Whenthe energy is small, the area of the particle's

Fig. 13. Equipotential lines in a binary's orbitalplane. The ratio of the stellar masses is 0.4

motion is small too. By "launching" particles with increasing speed we increase theirenergy and enable them to travel farther.When the energy reaches a certain value,the areas meet at the Lagrangian point L 1making a surface which looks like figureof eight. In memory of the French astronomer and mathematician Edouard Roche,it is called a Roche lobe.

If a test particle finds itself at point L1,

it can reach the neighbouring star withoutA loss of energy. Part.icles with great ener-


gies can go beyond the Roche lobe for theyno longer belong to a single star. The resultant of all the forces acting on a test particleis zero at Lagrangian points. We may usethis fact to find the position of these points.The solution is simplest when the starsare of equal mass. We suggest you solvethis problem on your own.

For an eccentric orbit, the motion of atest particle cannot be described by a potential function and we can no longer assume that a particle with some energywill necessarily move in the vicinity of onestar. True, we can calculate the particle'strajectory using a computer, hut a qualitative analysis is impossible.

Stars' Shape

A student who was to speak about the interior of a star but who had only a vagueidea of the subject began by saying, "Well,a star is a hot gaseous sphere. The formulafor the area of the surface of a sphere isS = 4nr2 and that for the volume..."

The student was right in saying thatsingle stars are spherical, the reason beingthat this is energetically favourable. Star'smatter, like a fluid, will fill any vesselinto which it is poured and its surface willbe an equipotential one. The star itselfcreates the field which forms its matter.I. have already said that the potential of

Ch. L What is a Binary Star? 4f

a point mass is inversely proportional todistance (formula (1)) and at the pointr = 0 the potential is at "minus infinity".Such a situation is impossible in nature, atleast in ordinary stars. The potential of astar is more like a wineglass without a stem

na r

Fig. 14. A star's gravitational potential does notgo to "minus infinity"

(Fig. 14). If r = 0, the potential is small,yet finite. On the surface r = const, i.e.the star is a sphere.

Now let us look at the shapes of stars ina binary system. Imagine the binary'spotential well is filled. with a fluid, as wedid for a single star (see Fig. 11). The fluidspreads out so that the star's surface is anequipotential one. The star will stretchalong the line connecting the binary's components.


You see, the problem is not as hard a nutto crack as you might imagine. Yet in the18th century the argument on another, moresimple, issue was so heated that a specialArctic expedition was organized. The storyof this endeavour is told by the outstanding

North

SouthFig. 15. Newton's thought experiment

astrophysicist Subrahmanyan Chandrasekhar in his monograph about stars' shapes.

In 1687, Isaac Newton published hisfamous Principia. I n it, he explainedKepler's laws by the law of universal gravitation and answered many other specialquestions. One concerned the Earth's shape.Newton believed that our planet is flattenedat the poles due to its rotation. To provehis hypothesis he showed his wit. Suppose,he suggested, we dig two wells leading tothe Earth's centre (Fig. 15), one along the

Ch. 1. What is a Binary Star? 43

rotation axis, the other in the plane of theequator and till the wells with water. Inorder for the water to stay in equilibriumand not splash out, the weight of the waterin the both wells must be the same. Buta body's weight is its mass times the effectiveacceleration, the latter being the differencebetween the acceleration due to gravityand the acceleration due to the centrifugalforces. At the equator, the accelerationdue to gravity would be slightly reducedby the centrifugal acceleration. In orderfor the water weight in the equatorial andpolar wells to be the same, the water levelin the equatorial wel l must be higher.

This conclusion, however, ran counter tocontemporary astronornical data. The famous French astronomer Dominique Cassinihad measured the meridian arc lying wi thinFrance and considered that his data demonstrated that the Earth was stretched alongthe poles. It was difficult to argue witha man who was the first director of the Parisobservatory and the discoverer of four ofSaturn's satellites and the famous darkdivision (Cassini division) in Saturn's rings.

Who Was Right?

Several generations of Cassinis and Newton's successors kept up the argument.Only in 17;~8 a speci a J exped i tinn Jeff by the

In the World of Binary Stars

French scientist Pierre Maupertuis went toLapland and discovered that Newton wascorrect. Voltaire taunted the Arctic voyagers, asking why they had to go to Laplandto learn what Newton had known withoutlooking out of his window.

You don't have to go to the Arctic tolearn what a star looks like. What you doneed is a telescope. Like the Earth, a starin a binary system is also oblate. Qualitatively, the situation is the same but ina binary things are complicated by thestars' motion about their centre of mass,which does not coincide with the stars'centres. Besides, a star in a binary systemis attracted by the nonuniform gravityfrom its neighbour. We have already seenthat the equipotential surface is a spotwhere the velocity of a particle with a specified energy turns to zero. In order fora surface to be an equipotential one, thestar must be fixed in a rotating referenceframe. This is only possible when the star'speriod of rotation equals the binary's period. The astronomers then say that thestar rotates synchronously.

Figure 16 illustrates the shape of a starin a binary system. When it is small incomparison with the size of the Roche lobe,it looks like a triaxial ellipsoid Of, say,a Central Asia melon (Fig. 16a). Strictlyspeaking, this is not true because the star'sshape is asymmetric. This is more obvious

Ch. t What is a Binary Star? 45

when the size of a star is comparable withthe size of the Roche lobe.

A star filling the Roche lobe looks different (Fig. 16b). In such a star a "spout"appears near the inner Lagrangian point

(6J

Fig. 16. The shape of the stars in a binary

and matter may be transferred through this"spout" to its companion without any lossof energy.

This is a surprising and remarkablefeature of a binary star. At first sight, onemight think this is abstract situation andwhy the star should be as large as a Rochelobe. In nature, however, nearly all binarysystems sooner or later turn into such"freaks".

Formation of Binaries

The history of a star begins with its condensation out of gas and dust due to gravitational instabili ties. This idea was firstapplied to the Sun by Immanuel Kant in1755 and was later supported by PierreLaplace. The birth of a star is hidden under


"a veil" but has nothing to do wi th "devilry". The "veil" is a result of the opacity ofthe con tracting cloud of gas to electromagnetic waves. You may still run intoan opponent of this hypothesis who willpoint out that, while no astronomer hasever seen a contracting gas cloud, explosionsmay be seen every day. I-Iere the stronginfJuence of selection is at work, Stars areborn \vhere there is a great deal of gas anddust, namely, where the visibility is worst.

A rigorous mathematical explanation asto what contracts a cloud of gas was provided by the English astronomer JamesJeans at the beginning of the 20th century.Let us suppose that an infinite space isfilled wi th a homogeneous gas and perturbthe gas parameters in some limited area,for example, by slightly compressing thegas. If there were no gravitation, the compression would be followed by an expansionperturbing neighbouring areas of gas. Awave of perturbations (sound) would runthrough the gas. I n in terstellar space, gasclouds are of great size, and gravitationalforces become as importan t as the forces ofpressure. Suppose we lind a spherical areain the interstellar space and compress ita little. The force with which the cloudattracts i tself is proportional to GM2/R2,where AJ and R are the mass and the radiusof the cloud respectively. External forcesare ignored for they compensate each other.

cu. 1. \Vhal. is a Binary Star? 47

Hence, the pressure of the gravitationalforce is proportional to (pR)2.* The pressureof an adiabatically compressed monatomicgas depends on the density as P oc pS/3.From this it follows that for very largeclouds (at some R) the gravitational forcewill exceed the force of the gas pressureand the cloud will start contracting. Thisprocess is referred to as Jeans instability.

The minimum size from which a cloudbecomes unstable with regard to compression is termed the Jeans radius. Approximately the Jeans radius may be found byequating the gas pressure in a sphere withradius R to the pressure of the gravitationalforce. The gas pressure is calculated by theequation of state for an ideal gas. We determine the pressure of the gravitational forcesby dividing the sphere into two equal partsand calculating the force of attraction between the two halves. The result is

!/I,T GM2P-,....-= 41tR4 ,

from which we get an approximate expression for the Jeans radius:

• In fact, given M = pR3, we can represent thepressure of the gravitational force as Pgr ec(GM2/RZ)/R2 ex. p2R2.


Here T is the gas temperature, !fl is the gasconstant, and ~.. is the relative molecularmass of the gas.

Of course, the gas is not uniformly distributed throughout the Galaxy. In thelate 1970s, it was discovered that almostall the gas is concentrated in gigantic molecular clouds with masses hundreds of thousands and millions times greater than thatof the Sun. We might ask why a single clouddoesn't produce a single giant star. Thisdoes not happen because the clouds arealso inhomogeneous and the stars are formedin the densest central parts of the clouds.Besides, the gas in the clouds is not at rest;it moves chaotically and this motion iscalled turbulence. Different parts of thecloud rotate in different directions. Eddiesoccur at all levels, from the cloud as awhole to small parts.

So binary star systems might arise dueto turbulence. A computer simulation ofthe process is shown in Fig. 17. First thecloud is compressed along the rotationaxis, and then it turns into a torus. Thetorus breaks into separate blobs whicheventually grow into binary stars. In fact,star systems evolving in this way can contain more than two stars (multiple stars)and there are a great number of multiplestars in our Galaxy. This book is aboutbinaries, although our discussion refers totriple systems too. The simplification is

~..'

';;:::1;' .:\.~';.~; .:.•••••.....~ .. • • ', t •

"':. :\~ ...::~.~~ _. ~.":'.; .. :~':" "~.'"...~.. :', ,- : . . ,'I ';,

...'s., .......... ...t • ~ •••. ' .. ~~. ....... "... ,-

t •••

,:'"'~.. .:

..:;~(~..:~ :'<'j:.':-:.}. . ..e .

,

.- .',:!',\ •• ·At

••:-.. , .~ •. ,...

t. ...: . .~ . : ~

17. This is, probably, how a binary is born,f a rotating cloud. A view from the poles (left)from the equator of the rotation (right)


justified because we are interested in multiplicity only from the viewpoint of theeffects the stars have on each other. Natureis such that any multiple system can bereduced in this sense to a binary. Threestars cannot coexist as equal companions.A triple system sooner or later ejects onemember provided the distances betweenthe stars are comparable. It happens dueto a cumulative effect which is well-knownin fluid dynamics. The effect is used in modern antitank projectiles. I t can be illustrated by the Pokrovsky experiment.

Take a test tube full of water. If youlet it fall several dozens of centimetresvertically onto the floor, a spout of waterwill rise several metres into the air (Fig. 18).I t seems like the effect violates the law ofenergy conservation. In reality this isnot so. What does happen is that the energyand momentum are redistributed in thewater. Most of the water remains in thetest tube and does not splash out at all;however, it gives a small portion of thewater its kinetic energy, thus pushing it upmuch higher than the test tube initially was.

You may have come across the same effectin the underground. The designer of theautomatic coin changer* must have con-

• All journeys on the underground in the USSRcost five kopecks regardless of distance. Machinesare provided to give five-kopeck coin change forother denomination coins.-Eds.

CIl. 1. What is a Binary Star? Sf

sidered the height the coins have to fallto the collection tray. A single coin cannotjump over the wall of the collection traybecause it lacks the initial potential energy.However, once in a while a coin falls tothe floor. The coin has taken momentum

lI»"D

h

Fig. 18. The Pokrovsky experiment illustratingthe cumulative effect

from its neighbours. The more coins fallat one time, the more probable it is thatone will fall to the floor.

As a result of the cumulative effect, asystem either becomes binary or remainstriple, with one of the stars moving along anorbit such that the distance between it andthe other two is much greater than thedistance between the other two stars. The

52 In the World of Binary stars

gravitational attraction between the twoclose stars is then the most important factor.

Fig. t9. As a rule, the distance between two of thestars is shorter than the distance to their thirdcompanion

For this reason multiple systems can beconsidered to be binaries (Fig. 19).

Chapter 2

Algol Paradox

I shall begin my story about the evolutionof binary stars by telling you about a starwhich is well-known to any student of

A classical light curve

astronomy. This star is in the constellationPerseus and is called Algol or ~ Persei. Thefact that ~ Persei is a variable star was dis-


covered by the I talian mathematician andastronomer Giovanni Montanari in 1669and for several centuries Algol was a typicalexample of an eclipsing variable star. Variations in brightness similar to those characteristic of Algol have been recorded inmany hundreds of other stars, making Algola generic term for a star type. Generationafter generation of astronomers observingAlgol found nothing surprising about it.Then all of a sudden (I mean "sudden"historically speaking, of course) they cameto realize that, although they had knownall about it for ages, something was wrong.What had seemed normal now appearedboth mysterious and surprising. When youhave no knowledge, you are never surprised.Astronomers had been scrutinizing the lawsgoverning star life when they realized thatan Algol-type star could not exist becauseit violated the laws they had so thoroughlystudied. Like any true paradox, the Algolparadox proved exceptionally helpful andit became a key to the mystery of binarystar evolution. To convince the reader thatthis is so, I shall have to start from the verybeginning.

Whose Light Is "Louder"?

In astronomy, a star's brightness is measuredin dimensionless magnitudes. Magnitude isintroduced by choosing a reference star

en. 2. Algol Paradox 55

whose brightness corresponds to the zerothmagnitude (m = 0). The brightness of anyother star is then given by

m= -2.51og ~ • (3)

where F and Fo are the brightnesses of starsof the mth and zeroth magnitudes respectively.. tJsually the energy of a star is characterl~ed by its luminosity, i.e. the amountof light falling from the star per unit areaper llni t time. I t follows from the defini tionof magnitude that a star of the 5th magnit,!,de is 100 times weaker than a zeroth magnItude star.

Y.ou may wonder what point there is inhaVing magnitudes if there is a physicalcharacteristic such as luminosity. We needmagnitudes because the luminosities of starsvary very widely. For example, the brightests~a~) I.e, the Sun, and the weakest starVISIhle to the naked eye differ in brightnessby ~ factor of more than 1012 • In terms ofmagnitudes, this corresponds to slightlymore than 30m, which is much easier tohandle.

Nature took tHis into account when itdesi.gned a human I eye. I t provided us witha sc.rt of a "computer" which can take logs.In llhysiology, this computer is called theW~her-Fechnerlaw: our sensations grow in~rlthmetic progression when stimulation1DCt'eases in geometric progression. The


same thing happens with our ears. I t isbecause of the Weber-Fechner law that soundintensity is measured in decibels. True, incontrast to magnitude, the definition ofa decibel contains a factor of 10 in frontof the sound intensity logarithm insteadof -2.5. In principle, however, there is nodifference between magni tudes and decibels.If you like, you could measure sound intensity in magnitudes and star brightnessin decibels. However, magnitudes havebeen used in astronomy since Hipparchus(2nd cent. B.C.). To get some idea of thebrightness of a zeroth magnitude star lookat Vega in summer. Its magnitude is closeto the standard one: m = 0.14.

Now back to Algol.

Light Curve

Late in the 18th century, the English amateur astronomer John Goodrike first noticedthe strictly periodic variations in Algol'smagnitude. The variation occurs with aperiod of 2 days, 20 hours, 49 minutes.A periodic process is best described bya phase. A phase is a time expressed inparts of period P. In practice, the phase iscalculated by assuming that at some moment in time the phase is zero. In starobservations, the initial moment (or zerophase time) is when a star's brightness isat a minimum. Then the observation time

Ch. 2. Algol Paradox 57

minus the initial moment is divided by theperiod. The fractional remainder is calledthe phase. The brightness of a variable staris usually measured by comparing it toa nearby constant star. A graph of brightnessagainst a phase is called a light curve.

Algol's light curve, as discovered byGoodrike, has two minima per period: a

o

Fig. 20. Algol's light curve

Phose1.0

"deep" primary minimum at phase zero anda "shallow" secondary minimum at phase 0.5(Fig. 20). We can understand this lightcurve pattern only by assuming that Algolis a binary. (The phenomenon is illustratedin Fig. 20.) Algol's components revolveabout each other with a period of 2.9 days;thus the two stars periodically eclipse eachother. True, they only do so partially.A question arises as to why one minimumis deeper than the other, because each star


eclipses the same area of its companion,namely, the area of the overlap betweenthe two circles. So what do we actuallymean by the depth of an eclipse and how dowe measure it? When there is no eclipse,we see both stars and the brightness of thesystem is the sum of the brightnesses of itstwo components. When one star "covers"the second one, the binary's brightness isreduced by the amount of light radiated bythe eclipsed part of the star. To calculatethis amount we must multiply the energyproduced per unit area by the area of theeclipsed surface. I t is clear that the difference in the depths of the minima is due tothe difference in the amount of energy radiated per unit area of the two stars. Asa first approximation, we may assume thatthe stars radiate energy like blackbodies.The latter may be represented by any bodywith constant temperature. Blackbodies aredescribed by the Stefan-Boltzmann law,according to which the radiation from unitarea is proportional to the fourth power ofthe body's absolute temperature:

F = oTt.

Here 0' == 5.67 X 10-8 J.m-2 . K-4· s- 1 ,

which is the Stefan-Boltzmann constant. Infact, we need this law only to say a verysimple thing: the hotter the body, thebrighter it is. Now one thing is certain:Algol is composed of two stars with different


temperatures and at the zero phase thehotter star is eclipsed.

Algol-type light curves are characteristicof variable stars and we know of severalthousands of such stars. In each case, thereare two minima with nearly constant brightness between them. Why do we have to say"nearly" constant? Since we see two wholestars between the eclipses and the system'sbrightness is the sum of the stars' brightnesses, one might think that the light curvemust stay constant. However, after the

Reflection effect

primary minimum the system's brightnessslowly rises until phase 0.5 and, but forthe second eclipse, there would be a maximum here (see Fig. 20).

The growing brightness is explained bythe reflection effect. Keep in mind that


one of Algol's stars is hotter than the other.The hotter star illuminates the closer sideof the colder star and thus makes it a littlebrighter. I t is as if the hot star is refleetedin its colder companion. In fact, we havehere a case of reemission with a change inthe wavelength of light. By the way, thereflection of visible light in a mirror is alsoreemission. Here the light is reemitted bythe electrons in the thin layer of metal deposited on the glass. However, in the process the wavelength remains constant. Ifyou see redhair in the mirror, then you definitely have redhair,

Let us once more follow the phase dependence of the reflection effect. At phasezero, the colder star eclipses the hotterone. This means we are seeing the near andcolder side of the darker star. Later in theorbit (increasing phase), we see more andmore of the illuminated side of the colderstar and the whole system slowly becomesbrighter. By phase 0.5 we see the hottestpart of the colder star. Now the reflectioneffect is at a maximum. Then the system'sbrightness symmetrically falls to phase one.In the Algol system, the reflection effectis small and insignificant, the eclipsesbeing really important. Now imagine asystem in which its components do noteclipse each other because either there isnothing to eclipse or nothing to eclipse with.This is quite possible. Then the reflection


effect may be the only reason for variationsin the brightness of a binary. Below weshall come across binary star systems inwhich the reflection effect is many timesstronger than it is in Algol.

What Can an Algol-TypeLight Curve Tell Us?

What information can be obtained froman Algol-type light curve, i.e, the one consisting of two minima? Suppose we observea binary system edgewise, and our line ofsight lies in the orbital plane (Fig. 21).For simplicity let the orbit be circular. In

A D--~ ...---

Ad sc CD

Ilf @ @I I I

I: II! I

Fig. 21. A central eclipse

this ease, eclipses will be central and oneof the components will be wholly eclipsed.The binary's plane is at an angle i = 90°to the plane of the sky. Then th~ light curve


will be trapezoid-shaped. Let brighter star 1be the larger. At the primary minimum, thebrighter and larger star is eclipsed. Theside AB of the trapezoid shows the starbeing gradually covered by star 2. Atpoint B the edges of stars 1 and 2 touch (aninternal contact) and while star 2 "travelsacross" the disk of star 1, the system's brightness remains the same. Side CD correspondsto star 1 gradually appearing from behindstar 2. Side CD is proportional to the radiusof star 2. The duration of the eclipse (fromA to D) is proportional to the sum of thestars' sizes, and the duration of the flatbottom Be is proportional to the differencebetween the stars' sizes. We have two equations and two unknowns, Thus the lightcurve makes it possible to calculate starradii. You may wonder how we can measuretime and obtain distance. We are alsosupposed to know velocity from the lightcurve. Truly, we cannot determine starsize. We can, however, determine it inunits of the radius of the binary's orbit (a).I t takes the star a period to travel 2na.Since we know the period, we may learnwhat fraction of the orbit's total-length iscovered by the star during the eclipse.

If the line of sight does not lie in thebinary's orbital plane, the duration andkind of eclipse will be determined hoth bythe relative stars' sizes and by angle i,The shape of the light curve wil] also be


different. At a certain angle, the smallerstar is not wholly eclipsed, which meansthere will not be a flat bottom in the secondary eclipse. The form of eclipse allows usto determine the angle at which a binary'sorbital plane is inclined. At a small angle i,

Fig. 22. Are there eclfpses?

there might be no eclipse at all. The minimum angle can be obtained geometricallyfrom the relative sizes of the binary'scomponents (Fig. 22).

The light curve thus permits us to determine a binary's period and the relativesizes of the stars. So now let us discusshow we measure the absolute dimensionsand, more important, the masses of thestars. What magic "ruler" could be putbeside the binary's orbit and stars? Nature


has taken pains to place such a ruler insideevery atom whether it is on the Earth orin a distant galaxy.

Doppler ERect

Accounts of the Doppler effect often beginwith a joke about a physicist who drovehis car past red traffic light. He tried toexplain to the policeman who stopped himthat he had mistaken a red light for a greenone because of the Doppler effect. Thepoliceman, however, knew some physics tooand booked the physicist for speeding. Itseems like the joke was duly appreciatedby the traffic police as they use now a speedgun based on the Doppler effect. Now weshall see how it operates.

Suppose we have a source and a receiverof periodic pulses. If the source is at rest,the period of the signal arriving at thereceiver will not be changed and will equalPsource. Now suppose the source is eitherapproaching or departing from the receiverat a speed v. During the period Psourcethe source will cover a distance vPsource.This means that the next pulse will have totravel a shorter (longer) distance. The pulsewill arrive earlier (later) by the time inwhich the signal travels the length of thesource's shift, i.e. vPsource/c, where c isthe signal's speed. The period of the signalsarriving at the receiver will change by

Chi 2. Algol Paradox 65

P = P r ec - Psource = ~P. In fact, thisrelative change in period is what we callthe Doppler effect:

d}>/ Psource -== ulc. (4)

-----=--:-----~

By comparing the periods of the transmittedand the received signals, we can obtain

the speed at which the source is either moving away or towards the receiver. Thepolice's' speed-gun is both a source and a

5-0667


receiver. It both illuminates vehicles andpicks up the echo. This makes it easy todetermine the speed because Psource isknown.

The Doppler effect is used in astronomy tomeasure velocities. The light coming fromstars is electromagnetic waves, which arespecial objects composed of oscillating magnetic and electric fields. An astronomerworks in terms of wavelengths rather thanperiods. Remember that a wavelength isthe distance covered by a wave in one period: A == cI'. The relative change in wavelength is also determined by the ratio ofthe velocity of the light source to the velocity of light. The Doppler effect depends solelyon the component of the source's velocity along the line of sight. This component is termed the radial velocity Urad:

~A __ urad

"-source - C(4a)

So how does the astronomer know thewavelength Asource of the light actuallyradiated hy a star? Star light contains a set,or spectrum, of electromagnetic waves ofvarying wavelength. Generally speaking,the spectrum of radiation emanating fromthe depths of a star to its surface (photosphere) is continuous, the amount of energyproduced by a star in each wavelength varyIng gradually with the wavelength (Fig. 23).


We have already mentioned a blackbody,i.e. a body in thermodynamic equilibrium.To obtain a blackbody, we mustwrap it ina heatproof and lightproof blanket. I t isnecessary for the substance and electromagnetic waves to be in thermal equilibri-

A_ ..... -_.

._----

Fig. 23. The formation of absorption lines in acontinuous spectrum

urn, in which, on average, the energy is nottransferred from the substance to the radiation and vice versa. Bodies which have thesame temperature are also in thermal equilibrium and so in this sense we can speakabout the temperature of radiation. A blackbody's spectrum is described by a formuladerived at the turn of the 20th century byMax Planck.

Since a star's depths are very opaque, ittakes electromagnetic waves a long time


to be absorbed and radiated before theyarrive at the star's surface. During thistime the substance and the radiation cometo a quasiequilibrium, and therefore thespectrum of the waves leaving the photosphere is very close to that of a blackbody.This spectrum is as regular "as it is useless.

Fig. 24. An electron can have only discrete valuesof energy in the potential well around an atomicnucleus

If you receive such light even if its wavelength has changed due to the star's motion, it will not help you determine thestar's velocity.

Fortunately, nature is not perfect andtherefore things may be learnt. Stars havetheir own atmospheres. By definition, a star


is a "hot gaseous sphere". Although hotinside, stars are surrounded by space witha temperature of -270°C. A star's atmosphere is generally colder than the star.Practically unobstructed, radiation passesthrough the atmosphere and thus does nothave enough time for attaining equilibriumwith it. This "spoils" the spectrum withirregular spectral lines.

You must bear in mind that light is notonly an electromagnetic wave but also aparticle. Light is absorbed and emittedin discrete bundles called quanta. Everyatom can emit quanta of a particular wavelength. Each quantum corresponds to atransi tion of an electron from one discretelevel to another (Fig. 24). Max Planckproved that each quantum's energy is determined only by its wavelength:

(5)

where h is 6.63 X 10-34 J. s (the Planckconstant). For example, a green quantumhas energy of approximately 2.5 eV. (1 electronvolt (eV) is the amount of energyacquired by an electron when it passesthrough a potential difference of 1 volt.)The atoms of a star's atmosphere, whichis colder than the star itself, only absorbquanta corresponding to certain energytransitions, i.e. certain wavelengths. As

70 Tn the World of Binary Stars

a result, in the star's spectrum there willbe a lack of light at some wavelengths, andwe get dark absorption lines. When wesaid a "cold atmosphere" we somewhatsimplified the situation. In fact, starshave no solid surface, and, therefore, theydo not have what we would call an atmosphere. The atmosphere of a star is thatlayer of it in which spectral irregularities(lines) are formed. The shape of the spectrallines varies with the temperature of theatmosphere at different depths, If the temperature is lower in the layers closestto the observer, then dark absorption linesappear. The reverse of a dark-line spectrumis a bright-line, or an emission, spectrum.In the overwhelming majority of stars thetemperature in the outer layers first fallsand then again rises. A rise starts in a layerwhere the density of substance is smalland quanta pass freely to the observer.This means that in most cases the opticalspectrum of a star has dark absorptionlines.

The wavelength of a line is governedsolely by the laws of electron-nucleus interaction. These laws are the same everywherein the Universe and so by measuring thewavelengths of the quanta emitted orabsorbed by a particular chemical elementwe can find "'rad- Then comparing the star'sspectrum with a laboratory spectrum permits us to calculate the star velocity.

Ch, 2. Algol Paradox 71

Radial Velocity Curve

In practice, a star's spectrum is obtained byplacing at the focus of a telescope a spectrograph, that is, an instrument which splitsthe light like a prism but much better.

The telescope is needed to collect morelight from a faint star. The spectrum iseither photographed or recorded digitallyand stored in a computer memory. Thestar's spectrum can then be compared witha laboratory spectrum of a chemical element.

We can find the velocities of binary starsby taking their spectra at different momentsand orbital phases. A velocity-phase curveis called the radial velocity curve. Theprojection of a star veloci ty on a line ofsight varies periodically as the star movesalong its orbit. Note that in a binary thevariations occur strictly in antiphase. Thespectra show the lines of the components"moving" in antiphase (Fig. 25).

Consider the simple case of circularorbits. Suppose the plane of the binary liesalong the line of sight (see Fig. 25). Weassume that star 1 is hotter and more massive. At the time of the primary minimumthe projection of the orbital velocity on theline of sight is zero. Then star 1 startsapproaching us, while star 2 starts movingaway. The first star thus has a negativespeed and the second a positive speed.


Radial velocity curves are thus like t\VOsinusoids in antiphase. The amplitudes ofthe sinusoids are the orbi tal veloci ties.However, sometimes the sinusoids meet offthe x axis.

In this case, the system is moving as awhole with respect to us. The velocity at

Towards thf o6sl'rVi'r

Ph,]s~

~ .~~ ~ 1---------.......-.....-0.--.--._~ ~

~ ~~

violt't ~

[ Jr'Dl N V

-- = --

-

Fig. 25. A periodic "moving" of lines in a binary'sspectrum (the Roman numerals show how the starpositions correspond to spectral lines)

which the radial velocity curves of the components cross is known as gamma-velocity.It is a projection of the velocity of thebinary's centre of mass on the line of sight.

Let us see how much information we canget about a binary star's parameters if weknow the radial velocity curve. First, wecan determine the absolute dimensions ofa binary's orbit from its orbital velocityand period. The wavelength of the lightquanta emitted by atoms is the ruler weused to measure the dimensions of a binary.Knowing the semimajor axis and the period,we can use Kepler's third law (formula (2»


to calculate the sum of the masses of thebinary star. lsi t possible to obtain theindividual masses of the stars? Yes, it is.Remember that the ratio of the stars' orbitalvelocities is equal to the inverse ratio oftheir masses. I t follows that we can use theratio of the amplitudes of the radial velocitycurves to find the ra tio of the star masses.

Thus the radial velocity curves and lightcurves enable us to find both the size of thebinary's orbit and the sizes and masses ofthe individual stars. However, this is possible if the lines of the both stars can beseen in the spectrum and if a binary systemis visible from the edge. Otherwise, additional (indirect) information is necessaryto estimate the parameters of the stars whichmake up a binary.

Often we only see the lines of a binary'sbrighter star in the spectrum. We, therefore,have only one radial velocity curve, i.e,that of the brighter star. One curve is notenough if you are after the mass of thestars, but we make the best of it.

To see what we can learn from a singleradial velocity curve, let us first solve avery simple system of four equations. Weassume that the stars' orbits in the binaryare circular. The sum of the radii of thestars' orbits is obviously equal to the semimajor axis of the binary. This gives us thefirst equation:

r1 + r2 = a.


The second equation arises from a conditionfor the distance from the binary's centreof mass to each of the stars:

M1r1 = M 2r 2 •

Kepler's third law (2) is our third equation:

a3 Gpi= 4n2 (M t+A12) ·

The radial velocity curve of star 1 yieldsa sinusoid's swing (amplitude). We divideit by two and label it K 1 • This value iscalled the semiamplitude. Let us considerits physical meaning. According to theDoppler effect, the shift of a wavelength isproportional to the projection of the star'svelocity on the line of sight (a radial velocity), therefore it is clear that K} is a projection of the orbi tal velocity of star 1 onthe line of sight. In a uniform circularmotion, the orbital velocity is equal to thecircumference of the circle divided by theperiod. Hence,

K 21ta..i = p- SIn i.

This is our fourth equation. By substitutingone equation into another and by eliminating r1 r-, and a, we get one equation infive variables, namely, the masses of thestars, the inclination of the binary's orbit,its orbital period, and the semiamplitudeof the radial velocity curve. Now we sim-

f:h. 2. Algol Paradox 75

(G)

plify t.he equation so that we have known ordirectly measurable values on its righthand side. The rest of the terms calleda mass function remain on the left-handside:

f . (AI) == J~f~ sin3i PK=:

1 (M 1 +M2 ) 2 2nG ·

To be more exact, /1 (M) is a mass functionas determined from the lines of the first star.From symmetry, it is easy to see thatshould we change index 1 for index 2, wewould get the same combination, with M 2

replaced by M]. This would be a massfunction as obtained from the spectral linesof star 2. Note that the mass functionin (6) is measured in the uni ts of mass.Usually it is measured in terms of the massof the Sun. To plunge deeper into "physics",let us study how the mass function dependson the semiamplitude of a radial velocitycurve. Assume that the binary's period isconstant while the semiamplitude K I increases. Star 1 will thus be moving faster.I t moves in the gravi tational field ofstar 2 and its mass must grow. Hence thenumerator of formula (6) contains thesecond star's mass raised to the third power.Of course, in reality the semiamplitude K I

is a constant. Our thought experiment isa search of different binaries with the sameorbital period and angle of inclination. Itshows that the mass function con tains some,


though indirect, information concerning themass of star 2.

Note that the mass function is determined spectroscopically, but it does not allowus to find the separate masses of the starsor the angle i. We need addi tional information, say, from the light curve. Thenwhy is the mass function so important?

The mass function has a valuable property. If we divide /1 (M) by the mass ofthe second star:

fdM}/Mz = ( Ml~2M2 Vsin" i,

we see that the right-hand side is smallerthan or equal to one, which means that

M2~ 11 (M).

This inequality is of great practical value.Suppose you are an astronomer, and onestarry night you obtain a spectrum ofa mysterious star. In fact, at first you seenothing mysterious. You photograph thespectrum, have a good sleep, develop theplate, and study it thoroughly. You haveobtained an ordinary spectrum of a singlestar. However, you do not give up. On thenext night (thank God, it is as starry) youmake another photograph of the spectrumrepeating your previous procedure. Nowyou have two spectra of the same star. Youstart comparing them and discover thatsome of the lines in the second spectrum have

Ch, 2. Algol Paradox 77

definitely shifted with respect to the linesin the first spectrum. I t becomes clear thatyou have discovered a binary star. None ofthe catalogues lists the star as a binary,which you know for sure because anyastronomer first looks through the cataloguesand the literature for information aboutthe object he or she is about to observe.

So the star is binary. Yet you may askwhy the second star's lines are not seen inthe spectrum. Clearly, the second star ismuch fainter and cannot be seen against thebackground of its brighter companion. Thismakes the object even more puzzling. Youspend another month and obtain a lightcurve which is pure sinusoidal with a periodof 1.2 days and a semiamplitude of K1 =400 km/s, Substituting these values into (6)you find the mass function /1 (m) = 8M0'Now you know that, whatever the inclination of the binary's orbit and whateverthe mass of the visible star, M 1 , the mass ofits invisible companion is more than eighttimes the mass of the Sun. Sometimessuch information is extremely important.In Chap. 7, I hope to convince you ofthis.

Why Are Stars So Different?

If you use a small telescope to observe thebinary star system ~ Cygni, you will seefor yourself hew different stars are. I t is one


of the most astonishing sights an amateurastronomer can ever see. The telescopewill reveal a pair of stars, one blue and oneorange. Why do they differ in colour?Objects around us are differently coloured,and this is because of their different chemical compositions, properties of their surfaces, etc. Yet we know that colour alsovaries with temperature. For instance, whenheated metal first becomes red hot andthen white hot. So we want to know whetherthe difference in star colours is due to thedifference in temperature or chemical composition.

We turn to spectral analysis to answerthis question. We use a spectrum of a starto determine chemical elements corresponding to the spectral lines. We find thatlines of different chemical elements existin the spectra of differently coloured stars.For example, the strongest lines in the visible spectrum of the Sun are those of calcium. In blue stars, calcium lines are missingwhile hydrogen lines dominate. The mainlines in white stars are those of helium. Thismeans star colour is due to its chemicalcom posi tion.

However, this is not exactly so. In fact,the intensi ty of spectral lines indicatingthe presence of a particular element in thespectrum of a star depends greatly on thetemperature of the star's atmosphere. Inyellow stars, like our SUIl, the strongest


lines of the visible spectrum are those ofsingly ionized calcium ions (that is calcium atom minus one electron). In blue starspectra, however, no calcium lines can befound because the stars are hotter andcalcium is almost totally ionized in theiratmospheres, i.e. nearly all its electronshave been torn off. When heated, atomswith weakest binding ionize first, and consequently their lines cannot be observed inthe spectra of very blue stars, even thoughthe chemical com posi tion of all stars isabout the same. In terms of mass, starscontain 70% hydrogen, 29% helium, therest being heavier elements.

To describe star colour or temperature,astronomers use spectral classification. Theseven classes are designated by the letters

OBAFGKM,

with the classes arranged in descendingorder of temperature, the hottest starsbeing of several tens of thousands of kelvins(0 and B) and the cooler stars being ofseveral thousands of kelvins (K and M).These classes can be memorized wi th thehelp of mnemonic phrases, such as, "0 BeA Fine Girl, Kiss Me!"

Initially, seven letters were thoughtenough. Later, however, a more detailed classification was required. Now each spectralclass is subdivided with the numerals 0through 9 used to indicate subclasses. So


the refined classification is ...B9, AO, At,A2, ... A9... The spectral class of the Sunis G2. There are millions of such stars inour Galaxy.

Stars range widely in temperature fromseveral tens of thousands of kelvins toseveral thousands of kelvins. The spectralclassification allows us to determine thetemperature of a star's surface. However,the stars differ even more markedly inluminosity. Note that a star's luminosity Lis defined as the amount of energy it produces per unit time. Infact, luminosity is therate at which energy in the form of lightis radiated. The Sun's luminosity is 4 X1028 W. There are stars which are a millionof times more powerful and thousands oftimes weaker than the Sun. Hence, there isnothing special about the Sun's luminosity.

Astronomers use absolute magnitudes todescribe the luminosity of a star. Thelatter is the magnitude the star would haveif it were located 10 parsecs (32.6 lightyears) away. In this case, its apparentmagnitude would be absolute. The Sun'sabsolute magnitude is 4.7. But the stars areat different distances from the Earth, theirapparent brightness yields no informationabout their luminosity or absolute magnitude, so how has it been possible to discover that stars have such a wide range ofluminosity?

Ch. 2. Algol Paradox 8t

First, we can use binary star systems, inwhich both companions are equidistantfrom us. Therefore, if one is brighter, it iscertainly more powerful.

In a binary, only two stars may be compared, while in a star cluster several thousand stars can be compared. The closeststar clusters to the Earth are the Pleiadesand the Hyades, each consisting of hundreds of stars approximately equidistant fromthe Earth. This fact gives astronomersa good opportuni ty to study the differencesin luminosities. The first astronomer whoused this opportunity was the Dane EinarHertzsprung. At the beginning of the 20thcentury, he observed the Pleiades and theHyades and plotted two similar diagramsshowing the apparent magnitudes of thestars as a function of their temperature.(In fact, he did not use the temperature,but the directly observed degree of bluein their light.)

A few years later, the American astronomer Henry Russell independently beganplotting such diagrams for stars located atknown distances. Today these diagrams arecalled Hertzsprung-Russell diagrams. Hundreds of different sorts of diagrams havebeen plotted by various astronomers overthe last hundred of years, but the Hertzsprung-Russell diagram (Fig. 26) hasturned out to be most useful.

Even the first Hertzsprung-Bussell (H-R)

6-0667


diagrams showed that the stars are notscattered randomly but instead are groupedalong certain lines. The overwhelming majority of stars are grouped along a diagonalline called the main sequence. By the way,that's where our Sun is to be found. The

Spectral ctasso 8 A F o I( H MB

I 1 I t I I ,-

..:~. '".: .-..-5~

+10-

+15-

. ...'•••., <# ... ,-

..;:~::.,. ;.::; :'i.~.I"fl".s..,.~~;:"'I~I'I":""" '. • ••

~ :;~~:. ::::: ·:.::::n\~;·," f.'::~~ •,;; •

. .!:I .':. ...'f'

.... r

...... .,"c _

: ~ -

Fig. 26. The Hertzsprung-Russell diagram

most luminous stars form a thick horizontalband. Altogether stars form a slightlydistorted Y-shaped curve. That stars "flock


together" proves that there is a certain,though complex, dependence between spectral class and luminosity.

The reason for this mysterious diagrambecame clear after a breakthrough in ourunderstanding stellar structure and stellarevolution. This theory, which we shall studyin detail in the next chapter, explains whymost stars are grouped along the mainsequence and why the main sequence existsat all. The band is a locus in which starsspend most of the time. The less massivea star is, the colder and dimmer it is.Stars, however, do not stay in the mainsequence for ever. Sooner or later, theyleave it, to become subgiants, then tobecome giants, and so on.

The theory of stellar evolution showsthat more massive stars spend least time inthe main sequence. Stars like the Sun stayon the sequence for billions of years,while blue O-B stars stay for a periodhundreds of times shorter. In otherwords, the more massive a star is, thefaster it burns out. The main sequenceitself is a sequence of stars with differentmasses (blue stars are more massive thanred ones). The masses of the coldest starsare about a tenth that of the Sun, whilethe hottest stars are dozens of times moremassive than the Sun. So the differencesbetween virtually all the stars in theGalaxy are caused by their different masses.

6*


Finally, a ParadoxThe theory of stellar evolution and structure has been brilliantly confirmed bymeasurements of star masses. We have already seen how the stars in a binary may be"weighed" using radial velocity curves.Many hundreds of measurements have unmistakably shown that the more massivecomponent of a binary is always bluer.That's what we call a triumph of theory.

In the mid-1950s, however, astronomersbegan to realize that this concord between

theory and observations was not ideal. Thedisillusioning cracks started from Algol,the most ordinary binary star imaginable.The Soviet astronomers AlIa Masevich and


Pavel Parenago were puzzled by the Algolparadox and analyzed its light curve andradial velocity curve. They discovered thatthe more massive component was on themain sequence, while its less massivecompanion had left it and turned intoa subgiant. Obviously, both stars had beenborn at the same time. So far there was noexplanation of the phenomenon.

The theory seemed so flawless that no onewanted to ruin it because of one little star.It soon became obvious, however, that theAlgol paradox was endemic among thebinaries. There was only one thing to do,namely, to admit that binaries did notevolve in the same way as single stars,which were described by the theory of stellarstructure and evolution.

The only key to the Algol riddle was theassumption that star mass is a variablein a binary star system. Suppose that a lessmassive star in Algol used to be the moremassive one. Now we suppose it left themain sequence, lost some of its mass forsome reason, and then became lighter thanits neighbour. That would resolve the Algolparadox. But why does a star begin to loseits mass? The explanation was offered by theAmerican astronomer J. Crawford, whosuggested a .seenario for a binary's evolution with a change in roles.


Changing Roles

The theory of a single star's evolutionpredicts that a star departing from themain sequence expands. This is why oneof the stars in a binary starts losing mass.Suppose we have a binary star composedof t\VO stars on the main sequence. Letstar 1 be more massive than star 2. Initiallythe stars evolve independently (Fig. 27a).Now suppose star 1 is the first to leave the

fa'

(hJ

ie)

Fig. 27. Changing roles

main sequence and it starts expanding. Ata certain moment the first star will fill itsRoche lobe. Then the substance of star 1will flow to star 2 via the inner Lagrangian


point (Fig. 27b). Suppose the outflow ofmass is such that the remainder of star 1became less massive than star 2. Thus theflow of mass has reversed the stars' roles.The resultant system has a more massivestar on the main sequence and a less massivestar which has expanded to the size ofa subgiant. Obviously the same occurred inthe Algol system.

By the mid-1960s, astronomers had shedmore light on this scenario. At the time,nobody realized that the scenario containeda "delayed-action mine". I t turned out thatthe change in roles accounts for the existence in the Universe of some very unusualobjects. The discovery and investigationof these objects in the 1970s gave birth toX-ray astronomy, a brand new science.

Binary stars whose evolution is accompanied by an exchange of mass are called closesystems. The evolution of close binariesis still not completely understood, withthe final stages in the evolution of binaries being especially vague. The whole situation is very interesting. The older a binaryis, the less we know about it. The reasonfor this is that at later stages exotic objectsappear that only recently have become available to observations.

I would like to invite you for a timejourney from the birth to the death ofa binary star. I shall try to explain whatis happening if I can.

Chapter 3

Living Apart Together

We shall start our time journey throughthe stages of a binary's evolution from

Livtng apart together

the very beginning. At this point, thestars are much smaller than the distancebetween them. This means that they are

Ch. 3. Living Apart Together 89

deep in their corresponding Roche lobes anddo not affect each other. Both stars evolveas if they have no companions. Binary ornot, they live the life of a single star.

In 40 Minutes Only

Even though the Sun is vast (its radius is700000 km), it only creates a gravitational potential at its surface 3600 times thaton the Earth's surface. Do you rememberformula (1) for the potential of a pointbody? It remains valid outside a sphericalbody of random size. The potential ismeasured in the units of energy per unitmass, which are the dimensions for thevelocity squared. The potential (Itsabsolute value) is, in fact, equal to the kineticenergy of a test particle having a zero totalenergy. Zero energy particles are thosewhich fall freely from infinity. Freely falling onto the Earth's surface, a body wouldhave a speed of 11 km/s, while on thesurface of the Sun it would have 617 km/s.The square of the ratio between these twospeeds is about 3600. The SUD has a highervalue because of its mass (the Sun's mass isMe = 2 X 1030 kg and the Earth's massis MfB = 6 X 1024 kg).

Gravity seeks to compress the Sun, butthe Sun does not shrink because the inwardpressure due to gravity is strictly balanced by the outward gas pressure forces


of the Sun. If at some moment the gaspressure disappeared, it would take the Sunonly 40 min to collapse to a dot. Yet theSun has been shining for billions of yearswithout any significant change in size. Thetime it would take the Sun or other star tocollapse is called dynamic time. I t isapproximately equal to the ratio betweenthe star's radius and the escape velocityon its surface:

t D ~ RGlvII. (7)The collapse time (i.e, 40 min for the Sun)

may he calculated very accurately. Assumethat the pressure inside the Sun has disappeared and the Sun begins to "fall intoitself". To see how long this will take,recall that the gravity outside a sphere isthe same as the field of a point with thesame mass at the sphere's centre. So, weneed to calculate how long it takes a testparticle to fall to a point whose mass equalsthat of the Sun from a distance equal to theSun's radius. The motion of such a particleis called free motion in a potential well,which can he described by Kepler's laws.It (motion along a straight line) can herepresented as moving along an ellipse withan eccentricity of one and a semimajoraxis of half the Sun's radius.

I t is obvious that the time we desire ishalf the period of revolution along a collapsed ellipse, the period being determined


by Kepler's third Iaw (formula (2)). Assuming that M 1 == M0 and M 2 = 0, a =R(i'/2, we obtain

P 1t V R0 ·to=T=Y 2GM ~40 min.o

You see that the estimate (7) does not differmuch from the accurate result. The finenesswith which gravity and pressure are balanced in the Sun can be seen from the ratioof the dynamic time to the age of the Sun,Le. approximately 10-14• You might getthe impression that the Sun is fragile. Suppose the pressure and the gravity were several percent off-balance. Clearly, the Sunwould collapse in a few hours. This, however, does not happen. Now let us see why.

The balance in the Sun and other stars isstable. We know that a stable equilibriumalways means that the system's energy isat a minimum. The star's energy is composed of its potential gravi tational energyand the kinetic energy of its particles,i.e, ions and electrons. All the kineticenergy in stars is concentrated in thechaotic motion of particles. (The rotation ofthe Sun adds very little to the kinetic energyof its particles; this is why the Sun isround.) The particles' chaotic motion isthermal; therefore, the Sun's kinetic energyis i t8 thermal energy. The Sun's totalenergy E is the sum of the thermal energy K


and the potential energy U:

E = K + U.

Now we shall see how stable a star is bycompressing it. Figure 28 shows how thestar's total energy varies with radius. On

E

Emln ---

R

Fig. 28. A change in the star's energy during compression and expansion

being compressed, the star's thermal energyrises faster than its gravitational energyfalls. True, the star's total thermal energy Kis proportional to the star's mass times itstemperature: K ex /11 MT, where __1? is thegas constant. We shall assume that thestar's compression is so fast that it does nothave time to give off heat. Such compression is called adiabatic. When adiabatically compressed, the temperature of a monatomic gas depends on the volume as T ocV-2/ 3 • But the star's volume is V ex: R3;


therefore, T oc R-2. The thermal energygrows during the star's compression as f/R2.In contrast to this, the star's gravitationalenergy is proportional to the potential,i.e. it grows as fIR (see formula (1)). Thegraph in Fig. 28 is the sum of two hyperbolas, a positive quadratic one and a negative first-order one. The energy minimum(E min ) corresponds to the star's equilibrium radius R•.

I have already mentioned that duringa star's compression its thermal energyrises faster than its gravitational energydoes. Therefore, the pressure forces growfaster too. In other words, the star tendsto expand. Like a spring, the star will varyin size about some equilibrium, and theperiod of the variations will be of the orderof its dynamic time. For the majority ofstars, this time varies between dozens ofminutes and several hours.

Getting Hot by Cooling

In equilibrium, the star's energy, like thetotal energy of a binary, is negative. Thisis a common property of any gravitationallyrelated system. The thermal energy K inequilibrium is always of the same orderas the gravitational energy (but with theopposite sign). This accounts for the astonishing property of a star, namely, its nega-


tive heat capacity. If you heat any normalobject, in other words, if you increase itsenergy, it will get hotter because it hasa positive heat capacity. I t is the otherway around with a star. An increase in thestar's total energy (heating it up) lowersits thermal energy, i.e. its average temperature. Indeed, if you increase the star'senergy (heat it up), it must expand andattain a new state of equilibrium. This,however, will be accompanied by a fallin the absolute value of the gravitationalenergy (the radius has grown) and a fall inthe thermal energy. The star will becomecooler! The reason is that an increase inenergy causes the star to expand and dowork to overcome gravity. That's whataccounts for the cooling.

This does not contradict the laws of thermodynamics. I t might seem unusual to usbecause on Earth we are used to dealingwith objects whose equilibrium is supportedby short-range forces. As a rule, these arethe forces of molecular attraction (in fact,they are ordinary electrostatic forces actingbetween charged parts of molecules eventhough, on average, the system is neutral).Considering some small segment. of a body,we ignore its interaction with other partsnot touching it. We can do this becauseboth negative and positive electric chargesexist in nature and they screen each other.In contrast, gravitational charges or masses


always have the same sign since there isno antigravi ty.

That is why gravity is long-range. Anypart of a star "feels" both the attraction ofits immediate neighbour and that of allthe other components of the star. Thus,

A star's negative heat capacity

a common gravitational field creates a poolof negative energy. When a star expands,this reserve turns into an insatiable consumer. When the star becomes colder andsmaller, it is a heater. Look at the starsthat are shining. Now you know that thismeans they must shrink to lose their energy.


As Little As 30 Million Years

The total thermal energy in a star is approximately equal to its gravitational energytaken with the opposite sign, i.e. it is ofthe order of GM2/R. The thermal energy ofthe Sun is 4 X 1041 J. Every second theSun loses 4 X 1026 J of energy. This meansthe Sun should run out of energy in 30 million years. The time it takes a star to loseits energy is called its thermal time

tT ~ K/L. (8)

The Sun and the Earth, however, haveremained nearly the same for severalbillions of years. Consequently, the Sunmust have an inner source of energy tomake up for what is lost. We have alreadydiscussed the gigantic pool of gravitationalenergy in the star, but it is not inexhaustible. The Sun must shrink to half its sizeevery 30 million years. Some other sourceof energy must exist.

The source was discovered by Sir ArthurEddington, an English astronomer andthe father of the modern theory of stellarstructure. He worked at Cambridge University, as Newton had done some 200 yearsearlier. Eddington suggested that nuclearenergy is produced inside the Sun. This ideais much older than the H-bomb.

Let us look at where nuclear energycomes from. Atomic nuclei are a mixture of

Ch, 3. Living Apart Together 97

neutrons and protons (nucleons). The totalenergy of a nucleus is made up of the potential energy of the nuclear interaction, theelectrostatic repulsion of protons, and thekinetic energy of all the particles, Figure 29 shows a graph of the energy per one

40 80 120 150 200Moss number

Fig. 29. The average binding energr per one nucleonin the nuclei of different chemica elements

nucleon against the mass number. The average energy per one nucleon in a nucleus, iftaken with the opposite sign, shows howclosely the particles in a nucleus are associated (or, in other words, how deep thepotential well in which they found themselves is). The graph indicates that thestrongest bonds exist in the elements neariron. The lighter elements only have 8 fewnucleons in a nucleus and the nuclearinteraction energy is proportional to thesquare of the number of particles. (Eachparticle interac ts wi th all the rest! Bythe W8)T, this is why gravitational energyis proport ional to the square of mass.)

7-0667


The amount of energy per particle growswith increasing the number of nucleons.This is true as long as the nuclei are light.Nuclear forces are extremely short-range.They act at a distance of 10-13 em, but, asthe distance grows, fall to zero. For thisreason in nuclei with many particles nucleons only interact with their immediateneighbours. By contrast, electrostatic repulsion, whose energy is positive and whichis long-range grows continuously and isdirectly proportional to the square of thenumber of nucleons (protons and neutronsare present in nuclei in approximatelyequal numbers). Therefore, the nuclei ofelements heavier than iron have smallerbinding energies. The graph shows that it isenergetically more favourable for lighterelements to fuse into heavier ones and forheavy elements to break into lighter ones.(The fission of heavy elements is used innuclear power plants.)

The fusion of light elements into heavierones is accompanied by a terrific release ofenergy. For instance, four nuclei of hydrogen (protons) have a mass of 6.69 X 10-27 kg,while a nucleus of helium weighs 6.65 X10-27 kg. This mass defect is explainedin relativity theory and, according toEinstein's mass-energy relation, a body'stotal energy is related to its mass thus

(9)


The binding energy per one nucleon inhelium is greater, so its potential well isdeeper and its total energy is smaller. If wecould convert 1 kg of hydrogen into helium,the liberated energy would amount to 6 X1014 J. This is about 1 % of the totalenergy of the fuel consumed. Here is anenergy reserve.

When Eddington suggested that a star'senergy is nuclear in nature, his hypothesiswas given a "death blow".

But for Quantum Mechanics

I-lis critics, staying wi thin the frameworkof existing laws, "proved" that fusion isimpossible at the centre of the Sun. Figure 30 gives a graph of how the interaction

Fig. 30. Interaction energy between two protonsagainst the distance between them

energy of two protons is related to thedistance between them. At greater distances, the interaction is governed by theelectrostatic repulsion of two positively

7*

100 In the. \Vorld of "Binary Stars

charged particles. The interaction energyis positive and, as the particles get closertogether, grows as 1/R. At a maximum,this energy is approximately 1000 keYThen, at about 10- 13 em, the nuclear interaction creales an area with negative ener-

Quantum tunneling effect

gies, which corresponds to the bound state.To get into the negative energy area, however, a barrier of 1000 keV must be overcome,

We can estimate the temperature at. thecentre of the Sun by assuming that thethermal energy and the gravi tational energy are about the Sal11C (.fiAtT ~ Gll12/R),

and the resultant ligure is ten millionkelvins. The average energy of protons atsuch a temperature is about 1 keV, which isa thousand times too small for helium fusion. The critics maintained that it was too


cold at the centre of the Sun for fusion.This did not embarrass Sir Arthur Eddington, who stubbornly suggested they shouldlook for a hotter spot. What was oncebelieved to be obstinacy is now calledin t11 i tion,

The advent of quantum mechanics endedthe argument. Developed in 1926, quantummechanics was soon applied to astronomyand revealed the astonishing properties ofthe microcosm. One such property is tunneling, i.e. particles' penetration of a potential barrier. Elementary particles appear tobe able to penetrate a barrier even if theirown energy is smaller than that of thebarrier. Suppose a high jumper were to rununder the bar, the judge would hardly belikely to say the jump was fair. The laws ofnature are not so "strict".

Clearly, the 1110St visible quantum effectis the shining of the stars. What would happen if there were no quantum mechanics?

Nuclear Evolution

The conversion of hydrogen into hnlium isirreversible, and the reserves of hydrogenin C\ star arc limited. Moreover, fusion reactions require high temperatures and densities to proceed. At the centre of the Sun,the density is 100 g/cm", Thus only thehydrogen in the star's central core, some10% of the star's total mass, call go to fuel

102 In the World or Binary Stars

(10)

it. Now let us see how long it will take theSun to run out of its nuclear fuel.

The Sun's total energy is M Cft2 = 1047 J I

nuclear energy E n uc is about 1 %, i.e. about1045 J. Given that not all the fuel mayburn, we get 1044 J. By dividing thisquantity by the Sun's luminosity L == 4 X1028 JIs, we conclude that it will taketen billion years for the Sun to burn out.This is consistent with geological dataconcerning the Earth's age. On the otherhand, it means that stars are not eternal,they evolve. Their nuclear evolution isdetermined by the gradual exhaustion ofthe light elements. This period is calledthe nuclear time and it is given by theformula

Enuc 1010 ( 1ft ) - 2tnuc ~ -L- ~ ~ years.o

The nuclear time versus the mass ofa star may be calculated if we rememberthat the star's nuclear energy is E n uc ocMc2 , and that luminosity behaves approximately like L ex: M 3 • I would like toemphasize here that the right-hand sideof (10) is only a rough approximation. Theresult is that the larger a star, the fasterit burns itself on t.

Now let us return for a while to theH-R diagram. Most stars are grouped alongthe main sequence. These are stars at whosecentres hydrogen is burning. I say hydro-

Ch, 3. Living Apart Together 103

gen and I mean it. The heavier elementsrequire higher temperatures to start fusing(their potential barrier is higher) and thisis accompanied by the departure of thestar from the main sequence.

Let us compare the star's nuclear timeand its thermal time (see formula (8)).An approximate relation between the thermal time and the star mass is

tT ~ 3 X 107(:: ryears. (11)

cThe dynamic, thermal, and nuclear times

determine the star's evolution. Becausethe dynamic time is much less than thermaland nuclear times, the star always has timeto attain hydrostatic equilibrium. Furthermore, because the thermal time is lessthan the nuclear one, it also has time toattain thermal equilibrium. In other words,there is always an equilibrium between theamount of energy emitted at the centre perunit time and the amount of energy radiated by the star's surface (a star's luminosity). Every 30 million years the store ofthermal energy in the Sun is renewed. Butthe energy in the Sun is transferred byradiation, in other words, it is carried byphotons. A photon that is born in a fusionreaction at the centre appears on the surfaceafter about 30 million years.* If the ther-

• Although it will be quite another photon (seebelow).


mouuclear energy supply were stoppedtoday, the Sun would continue to shinefor many more milltons of years,

Burning hydrogen not only producesphotons it also yields neutrinos. The latterleave the Sun at the velocity of light,that is, it takes 700000 km : 300000 km/s =2.3 s (the Sun's radius in light seconds).

Fig. 31. The trajectories of a photon (y) and a neutrino (v) emitted Irom the centre or the Sun

How COOlC that it takes a photon, whichalso travels at the velocity of light, asmuch as 30 million years to cover the samedistance? Clearly, because the photon isrepeatedly absorbed and reemitted. Itstrajectory is thus so complex that it has totravel 30 miflion light-years, i.e. the distance to distant galaxies (Fig. 31). Over sucha long period the radiation attains thermal


equilibrium with the matter in which it ismoving. Therefore, a star's spectrum isclose to the spectrum of a blackbody.

The theory of the structure of mainsequence stars worked out by Eddington in

0.1 t tostars moss,M/M

0

Fig. 32. The observation of binaries made it possible to plot the mass-luminosity relation for mainsequence stars

the early 19308 fully explained their observedproperties. The H-R diagram, however,features luminosity-spectrum dependencewhich, in turn, is a function Qf star masses.This means that a final verification of thetheory demanded that the stars be weighed.That's when binary stars were handy.Photometric and spectral observations ofbinaries made it possible to determinethe masses of many hundreds of stars andtheir relationships with luminosity (Fig. 32).This relationship revealed consistency be-


tween observation and theory. However, thetheory failed to explain why some stars inthe H-R diagram do not lie on the mainsequence.

Fleeing the Main Sequence

The line along which a star moves in theH-R diagram is called an evolutionary

Colour

Oh, flocks of stars! Like sheep you roam.Who is your shepherd? Where is your home?

track. The first astronomer to introducenuclear reactions into the calculations of


stellar structure was an American namedMartin Schwarzschild. His calculations explain why the stellar groups wander toand fro.

Initially, a star starts crawling up themain sequence. At this stage, the star'sluminosity and its temperature vary withgradual changes in the chemical composition of its core. Hydrogen turns into heliumuntil it burns out completely. First ithappens in the core where the density andtemperature are at a maximum. Thus a helium core is formed. Although the temperature in the core is not high enough tosustain the burning of helium, hydrogenkeeps on burning in a shell surrounding thecore. Such a process is called burning ina shell source. The helium core thus findsitself inside the source of energy; the temperature in the core becomes constant, andsuch a core is said to be isothermal. As theshell source is formed, the flow of energy tothe surface of the star increases. The envelope around the shell source becomes turbulent like water boiling in a kettle.

The analogy with a kettle is more thansuperficial. At a certain moment, the waterin a kettle starts bubbling because thewater's heat conductivity is too small totransfer the heat supplied from the bottom.Heat is transferred in fluids by a mechanismcalled convection. The warmer and less densewater at the bottom rises ill accordance

108 In the \Vorld of Binary Stars

with Archimedes' principle. The instabilityof a denser fluid being above a lighter fluidis called Rayleigh-Taylor instabili ty.

The formation of the shell source leads tot.he formation of a convective envelope.

tog (llle)

5.0

5.0

4.0

.-...-- ~2110

-..--15110

- ....--10110

4.8 4.5 4.4 4.2 3.8togT

Fig. 33. Stars of different masses leaving' the mainsequence

The star expands. In the H-R diagram, thisis manifested as a departure from the mainsequence towards giants and supergiants(Fi.g. 3:l).

Have you ever seen peat burning ill a peatbog? You don't actually see any fire, onlysmoke. The peat burns in a small ringshaped segment. Gradually the segmentgrows wider and the yellow patch of fadedgrass inside the circle expands.

The same happens to a star. As thehydrogen begins to fuse, the mass of the

Ch. 3. Living Apart. Together 109

helium core increases (Fig. 34). The corebecomes too large and the expansion ishal ted and then reversed. What happens next

Isothermatfluel PIIS

-,IIIIIIIIII! I

R

\Shell source

Fig. 34. The structure of a supergiant

depends on the mass of the star. In a verymassive star, the temperature in the heliumcore rises until it is high enough for thehelium to turn into carbon. This occurswhen three helium nuclei (a-particles) fuse,


therefore, it is sometimes called the 3aprocess.

A star may have two or more shells. Instars with masses exceeding 10M0, thenuclear fusion even involves elements belonging to the group of iron. Such a star isa blue supergiant. I ts radius increaseshundredfold and in some cases becomesthousands of times as great as that of theSun. If you were to place such a star in theSolar System instead of the Sun, the Earthwould be deep inside it (the distance between the Sun and the Earth is equal to214 radii of the Sun). What is the finaloutcome of the stellar evolution?

White Dwarfs

Eddington's theory elegantly explained thepositions of the .stars in the If-R diagram.Given the mass and chemical compositionof astar, all the observed parameters including luminosity, radius, surface temperature, etc. could be obtained. However, onestarlet, namely, 40 Eridani B, seemed tospoil the whole picture. I t was located inthe H-R diagram below and to the left ofthe main-sequence stars. Even though it hasa high temperature, it was glowing toofeebly and was thus too small. Eddington'stheory could not explain this.

The puzzle was solved by the Englishphysicist Ralph Fowler in 1926. Those were

Ch. 3. Living Apart Together iii

years of great discoveries in physics. Between 1925 and 1927, the German physicistsWerner Heisenberg and Erwin Schrodlngerdeveloped quantum mechanics. In 1927,Heisenberg enunciated the uncertainty principle, which stated that it is impossible toknow both the coordinates and velocity ofan elementary particle at the same time. Atthat time, Wolfgang Pauli proposed hisfamous exclusion principle, which statedthat no two electrons in an atom may be inthe same quantum state. Enrico Fermi andPaul Dirac studied the fundamental properties of matter using quantum-mechanical principles. These discoveries showed thatunder certain conditions matter starts behaving in an unusual manner.

Eddington's theory starts from theassumption that a star consists of an idealgas, i.e, matter satisfying the equation

P=p91:, (12)

where P and p are the pressure and densityof the matter respectively, and ~t is itsrelative molecular mass. The relative molecular mass is the average mass of one particle expressed in masses of an atom ofhydrogen. This definition is also true forplasma, which has no molecules whatever.Accurate measurements show that the relative molecular mass of fully ionized hydrogen plasma is 0.5. The mass of a pair of

1t2 In the World of Binary Stars

particles (an electron and a proton) is almostas much as that of a proton (the electron hasa mass 1800 times less than that of a proton).

According to the classical definition, anideal gas is one in which particles are muchsmaller than their mean free path. Have youever asked yourself why the densi ty ofobjects around us is close to 1 g/cmS? Thedensity of water, for example, is equal to1 g/cm", Let us try to answer this question.

If we cool bodies around us, their densities will not be significantly changed. I t canonly mean one thing: these bodies are already so cold that the particles they are madeof touch and further cooling will cause nofurther contraction. The world we inhabitis coldl The density of bodies we comeacross in everyday life is the densi ty of theatoms they are made of. A hydrogen atomhas a mass of 10-2.& g and a diameter10-8 em. Dividing the mass by the cubeof the dimension, we get 1 g/cm",This means that the atoms and moleculesin solids and liquids are packed so closelytogether that they are far from being idealgases. Now let us turn to the density ofthe Sun. If we divide the mass of the, Sun(2 X 1033 g) by its volume (4/3) n (7 X1010) 8 em", we get 1.4 g/cm", That is, theSun is denser than water and yet it is verymuch an ideal gas. Why?


Let us recol lect the definition of an idealgas: particles' mean free path is much greater than their dimensions. The mean freepath is the average time between successivecollisions multiplied by the speed of a particle. As the density grows, the time between the collisions decreases and so themean free path becomes shorter. But ifthe density increase is accompanied by anincrease in the speed of particles, the meanfree path may even become longer. Weknow that the speed is proportional totemperature and this is the key to thepuzzle! The Sun's matter is so hot thateven though it has a density exceeding1 g/cm", it remains an ideal gas (or, tobe more exact, a plasma). Clearly, if thecontraction of the Sun is not accompaniedby a significant rise in its temperature, itsmatter wi ll sooner or Jater stop being anideal gas.

If we admit that t.he Sun could be COD1

pressed, its matter will cease to be idealsooner for qui te another reason. I t is worthyof note that as early as tho end of the19th century the Iarnous Russian inventorand rocket expert Konstantin Tsiolkovskysuggested that if the Sun contracted, itsmatter might. be converted into anotherstate that cannot be described by the idealgas equation. Considering radiation-induced star contraction, Tsiolkovsky admittedthat at some density further contraction

8-0667

i14 In the \Yorld of Binary Stars

would not be possible due to deviationsfrom the ideal gas condition. Surely, hecould not know that the reason lay in thelaws of quantum mechanics. I twas Fowlerwho first noted that if a star was contracted10-100 times, the electrons would be pushed apart due to Pauli's exclusion principle.At densities of 10s-108 g/cm", protons andions would be so close together that theiratomic levels would unite. Electrons alsounite to form a special degenerate gas.When Fowler used an equation of state fora degenerate electron gas, he discoveredtI18t such stars would be about 500010000 km in diameter, i.e. about the sizeof a planet. This explained the contradiction of 40 Eridani B and other stars of thistype. Such stars are called white dwarfs.

If a star'smass is Jess than 3M0, a heliumcore grows at its centre which contractsto become a white dwarf. The contractiongenerates gravi tational energy which heatsup the envelope; then the latter cools, i.e. itexpands and leaves the star in the formof a ring-shaped nebula. This may be theway planetary nebulae are born (Fig. 35).As a rule, at the centre of such a nebula,there is star which is as hot as it is cornpact; it is not unlikely that the star isa helium core being cooled.

In more massive stars, the temperature atthe centre is high enough to initiate thefusion of heavier elements. If the star's

Living Apart Together 115

lass is less than abou t 81\;/0 a core ofarbon and oxygen grows in its interior.'he contraction of such a core, provided itsnass is close to the Chandrasekhar Iimit,

~ig. 35. A planetary nebula in the constellation~quarius

s not as harmless as it is in the case ofl helium core. The carbon begins to fuseIt the star's centre. As a result, the burnng spreads slowly throughout the star.


This process is knowu in explosive chemistry and is called deflagration. At a certainmoment, the hydrostatic equilibrium isdisturbed and the star begins pulsatingwith a growing amplitude. This pulsationmay result in the explosion and completedisintegration of the white dwarf. Theenergy thus produced is comparable to thatof a supernova explosion. The star ejectsheavy elements into interstellar space wherenew stars or planetary systems form andwhich wi ll probably be able to support.Jife.

Filially, in stars wi til masses of (8-10) llfrv,the carbon is completely used up anda core consisting of a whole set of elemen tsincluding "Oxygen, neon, and magnesiumis Iormed , Later this core cools and becomes a white dwarf. The extra materialdisperses "quietly" ill the Iorm of a planetary nebula.

Neutron Stars

A degenerate gas possesses the remarkableproperty that its pressure is iudependentof temperature and varies only with thedensi ty as P ex: p5/3. However cold a whi tedwarf may be, it wi l l never contract.The white dwarf's equation of state revealsan unusual relationship between its radiusand mass,

Ch. 3. Living Apart Together it7

The pressure of gravitational forces isPg r ~ (GM2/R)/4nR2 ~ M 2/R4, while thepressure of tile gas is P ex: p5/3 ex: (M /R3)5/3oc.M 5/ 3/Rs. At equilibrium, both pressuresmust be equal; therefore, R ex: M-1, 3 . Itfollows that increasing the mass of a dwarfdecreases its radius and increases its density. A rise in density, however, is accompanied by a rise in the energy of the electrons. As is known, electrons in an atomseek to occupy the lowest levels and thesame happens in an electron gas. But ina degenerate electron gas all the lowestlevels are occupied, and thus are "closed"to the electron according to the Pauliexclusion principle. As the mass of a whitedwarf and, hence, its density increase, theelectrons will be "packed" more and moreclosely. Since there is no room at thebottom, the electrons will have to move toupper "shelves" where the energy is higher.Little by little, the electron energies become comparable with their rest energy mec

2•

Thus the electron gas turns into a relativistic gas which is much more compressible.Here the pressure is P ex: p4/3, i.e, 1J ex(MIR3)4/3 oc.M4j3/R 4. During compression,gas pressure grows in the same manner asthe pressure of gravitational forces. Thismeans that a white dwarf may only be inequilibrium at a certain value of mass;this critical value is approximately 1.5Mevand is called the Chandrasekhar Iimi l. It


was obtained by a twenty-year old Indianphysicist Chandrasekhar in 1931. For histheoretical study of white dwarfs he wasawarded the 1983 Nobel Prize in physics.

If a star's n18SS exceeds the Chandrasekharlimit, the electron gas pressure can nolonger balance gravity and contraction willcontinue. Independently of Chandrasekhar,the limit was obtained by the Soviet physicists Yakov Frenkel and Lev Landau.Landau wrote that a star with a mass abovethe critical limit will continue to contractuntil the nuclei come into contact makingup a single nucleus of enormous size.Landau wrote this article a year beforethe neutron was discovered. At the time,physicists did not know that protons andelectrons could fuse to form neutrons.Another year passed and American astronomers Walter Baade and Fritz Zwickyhypothesized that a supernova explosion isthe result of all ordinary star's collapsein10 a star consisting only of neutrons.Such stars were called ueu troll stars. Thedensity of these stars is close to nuclear density, i.e. 1013-1015 g/cm''. It means thesize of a neutron star with closely packedneutrons is (1015/ 1 ) 1/ 3 t imes smaller than theSun whose aver-age density is close to unity.It fnllows that the neutron star's radius isabout 10 km even though its mass exceedstha t (}f the Sun.

Neutron stars are the end result of the


evolution of stars whose initial masses aremore than 10M0. I t must be massive in orderto be able to sustain the fusion of the heavierelements after the lighter elements havebeen exhausted. In such stars everything upto iron is burnt out. Further fusion produces no more energy and, instead, starts toconsume it. Therefore, once the contractionof an iron core has begun, it cannot bestopped.

The gravitational energy produced feedsthe fusion of the heavier elements, thecontraction becomes disastrous, and collapse takes place. During the collapse, theamount of energy liberated is so great thatthe whole envelope is ejected with a velocityof several tens of thousands of kilometresper second, and this is what we observewhen we see a supernova explosion. Baadeand Zwicky's hypothesis was convincinglyconfirmed in 1968 when the Crab radiopulsar was discovered in what might be theremnants of a supernova explosion.

The radio pulsar's signals come to us asa sequence of very regular narrow pulses.The pulsar's light curve looks like an oldcomb with widely spaced teeth (Fig. 36).The teeth (pulses) may disappear but whenthey emerge they do so in accurately predictable places (moments). The equidistantteeth of a' comb are made by a machine, butwhy are a pulsars signals so very regular?The reason is that they are emitted by


a rotating neutron star. Indeed such a staris the only body that can rotate with a period of 0.033 s. Any other star would he torninto pieces by centrifugal forces.

The American astrophysicist ThomasGold first realized that radio pulsars are

L...~ .

~

~.~ ,~O 2 4 5 8

Iime,s

Fig. 36. The recorded radio signal from the pulsarPSR 0329 + 54, which was one of the first to bediscovered

neutron stars whose energy comes fromtheir rotation and whose magnetic field isa drive bel t which transfers the energy fromthe star. The magnetic field of a neutronstar, like that of the Earth, is dipolar. Inother words, there is a line passing throughthe magnetic poles. Fluxes of relativisticparticles and radiation are ejected alongthis line (Fig. 37). The pulsar "illuminates"space like a rotating searchlight. Periodically the beam strikes the Earth and we havean opportunity to receive it. However, thesignals must slow down the rotation rate,and this is exactly what is observed. Thepulsar's periods gradually increase (Fig. 38).I t is not yet clear, though, why neutron stars

Ch. 3. Living Apart Together

Fig. 37. The magnetosphere of a radio pulsar

121

0.089230tI)~...•~ 0.089220~C(

0.088210

1970 f975Years

Fig. 38. A change in the period of a radio pulsarin the constellation Vela


rotate so rapidly and possess such strongmagnetic fields.

The rapid rotation, a strong magneticfield, and an accompanying nebula are thebirthmarks of a neutron star. The onlydifference between them is that the nebulaedisperse after several tens of thousands ofyears and become invisible, while the rotation and the magnetic fields retain formany millions of years.

The pulsar's unique properties prove thatneutron stars are born when ordinary starscollapse. To be more exact, it happens totheir iron cores. This core, which grows at.the centre of a massive star, may havea mass exceeding the Chandrasekhar limit.I ts collapse is accompanied by a terrificoutburst of energy produced by gravitational forces. There is enough energy to shakeoff the star's massive envelope (and thusproduce the nebula, that is, the remnantsof the supernova), increase the magneticfield, and accelerate the rotation.

I t is not yet clear how the envelope isshaken off. As the star's iron core contracts,atomic nuclei are pressed together, thematter's neutronization sets in , and theprotons p + uni te wi th the electrons e: toform nell trons n, i. e.

p + --t- e- -+ n .+ v,

This reaction also produces neutrinos Y.

Noulri nos that take up the energy. The

Chi 3. Living Apart Together 123

density is so high that even neutrinos withtheir all-penetrating abili ty cannot finda direct way out of the star. They areabsorbed (like, for example, in reverse reactions) and give off their momentum. A PO\verful neutrino pressure emerges. Astrophysicists believe that it is this pressurethat pushes away the star's envelope. Thisview is confirmed by the existence of neutron stars.

In some cases, however, the envelope isnot shaken off. You may wonder whetherit is possible for a massive neutron star tobe formed. The answer is no. Massive neutron stars do not exist.

Black Holes

Just as white dwarfs cannot have a massexceeding the Chandrasekhar limit, neutron stars cannot be infinitely large. Thiswas first discovered by the American physicists R. Oppenheimer and G. Volkoff in1939. Unfortunately, the accurate valueof the Oppenheimer-Volkoff limit is stillunknown. because we do not know howmatter behaves at densities exceeding nuclear density. This limi.t is now estimated tobe (2-3) MG. In a more massive star, nopressure can withstand gravity. The starcol lapses and a black hole appears.

Early in the 1Uth century the greatFrench scientist Pierre Laplace used New..


ton's law of universal gravitation to inventwhat was later called black holes. Simplereasoning brought him to an astonishingconclusion.

If you want to get a freely moving bodyoff the surface of an attracting body youmust attain its escape velocity

.. /' GMv== V 2jfe

For the Earth, it is 11.2 km/s, for the Sunit is 600 km/s, If we start compressingthe body, the escape velocity will grow.At a certain radius, it will reach the velocityof light. This critical radius which is nowcalled the gravi tational radius is

2GMR g r = - 2- . (13)

c

For the Sun, R g r is equal to 3 km. A bodywith such a radius cannot shine becausethe light cannot leave its surface.

In fact, our calculations are wrong. Ifthe body's size is of the order of thegravitational radius, Newton's theory canonly give an inaccurate picture of thephenomenon. A rigorous solution of theproblem of the gravi tational field for sucha body demands that we solve the equationsof general relativity. This ","as done in1916 by the German astronomer KarlSchwarzschild, the father of Martin Schwarz-schild, who first calcula led the nuclear


stellar evolution. I t is interesting thatthe rigorous solution contains a quantityhav ing the dimensions of distance and theformula for it coincides with formula (13).That's where the similarity with Newton'stheory ends. In Schwarzschild' s rigoroussolution, something more subtle happensat the gravitational radius than mere equating the velocity of light and the escapevelocity.

You do not need to develop escape velocity to leave a body, this requirement isonJy necessary for a freely moving body.You could leave the Earth at a snail'space; what you would need is an enginethat never stops working. Similarly, youcould flee a Laplacian black hole with thehelp of, say, a ladder.

But, however hard you try, you willnever be able to escape from under thegravitational radius. No engine can helpyou out, because gravity rises to infinity.The road to a black hole is like a one-way"anisotropic" road in the novel Hard to BeGod by A. and B. Strugatski.

This is not the only paradox of relativitytheory. For a distant observer on Earth, theblack hole never forms completely. Thereason for this is the Jack of tirne. 1n a stronggravitational Held, time itself slows down.If you were to look at a clock on a collapsing star, it wi ll be slower than our (laboratory) clock. As you approach the gravita-

126 In the World of Binary

The road to a black hole resembles all "anisotroad

ci, 3. Living Apart Together 127

tional radius, the time dilation grows toinfinity. The star will contract to the sizeof the gravitational radius over an infinitelylong period of time. This is why black holesare sometimes called frozen stars.

Things would be different if we wereunlucky and found ourselves on the surfaceof a contracting star. We would reach thegravitational radius in a matter of seconds.Even then nothing terrible would happen(provided we could survive the tidal forces).But as soon as we cross the gravitationalradius, we would be hopelessly lost fromthe rest of our Universe. Our cry for rescuewould accompany us in our fall into theblack hole. Now it is easy to see why thesurface with radius R g r is called the eventhorizon.

Ancestors and Descendants

Let us make a brief summary. A star's futuredepends on its initial mass. Less massivestars (lighter than 2-3 masses of the Sun)give birth to helium white dwarfs (Fig. 39);the greater the mass, the heavier the core.(When I say "heavy", I mean both the massand the mass number of chemical elements.)Stars with masses of (8-10)AJ0 producewhite dwarfs composed of carbon, oxygen,etc. The masses of white dwarfs are less thanthe Chandrasekhar limit (about 1.5AJ0).It is interesting that in a carbon core

128 In the World of Bi

wi th a mass close to the Chanelimit the liberation of a great atenergy can completely destroy j

.4 lJl'Scl'ndanl s "

Mass,M/M 0

1.5 rv2

OJ:r:

wott»dwarfs

oI

u

oI

Nelltronstars

1 3 8 fO ""JO ocMass,M/MQ

"Ancestors "

Fig. 39. Ancestors and descend an ts

If this happens, all that remairstar is a supernova explosion andn-bula.

IrUit ceres grow in stars with mr


10M0 . When the star collapses, the resultis a neutron star. A neutron star's masscannot exceed the Oppenheimer-Volkoff limit «2-3)MG). Finally, especially massivestars may turn into black holes.

These considerations allow us to cl assifystars into massive stars (~ 10M·f=) whichproduce neutron stars and black holes, andlow mass stars which produce white dwarfs,

Such is the Iifestyle of single stars orbinary stars whose components are farapart. What happens to binaries in whichthe two components are close neighbours?Or in systems where mass exchange is possible? Evidently, the same thing, onlywith one reservation. The situation is radically different when one of the stars Hils itsRoche lobe and when mass exchange begins.The star losing its mass may become theless massive component, When the exchangestarts, however, the star already "knows"what kind of descendants it can hope for.The type of compact star is determined bythe star's initial mass, I t is called the ancestor star.

Mass exchange in a massive Liuary savesit from destruction. The exchange, however,does not happen instan taneously, whichmeans there must. be binaries in our Galaxyin which the exchange is underway now.The discovery of such systems would bethe best proof tha t the role change actuallyoccurs the way we think it does.

9-0667

Chapter 4

First Exchange

The Algol paradox indicates that starmasses in binaries can change dramatically.This does not sound surprising today. Thetheory of stellar evolution demands thatstars expand. On the other hand, celestialmechanics requires that matter leaving theRoche lobe is lost from the star for good.It is only logical that the mere combinationof these two facts means that mass exchangebetween stars is a reality.

However, the Algol paradox and our theoretical arguments do not yet prove thatmass exchange is a fact. What we need issubstantial evidence. Astronomers have nowmanaged to find undeniable evidence bycatching a star "red-handed" in the processof mass exchange.

John Goodrike Again

On September 10, 1784, John Goodrikediscovered Ii Lyrae, the second eclipsingvariable star. Like Algol, this star" is famous for its light curve. In contrast toAlgol, though, the light curve of ~ Lyraedoes not have even roughly flat parts.

Ch. 4. First Exchange 131

I t winks allover the place (Fig. 40). Between two successive eclipses its light curveis not constant, and when it attains maximaat phases 0.25 and 0.75, it falls rapidly.The flat part of the light curve correspondsto the moment we see both components of

1.0

0.0 0.5Phose

Fig. 40. The light curve of p Lyrae recorded at theTien Shan station of the Sternberg State Astronomical Institute

the binary simultaneously. The light froma system is the sum of the light or, to bemore exact, energy fluxes coming fromthe both stars. The period of variation of ~

Lyrae is 12.9 days, Le. more than fourtimes longer than that of Algol. I t seemsas if this system must be wider and its eclipses must be shorter. In fact, the stars of ~

Lyrae are both heavier and larger.

9*


Do you remember Newton's argumentwith Cassini about the shape of the Earth(see Chap. t)? The rotation of a singlestar about its ax is flat tens itat the poles.However, the revolution in a binary systempulls the stars along the line which holdsthem together; therefore, the hinary's cornponents are melon-shaped, i.e, triaxial ellipsoids. As a resul t of the orbi tal motion,

The ellipsoidal effect

the stars show different sides to the observer, which is why we see variations in thebinary's light curve. This phenomenon iscalled the ellipsoidal effect. We may obtainthis effect by examining a binary in whichone star is shaped like a melon and theother one is extremely small (Fig. 41).

For simplicity, we assume that the binary's orbital plane is inclined at i = 90°.At phase 0, we see the star from its face and


its light curve is at a minimum and theradial component of the star's velocity iszero. Then the star turns and the visiblearea gradually grows. The light curve

0.5•

0.75-

,o

JV0I ! I I I

o 0.5 1.0 Phose

~ t/~ 0 Phose

Fig. 41. A change in the light of a star caused by theellipsoidal effect

increases and reaches a maximum at phase0.25. At this moment, the star is seen fromthe side, its velocity is at a maximum andis directed away from us. In a quarter ofthe orbital period, the light curve againdecreases to a minimum. Then the whole


thing is repeated. In one period, the lightcurve rises twice to a maximum and twicefalls to a minimum. For this reason,such a light curve is called a double wave.

Note that the reflection effect results inone wave per period and a maximum of lightat phase 0.5. The ellipsoidal effect bringsabout a double wave and maxima at phases0.25 and 0.75.

Yet the model curve in Fig. 41 is different from the observed curve (cf. Fig. 40).The depths of the minima especially aredifferent. At the primary minimum, thebinary's light curve falls to a magnitude of4.2, while at the secondary minimum it fallsto 3.9. But this may easily be explained bya partial eclipse of one of the binary'scomponents. To sum up, the light curve of ~

Lyrae is a combination of two effects, eclipses and the ellipsoidal effect. The latter isgoverned by the degree to which the starsin a binary are flattened. The maximumelongation is achieved when the star fillsthe Roche lobe completely. Therefore, variations in light due to the ellipsoidal effectcannot exceed several tenths of a magnitude.

As I mentioned in Chap. 1, the shape ofa star in a binary corresponds to an equipotential surface. If a star approaches the sizeof the Roche lobe, then its shape willresemble a pear rather than a melon. Theside facing its neighbour differs in shape

Ch. 4. First Exchange f35

from the far side. This means that therewill be a difference in brightness betweenthe two sides. The star's spout (the areanear the inner Lagrangian point) is colderthan other parts of the surface. Therefore,the minima on the curve of the ellipsoidaleffect are always different. (And what aboutmaxima?)

By measuring the amplitude of the ellipsoidal effect, we can determine the extent towhich the Roche lobe is filled by a star ina binary. In other words, we can determinethe relative size of the star. An analysis ofthe light curve of ~ Lyrae shows thata brighter component is close (or, probably,equal) in size to its Roche lobe. I t is reasonable to suspect it of mass transfer to theother star. If you study the spectrum of ~

Lyrae, the suspicion will grow into confidence.

The Spectrum of ~ Lyrae

The spectra of most stars, like that of theSun, contain dark absorption lines. Whenwe were discussing the Doppler effect, wesaid these lines resul ted from the thincold stellar atmospheres. You can observethese lines in the spectrum of ~ Lyrae too.However, there are also bright lines due todifferent chemical elements. These are knownas emission lines. Like the binary'slight curve, the emission lines and ab..


sorption lines are "moving" along the spectrum with the same period of 12.9 days.I hope you can see why. What is moredifficult to explain is that the emissionJines of different elements are "moving" indifferent phases, of which there are morethan two. Y 011 get the impression thatinstead of a good old binary there is a hordeof stars, which, of course, is impossible forhow could they all rotate with the samerate.

The absorption lines, however, all are"moving" in the same phase. Characteristically, the star's radial veloci ty at theprimary minimum becomes zero and thenthe velocity becomes negative. This meansthat the star, whose lines we are observing,was moving away from us before the primary minimum; then it began moving towardsus. In other words, the star is being eclipsed. This is only natural because it is thebrighter star, whose lines must be visible,that is eclipsed at the primary minimum.Until recently, however, nobody could havedetected the lines of the second star.

Researchers determined the spectral classof the bright component of ~ Lyrae fromthe absorption lines; the component happened to be of spectral class B8. Thebright star's mass could then be determined from the mass-luminosity relation (seeFig. 32) for main-sequence stars. I t is approximately 3M0. The mass function deter-

Ch. 4. Fir~t Exchange t37

mined by the radial velocity curve wassurprisingly large, approximately 12M0.Remember the remarkable feature of themass function (see Chap. 2): the mass of thesecond star cannot be less than the massfunction. I t means the second star is mnchmore massive. There was the puzzle: thesecond star is nearl y four times the sizeof the first and yet the binary's spectrumcontains none of its lines.

Consequently, the observation databrought the researchers to stop. Two mostimportant issues had to be tackled: whatwas the origin of the emission lines and whydoesn't the more massive star give anyspectral lines?

Gravity Wind

When we discussed the measurement ofthe stars' radial velocities (see Chap. 2),we saw how absorption lines appeared ina star's spectrum. Now it is clear that if weplace cold semitransparent gas between usand a hot body giving the Planck spectrum,the latter will contain absorption linesmaking it look "chipped" (Fig. 42a; seeFig. 23).

Now imagine that we gradually heat upthe gas hiding the star. In the heatedgas, atoms collide more frequently and witha greater vigour. The atoms are mutuallyexcited as the kinetic energy turns into

138

(a)

{bJ

(e)

(d}

In the 'Vorld of Binary Stars

~t .~IA--·_····-.·~ -- ..

~:--::.-:.~---._._--_... ")-

A··-=~;=-_c-_

...._------

Fig. 42. The formation of emission lines in a star'sspectrum


the energy of the electrons. Electrons whichhave received energy are "thrown" up tothe upper energy levels. However, theycannot reside there for long and in a while(to be more exact, in about 10- 8 s) theywill fall down the potential well of theelectrostatic field of the nucleus after givingoff a quantum of light. Since the gas is notvery dense, the quantum will leave withoutdelay and cool the gas by taking up some ofits energy. We shall make up for this loss ofenergy by further heating. What will happen next?

In accordance with the quantum mechanics laws, a hot gas radiates only the samelines it absorbed from the star's radiation.It is clear now that, as the gas gets hotter,the "chips", i.e. the absorption lines, willbecome filled with radiation and becomeless deep. If we increase the incomingthermal energy and thus continue to heatthe gas, the "chips" will be completelyfilled and we get a spectrum without anylines at all (Fig. 42b).

Now we take the star away. The continuous radiation will disappear and thespectrum will turn into an irregular fenceof bright emission lines replacing the absorption lines. In this way, a transparent gasheated to a temperature which still permitsthe existence of neutral atoms (10000 K)produces a spectrum that is almost completely composed of a fence of lines. True, at


a temperature of 10000 K free electronsalso exist. A free electron, i.e. one outsidean atom, can emit a quantum of any frequency. At a temperature of 10000 K,however, the electron's energy is so smallthat, as it comes a little closer to an ion,it is captured (or recombines), and later itemits the whole of its energy only on thefrequencies of the lines. Therefore, thecontribution of the continuous spectrumwill be small (Fig. 42c).

I t is obvious that such a spectrum differs greatly from the Planck spectrum ofa blackbody. It is easy to see why. The gasunder study is optically thin. A new photonundergoes only several scatterings and absorptions before it leaves the gas. Theradiation does not have enough time to attainthe temperature of the gas, which is whyits spectrum looks like an irregular fence.

Try to imagine another thing (Fig. 42d).Suppose we put the star back to its initialposition and make the gas move with respectto the star. For instance, suppose the gas ismoving towards us. Due to the Dopplereffect, all the emission lines produced by thegas will be shifted to the violet end of thespectrum, while the absorption lines wi llstay where they are in Fig. 42a. The star'sspectrum thus has both emission and absorption lines for the same atomic transi tion ~

If the hot gas moves in a circle, there will bea periodic shift in the emission lines with

ch. 4. First Exchange t4t

regard to the absorption lines in the spectrum.

This accounts for the "moving" of thelines in the spectrum of ~ Lyrae. Only theemission lines of different elements movedifferently depending on a phase. The radialvelocity curves for the different lines variesin amplitude between 80 and 360 km/s,which is due to the presence of severalstreams of optically thin gas, each withits own temperature. The lines of the easilyexcitable elements will form in the colderstreams while the lines of the elementswhich need more energy to be excited willform in the hotter streams. I t might be thesame stream which varies in velocity andtemperature.

Where do gas streams come from? As thecurve of ~ Lyrae shows us; the brighter staris close in size to the Roche lobe or, probably, fills it completely. What we haveto do is to put two and two together. Thematerial in the brighter star near the star'sspout (the inner Lagrangian point) is captured by the gravitational field of the darkerstar. An accelerated gas stream stretchesout and rushes towards the darker star. Wecall this a gravity wind, for what is a windon the Earth if it is not a stream of gas causedby a difference in pressure. Similarly, ina binary the wind is driven by gravity,which is the kind of wind "blowing" in the~ Lyrae binary (Fig. 43).


Fig. 43. Gas streams in the ~ Lyrae system

A n exchange

ch. 4. First Exchange 143

In this way, the familiar variable starJohn Goodrike discovered at the end of the18th century has proved to astronomersthat mass transfer really happens in theworld of binaries. The Algol paradox gaveus a "hint", while the emission lines of ~

Lyrae are the substantial evidence.Another convincing example of an intense

mass exchange is given by RX Cassiopeiae,a binary, which was thoroughly studied byDmitry Martynov, one of the first Sovietastronomers to study binaries. The list ofexamples is still incomplete.

A Peculiar Device

I t is well-known that the liquid levels incommunicating vessels are the same. Thisphenomenon can be seen in a laboratory device which looks like two vessels linked bya connecting tube. For convenience, thetube should be at the bottom to make thecommunication between the vessels independent of the quantity of water pouredinto them. Suppose an eccentric or incompetent glass blower had fixed the connectingtube at the upper end of the device. Beforeyou smile, let me tell you that the devicedemonstrates how mass transfer occurs ina binary. If the vessels are different inheight, so much the better; the narrowervessel should be a little higher (Fig. 44).The connector acts as the spout near the

t44 In the World of Binary Stars

inner Lagrangian point. The upper, edges ofthe vessels are like the outer Lagrangianpoints (L 2 and L 3 , see Fig. 13). Note thatthe point L 2 is closer to the less massivestar and the level of its potential is lowerthan at the point L 3 • By pouring water

Fig. 44. An experiment illustrating mass transferin a binary

into the left-hand vessel (see Fig. 44) wesimulate the nuclear evolution of a binary.As long as the water does not reach theconnector, the vessels are disconnected andthe water levels are different. A gradualrise of the water in the left-hand vesselsimulates the expansion of the more massivestar during nuclear evolution. The secondstar, which is lighter, remains practicallythe same. As soon as the water reaches the


connector (the Roche lobe is filled), thevessels become connected and the waterlevels in each vessel begin to equalize. Ifthe incoming stream of the water does notstop, the water levels will equalize andeventually the water will fill the right-handvessel, which is lower, and splash onto thetable. In other words, material will leavethe binary system. I t is clear that, howeverperfect, two vessels are an inadequate analogue of a binary system. Mass exchange isa complex process and its gas dynamics isstill a subject for research. A laboratory device can give us a correct idea of how it allhappens, but a more detailed study willtake more time.

Suppose a star Iills its Roche lobe. If yougive a particle a push near the Lagrangianpoin t, it will sl ide in to the other star'sRoche lobe. Note that at a Lagrangian pointthe resultant force of three (not two!)forces is equal to zero. Two of the forcesare the stars' gravity and the third is centrifugal (don't forget we are inside a rotatingcoordinate system). If the particles slidingfrom the Lagrangian point could avoidmutual interaction, each would move alongits own intricate trajectory (see Fig. 10).But star material is a gas whose particlescollide. Random motions are neutralizedin collisions and disappear turning intoheat. What remains is the motion of particles from one star to another. Thus the gasJO-OfHi7

In the Wortd of Binary Stars

forms a stream, otherwise known as a "gravity wind". In order for the stream to besufficiently dense the star must overfillits Roche lobe a little [i.e. the water levelin the left-hand vessel TIllist be a Ji t tJe higherthan the connector).

What is the gas stream np to? A qualitative picture of its motion is as follows.A gas stream wi th a thickness of abou t0.1 of the radius of the outflowing star cannot find its way directly to the other star.This is easy to see if we adopt a referenceframe fixed to the second star. The ou tflowing star and the gas stream revolveabout us with a binary star's orbital period.I t follows that the material of the streamhas an angular momentum in relation to us.In other words, it strikes but misses, Theaccumulation of matter due to gravity iscalled accretion and the star on which thematter accumulates is called the accretingstar. The gas stream is turned away by theaccreting star's gravitational field. If thesecond star is far from Iilling its Roche lobe,the gas stream does not hit it directly butwraps around it in a gaseous ring or a discshaped envelope. At different distances,the rotation in the envelope occurs withdifferent angular velocities. Such a rotationis known as differential. Friction betweenneighbouring layers causes an exchangeof angular momenta. The ring turns intoa disc. The inner layers give their angular

Ch. 4. First. Exchange 147

momentum to the outer layers and so approach the accreting star. Eventually, thematter settles on the star's surface.

A Strait-Jacket for a Star

A star that gives off part of its mass isclose in size to the Roche lobe. The star'ssurface is, in fact, the surface of the Rochelobe. The si tuation , it seems, should changeshortly. A star which has lost part of itsmass and become smal ler should shrinkto maintain hydrostatic equilibrium. Thesmaller the mass of a nondegenerate star,the shorter its radius. The star adjuststo fat a new mass almost instantaneously.For example, it would take the Sun only20 min. That is when a mass transfer wouldseem to come to an end.

This is not always so. As the star losesits mass, the relation between the, massesof the binary's components changes andso does the size of the Roche lobe. Don'tforget that the star nuclear evolution whichis accorn panied by an increase in the star'sradius causes the star to fill the Roche lobe.I t is the combined effect of these threeprocesses that governs the rate and durationof matter transfer.

Will the Roche lobe size vary much? Forsimplicity, let us assume that one star'smatter is completely transferred to theother star. This siInplification allows us


to solve the problem without resorting tohigher mathematics. All we need is the lawof the conservation of the binary's orbitalmomentum because the binary does notlose mass. If a mass exchange does notcause a loss of matter by the binary, thenthe exchange is said to be conservative.

Assume that the matter is transferredfrom the more massive star to the lessmassive companion. Now suppose we transfer one gram of matter from the larger starand put it carefully on a smaller star. Thelatter has the greater orbital momentum,which makes it chief (see Chap. 1). Thistransfer reduces the average (per unit mass)orbital momentum of the less massive starmoving ita Ii ttle closer to the larger one.It follows that the mass transfer from thelarger star to the smaller one reduces thedistance between them! The binary shrinksand the Roche lobe becomes smaller too.However, a star that has lost mass willcease to be in thermal equilibrium. Itradiates energy which does not correspondto its mass. The thermal equilibrium isestablished much more slowly than hydrodynamic equilibrium but more quickly thanthe star's nuclear evolution (see formulas(7), (10), (11»).

The reduction in the size of a star is,therefore, made up for by the reductionin the size of the Roche lobe. The star losesmass for a period of the order of its thermal

Ch. 4. First. Exchange 149

timescale remaining for that time in the"strait-jacket" of the Roche lobe. For massive stars this period is about 104-105 years.The rate of mass exchange in such systemsis about 10-4-10-3M0 per year. This is tenbillion times more than the rate with whichthe Sun loses mass due to the solar wind.In contrast to the solar wind, the gravitywind looks like a hurricane.

Now suppose the less massive star fillsits Roche lobe. In this case, matter witha larger angular momentum settles on thestar with the smaller angular momentum,thus increasing the latter's average angularmomentum. The binary "spreads out", andthe Roche lobe expands. If the mass transferis too intense, the Roche lobe will tear offfrom the star and the mass exchange willcome to an end. I t will resume once thenuclear time expires and the expandingstar again fills the Roche lobe. In fact, themass transfer may not be quite so intense,and the star's size may be kept equal tothe Roche lobe size. The star again findsitself in the "strait-jacket" of the Rochelobe. However, the exchange proceeds onthe nuclear timescale rather than on thethermal timescale (see formula (10»).

The Second Puzzle of P Lyrae

P Lyrae has another puzzle. The emissionlines convinced us that a mass exchange isa reality, but it wasn't clear why the more

t50 In the \,r orld of Binary Stars

massive star doesn't show up in the binary'sspectrum? The absorption lines in thesystem belong to the less massive hot starof spectral class B8. The mass functionestimated from these lines puts a lowerlimi t on the mass of the invisible satelliteof 12At10. This is nearly four times theID8SS of the 88 star. I t is impossible forsuch a star not to show itself in the spectrum.Note that the luminosity of a regular staris approximately proportional to the cubeof its mass (see Chap. 3). With a mass ratioof 4 to 1, the luminosity ratio should be64 to 1 in favour of the more massive star!

How do we account for this? The Algolparadox brought us to the idea that a changein roles was possible. Where will the paradox of p Lyrae take us? The most radicalastronomers were prepared to believe thatthe more massive star in this system isa black hole! We don't. see it because itis black. This is absolutely wrong, andas we are going to see, a black hole in sucha binary would be much brighter than many"whi te" stars.

Another hypothesis is more prom ising.The mass-luminosity relation is applicableto main-sequence stars, i.e, stars in hydrostatic and thermal equilibrium. Is it possible that the second darker star in the~ Lyrae system has yet to "settle down"after the violent transfer of matter fromits neighbour? Before the exchange it was

Ch. 4. First. Exchange 15t

less massive and emitted little energy. Nowit is more massive and although energygeneration in the centre has accelerated weshall only learn about it a million years(the thermal time) from now. There's a darkmassive satellite for you! I t is like a hugedinosaur which tries to scratch its ear witha tail that was bitten off long ago.

After the Exchange

A star which filled the Roche lobe rapidly"loses weight". The calculations concerning

100000 tOOOOTl'mperoturf', K

Fig. 45. The evolution of stars after a great lossof mass (* shows the initial positions of the starson the main sequence; .. labels the positions ofpurely helium stars)

the evol ution of such stars were made inthe 1H60s. The results in the form of evolutionary tracks on the H-R diagram areshown in Fig. 4:), As the star loses mass,


deeper and deeper layers are exposed. Thelonger the star has lived before filling itsRoche lobe and the more massive it is, themore advanced its nuclear evolution is andthe heavier are the elements that haveburnt within it. Mass exchange in a binarystrips away the upper, unburnt hydrogenenvelope. What remains is a star with ananomalous chemical composition and consisting almost wholly of heavier elements.Mass exchange terminates when the hydrogen finishes and helium starts burning inthe star's centre. The star's radius reducesdramatically and the star is torn off fromthe Roche lobe. This is how helium starsare born.

I t is possible that the stars discoveredby the French astronomers C.J .E. Wolf andG. Rayet about 100 years ago were born inthe same way. The spectra of these starsare quite astonishing (Fig. 46). They consist of not a fence but a forest 0"1 emissionlines of helium and other heavy elements.Some of the lines are as wide as severalnanometres. This indicates a rate of expansion of several thousands of kilometres persecond.

In the early 19208, the English astronomer Carlyle Beals surmised that the properties of Wolf-Rayet stars could be explained by a powerful isotropic flow ofmatter from the star's surface (Fig. 47).The Doppler effect broadens the lines in the


spectrum. Thus the outflowing matterscreens the light from the star itself. Thereare no hydrogen lines in the spectra ofWolf-Rayet stars.

To explain why stars are different (seeChap. 2), I have already emphasized that

4-.()

~20 ""0 "50 480Wovtltl7/th, nm

ao =+c.J

2.0

1.0

~ 0.0"..a....~::~ 3.0

2.0

Fig. 46. Two typical spectra of Wolf-Rayet stars

it is their temperatures and not the chemical composition of their atmospheres thatmatters. If we keep this in mind, it must

f54 In the World of Binary Stars

mean that there are no hydrogen lines inWolf-Rayet stars because the situation isunfavourable for exciting these lines. Butit is possible that there are no hydrogenlines solely because there is no hydrogen.

A helium star of the same mass as a hydrogen star is smaller. A helium star with

Fig. 47. The surface of the Wolf-Rayet star visiblein lines is much larger than it actually is

a mass of lONl0 should have a radius of(2-3) R8 , i.e. about 3-5 times smallerthan that of a hydrogen star. How do weestimate a star's radius? Not much of itssurface can be seen, nor does its spectrumyield much for it differs too greatly fromthat of a blackbody. If a WoJf-Rayet star


is a part of an eclipsing variable s ystem,then its size can be estimated from itslight curve.

Fortunately, a number of such eclipsingvariable systems can be observed. The best-

I

1.0II

o

..~• ••, i .... :.. . ... -: .. : .~:., :. .. -. .: .~.. .... ..

0.5Phose

Fig. 48. The light curve of V 444 Cygn i

studied one is V 444 Cygni (here V standsfor "variable"). I ts light curve is shown inFig. 48.

Since a Wolf-Rayet star is wrapped intoa large semitransparent envelope, its sizecannot be determined by a light curveusing traditional methods. In the early1970s, Soviet astrophysicists used a newmethod to reconstruct the parameters of theV 444 Cygni system from its light curve.It turned out that the Wolf-Rayet star hasa radius of (2-3) R00 This means thatWolf-Rayet stars are helium stars. Anotherthree eclipsing' binaries with Wolf-Rayetstars were examined in later years and thisresearch has proved the helium hypothesis.


As heavy elements burn out, a star evolvesever more rapidly. The lifetime of abeli urn star or, in other words, the nucleartime for helium is about 100 times shorterthan that for hydrogen. The second star,however massive it gets due to a change ofroles, is still on a "hydrogen diet" and therefore cannot catch up with the first one. Ittakes a helium star inhabiting a massivebinary system several hundreds of thousandsof years to explode and thus give birth toa neutron star or a black hole. In less massive binaries, white dwarfs appear quietly.

The Story Will Be Continued

In Chap. 2, I said that the mechanism ofa role changing is like a mine provided witha delayed-action fuze. This mechanism isvery important in the evolution of a binary.Suppose there is no mass exchange at alland the more massive star stays as such.If its core collapses, practically the wholeof the binary's mass will be ejected fromthe binary because neutron stars are 10-20times lighter than the massive star. Thusthe system is doomed to the loss of morethan a half its mass and to disintegration.

Therefore, but for the change of roles, thesystem would cease to exist as a binary. Itis different when the exchange results inthe explosion of the less massive star whichcannot shake off more than a half the bi-


nary's total n18SS. This leads to a newtype of binary in which an ordinary starcoexists with a neutron star or, perhaps,a black hole. Note that any massive systemliving according a scenario with a rolechanging will end up with such an exoticcombination. How many stars of this typeare to be found in the Galaxy? Can we discover them? These questions intrigued astronomers in the 1960s. At the time, few suspected that scores of them were abovetheir heads.

How to Count Stars

Imagine you are queuing for an interestingexhibition at a museum. The line is longand is moving slowly. You are cold andbored, and having nothing better to do,you wonder how many lucky ones havealready got inside the museum. You havebeen waiting for a long time and see thatthe situation is stable. In this case, stability means that the number of peopleleaving the museum is equal to the numberof people entering it. If this assumptionwere not true, the museum would be eitherempty or overfull.

So how do you count the people insidethe museum? The simplest way would beto wait until you get inside and quicklycount them all. But it is not a good plan.First, it is technically impossible; second,


you have come to see the exhibition, not tocount people.

There is another way. You estimate theaverage number of people entering themuseum. Suppose it is 100 people per hour.(The number itself "warms you up".) Towhi le away the time, you go to the exit tofind out how much time it takes people tosee the exhibition. By averaging their replies, you find that they spent, say, fourhours. What you know now is enough toget an answer to your question. The numberyou need is the number of people per hourtimes the duration of their slay inside:

100 people per hour X 4 hours 400 people.

I t is easy to see that this solution requiresno informat.ion about the structure of themuseum. The only important assumptionwe made is that the flow of visitors hasstabilized. "Ie call this ithe steady-statecondi tion.

The same technique is widely used inastronomy. Suppose you know how manystars of a given 1118SS are born in the Galaxyper unit time. This quantity may be termedthe star birth-rate. I t is dependent on themass of the stars and is called the Sal peterfunction. Edwin Salpeter first derived thisfunction in the 1950s by the application ofstatistical methods to a great number ofobservations. For our Galaxy the numberof newly born stars N (M) with a mass

Ch, 4. First Exchange t59

exceeding M is

N (AI) =0.7 (JlI/I!ltf0)-1.3J stars per year. (14)

According to this formula, the number ofstars with a mass greater than 10Af0that appear annually is 0.0:1, Le. one starevery ao years, while the correspondingnumber for stars with a Jl18SS of more than]l,J0 is 0.7 per year. It is clear that thebirth-ra te faJ Is rapidly wi th mass,

Now it is not dillicul t to count how manystars wi th a certain mass exist in the Galaxyat one time. Assume that we are interestedin main-sequence stars with masses exceeding 10M0' The lifetime of 8 star on themain sequence depends on the nuclear timefor hydrogen (formula (10». For stars witha mass of about 10M0 this is 10 millionyears. We multiply this by the star birthrate (0.03 per year) and see that there areapproximately 300000 such stars.

I t is easy to calculate the number ofsupergiants too. They have masses in excess of 10M0 and an average lifetime100 times shorter than the nuclear time.This means that in the Galaxy there areabout 3000 O-B supergiants, i.e. supergiants of spectral classes 0 and B. \\ie have,it is true, started from the assumptionthat the situation in the Galaxy is steadystate.

Now we can find the number of binarieswi th rela tivistic stars (nell troll st ars and


black holes) in the Galaxy. Due to thechange of roles, practically every closebinary gives birth to such a binary system.How long does such a couple live? Wesurmise that it lives for as long as thenormal star is on the main sequence. Theexchange increases the mass of a normalstar from 10M0 to, on average, 20M0;therefore, we should use nuclear time for20Me, i.e. about 5 X 106 years.

Such stars are born in the Galaxy every30 years, half of them being binaries.Don't forget that a binary is a two-stararrangement, and we must half 30 years.Eventually, we conclude that a binary weare interested in appears every 50-100 years.We multiply this birth-rate by 5 X 108

and find that there must be 50-100 thousandmassive stars with relativistic componentsin the Galaxy.

Well, the story is really to be continued.

Chapter 5

Massive X-ray Binary Stars

The only possible outcome of stellar evolution is the depletion of nuclear sourcesof energy. Sooner or later, any star willeventually turn into a compact star, I.e.a white dwarf, a neutron star, or a blackhole. Massive stars, i.e. the ones whosemass exceeds 10M0, evolve into neutronstars and black holes. These compact starsmay be called relativistic. Their total energy (with their rest mass taken into account)Mc2 is comparable with gravitational energyGM2/R. Nearly all the energy in a blackhole is concentrated in the gravitationalfield, while 10-30 % of the energy in a neutron star is gravitational energy.

Al though it became clear in the 1930sthat neutron stars and black holes couldexist, no attempt was made to discoverthem experimentally during the next 2030 years. I shall now explain why the situation changed in the 19608.

"M with a Dot"

The reason is to do with the size of neutronstars and black holes. Although havingthe mass of a star, these objects have radiiof only a few dozen kilometres.

According to the Stefan-Boltzmann law,

11-0667


the luminosity of a hot body is proportionalto the area of the radiating surface andthe fourth power of its temperature. Thearea of the surface of a neutron star witha radius of 10 km is (700000/10)2, or about5 billion times less than that of the Sun.Hence, a neutron star having the same temperature 8S the Sun would be 5 billiontimes weaker. This difference in brightnesscorresponds to 24 magnitudes (see formula(3». The absolute magnitude of a star, i.e.the magnitude it would have if it were10 parsecs (32.6 light-years) away, wouldbe 29m• Note that the best telescopesavailable today are limited to magnitude 25.You would have to be very optimistic tolook for such a faint star. The situationis still worse with black holes, which notonly do not radiate anything at all, butalso absorb everything within the reach.Nobody wanted to undertake something aseffort-consuming as it was hopeless. Therewere only a few papers on neutron starsand black holes from around the worldbefore the 19608. I t is surprising they appeared at all. For example, in 1949, theSoviet astrophysicist Samuil Kaplan wrotea paper on the motion of a test particle inthe gravitational field of a black hole. Itis interesting that at the time he was working at the observatory of Lvov Universitywhose telescopes were much worse thanthe contemporary standard.

Ch. 5. Massive X-ray Binary Stars 163

The situation changed dramatically inthe 'early 1960s when quasars and the firstX-ray sources were discovered. Quasars(quasistellar objects) look in the visibleregion of the spectrum as stars, but theirluminosities are incredible. They are1013_1010 times brighter than the Sun andmany thousands of times brighter thanwhole galaxies. Galaxies radiate the energyproduced by nuclear sources in stars andare much larger than quasars. This meansthat quasars must be powered by someother, more efficient energy sources.

In 1964, the Soviet physicist Yakov Zeldovich and the American astrophysicistEdwin Salpeter working independently discovered a source of energy which is hundreds of times more effective than fusionreactions. To realize this method, you needa relativistic star whose gravitational fielddoes the work, and the surrounding matteris used 8S a fuel.

Suppose the surrounding matter fallsfreely onto the star's surface. Moving likea test particle with zero energy (see Chap. 1),this matter will fall onto the surface withthe escape velocity. The kinetic energy ofparticles per unit mass equals in this casev2/2 = GM/R. Assume that all kinetic energy turns into heat when the matter hitsthe surface and is radiated into the surrounding medium. I t is convenient tomeasure the amount of matter attracted11*

tM In the \Vorld of Binary Stars

by gravi tation (i.e, accretion) by the quantity of matter falling onto the star's surface per unit time. This quantity is otherwise known as the accretion rate, and itshows how rapidly a star's mass changesin time. The rate of change of some quantityis called a derivative. To distinguish between the derivative and the quantity,a dot is placed above the letter designatingthe quantity. (This notation and the ideaof a derivative itself is due to Isaac Newton.) Thus, the derivative of distance X

is written i: This is a speed. The rate atwhich a star either loses or gains mass is.designated M. This notation is so wide-spread that the temptation to use it in atext for the layperson is too great, but itdoes not demand any special knowledge.I t is. enough to understand that a letterwith a dot shows the rate of mass change.

So each second Ai .kg of matter falls ontothe star's surface. Each kilogram has a kinetic energy of GMIR., where R. is thestar's radius. If all this energy is radiated,the star's luminosity will be

(15)

For neutron stars, GMIR. ~ (0.1-0.3) c2.t

i.e. L ~ (0.1-0.3) Mc2• Consequently, 1030 % of the total rest energy of the fuel is

Ch. 5. Massive X-ray Binary Stars t6S

radiated. This is about 100 times morethan the fraction of the rest mass producedby the burning of hydrogen into helium.

But what happens around a black hole,which has no hard surface? The fallingmatter has nothing to hit. The answer wasfound by Zeldovich. We "make" the matter

__U..,...-Q.

'\..~v..,cAccretion ~ . ~, Vo«C------.-- -e------------------oris /l'\::s;. Do

-------....~~

Fig. 49. A gravitational "accelerator". Collisionwith a relativistic star will liberate up to 10%of the objects' rest energy

collide with itself! Imagine two particlesfalling onto a black hole symmetrically(Fig. 49). They accelerate in the hole'sgravitational field and eventually collideon a certain axis called an accretion axis.The collision must occur as close to theblack hole as possible. Then the velocityof the colliding particles will approach thevelocity of light. The collision will release1.0 % of the rest energy of the particles. Itshould be noted, however, that the collisionshould not occur too close to the black hole.The energy gain in this case will not be

t68 In the \Vorld of Binary Stars

impressing because the hole itself will soakup most of radiation.

, 11\\\lljl~~r 1111~IJj~/~1111 ~/'I{Pi

~

The di.8coverg 0/ X -rag stars

In his 1964 paper, Zeldovich stressedthat a very powerful source of energy would

Ch. 5. Massive X-ray Binary Stars f67

be a binary where the second star suppliesthe matter fuel for the gravitational machine.

A combination of the idea of accretion bya relativistic star and the concept of a binary system stimulated both theoreticaland experimental thought.

"Brighter 'Than a Thousand Suns"

All ordinary stars lose matter irrespectiveof whether they are a part of a binary system or not. The first space flights showedthat the Sun is a source of a continuous flowof charged particles travelling at severalhundreds of kilometres per second. Thisflow of plasma is called the solar wind.

The same quantity M may be used todescribe the power of this flow. The rateat which matter flows from the Sun isabout 10- 14M 0 per year, so it halves themass of the Sun over a period of 1014 years,i.e. much greater than the age of the Sun.So we conclude that the solar wind doesnot affect the Sun's evolution. Similarwinds emanate from other stars. The largerthe star, the stronger its wind. The flowrate for a blue supergiant is 10-6M 0per year, while that of a Wolf-Rayet staris approximately 10-&Me per year.

Let us determine the power of a gravi-


tational machine if its efficiency is to%:

L "" O.1,ilc2 ~ 15000L0 10 8M !If • (Hl)- 0 per year

I t is easy t.o see that if even only t/1000thof the matter in the wind flowing from ablue supergiant falls onto 8 relativistic star,the new source will flare up "brighter thana thousand suns".

Ifow can a tiny (by area) neutron starradiate SUCll great amount of energy?Obviously, only due to a high temperature.\\'e have mentioned that the area of a neutron star is 5 billion times smaller thanthat of the Sun. According to the StefanBol tzmann law, a star which is 1000 timesbrighter than the SUIl must have a temper-ature (5 X 109 X 1000 ~ 1500 higherthan the Sun's, The temperature of a IU'11

trou star must reach several dozens ofmill ions of kelvins. Photons emitted by abody with such 8 temperature have Cnl\~gyof abolll1000 eV This is in the X-ray rangeof the spectrum, and the Earth's atmosphereis opaque to quanta wi th such energy.

The atmosphere consists of neutral atoms,Neutral atoms can absorb light. only innarrow Jines or t.he spectrum, in a transitionfrorn one level to another. This is only trueun ti 1 the energy of the quantum en teringthe a tmosphere is not sufficient for ionizingthe constituent atoms, The ionization energyfor ditlerent atoms amounts to dozens of

Ch. 5. Massive X-ray Binary Stars t59

electronvolts, llowever, an X-ray quantumhas hundreds of times more energy. So asthey pass through the air, X-rays ionizeits atoms and are absorbed by it. Thatis why you must go beyond the dense layersof atmosphere to register X-rays.

In 1962, a group of American astrophysicists headed by Riccardo Giacconi put equipment. for observing X-rays into orbit aboardthe Aeroby-150 rocket. The equipment consisted of three Geiger counters capable ofdetecting photons with energies varyingfrom 1600 to 6200 eV.·

The history of X-ray astronomy and thatof atomic physics are closely interwoven,both theoretically and experimentally. TheGeiger counter is little more complicatedthan the electrometer, which helped discovercosmic rays. The operational principle ofthe Geiger counter is based on the abilityof X-ray photons to ionize gas.

Imagine 'an X-ray quantum being absorbedbetween two electrodes to which a highvoltage is applied (Fig. 50). The quantumspends its energy to ionize an atom. Thefree electron then accelerates in the electricfield and ionizes other atoms, Thus thenumber of electrons starts snowballing anda current emerges, This counter is calleda proportional gas-discharge counter. I t is

• X-rays radiated by the SUIl were observed in 1948using a rocket which climbed to an altitude of200 km.


proportional because the output current isproportional to the energy of the absorbedphoton. (The more energy a photon has, themore free electrons are produced.) The coun-

Fig. 50. A Geiger counter block diagram

ter is placed in a protective housing witha window to enable it to work in a particular direction.

As 800n as a rocket carrying equipmentis beyond the atmosphere, it starts sendingsignals showing evidence of variable X-rayradiation. The period of the X-ray radiationwas exactly the same as the period of therocket's rotation about its axis. The rotationof the rocket made it possible to determine,however roughly, the direction towards thesource. The radiation was coming from theScorpius constellation. In this way, astronomers discovered the first X-ray sourcebeyond the Solar System, namely, ScorpiusX-1 (here X comes from "X-rays"). TheX-ray radiation was so powerful that ittook astronomers some time to dare suggestthat it was, in fact, located beyond the

Ch. 5. Massive X-ray Binary Stars 17t

Solar System. If we surmise that the distance between the Earth and Scorpius X-1is 1000 light-years, then its luminosity willexceed that of the Sun by dozens of thousands of times.

In 1967, Yakov Zeldovich, Igor Novikov,and (independently) Iosif Shklovsky suggested that Scorpius X-1 is a binary systemin which a neutron star is intercepting thematter flowing from a normal star and radiates in the X-ray range. It was a hypothesisunsupported by rigorous proof.

While X-ray astronomy was taking itsfirst steps, radio astronomers ran into neutron stars (we discussed radio pulsars inChap. 3). The discovery whipped up workby theorists. In 1967-1971, a large group ofyoung Soviet astrophysicists inspired byZeldovich launched 8 wide-scale assaultagainst the problem of accreting relativisticstars.

In 1969, the formation of the spectrum ofan accreting neutron star was studied forthe first time. Neutron stars were predictedto radiate X-rays, and, as a first approximation, a neutron star without a magneticfield was considered. Radio pulsars, however, are direct evidence that neutron stars,in fact, possess magnetic fields which playa vital role in matter accretion.

The accreting matter in a binary systemis plasma, which is an excellent conductorof electricity. At the accretion rate typical


of a binary, a magnetic field affects themotion of the matter over several thousandsof kilometres, i.e, over distances severalhundreds of times greater than the size ofa neutron star itself. When affected by amagnetic field, matter moves anisotropically, which means that a neutron star willradiate anisotropically too. The star's ownrotation "rill cause the radiation to fluctuate!

At almost the same time, Zeldovich'spupil Viktory Shvartsman boldly suggestedthat X-ray pulsars should be found in closebinary systems. His argument was that ifa radio pulsar (for instance, the radio pulsar in the Crab Nehula) were placed in abinary system, the pulsar would radiate(eject) powerful fluxes of electromagneticwaves and relativistic particles (Fig. 5ia).The pressure produced by the ejected fluxeswould be so high that all the matter in thestellar wind would be "swept out" of thebinary, thus making accretion impossible.

This cannot go on for ever. The pulsarwill lose its rotational energy: and willrotate more slowly. As this goes on, itsluminosity and the pressure ejecting thestellar wind will lower. Obviously, therewill come a moment when the radio pulsarwill "fade" to a degree at which the accretingmatter attracted by gravitation will rushto the neutron star. Although the pulsar'sradiation has "faded" accretion is still im-

Chi 5. Massive X-ray Binary Stars 173

possible. The reason for this is the rapidlyrotating magnetic field of the neutron star.

Fig. 5f. Three states of a neutron star in a binarysystem: (a) an ejecting pulsar; (b) a "propeller";(c) an accreting neutron star

Like an enormous propeller, it scatters thematter and does not let it fall (Fig. 51b).This effect was later 'called the propeller


effect by Andrei Illarionov and RashidSyunyaev. Whilst repelling matter, thepulsar continues decelerating until accretionis possible (Fig. 51c). The matter rushes tothe neutron star's magnetic poles, and theresult is a terrific outburst of energy. AnX-ray pulsar powered by accretion flares up.

The scenario seemed too bold, and manyexperts were critical of it. At the same time,a young astrophysicist named Nikolai Shakura studied disc accretion on a black holeand proved that a black hole in a binarysystem must be an X-ray source. Withina year these results were all brilliantly conurmed by observations conducted fromUhuru, an American X-ray satellite.

The breakthrough made by Soviet astrophysicists prior to the launch of Uhurucorrectly explained nearly all the propertiesof the X-ray sources discovered in the 19708.Suffice it to say that all the research onaccreting neutron stars before the launchof Uhuru- was done in the USSR.

Uhuru

On December 12, 1970, Uhuru, the firstX-ray satellite was launched from Kenya.The Swahili word "uhuru" means "liberty".The satellite carried X-ray counters to record an X-ray source 10000 times as weakas Scorpius X-i. This satellite allowedhumanity to see clearly in the X-ray range.

Ch. 5. Massive X-ray Binary stars 175

Within a few years of observation, thissatellite discovered more than 300 X-raysources. The brightest X-ray pulsars turnedout to be of especial interest to the experts.As predicted by Soviet astrophysicists,some of them were fluctuating. One of thefirst pulsars to be discovered was CentaurusX-3. Its fluctuation period is 4.8 s, The

Time

Fig. 52. Recording of X-ray pulses coming fromthe pulsar Centaurus X-3 (the modulation is causedby the receiver's rota tion)

light curve of an X-ray pulsar greatlydiffers from that of a radio pulsar (Fig. 52).The light curves of radio pulsars resemblea broken comb wi th very fine pulses. Thepulses of an X-ray pulsar are less distinctand look like the edges of an accordion'sbellows. This means that the "lighthouse"beam of an X-ray pulsar is much wider thanthat of a radio pulsar.

Soon it was noticed that the pulses comingfrom the pulsar were coming at varying


frequency. I t looked as if the bellows werebeing contracted and expanded in the handsof an invisible accordion player. This

o

. .

a 0.5 f.0Orbital phose L

Fig. 53. Light and radial velocity curves of thooptical component of Centaurus X-3

"melody", which had a period of 2.087 days,was soon explained. A pulsar's radiationis a periodic process and so subject to theDoppler effect. When the pulsar approachesus. the pulse period is reduced (the bellows


are contracted) and when it recedes, theperiod becomes longer.

Thus the binary nature of the X-raypulsar was proved. Moreover, every 2.087days the X-rays disappeared. When the pul-

"Relations" between radio and X -ray pulsars

sar's X-ray light curve and radial velocitycurve were compared, there was no doubtthat the pulsar was periodically eclipsed bya second large star (Fig. 53).

X-ray equipment cannot yield accuratecoordinates of the source, only a certainarea (the error box) within which the sourceis located. Before long, a star whose lightcurve varied with a period of 2.096 dayswas found near the error box of CentaurusX-3. The minor difference could be explainedby the inaccuracies in the determinationof the optical light curve. Initially this

12-0667

Table 2. X-ray pulsars ....-J00

CharacteristicName

Fluctuation Orbital Normal time of Luminosity,period, S per iod, da y8 component acceleration. L1L0

years

A 0538-66 0.069 16.66 Be 200000

SMC X-1 0.71 3.892 BO I 1400 150000 .....=

Her X-1 1.24 1.7 HZ Her 340000 2500 :;.H 0850-42 1.8

(tl

".

4U 0115+63 3.61 24.31 B 30000 2500 -<0~

V 0332+53 4.4 34 Be 200 6:

4.84 2.087 34000

Cen X-3 06 II-III 12000 .....

IE 2259+59 6.98 0.03 2000tX'S·

7.68 5000 2000~

4U 1627-67 0.0288 KZ Tra ~

28 1553-54 9.26 30.07 (?) enit

LMC X-4 13.5 1.408 > 1000 88000 ""ttIJ

~ 28 1417-62 17.6 > 15 :G tOOOO("")

:=r-OAO 1653-40 38.2 200 ~100 ~

A 0535+26 104 1t 1 (?) 09.7 lIe 1000 5000 :::~

GX 1+4 122 M6 IIIe ·47 10000fIJ

> 15 rIJ<.4U 1230-61 191 ('D

GX 304-1 272 135(?) BO-B5 500 ~.,~

Vela X-I 283 8.965 O-B 3000 350 '<

4U 1145-61 292 187 (?) >1000 75t:I:J5·

HE N 715 >300 =tE 1145.1-614 297 >12 750 ~A 1118-41 405 Mira-type 1200 00

S"4U 1907+09 438 8.4 O-B ....

rJJ

4U 1538-52 529 3.73 BO I ~500 1200GX 301-2 696 41 A 977 ~100 2500X Per 835 580 (?) 09 1400 2.5

~

-:Ico


explanation was accepted. As the numberof observations grew, however, the difference between the X-ray and optical periodswas neither eliminated nor reduced. Theorists would have had to burn a great dealof midnight oil had it not been for the

1970 '9744.845 "

.~.

~ 4.84 .-

~~ 4.835 ~

t=::.~ 4.830-'io...Ie:::t">~ 4.825~

1878 1982 1985

4.820L--- ~

tlaservation time 1 years

Fig. 54. A change in the period of the CentaurusX-3 pulsar

Polish astronomer Wojciech Krzeminski.In 1974, he found a star which, althoughlocated not far from the error box determined by Uhuru , had exactly the sameperiod as the X-ray pulsar. The pulsar'sneighbour turned out to be a blue supergiantwith a mass of about 20Aif0. It was excellen t proof of the theory: a massive binary,a blue supergiant with a strong stellar windand a neutron star, the pulsar. All in all,about 20 X-ray pulsars, which for. the most


part are components of a binary, had beendiscovered (Table 2).

When the first X-ray pulsars were discovered, their discoverers were uncertainabout what powered them. Many were impressed by radio pulsars. Soon, however,everything became clear. In contrast toradio pulsars, the periods of X-ray pulsarswere not increasing, indeed, they were decreasing (Fig. 54). The only mechanismthat could offer an adequate explanationto the luminosity of X-ray pulsars wasaccretion.

In the Stellar Wind

The optical light curve of Krzeminski'sstar (see Fig. 53) is a double wave with anarnpli tude of abou t 0.1 m. This is caused bythe ellipsoidal effect and a weak eclipse.However, this sm al l neutron star can hardlyeclipse a supergiant. Since the neutron staritself does not actually radiate ill theoptical range, tho system IURy not be eclipsing in visible light. Sti ll there is a weakeclipse: the secondary m i nimum is deeperthan the primary one. The normal star isprobably eclipsed by :1 semitransparentaccretion flow around the neu tron star. I tappeared possible to detorrn ine many of thebinary's parameters using the visible andX-ray light curves and the radial velocitycurves found from the spectral lines of the


optical star and the change in the periodof the pulsar. The binary is schematicallyshown in Fig. 55.

The optical star is very close in size tothe Roche lobe but does not fill it. The same

Fig. 55. The binary system of the Centaurus X-3pulsar

thing occurs in some other pulsars. I t seemsthat this is a direct indication that thereis matter transfer through the inner Lagrangian point ("gravity wind"). However,this is hardly so.

Later astronomers discovered binary systems with bright X-ray pulsars in whichoptical stars were far from filling the Rochelobes. I t is probable that in all these systems the main role is played by the stellar


wind. A neutron star captures some matterfrom the stellar wind by its gravitationalfield (see Fig. 55). I have already shown thata neutron star needs only capture one thousandth or one ten thousandth of the matterflowing from a normal star to attain the

I n the stellar wind

observed luminosity of pulsars, that isseveral thousands of times greater than theluminosity of the Sun. It is possible thatin some cases the stellar wind is supplemented by a weak "gravity draught".

To see what then happens to the capturedmatter, we take an imaginary all-wavelength telescope and observe such a binaryat a growing magnification. The size ofa binary, i.e. the size of the neutron star'sorbit within the system, is of the order of10 million kilometres. The size of the nor-


mal star is of the same order. The size ofthe "sail" with which the neutron star captures the stellar wind is about 100 timessmaller but is still hundreds of thousandsof kilometres. From the side that is away fromthe wind a cone-shaped shock wave occurs.Shock waves appear when a fluid is stronglydisturbed. Imagine a body moving in agaseous medium. On the body-gas interface,the gas is slightly denser. This denser layerbegins travelling throughout the gas withthe velocity of sound. If the body moves ata supersonic velocity, it will go fasterthan the "perturbation", which is concentrated along the cone-with a vertex near thebody's surface, leaving them behind. Outsidethe cone the gas is at rest, for it does notyet "know" anything about the body. Insidethe cone the gas is entrained by the body.In other words, on the cone's walls thereoccurs a sharp increase in the velocity and,consequen tly, in the pressure of the gas. Bypassing through the cone's wall, the gasis subjected to a kind of a blow. Since theneutron star is also moving, the wake turnsalong the vector of the difference in veloci tybetween the neutron star and the stellarwind. The matter captured by the neutronstar rushes towards it. Yet it does not falldirectly onto the star. This is because ofthe angular momentum caused by theorbital motion. In some cases, the angularmomentum is so substantial that ail accre-


tion disc appears around the neutron star.In other cases, the trajectory of the accretingmatter is only a little twisted.

How an X-ray Pulsar Works

The accreting matter is actually a plasma,which conducts electricity very well. Anideal conductor can prevent external magnetic fields from penetrating. This ability

Fig. 56. A plasma's diamagnetism

is known as diamagnetism. A plasma'sdiamagnetism may be thought of in thefollowing way. Suppose an external magnetic field has penetrated the plasma(Fig. 56). A plasma contains many freeelectrons. When acted upon by a Lorentzforce, they all start rota ting in the magneticfield in the same direction. Each electroncreates its own small circular current. Byadding with its neighbour, it cancels out.The surface current is the only one that


remains. This current creates a magneticfield equal in magnitude and opposite indirection to the external field. The totalfield inside the plasma is, therefore, zero.The plasma as if pushed away the magneticlines of force.

For the accreting plasma, the neutronstar's magnetic field turns out to be "exter-

= .. : ' : ,-,:,- .: .

Fig. 57. A neutron star's magnetosphere for a caseof rad ial accretion

nal" but staying inside it! As it falls ontothe neutron star, the plasma compressesits magnetic field (Ftg. 57) which, in turn,resists this compression. The pressure of themagnetic field rapidly increases as thedistance to the neutron star decreases. Ata distance of several thousands of kilometres, the magnetic field's pressure becomes

Ch. 5. Massive X-fay Binary Stars 187

equal to the neutron star's gravity. Herethe neutron star's magnetosphere forms.

Now we shall see how matter penetratesto the neutron star's surface. I t happensdue to various hydromagnetic instabilities.The plasma cannot go through the lines offorce but it can pull them apart and squeezein. If there were no accretion disc, theplasma would accumulate on the neutronstar's magnetosphere. The magnetic fieldis like an elastic "fluid" on which another"fluid", i.e. the plasma, lies in the gravi tyof the neutron star. As the plasma cools,it becomes thicker and "heavier". This isknown as Rayleigh-Taylor instability andis what happens in a kettle or in a star'sconvective envelope. The heavy fluid (plasma) seeks to switch the roles wi th the magnetic field. The plasma pulls apart the linesof force and infiltrates into the neutronstar's magnetosphere in the form of enormous drops which can be several hundredsof kilometres in diameter. Then the dropsbreak in to Iragments, As a resul t, the electrons and ions start being affected by themagnetic field across which they cannotmove. Thus they slide onto the neutronstar's magnetic poles along the lines offorce. The plasma eventually collides withthe neutron star's hard surface at100000 km/s and is heated to a temperatureof several billions of kelvins. We neededa "hotter spot" and we got it! Here within


a small spot of less than one square kilometre, half of the X -ray pulsar's totalenergy (for there are two poles) is released,with thousands of times the luminosity ofthe Sun.

If the neutron star is surrounded witha disc, the Rayleigh-Taylor instability does

Fig. 58. A magnetic field '8 structure in the caseof disc accretion

not "work" because matter rot.ating arounda star is weightless. I t is replaced, however,by the ruagnetohydrodynamic instability.As shown in Fig. 58, a thin accretion disc"constricts" the magnetic field, which isespecially strong at the internal edge ofthe disc. The lines of force seek to straightenout, which requires a change of positionbetween the plasma and the field. As a resul tof such instabi li ty, smal 1 blobs of the


plasma enter the neutron star's magnetosphere to be broken into fractions and "frozen" into the field's lines of force. As iftransported by rail the plasma rolls ontothe neutron star's magnetic poles.

The surface of the disc is covered withwaves due to what is known as the KelvinHelmholtz instability, something we comeacross every day. It ri pples water, wavesflags, and causes sails and curtains to flutterin the breeze. The origin of this instabilityis clear. Suppose we have two fluids, onestatic, the other flowing parallel to the flatinterface between them (we shall assumethat there is no gravity). The interfaceremains the same because all the forces arein equilibrium. Now we need only bend theinterface 8 little. The pressure in the fluid atrest will not change; hence, the force actingfrom the static fluid on the interface willnot change either. However, the pressureproduced by the flowing fluid will decrease.The fin id flowing over the "hill" will beacted upon by a centrifugal force. Thus the"hill" grows, which means that the interfaceis unstable. The role of a flowing fluid ina river is played by the wind, and the heightof waves is restricted by gravity.

The neutron star's rapidly rotating magnetic field acts as the "wind" on the surfaceof an. accretion disc. The "waves" can riseto an altitude of several dozens of kilometresand plasma splashes get mixed up with the

tOO In the World of Binary Stars

magnetic field. The resulting "seascape"is one no marine painter has ever seen orpainted.

Most of the pulsar's energy is producedat the magnetic poles in a strong magneticfield of the order of 1012-1013 G. Recall thatthe Earth's magnetic field is about 1 G.The radiation in such a strong magneticfield is anisotropic, and the neutron starilluminates the Universe with two X-raybeams, like a gigantic rotating lighthouse.Periodically, a beam strikes the Earth, andthat's when we see the pulsar.

Why Neutron Stars?

Most X-ray pulsars have periods exceeding100 s. What makes us believe these areneutron stars? No normal star could rotateat such a high rate because if it did thecentrifugal forces would tear it into piecesno matter how large it is. Yet there arewhite dwarfs whose size (only several thousands of kilometres) allows them to rotatewith periods up to several seconds.

Initially it was thought that X-ray radiation could only be generated by b1ackholes and neutron stars. In the mid-1970s,however, astronomers discovered fluctuatingradiation from white dwarfs in systems ofthe type Al\tl Herculis, polars, which we shalldiscuss in the next chapter. I t follows thatneither the rotation rate nor the X-ray


radiation is characteristic only of a neutronstar.

Still it is possible to identify a neutronstar unmistakably.

Here the fact that X-ray pulsars acceleratecomes to rescue. I t was, in fact, this property that "buried" the radio pulsar modelof energy generation. I have pointed outmore than once that the matter capturedby an accreting star in a binary systempossesses an angular momentum with respect to the system. The matter acceleratesthe star's rotation by falling onto it. Thatis what accelerates pulsars. The exact valueof the accelerating impulse of the forces isdependent on many unknown parameters,e.g. the neutron star's magnetic field. Thereis, however, a restriction on the acceleratingeffect of falling matter, a restriction freeof any supposition concerning minor detailsand which is dependent only on the star'srotation period .and the accretion rate.Qualitatively, it is clear. If the matter hadan infinitely large angular momentum, itwould not fall onto the neutron star'ssurface and would not give up its momentum.The pulsar's period is known while theaccretion rate may be obtained from itsluminosity using formula (14). The smallerthe star, the higher its acceleration fora given accelerating momentum. Neutronstars are hundreds of times smaller thanwhite dwarfs. This means that the ultimate


acceleration for neutron stars is much higherthan that for white dwarfs.

The annual change in the pulsar's periodis generally used as the measure of the pulsar's rotation acceleration. When the pulsars' annual acceleration was compared withthe top limi t established for neu tron stars

Observed parameters 01pulsars

Fig. 59. The rotation acceleration of X-ray pulsarsproves that they are neutron stars

and white dwarfs, it became evident thatthe pulsars (those whose acceleration hasbeen measured) had been accelerating fasterthan white dwarfs were expected to do it(Fig. 59). Then it became clear that thesewere neu tron stars.

"Discology"

The possibility of black holes is a significantresult of Einstein's theory of relativity.Scientists are too impatient to wait until

Ch, 5. Massive X-ray Binary Stars 193

it is possible to create a black hole in thelab, so they are searching for natural ones.Where should we look for them? How do wetell a black hole from some other heavenlybody? In most cases, it is easy to distinguishit from a normal star (not always! seeChap. 7). But how to distinguish it from awhite dwarf or a neutron star? The latteris especially "unpleasant" in this sense. Thesize of a neutron star is only 2-3 gravitational radii. At these distances, a neutron starand a black hole do not differ in gravityat all. On the one hand, we know that ablack hole has an event horizon, and everything tha t is behind it is lost for good. Onthe other hand, we know that for a distantobserver on Earth a collapsing star neverfalls beneath the event horizon because ofgravity-induced time dilation.

Is itat all possible to trace a bl ack hole?The only characteristic feature of a blackhole is its mass. We know that a neutronstar cannot exceed the Oppenheimer-Volkofflimit, while a black hole's mass is unlimited. The problem is that the exact value ofthe Oppenheimer-Volkoff limit is so farunknown, Advocati diaboli argue that theIimit may he 8JltI0t but it is more probablethat it lies between 1.5M0 and 3M0'

I n the mid-1960s, Soviet astrophysicistssuggested that we should identify blackholes by mass, which may only be estimatedif the hole is a part of a binary system. If

13 -0667

194 in the World of Binary Stars

you run into a dark massive satellite, itis probably a black hole. This eliminationmethod is very time- and eflort-consuming.

Clearly, a black hole hunter has to knowa more distinguishing feature. In 1972,Shakura published a paper in which he provedthat the accretion disc around a black holemight be an X-ray source. I-Ie described thecollision of matter with itself in the discin the following way. The matter in thedisc is engaged in three motions. The firstone is a chaotic whirling known as turbulence. This occurs in rapid shifts of a fluidas happens, for example, in the turbulentwake of a motor-boat. In the disc, however,these motions are averaged. The secondmotion is the rotation of the matter asa whole around the attracting body according to Kepler's laws. A combination of thesetwo motions initiates the third. Neighbouring rings rotate with different angularvelocities and, owing to this, turbulent vortices in them collide and exchange momenta.As a result of this rubbing, the inner ringsgive their angular momenta to the outerones. Kepler's second Jaw is violated for theparticles of the inner ring and they slowlymove towards the disc's centre. A "Cassinidivision", however, does not appear becausethe centre-bound matter is replaced byanother portion of matter, which gives itsangular momentum to the ring that isfarther from the centre. The result is a


general radial motion of the matter to thecentre accompanied by an outward] y d ireeted transfer of angular momentum.

Such turbulent discs were first studied bythe German physicist J\ arl von Weizsackerin 1944, when he did research 011 the originof the Solar System, In 1964, the Sovietastrophysicist Vi taly Gorbatsk y examinedturbulent disc-shaped envelopes in binariesexchanging mass. The EngJish astrophysicist Donald Lynden-Bel l first studied discaccretion 011 a gigantic. black hole witha mass of about 106

..11f,'V.This work was the birth of the new science

0)) accretion discs (i t is called in jest discology). For a long time, the turbulence wasa stumbling block in discoJogy. We did notknow anything about the extent and natureof the turbulence, and it was impossible tomake progress without this knowledge. Turbulence is, in fact, one of the most enigmatic phenomena in modern physics. Physicists have developed general relativity theory, relativistic quantum theory, and nowthey are venturing toward a "grand unification" of all physical fields, but they havefailed to answer fully why at a certain speedturbulence emerges in a fluid.

This poor understanding of turbulencehindered the developmen t of discology.Probably, no progress would have beenmade if in 1972 Shakura had not founda way out by including our lack of know-

13*


ledge abou t turbulence into one unknownparameter, i.e, some dimensionless number a. (The disc accretion model is, therefore, sometimes called the ee-model.) In1973, Shakura and Syunyaev used this simplifying assumption to solve the disc accretion equations, which stimulated the development of discology. Since then hundreds of papers on accretion discs havebeen published every year. Today, anattempt has even been made to treat theorigin of the Solar System in the frameworkof the standard disc accretion theory.

What does an accretion disc around ablack hole look like? Matter moves to thecentre of the disc along a steep spiral path.As it approaches the black hole, both therotation frequency (strictly according toKepler's third law) and the difference inangular velocities of the neighbouring ringsin the disc increase. The greater frictioncauses higher temperatures. At a distanceof several dozens of gravitational radii, thetemperature rises to one hundred millionkelvins (see formula (12)). Nearly all theenergy of the accretion disc is released inthis zone (Fig. 60). When we rub our hands,muscular energy is released, while in theaccretion disc the work is done by the blackhole's gravity. The disc's inner boundary islocated three (and not one) gravitationalradii away. I n the case of a neutroll star,the disc is destroyed by the magnetic


field. Now we shall see what destroys a blackhole's disc so far away from it.

In fact, the disc is not destroyed at a distance of 3R ttr • From this distance on, thematter falls freely, i.e, without friction,into the black hole. I t turns out tha t circular orbi ts are not stable at distances less

Fig. 60. Accretion disc around a black hole

than three gravitational radii. At shorterdistances, the black hole's gravity growsmuch faster than 1/R2, and centrifugal forcecannot withstand it. Therefore, when mattercomes to within three gravitational radii,it is torn off from the disc and falls intothe black hole without being heated andradiating anything.

The disc's inner parts are extremely unstable. The disc disintegrates into spots orseparate hot areas which revolve around theblack hole at a terrific rate. The characteristic period of revolution is one millisecond.An outsider observing these spots appearand disappear will get the impression thatX-ray sources' periodicity is unsystematicand chaotic. This is exactly what was observed in Cygnus X-i.


Cygnus X-t and Other Black Holes

Cygnus X -1, Ji ke X -ray pulsars, is one ofthe brightest X-ray sources in the Galaxy.Unlike an X-ray pulsar, however, CygnusX-1 shows no periodic variation, instead. itfluctuates chaotically with extremely shortperiods (about 10-3 s). This variability wasobserved by American astrophysicists in the

Seconds

Fig. ()1. A recording 01" the X-ray flux from Cygnus x-i

early 1970s (Fig. nn, u seemed evidentthat such a powerful X -ray source couldonly exist in a binary system, but it wasimpossible to identify the source as a binarydue to the absence of X-ray eclipses. Wewere unfortunate because the angle of thesystem's inclination i is too small. Astronomers looking in the error box for somesuitable optical star which could he thesecond component of the binary were attracted by a variable no supergiant, namely, V 1;\57 Cyglli. Thov fOl1lHI out that


the emission lines in its spectrum were"moving" with a period of 5.6 days. Soonthe Soviet astronomer Viktor Lyuty discovered that V 1357 Cygni's light intensity

A binary with a black hole

fluctuated with the same period. This variable star's light curve is essentially a doublewave and is given in Fig. 62. It resemblesthe curve of ~ Lyrae but differs in that inV 1357 Cygni both minima are of the samedepth. This is a pure ellipsoidal effect.When the Soviet astronomers carne to this


conclusion, they surmised that the "el lipsoidality" could be explained if the secondstar's mass exceeded 5M0- You shouldbear in mind that the ellipsoidal effectmakes it possible to estimate relative"

oo

o 0.5Phase

Fig. 62. The optical light curve of V 1357 Cygni

masses. The supergiant's mass, however, wasfound from its spectrum to be about 20M0.

The radial veloci ty curve was soon refinedand it turned out that the mass functionin tha t system is

M:~ sin3 i

(Mo\M x )2 =0.2M0 ·

Substituting the supergiant's mass M"o =20M0, we find that the mi ni rnum massof the in v i si bIe co m pa n ion (a tan a ng Iei = 90°) would be 5M0.* However, the

• It is not easy to solve such a transcendentalequation. however, we can use the Newtonianmethod or successive approximations. We write theformula as M x = [0.2 (Mo -{- M x)2jlJ3, then equate


angle is much smaller because there areno X-ray eclipses. Hence, the mass of theX-ray star is between 1211f0 and 15M0.

Remember the story concerning the optical component of the X-ray pulsar Centaurus X -3? You will probably understandwhy astronomers are anxious to know whether the supergiant is simply projected ontoCygnus X-1 by chance. The situation isaggravated because Cygnus X-1 was fora long time the only worthy candidate forblack hole status. Once in a while othercandidates appeared but they either hadregularly fluctuating X-ray radiation ortheir optical identification was wrong.

In 1983, astronomers discovered anotherprobable black hole, LMC X-3. It was foundin a neighbouring galaxy, namely, theLarge Magellanic Cloud (LMC). It was theUhuru satellite that recorded this X-raysource. The LMC X-3 source is extremelyunstable, it "switches on" and "switchesoff". Like in Cygnus X-1, no X-ray eclipseswere discovered. In 1983, American astronomers noticed radial velocity variations-in the probable optical component, a B3star, wi th a period of 1. 7 days. The semi-

It'Jx on the right-hand side to zero, find Mx on theleft-hand side using a calculator, and substitutethe quantity back on the right-hand side. By doingthis repeatedly, you will get an answer to theaccuracy of the number of decimal digits in thecalculator.


amplitude of the radial velocity variationsis about 235 km/s, which corresponds toa mass function of 2.3M0. If we take intoaccount that the optical star mass is noless than 3Mev , the invisible component'smass must be no less than (3-4)M0' whichalready exceeds the probable Oppenheimer-Volkoff limit.

True, there may be mistakes in theoptical identification. Although out of theten candidate stars in the error box of theX-ray source, only the B3 star belongs tothe Large Magellanic Cloud, there is noindication that the X-ray source itself belongs to the LMC. The research is continuing.

Back to the Scenario

In the early 1970s, the Soviet astrophysicists Alexander Tutukov and Lev Yungelsonand a Dutch astrophysicist Edwin van denHenvel working independently associatedmassive X-ray binaries with a special stagein the evolution of close binaries. Immediately after the formation of a relativisticstar in a binary system, the normal star isstill on the main sequence (Fig. 63a). Thestellar wind of the normal star is stillweak, and so the accretion rate onto therelativistic star is low and its X-ray radiation is insignificant. When the norma) starleaves the main sequence to become a blue


giant, the intensity of its stellar wind risesdramatically to about 10-6Jf0 per year.This marks the birth of a powerful X-ray

(a)

(bJ

(cJ

Fig. 63. Three stages in a binary's evolution afterthe first explosion

source with a luminosity many thousandsof times that of the Sun (Fig. 63b).

This is consistent with observations ofX-ray systems. Many components of X-raysources are stars that have left the mainsequence.

There ex ist , however, massive binariesill which the normal star is not a supergiantand still t.he X-rHY source is very bright.These belong to the type of the pulsar

204 In the 'Vorld of Binary Stars

A 0535 + 26. (Here A stands for Ariel,the European satelli te which discoveredthe X-ray source; then come the source'scoordinates on the celestial sphere: ex =05h35m ~ = +26°.) Still the opticalstars in these systems are extraordinary.They are all rapidly rotating Be stars.*The matter flow from these stars is verynonstationary and anisotropic. Inhomogeneous blobs of matter are torn off fromthe equator of the rapidly rotating star andcaptured by the relativistic star's gravity.These sources "work" only from time totime, which accounts for their name "transient".

Today astronomers have not yet foundthe place of these stars in the evolution ofbinaries. I t is possible that Be stars acquiretheir rapid rotation as a resul t of the firstmass exchange, during which the normalstar receiving the matter is accelerated inthe same way as an X-ray pulsar. Followingcollapse, the relativistic star "takes back"the matter it "lent" its partner. This mayoccur in the form of a stellar wind or astream flowing from the equator.

However, there might be another solutionto this problem and this is the second massexchange through the inner Lagrangianpoint. Sooner or later, the second star must

• Here Be stands for a B class star with emissionlines.

Ch, 5. Massive X-ray Binary Stars 205

(17)

fill its Roche lobe too (Fig. 63c). The mattertransfer will occur on the thermal timescale(the matter will flow Irom the larger starto the smaller one). The rate of accretiononto the relativistic star will be (see formula (11))

• ( M )3M ~ 3 X 10-8 M M0 per year.o

If optical star's mass is M = 20M0 , themass-flow rate will reach 10-'M0 per year.If all this matter falls onto the relativisticstar, then, according to formula (16), thestar must become a source with a luminosity a billion times that of the Sun. I onlywant to say here that this is impossible, butI leave the details to the next chapter.

Now I would like to draw your attentionto another factor. In a moderately massivebinary the mass-flow rate is not high, andthe relativistic star's luminosity will beless than 10000 that of the Sun. I t followsthat in such binaries an X-ray source flaresup as soon as the normal star fills the Rochelobe and the second mass exchange begins.This is, probably, the only chance for a moderately massive system to become abright X-ray source. A low mass star's stellar wind is thousands of times weaker thanthat of a supergiant. Strong accretion fromthe stellar wind as observed in massive Xray system is only possible in systems witha red giant (giant by size).

Chapter 6

Dwarf Binary Stars

I have already said (see Chap. 3) that neutron stars have massive ancestors, and thatstars wi th masses less than 10N10 turn i ntowhite dwarfs. Given this fact and given theconservative scenario (there is no loss ofmass from the binary's during the firstexchange), one might say that a low massbinary cannot evolve into a neutron star.In nature, however, there is no such rule,and neutron stars are often found in JO\V

mass binaries, as white dwarfs are.Looking back at the late 1960s, the study

of variable stars seemed a wonderland. Articles, books, and catalogues swarmed withthe types of variables whose diversity terrified the theorist. There were novae, nOV8

like stars, recurrent novae, dwarf novae,flare stars, cataclysmic stars, eruptive stars,etc. Some stars were known simply as irregular variables, while some stars of thesame type were often called different things.Any attempt at classification seemed hopeless.

Today we know why many stars behavestrangely. We do not know everything butwe know a lot.

Ch. 6. Dwarf Binary Stars 207

Pussy, Pussy, Hercules X-I

The X-ray pulsar Hercules X-1 is one ofthe best-studied X-ray sources. I ts radiation is periodic hut has three different periods. This remarkable property has earnedit the name a wonder clock. First, it pulsates with a period of 1.24 s. The X-raylight curve with this period is given inFig. 64. Second, it has a period of 1. 7days. This light curve has one Il-shapedminimum.

Physiologists say that cats sleep for18 hours a day. In other words, they areasleep for 75% of their lives. Hercules X-1is like a cat. Every 35 days, Hercules is at"work" for 11 days, and for the next 24days its X-rays do not hit the Earth at all.This is the third period.

The shortest period of 1.24 seconds is theperiod of the neutron star's rotation aboutits axis. The star's pulsation light curvein this case has t\VO maxima.

The period of 1. 7 days is, undoubtedly,the binary's orbital period. The short period of 1.24 s changes with the same perioddue to the Doppler effect. The 1.7-day changein the X-ray curve is the result of theeclipse of the X-ray source by normal star.I t is interesting that the eclipse begins veryabruptly. Recall that the duration of thedecreasing part in a light curve is proportional to the size of the eclipsed star (see

208

150


1.24s

~f50~~~a. 100.....:::s~ 50.~

~I

~

150

fOO

1.7days

...... .

35 days

51)

o 0.5Phose

Fig. 64. Three periods of the X-radiation fromHercules X-1


Fig. 21). So here we deal with a star that isactually a point object.

The star that periodically obscures theX-ray pulsar was found by the Soviet astronomer Nikolai Kurochkin. Stars in X-raydetectors' error box included the variablestar HZ Herculis of magnitude 12. Priorto the discovery of Hercules X-1 it had beenlisted as an irregular variable, whichmeans that its light changed chaotically. Apparently, no astronomer had ever lookedat its variability attentively. When theold plates stored at the P.K. Sternberg State Astronomical Institute were processed,it became clear that the light from HZHerculis changes smoothly with the Xray period of 1.7 days (Fig. 65). I t was revealed that the visible light of HZ Herculischanges 13-fold (about 2.8m) , and duringthe X-ray eclipse, the system's opticalbrightness falls to a minimum. As inp Lyrae, the light curve has no flat partsbut, in contrast to ~ Lyrae, it is not a double wave. I t is rather a "single" wave with amaximum at phase 0.5. You probably remember that we have already COOle acrossthis phenomenon in another classical star,namely, Algol, whose light curve tends torise slightly by phase 0.5. Now there wasno doubt that the whole light curve of HZHerculis is the result of the reflection effect.

In 1967, Shklovsky emphasized that avery strong reflection effect must be inherent

14-0667

210 Tn the World of Binary stars

in X-ray binary systems. This phenomenonis

p=l7.75

• - '1972x - 1944 - 1950

An X-roy curve (a schematic view)

o 0.25 0.5 0.75 co 9'IIE 1.7davs )IIOrbital phase(t=t.7days)

Fig. 65. A change in the brightness of HZ Herculis

the only difference being that in an X-raybinary the star is warmed up by intercepting the pulsar's hard X-radiation (Fig.

Ch. 6. Dwarf Binary Stars 2'lt

6.6). The X-ray luminosity of the pulsar Hereules X-1 is about 100 times greater thanthe total optical luminosity of the normalstar HZ Herculis. Intercepting only part

Fig. BtL A normal star's interception of X-rad iation

of the X-ray radiation, the side of HZ Herculis which faces the pulsar becomes sixtimes as bright as its opposite side.

The star's spectrum confirms this conclusion. At the minimum, HZ Herculis is astar of spectral class F and at the maximumof class AD. According to the mass-luminosity relation, the optical star's mass is(2-2.5) M0. The duration of the X-ray eclipse allows us to estimate the size of HZJlerculis, which is very close to the Rochelobe. This system is, probably, undergoing

14*


the second mass exchange: the normal starhas filled its Roche lobe and matter is flowing through the inner Lagrangian pointin the form of a gravity wind. The gas streamforms an accretion disc around the neutron star. An observer near the accretiondisc would see the following sight. At the"horizon", there is the optical component,i.e. the star HZ Herculis and a stream ofgas flowing from it. As the disc approachesthe neutron star, magnetic forces start bending its surface, which is covered with "ripple" due to Kelvin-Helmholtz instability.Eventually, the disc is broken up by themagnetic forces at a distance of severalthousands of kilometres. The matter isthen "frozen" into the magnetic field andflows down onto the neutron star's magnetic poles, which produce two X-ray beams.

I t is not yet clear, though, why HerculesX-1 is as sleepy as a cat. By the way, thefact that for us on Earth the radiation disappears for 24 days every 35 days does notindicate that the pulsar takes a 24-day holiday. Even when the pulsar is "off", its optical light curve remains the same, that is,the reflection effect does not disappear. Something seems to be obscuring the neutronstar from us, but this 'something cannot screenall the X-radiation falling onto its companion, HZ Herculis. The accretion discaround the neu tron star, probably, serves as ascreen. Since the binary's orbital plane lies


actually in the line of sight, the disc whichwobbles slightly with respect to this planeblocks the pulsar's radiation from us for

Hercules X-1

some time. The accretion disc's shadow onthe companion star is not large and has virtually no impact on the reflection effect.

The HZ Herculis-Hercules X-1 systemis a low mass binary. You may ask how aneutron star, whose ancestor must have had amass of more than 10M0, could have evolvedin such a low mass system. A probableanswer was offered by van den Heuvel in1977 (Fig. 67). In a nutshell, he suggestedthat the first mass exchange was not conservative.

The reason for the system's great loss ofmass lies in the large initial difference between the star masses. Note that the outflowof matter from a more massive componentto a less massive one occurs on a thermal


timescale. Because of the large difference inmass, the less massive star's thermal timeis tens and hundreds of times larger thanthat of its more massive companion. The

15He

(a) t-O, P-5days

(b) t-8.'p106yeQrs,p-5days

... -- ...... ,

t.Jl'1e ~2.5H0 ') (dJ t-10 7ytors, P-5. 7doysNeutron star'........ - .... .-.~

1.3H(J)~ 2.5110 (e) t-5",f08Yf Ors,P- t i days

Fig. 67. The evolution scenario in which a massivebinary evolves to become a system like HZ Herculis-Hercules X-1

accreting matter does not have enough timeto attain thermal equilibrium and settleon the star. Let us assume that such a rapidmass transfer results in the formation of acommon envelope with mass flowing from


the binary. If so, what is happening is nota mass exchange but a total loss of massfrom the system. The low mass star remainsalmost unchanged, while its more massivepartner collapses into a neutron star. Thesystem's total mass does not exceed(3-4)M0When the envelope is pushed away, the binary loses both material and angular momentum, i.e. the stars move closer to each other.

Finally, due to its nuclear evolution, theless massive optical component will fillits Roche lobe and thus initiate the second exchange, in which abright X-ray pulsar will flare. According to van den Heuvel, this is how a system such as HZ Herculis-Hercules X-1 appears.

Today we know of several systems likeHZ Herculis. They are not very numerousbecause when first formed the masses of abinary's stars are about the same. It isvery seldom that a binary is formed whosecomponents differ very greatly in mass.Van den Heuvel's scenario shows how aneutron star can be "smuggled" into a lowmass binary system, while a binary's legitimate child should have been a whitedwarf.

Novae and Recurrent Novae

The outburst of a nova is as impressive asa supernova explosion. A bright star maybe visible with the naked eye where sever-

'1142" • ~.. ': Fig. 68. The envelopeeiec~ed by a nova ex-p osion

f94~ •f944 II.

'945 ,f848

f95Q •fIJ5f ~

\'956 ef867 •1968 •


al days before there was nothing. This isexciting for someone used to an invariablecelestial sphere occasionally crossed byartificial satellites. This interesting physicalphenomenon has always fascinated stargazers and inspired poets. When the novaV 1500 Cygni began shining in the sky,astronomers looked through old plates todiscover that prior to the nova explosionthere had been a star 10-12 magnitudes weaker than the nova. For some reason, the starhad suddenly increased its luminosity bytens of thousands of times. Note that theexplosion of a supernova increases its energy output as much as a billionfold. Novaeare less spectacular, but seem as bright assupernovae because they are much closerto us.

A nova reaches peak intensity and thenslowly fades to its original brightnessover several months. Some years pass, andin the place of the explosion a vague spotappears, which indicates that the nova'sexplosion was accompanied by the ejectionof matter into space (Fig. 68).

Long ago astronomers noticed that therewere stars which exploded more than once.True, their explosions were weaker thantrue novae. Table 3 is taken from a book bythe well-known Soviet expert in variablestars Vladimir Tsesevich and it lists thecharacteristics of nova explosions. The explosions of recurrent novae are much weak-

218

Table 3. Novae


NameYear Magni- Orbital

ot dis- tude at period,coverr a::;'1- days

Luminositybefore andafter theexplosion,

L/L0

GK Persei 1901 0.2 0.685 1.2 251000DN Geminorum 1912 3.6 ? 3.3 100000V 603 Aquilae 1918 -·1.1 0.138 8.3 480000V 476 Cygni 1920 2.0 ? 1.6 690000RR Pictoris 1925 1.2 0.145 1.3 91200DQ Herculis 1934 1.3 0.194 0.06 '17400CP Lacertae 1936 2.1 ? 2.8 525000CP Puppis 1942 0.5 ? 0.06 440000V 1500 eygni 1975 1.8 0.14 250000

er than those of the novae and occur decades apart.

DQ Herculis (Nova I-Ierculis 1934) is oneof the best-studied novae and it exploded in1934. At its peak intensity it was one of thebrightest stars in the northern sky. Oldplates showed a star of magnitude 15 wherethe nova appeared. After its peak intensity the star faded, then suddenly increasedits brightness for 100 days and finallyfaded again to magnitude 15 (Fig. 69).

In 1954, the American astronomer MerleWalker made a discovery which shedmore light on the nature of novae. A thorough study of DQ Herculis showed thatthe star fluctuated in brightness with a period of 4 h 38 min (0.194 day). Its light


curve proved that DQ Herculis is an eclipsing variable star (Fig. 70) and its spectrumcontained bright emission lines of hydrogenand helium. I ts light curve and radial velocity curve indicated to astronomers that

~ 7~~ 9.~~ If

13

2;00o 40 120Time,days

Fig. 69. A change in brightness of DQ Herculis

the binary's total mass is smaller than themass of the Sun! This is a low mass systemcomposed of a cold red dwarf and a compact.hot star. The radiation lines were "moving" in the spectrum with the orbital period but did not coincide in phase withany star. This proved tha t there were largegaseous streams in the system.

In 1956, Walker made another importantdiscovery. It turned out that the system'slight "oscillated" by several percent witha definite periodicity (Fig. 71), repeating


J~5 ..... .-~.j, ( ..I

4.1J I !

4,4 ! j~ ,~

~I: i'-

~ "'.8 t :~ 12.08. f956 ..

~: I

5.2 .-':

0.8 0. 0.2 a4 0.6 0.8 0 0.2 0.4,. 417 38 mill .1IJr6ital phost'

Fig_ 70. The light curve of DQ Herculis

Time

Fig. 71. The oscillation in the light from DQHerculis with a period of 71 s discovered by Walkerin 1956 (before the discovery of radio pulsars)

Ch. 6. Dwarf Binary Stars 22t

itself every 71 s. Thus the first pulsar hadbeen discovered. I t was found for a whitedwarf and not for a neutron star!

In order to rotate so fast, the star had tobe either a white dwarf or a neutron star.The star could not be neutron for a numberof reasons. The pulsar's mass in this systemis much smaller than the Chandrasekharlimit. Yet while I say that the neutron starhypothesis is unlikely it is not impossible.In 1937, Landau proved that the neutronstar's smallest possible mass is much lessthan the Chandrasekhar limit. True, it isenergetically more favourable for a lowmass star to be a white dwarf than a neutron star, because the latter needs "lookingafter" and has to be "prepared" under special conditions. However, a whole set ofobservational data indicates that the staris a white dwarf.

The gaseous streams in a binary occurbecause the red dwarf star fills its Rochelobe. What is happening here is very muchlike what we have seen in the HZ Herculissystem, the only difference being that theneutron star is replaced by a white dwarf.

A white dwarf is about 1000 times thesize of a neutron star, which makes the accretion mechanism 1000 times less effective. When accreting onto a white dwarf,only 0.01 % of the rest mass, not 10 %,is released, i.e. less than in fusion reactions. Having no other energy source, the


white dwarf has to rely on accretion. By theway, these figures show that for the sameaccretion rate a neutron star must be1000 times brighter than a white dwarf.This agrees with what we observe: the luminosity of systems of the DQ Herculis typeis several times the luminosity of the SunOf, in other words. about 1000 times weaker than that of the HZ Herculis--Hercules X -1 and similar systems.

Now let us see what causes a nova or arecurrent nova to explode. You know thatwhite dwarfs are formed from not very massive stars and that the maximum temperature at the star's centre depends on the star'smass. There is a moment when there are nomore elements at the star's centre to support fusion at a given temperature and density. The isothermal nucleus contracts tobecome a white dwarf in whose interior nuclear reactions are impossible.

Nothing happens to a single white dwarfother than it gradually cools. Things aredifferent when it has a partner. The companion star flows onto the white dwarf andadds fresh fuel ready for burning. On theother hand, the accreting material becomeshotter because it hits the white dwarf's surface. For fusion to begin, it is necessarythat the density and temperature of thematter exceed certain critical values. Thetemperature in the inflowed layer varieswi th the inflow of energy. As a first ap-


proximation, we assume that the energy inflow is constant: and, therefore, the temperature is constant too. As for the density,it grows slowly with increasing the layer'smass. When the layer reaches a criticalmass, it explodes. Matter is ejected intospace at a very high speed. As the areagrows, the luminosity increases, and we observe a nova explosion.

A nova differs from a recurrent nova,probably, in that helium explodes in theformer and hydrogen explodes in the latter. The burning of helium requires a higher density and, consequently, more timeto explode. The explosion, however, is farmore powerful due to the greater amount ofmatter involved.

The idea of nuclear burning in a thinlayer on a white dwarf was discussed byscientists as far back as the 1940s, but itwas in the last decade that the relationshipbetween this phenomenon and accretionin close binaries was established. Roughly,we can estimate how often novae explodebecause it is assumed that they are also recurrent.

An analysis of nova explosions shows thatevery explosion ejects M total = 10-6M0.

If we assume that the accretion rate is

standard, i.e. Ai = to-lOMe per year,then the mass ejected in an explosion will

be replaced in about t ~ M/M = 105

years. On the other hand, hydrogen explodes


under milder conditions, i.e. more frequently.

Clearly, these are only estimates and ourunderstanding of what is happening in reality is still vague. For instance, white dwarfshave strong magnetic fields. We know thatDQ Herculis is a pulsar and its magneticfield (about 105-106 G) is strong enoughto make matter fall onto the poles. Thearea on which the matter falls, as well asthe temperature, depend on the magneticfield strength and the accretion rate. Thewhole picture becomes more complicatedbecause the material outflowing in different binaries has different chemical compositions, and dwarfs are not identical either.This is where the apparent diversity innovae and recurrent novae comes from. Thebehaviour between explosions is determined by the nature of the accretion, whichis also extremely varied.

U Geminorum Stars

In military terms, we could say that novaeand recurrent novae are like heavy artillery,while explosions of U Geminorum starsare like machinegun bursts. The explosionsare only several months apart and there isno fixed periodici ty, al though there is definitely repetition. Table 4 lists the characteristics of some U Geminorum stars. Allin all, we know of over 300 such stars.

Ch, 6. Dwarf Binary Stars

Table 4. U Geminorum Stars

225

Name

U GeminorumSS AurigaeI4~X Hydrae~S CygniRU PegasiZ Chamaeleoutis

Cycle Orbitalduration, dny~ period, days

103 0.17757 O.1BO

558 0.06850 0.27668 0.371

104 0.0745

The Polish astronomer Krzeminski (whocorrectly identified the X-ray pulsar Centaurus X-~i) calculated that the U Geminorumstar's light between explosions changedwith a. period of 4 h 14 min (0.177 day).The light curve is unstable (Fig. 72) andchanges shape, However, it reproduces itstypical shape partially or completely Iromone cycle to another.

We have Hot yet come across Stich an extraordinary light curve. We can "straightenit out" using the radial velocity curve. Atthe primary minimum (\f> = 0.5), the radial velocity is zero, which means that thelines are associated with the eclipse of ahot and bright area. The light changes aresketched in Fig. 73.

The light curve is peculiar because theminimum is not symmetric. At first, thestar gradually enters the minimum, At 8

certain phase, the light curve abruptly falls,

t 5-0667


indicating a rapid entry into the eclipse.You get the impression that the bright areais first obscured by one body and then bysome other. It is interesting that the shape

fa}

1.0o

N(JvemlJer January MarchIJl'(.fmIJer feIJrlJory

Time.(b)

1'.1.mt! E l .(J :~. xp aston.... -- .:.\~ .

~ fJ:~

~ 14.0 .,,::11 •• Calm state';f4.5 :~ .:. .

a..;: .... ,.\- e.L,~ 15.0 ~ • ,,~,-..-

Fig. 72. Achange in brightness of the U Geminoru mstar: (a) long-term change; (b) orbital change

of the eclipse's initial stage depends on thewavelength of observation. Only one explanation seems possible, that is, the start ofthe minimum (section 1.. 2) is an eclipseby a semitransparent body, the next stage

ell. 6. Dwarl Binary Stars

0.25I

DiS~.~~. Hot spot ~1)- -0.5 ~

~.~

~

227

t0.75" o fJ.5

Phl1SI'1.0

.7-

0.25I

r ,

I~Q 0.5 1.0

Fig. 73. A synthesis of the light curve of a U Geminorum star. Top, the light curve of the hot spoton the disc. For half a period, the spot is seenfrom phase 0.12 to 0.62,· while for the remainderof the time it is obscured by the accretion disc.In the middle, you see the light curve changecaused by the eclipse by the companion star;a minimum appears at phase 0.5. Finally, at thebottom, an eclipse by the semitransparent gasstream is taken into account; it reduces the brightness (section 1-2)

15*


(sections 2-3, 3-4) is an eclipse by anopaque body (see Fig. 73).

This can easily he explained in the following way. Imagine a system of two starsin which one star is surrounded by a disc.On the disc there is a bright hot spot, slightly to one side of the line connecting thestars. The spot is so bright that it actuallyaccounts for the brightness of the entire system. If we had only this spot and the disc,the light curve would be bell-shaped with atop to the left of phase 0.5. At the maximum, we would see the whole spot (frontview). The spot, however, is a little awayfrom the line connecting the stars; therefore,the phase is also a little shifted. Now weshall add the companion star. Now the eclipse will appear exactly at phase 0.5. Thispicture needs a final touch. Let the darkstar release a flow of semitransparent gastoward the disc. This will cause a sharpbreak (section 1-2). First the spot is obscured by the flow, then by the star. This explains why the minimum is not symmetric.

The spectra analysis shows that U Geminorum stars contain a red dwarf whosemass is several tenths of the mass of theSun, and a white dwarf surrounded by theaccretion disc. For some unknown reason,the accretion disc and the white dwarf, unlike DQ Herculis stars, emit almost nothing.


The origin of the bright spot is fairlyclear. The gas flowing from the red dwarf,which fills its Roche lobe, does not "wind"around the disc, but hits it at an angle.The angle of attack may be quite large, andthe result may be a shock wave at the collision site, which is a narrow area where thespeed of the gas decreases and its temperature increases dramatically. The stream'skinetic energy is converted into potentialenergy. .

It is easy to estimate the luminosity ofsuch a spot. If the speed of the stream relative to the disc is v, the kinetic energy of thegas unit mass in the stream is v2/2. On impact with the disc, nearly all the energyturns into heat and light. Therefore, eachkilogram of gas in the spot radiates approx-

imately v2/2 J. The stream carries AIkilograms per second, so the spot's totalluminosity is

• v2

Lspot = M T. (18)

In such dwarf systems, the matter in thestream may be moving' at several hundredsof kilometres per second, and the flow ratemay be 10-8-10-7M 0 per year. The spot'sluminosity would be comparable with thatof the Sun. The stars themselves in thesesystems glow tens and hundreds of timesmore feebly, so it is small wonder that thespot outshines the stars.


We are not certain why a white dwarfsurrounded by an accretion disc doesn'tradiate any energy between explosions.It could be that between the explosions thereis no accretion because in the discsof such systems there is no friction and theygradually increase in mass without accreting. When the disc density reaches a certainvalue, turbulence abruptly increases,and the accretion on the white .dwarf begins, which we describe as an explosion.The disc "flows" onto the white dwarf, friction disappears again, and the whole process is repeated. If so, why? Today thisquestion is unanswered.

Can we suggest that the absence of accretion between explosions is explained by thepropeller effect? If the white dwarf rotatesrapidly, its magnetic field will impede thefall of the matter on its surface. In thiscase, the matter will accumulate in the disc,the pressure in it will rise, and the disc'sinner boundary will move slowly towardthe white dwarf. The speed of the magneticfield's lines of force will decrease as theymove closer to the whi te dwarf because thelines of force rotate like a rigid body togetherwith the white dwarf. At some moment,the propeller effect disappears and the matter falls onto the white dwarf's surface.The result is an explosion. Then everythingrepeats again.

Indeed, calculations show that a white

Ch. 6. Dwarf Binary Stars 23t

dwarf in such a system must rotate at avery high rate, i.e. with a period from severalseconds to several hundreds of seconds.

Generally speaking, U Geminorum systems have another kind of explosion,which have a much greater output but aremuch shorter. These may be caused by athermonuclear explosion of the matter richin hydrogen and flowing onto the whitedwarf's surface.

In one star in this category, namely, SSCygni, astronomers have recorded an outburst of X-ray radiation pulsating with aperiod of 8 s. This is exactly the periodthe white dwarf should have in such asystem.

Polars

Another class of dwarf binaries called polars was discovered in the mid-1970s. Thefirst polar to be found was AM Herculis,the third remarkable star from the constellation Hercules after HZ and DQ.

This star has taught astronomers manylessons. The main one was that an accreting white dwarf might produce a hardX-radiation. True,. American astrophysicists first noticed in 1967 that the accretionon white dwarfs might be accompanied by aconsiderable Xvradiation, but it was takenfor granted for a long time that white dwarfscould not radiate X-rays. This belief was


refuted in 1976 when the American satelliteSAS-3 (Small Astronomical Satellite) discovered oscillating X-ray radiation witha period of 0.129 day (about three hours)emanating from AM Herculis. Opticallight from this star also varies between the12th and 13th magnitudes with the sameperiod.

Emission lines in the star's spectrumshowed evidence of strong gaseous streams,and the radial velocity's semiamplitudewas 240 km/s. Such a high speed as due tothe system's small size and not its greatmass. And the period is only :3 hours!

I shall give you an example Illustratingthe situation in this system. If we were toput an artificial probe into low solar orbit,it would go around the Sun in 2 h 40 min,The mass of the stars in AM Herculis isless than that of the Sun, and the binary'ssize turns out to be less than the Sun's radius.

The most surprising thing about thissystem was that the visible light emittedby AM Herculis was very polarized. Thisis why the AM Herculis stars are known aspolars.

I would like to remind YOU that what wecall visible ligh t is a set ~f electromagneticwaves. Each wave is composed of crossedoscillating magnetic and electric fields(Fig. 74). In each wave, the electric (or magnetic) vector's oscillation plane is fixed


and does not change. The combination ofwaves in which the oscillation planes ofthe electric vectors do not coincide iscalled unpolarized light. In contrast to this,light ;8 polarized if there is a plane inwhich most waves oscillate. Ordinary substances, like glass, allow light to travel

£

B

Fig. 74. In an electromagnetic wave, the electricvector E oscillates in a plane perpendicular tothe oscillation plane of the magnetic inductionv~ctor B

freely irrespective of its polarization. Innature, however, there are light-polarizingmaterials (polarizers) which only allowlight to pass if its polarization plane isorien ted in a certain way. This is becausethe crystals which make lip pol arizers havesome preferred directions (faces).

To illustrate how a polarizer works and'what polurization is, I shall give you anexample from everyday life. If yOH wantto enter the Moscow underground, you mustdo at least two things: take a five-kopeckcoi n ou t of your pocket and align it wi th aslot in the coin changer. Before yOH enterthe underground, the coins in your pocket


are unpolarized, i.e. they are oriented atrandom. The coin changer functions as apolarizer polarizing your coins as soon asyou en ter the underground.

This is a trivial example but it shows ushow to determine whether a star's light ispolarized. We put a polarizer at a telescope'sfocus. I t allows the light to pass throughonly if the waves' plane coincides withthe direction of the "slot", which blocks allother waves. By rotating the "slot" aboutthe telescope's axis, we see the star's lightchange. If there is no change, the light isnot polarized. In most cases, the lightgiven off by stars is unpolarized.

Things are different with AM Herculis.If we rotate the polarizer, we see its lightchange by more than 10%. Note that thepolarization also changes with the same period of three hours (Fig. 75).

The star's light curve and spectrum indicate that the main contribution to its luminosity is made by a white dwarf while thesecond star, a red dwarf, is, in fact, invisible. Let us see why the light emitted bythe white dwarf is so strongly polarized.The reason is that it is not the ordinaryradiation emitted by a plasma but the radiation of electrons moving in a strong magnetic field. This radiation is known as acyclotron radiation.* I t is generated at the

• So called because it was first recorded in cyclotrons, i.e. in charged particle accelera tors.

Ch. 6. Dwarf Binary Stars

x-raaiation

~ J4-VV~

0-- -

Visible light

~ O~Q-~-

~... o.5~

~ :===============:~ tutrariotet radiation.~ rf.o~~ 0.3

!otiial velocity~+200~ \ A J

!-4~ ~ ~-

235

Circular polarizationIf. 5 A-A~ 0 :::~=====:::;:=:;:;:==::::~ 7 li~neorpolarization _or: +v

~ -9 - -~ I I 1

0.0 0.5 1.0Or6ital phose

Fig. 75. Variations of different characteristics ofradiation from AM Herculis with orbital phase

surface of a white dwarf with a very strongmagnetic field, say, of the order of 108 G,which is about 100 times stronger than thatof the white dwarf in the DQ Herculis svs-


tern. The material accreted by the whitedwarf gets heated by hitting the surfaceand part of that energy is emitted in theform of X-radiation, i.e. following usualmechanism when electrons pass near ions(free-free radiation). The rest of the energyis emitted as the electrons revolve in thewhite dwarf's strong magnetic field. Cyclotron radiation has an optical wavelength,i.e, it is a visible light.

The way the mass flows in this system isalso unusual. Soviet astrophysicists discovered that if the whi te dwarf truly possessessuch a strong magnetic field then the latterstarts governing the motion of matter assoon as the matter leaves the red dwarf.In fact, the red dwarf lies within the whitedwarf's magnetosphere (Fig. 76).

The fact that the X-radiation and the visible light fluctuate with the orbital periodindicates that the white dwarf always showsthe same face to the red dwarf (like theMoon faces the Earth). This is hardly surprising since it has been calculated thatthe strong magnetic fields tie the stars in tosuch a close system that the stars quicklysynchronize their rotations. Now we knowof abou t ten such systems.

A high degree of polarization is not theonly peculiarity of these systems. They alsohave very characteristic visible light fluctuations. Every few months the systemfades tenfold or more.

ell. 6. Dwarf Binary Stars 237

A skilled interpreter of a system's lightcan determine whether it is a polar or not.Soon after the discovery of AM Herculis,the young Soviet astronomer Sergei Shuga-

Fig. 76. Matter flow in AM Herculis binaries

l'OV nominated another candidate for thepolar club, namely, the star AN Ursae Ma[oris, His only evidence was variations inthe star's brightness. Several monthspassed, and X-radiation from this starwas discovered.


Those Supernovae That AreSo Much Alike

Following Baade and Zwicky, we could associate the explosions of all supernovaewith the birth of neutron stars and, possibly, black holes. Stellar evolution theorysays that neutron stars' ancestors are massive stars.

Long ago astronomers noticed that, paradoxical as it might seem, supernovae often erupt in elliptical galaxies which haveno massive stars at all. We can solve thisproblem by assuming that the explodingstars are binaries. Supernovae which eruptin elliptical galaxies are very much alikeand in this they are peculiar. They evenconstitute a separate class known as typeI supernovae. It is typical that after themaximum they fade very quickly (Fig.77). In several weeks, the star becomes 6-8times fainter, then comes a period of aslow linear decline in brightness whichlasts for several hundred days.

There are also type II supernovae. Theirmaximum is followed by a flat part of thelight curve which lasts for several monthsand ends with a rapid decline (see Fig. 77).Supernovae of this type are more varied butall erupt in spiral galaxies, as a rule, inthe spiral arms, densely populated by massive short-lived stars. Type II supernovae


occur as a result of the collapse of massivestars.

Type I supernovae also explode in spiralgalaxies, but they do not tend to concentrate

-ISTn

~ -f7~ -15.~-1J

~-tI~ -9

~~ -ISm~ -17

-15-13-If-9

o 200 400

Days

Fig. 77. Typical light curves of type I and IIsupernovae

in the spiral arms. Statistically, type Istars are somehow connected with low massstars.

In 1963, the French astrophysicist EvrySchatzman noticed that white dwarfs inlow mass systems may collapse because ofthe accretion of matter flowing from thecompanion star onto them. A white dwarfwhose mass is initially below the Chandrasekhar limit grows to the limit, then loses


stahili ty, and starts collapsing. Let ussee how this happens. Calculations indicatethat the contraction of a white dwarf cornposed of helium or carbon is followed by arapid rise of temperature at the centre. Thetemperature is so high that the result is anuclear explosion and the complete disintegration of the white dwarf. This explainswhy the explosions of type I supernovae areso much alike; it is because the explodingstars have the same mass equal to theChandrasekhar limi t.

If this is true, the light curve of a typeI supernova is very elegantly explained.The nuclear burning of a white dwarf produces a large quantity of the nickel isotopeSIN i. This isotope is radioactive and it decays spontaneously into cobalt 56CO, which,in turn,' decays into iron 56}i'e:

56N i -+ 56CO -+ s6Fe.

Nickel's half-life (the time in which thenumber of radioactive atoms is halved) is6.1 days: while that of cobalt is 77 days.The energy from the cobalt decay is takenaway by electrons which later reemit thisenergy thus ensuring the luminosity of thesupernova's remnants which are ejectedinto space. The observed light curve is theresult of the nickel and cobalt decay.

At the same time, white dwarfs composedof heavier elements like carbon, oxygen,and magnesium (see Fig. 39) are not com-

Ch, 6. Dwarf Binary Stars 241

pletely destroyed and give birth to neutronstars.

It is clear that Schatzman's hypothesisis another way of "smuggling" neutron starsinto low mass systems. Like van den Heuvel's nonconservative scenario, it also implies the formation of low mass X-raysystems.

I wonder if novae, nova-like systems,U Geminorum systems, and polars are allstages to a supernova explosion. These aresystems in which there is accretion ontowhite dwarfs. I don't think we should rushto any conclusion but there is evidence thatat least some of these systems yield supernovae.

X-ray Bursters

In 1975, the American astrophysicist GeorgeGrindlay and his colleagues discovereda totally new type of X-ray source. A mysterious source was sending out short(100 s) bursts of soft X-radiation every fewhours (Fig. 78). (This is where the word"burster" comes from.)

Between bursts the radiation did notdisappear completely, hut it had a far harder spectrum. Astronomers hunting for theoptical "double" found that the X-ray sourcewas located in the direction of a globularcluster.16-0ob.,

242 to the World of Binary Stars

A globular cluster is a special kind ofstellar association with a very high densityof stars. On average, a globular clusterabout one light-year across contains up to a

Hours

Fig. 78. A burster's X-radiation

million stars. Outside globular clusters inthe Galaxy there is only one star in avolume that is 100 times larger.

Before long, bursters were discovered inother globular clusters. Given our knowledge of the distances to these globular clusters, it is possible to estimate the energygenerated by a burster. During a burst, theluminosity is some 30000 times the luminosity of the Sun. Between bursts the bursters fade to some constant level, which exceeds that of the Sun several thousandfold.Today we know of 20 sources of this type;some of them are not associated with globular clusters.

The I talian astrophysicist Laura Maraschifirst uncovered the secret of the bursters.


She noticed that the total radiation energyof a burster between bursts was 100 morethan its energy output during a burst. Itturned out that this magic number wascharacteristic of many bursters and thisnumber was the key to the' mystery.

Suppose we have two power stationswhich differ in efficiency. Given that theyburn the same amount of fuel, the ratiobetween their power outputs will, in fact,be the ratio between their efficiencies.The number 100 is, approximately, the ratio of the efficiency of the accretion on to aneutron star to the efficiency of fusion reactions.

The burster "operates" in the same manneras a recurrent nova, only the white dwarf isreplaced by the neutron star. Betweenbursts, a quasistationary accretion onto theneutron star takes place. The result ishard X-radiation, the "background". Unconsumed hydrogen and helium fall ontothe neutron star's surface from its partner.When the temperature and density in thesurface layer reach a critical point, explosive fusion reactions consume nearly allthe accreted matter which has fallen between bursts transforming into heavier elements within one hundred seconds. The explosion is accompanied by a rise in theX-radiation intensity. The burning takesplace at the layer's base, so the resultingradiation "thermolizes" and attains a black-

16*


body spectrum wi th the temperature ofthe envelope (less than 107 K).

A neutron star has an area about a million of times smaller than that of a whitedwarf; and, therefore, the temperature anddensi ty necessary for an explosion are achieved much faster than in the white dwarf.This is why "recurrent novae" on neutronstars burst once every few hours and notonce every 100 years.

The reason why bursters are 110t X-raypulsars lies in the great difference in thestrength of their magnetic fields. On an Xray pulsar, the magnetic field strength isas high as 1012-1013 G. Therefore, matterfalls only onto 1% of the whole surfaceconcentrating near the magnetic poles. Forthe same accretion rate, the temperatureand density of the matter in an X-ray pulsar is much higher than those in a bursterand, probably, higher than critical valuesfor initiating fusion reactions. For thisreason, the matter burns out without exploding. Since the fusion efficiency is100 times lower than the efficiency of accretion, the light from the fusion is hardly visible against the background of the accretion.

A burster's magnetic field is thousands oftimes weaker, its magnetosphere is veryclose to the neutron star's surface, and thematter falls isotropically. This is a difference between a burster and a pulsar. In


fact, even such a weak magnetic field caninfluence the homogeneity of the radiation;therefore, the burster's X-ray flux must"tremor" with a period equal to the rotation period of the neutron star. It is likelythat we should soon discover this phenomenon. I t has been calculated that burstersshould rotate with a period of several thousandths or hundredths of a second. Clearly,the accreting matter will accelerate theneutron star, while the weak magneticfield cannot practically impede it rotatingat a very high rate.

I t is also puzzling why astronomers failedto prove the binary nature of the burstersfor such a long time. No eclipses were noted, but this statistical paradox may, however, be explained. Imagine that a burster is a binary in which a neutron star isaccompanied by an extremely light (dozens of times less massive) normal star.Even when it fills its Roche lobe, it is muchsmaller than the binary's size (Fig. 79).I t is easy to see that the probabili ty thatsuch a small star will obscure the X-raysource is not high. On average, one in tensystems should be eclipsing. In 1982, onesuch system was found! Its period is only50 min (slightly longer than a school lesson).

Let us see how it was proved that a burster is a neutron star. I have already saidthat during bursts the X-radiation spectrum

246 In the -World of Binary Stars

is close to that of a blackbody, i.e. it obeysthe Stefan-Boltzmann law. Given the temperature and the luminosity, we can findthe area and, consequently, the radius of

Fig. 79. OutOO\V of matter in an X-ray burster binary s~stem

the radiating star. I t turned out to be about10 km, which had been predicted by neutronstar theory.

As to the origin of such a system, we maychoose between van den Heuvel's nonconservative scenario and Schatzman's mechanism, but which one? Observationaldata put specific restrictions on our ideas.Here we should mention the following puzzling fact: half of all X -ray bursters areconcentrated in globular clusters which on-

ChI 6. Dwarf Binary Stars 247

Iy contain 1/1000th of the sta.rs in the Galaxy. This is the bursters' third peculiarity.We need to explain why bursters are sounequally distributed between globularclusters and the rest of the Galaxy.

Neither of the two scenarios so far coveredfor a neutron star's formation in dwarfsystems seems to explain this disproportion.Hence a third scenario was necessary andthis was that a red dwarf is occasionally captured by a neutron star. The probabilityof capture is directly proportional to thesquare of the density of star distribution.In globular clusters, the density is hundreds of millions of times higher, therefore,the bursters form there. As for those bursters that occur elsewhere, they may havebeen ejected from globular clusters. I t isalso possible that these outsiders may havearisen according to either Schatzman's mechanism or van den Heuvel's scenario.

But for General Relativity

All the systems mentioned in this chapterare binaries in which the second mass exchange, implying the filling of the Rochelobe by the normal star, has occurred. Matter often flows from a smaller star to a larger one. This happens, for example, inX-ray bursters, U Geminorum systems, polars, etc. In Chap. 4, which was devoted tothe first exchange, we showed that a flow


from a smaller star to a larger one causesthe system to "move apart". The flow rateis the same as the rate at which the star expands to fill its expanding Roche lobe. Thisoccurs on the nuclear timescale. We knowthat in the systems described in this chapter a normal star's mass is about a tenth ofthat of the Sun, and its nuclear time is longer than the age of the Universe. On the other hand, observations of mass-flow rateindicate that typical durations of masschange in these systems are about 108

109 years, i.e. hundreds of times lesslThis do-es not mean that we have erred,

simply that we have to take into accountgeneral relativity. The Polish astrophysicist Bohdan Paczynski first noticed thatin very close systems with periods less thanten hours, gravitational waves must beim portan t. Just as accelerating chargesproduce electromagnetic waves, an accelerating mass must generate gravitationalwaves. These waves use much of the system's energy, and it contracts. It is thiscontraction that compensates for the binary"moving apart" due to the mass transfer.The flow continues on a scale of the energyradiation by the gravitational waves. Thecalculations based on a formula derived byEinsten for the binary's emission of gravitational waves are in conformity with theobserved typical mass transfer rate. Scientists are now building gravity wave detec-

ell. 6. Dwarf Binary Stars 249

tors and I hope gravitational waves willsoon be discovered.

The fact that such waves exist, however,is obvious today, since hundreds of thousands of the binaries in the Galaxy wouldbe unable to exist but for the gravitationalwaves.

The second mass exchange in low massbinaries leads to the formation of powerfulX-ray sources. We shall now turn to thesecond exchange occurring in massive systems in which flows of matter falling ontorelativistic stars are millions of times stronger.

Chapter 7

Cosmic Miracle

The Algol paradox and ~ Lyrae's spectralproperties indicate that the first exchangeresulting in a change in roles is a reality.Such an exchange allows massive systems tosurvive the first explosion and to continueevolving. This is confirmed by the existenceof such massive X-ray systems as Centaurus X-3 and Cygnus X-to In these systems, optical stars, i.e. blue supergiantswith masses in excess of (15-20)M0' are losing matter via a stellar wind at a highrate. A relativistic star capturing even asmall fraction of this wind will flare brighter than a thousand suns. The accretionrate onto a relativistic star amounts to10-1°-10-9 MG) per year. The accreting star'sluminosity is directly proportional to theaccretion rate (see formula (15». The nuclearevolution makes the supergiant fill theRoche lobe; thus the second exchange willbegin, and matter will flow from the moremassive star to its less massive companion.If this happens, the stars. as we alreadyknow, will get closer, which will not allowthe star to go under the Roche lobe. As a

Ch. 7. Cosmic Miracle 251

resul t, the star starts losi ng mass on thethermal timescale. For a massive star, thismeans dozens of thousands of years. If allthis matter were to fall onto the surface ofa relativistic star, the accretion rate wouldbe 10- 4M 0 per year, i.e. millions of timeshigher than in an X-ray system. We shallnot be so naive as to believe that under theseconditions the relativistic star's luminosity will be millions of times greater. Ourassumptions concerning the accreting star'sbrightness cease to be valid when we aretalking about enormous luminosities. Asthe luminosity and accretion rate grow, thepressure of the accretion-induced radiationmust grow too. If the radiation pressure exceeded the gravity matter could no longerfall. I t follows that an accreting star's luminosity cannot be infinite. A similar luminosity limit also exists for ordinary stars.I t was discovered by Eddington when he studied the equilibrium of high-luminositystars.

The force of the light pressure acting ona hot plasma is proportional to the amountof light falling per unit area, or illuminance.Illuminance is inversely proportional to thesquare of distance to a star. This means thatlight pressure falls with distance in the samemanner as gravity. If the light pressureexceeds the gravity at some distance, thenit exceeds it at any distance. Therefore,the critical Iuminosi ty (the Eddington lim-


it) does not depend on distance bu t onl ydepends on star mass:

L E ~ 33000£0 (: ) . (19)o

I n other words, a star wi th the mass ofthe Sun cannot be more than 33000 timesbrighter than the Sun, or else it would betorn apart by the radiation pressure.

An accreting star cannot be much brighter (say, 100 times brighter) than the Eddington Iimi t either. Since the star's luminosity is proportional to the accretion rate,the flow of matter cannot exceed the accretion rate corresponding to the Eddingtonlimit. The maximum accretion rate for arelativistic star of a star mass is of theorder of 10-8M 0 per year. On the other hand,an optical supergiant star, having filledthe Roche lobe. does not know anythingabout the Eddington limit for its partnerand may deliver 10000 times more matterthan its partner can possibly accept!

We still do not have a complete theoryfor a supercritical accretion rate and maybeobservations will help us because systemsof this kind must exist somewhere in theGalaxy at the moment. The second exchange lasts for about a supergiant's thermal time, i ,e. about 10000 years. I hopeyou still remember how to count stars. Allin all, there are about 1000000 massive binaries, each living 108 years. It Iollows that

-:11. Cosmic Miracle 253

there should be about 100 stars in the Galaxy involved in the second exchange. Evenif we adrni t tha t the second exchange stagefor the massive binaries is dozens of timesshorter than the thermal time, there muststill be at least one such system in theGalaxy!

Moving Four W8Y8 At Once

The discovery of the remarkable propertiesof the source SS 433 was the most strikingdevelopment in astrophysics in the late1970s. I ts spectral properties were astounding even for people with vivid imagination. Yet, not for all.

As you probably know, the Universe isexpanding. This fact was predicted by theSoviet physicist and meteorologist Alexander Friedmann* and established by theAmerican astronomer Edwin Hubble in1929. He discovered that distant galaxiesare moving away from us and that the greater the distance between the galaxies andthe Solar System, the higher the recessionvelocity. This is known as Hubble's lawand it allows us to determine how fardistant galaxies are by measuring theirrecession veloci ty.

• It shows that the weathermen are not alwayswrong. True, he gave no detail and only held thatthe Universe may be nonstationary, i.e. that iteither expands or shrinks.

254 In the VVorld of Binary Stars

In 1963, the American astronomer Martin Schmidt deciphered the spectrum of quasars. I t turned out that their spectral linesare shifted toward the red end of the spectrum by tens of nanometres, which, according to the Doppler effect, indicates thatquasars are moving away from us at tensof thousands of kilometres per second.According to Hubble's law, quasars are themost distant and luminous objects in theUniverse. During the following 15 years,hundreds of quasars have been discoveredand carefully studied, and all of them onlyshow redshifts.

Frequently great laws make us dogmatic.I t was taken for granted for a long timethat in the Universe there may he no large(tens of thousands of kilometres per second) shifts toward the hlue side of the spectrum. Let me illustrate the point with astory ahout a Moscow meeting of scientistson finding and contacting extraterrestrialcivilizations. One session was devoted tothe idea of "cosmic miracle" proposed byShklovsky. According to Shklovsky, thefact that we see nothing miraculous in spaceindicates that so far there are no extraterrestrial intelligence (at least, civilizations with whom it would be worthwhileto "chat a little"). This sounds convincing.To support its existence, a supercivilization would have to make some "superefforts" and would leave some trace in the

ch. 7. Cosmic Miracle 255

Universe. Then we would have observational evidence of "cosmic miracles" Of, inother words, objects or phenomena that areclearly defying the laws of inanimate nature. But we don't see any. True, the optimists maintain that we could not recognizea miracle. Would it be a square galaxy, ora UFO?

We already have a square or rather rectangular nebula (a disc viewed edgewise).I t is more difficul t to establish whether a"flying saucer" has been detected forthe lack of professional observationaldata.

In the heat of the argument someoneinvented the "miracle of miracles", namely, an object whose spectrum would containboth red- and blueshifts indicating motions at velocities of tens of thousands ofkilometres per second. I t seemed impossiblethat any object could be flying simultaneously to and away from us at such a deadlypace. Intelligent species would know this.And we realize that they realize that we...and so OD. I t follows that if there is a sighting investigators cannot attribute toknown phenomena, then intelligent species have left it there on purpose to let usknow... What we should do is to wait a little until such an object is discovered.

We did not have to wait long.Less than half a year later such an object

was discovered. I t seemed to stay in place,


move away and toward us, all at once.I t moved in all directions like someone spellbound in The Arabian Nights.

To Meet Again

In the early 1970s, American astronomersset out to look for celestial objects with

•

~~~

Fig. 80. A spectrograph with an objective prism

bright emission lines with a spectrographand an objective prism. This is an ordinary telescope wi th a large prism placed infront of the objective lens (Fig. 80). Ifthere were no prism, the photographic plateat the focus of the objective lens wouldshow just a sector of the starry sky. Theprism, however, splits a star's light into

Ch, 7. Cosmic Miracle 257

colours. The plate carries many pictures ofeach star in different colours, i.e. the stars'spectra. Although the spectrum of eachstar is not of the best quality, the advantage of the method is that many stars can beseen at the same time. A star's spectrum isits "fingerprints" which permit us to studyits properties.

You already know that bright emissionlines indicate some extraordinary phenomena. Today we have a catalogue of suchstrange stars.

In this catalogue, No. 433 is the variablestar V 1343 Aquilae. I t was enough tolook at the plate to see bright emissionlines of hydrogen and helium. I t took astronomers several years to find that in the direction of V 1343 Aquilae there are anX-ray source and the remnants of a supernova known as the radio nebula W 50. In1978, I talian astronomers scrutinized thespectrum of the SS 433 (here SS stands forthe initial letters of the names of the catalogue's authors) and were puzzled becausethe bright hydrogen and helium lines inthe spectrum were supplemented with unidentified lines. The I talian astronomerswishing to draw the attention of otherinvestigators to this star published theirdata in a special bulletin of the International Astronomical Union.

A response came from a group of American astronomers headed by Bruce Margon.

17-0667


They carefully investigated the object anddeciphered its spectrum (Fig. 81). Theunidentified lines turned out to be the samelines of hydrogen and helium but shifted toward the blue and red ends by severaltens of nanometres. I t looked as if every

nm

Fig. 81. The spectrum of SS 433

hydrogen transition were yielding threelines instead of one: the brightest line being located where it should be, while theother two lines shifted in opposite directionsby great "distances" corresponding to velocities of several dozens of thousands of kilometres per second. You can imagine howamazed astronomers were when they latersaw the shifted lines "crawl" to each other!

When this information was made public,the quickest comment came from MordehaiMilgrom, an astrophysicist known for his


unorthodox ideas. He suggested that thelines "moved" periodically and were bound tobe back where they had been first detected.

Soon American astronomers proved thathis suggestion of periodici ty of the lines

...f~ 00000

E:~

~.~ 30000~~

~

c::; 0 .......~-~------+- .........--'-----'~-~

~-30000

ooo~.~

t::~......~L....-,..j~_----<ClO~,.L- ~~ ___

Fig. 82. Radial velocity curves plotted for the twosystems of SS 433 lines

moving through the spectrum was absolutely correct. The period was close to 164days. The radial velocity curves constructed by the red and blue components wereclose to sinusoids changing in antiphase(Fig. 82).

17*


Not only were the spectra shifting by ahalf-period per phase, the curves hadthe same amplitude (about 40000 km/s)and were spaced wider apart along the yaxis making up a chain of alternating unequal links. The links' meeting points corresponded to the point where the radial velocities of the both shifted components coincide (see Fig. 82). The most remarkablething, however, was that at the meetingpoint there were two and not three lines.This is the case when it is enough to knowthe number of those meeting to know exactly what they are going to do.

All these remarkable properties helpedastronomers to realize why the lines "moved"and where the shifted lines had comefrom.

Why Two?

Now we shall try to understand why we seethe lines three times in the spectrum of theSS 433. Modern physics knows of threemechanisms by which spectral lines can besplit in an astrophysical situation. Theyare the Doppler effect due to motion, a gravity-induced shift, and the Zeeman effect.

The Zeeman effect occurs in a matterplaced in a magnetic field. If we put anatom in a magnetic field, the electron's motion will be go verned by both the nuclearattraction and the magnetic forces. The


splitting of the atom's energy levels is followed by the splitting of spectral lines occurring as a result of transition to these levels. When lines are spli t in a strong magnetic field, the relative shift of the lines,8"A/'A, depends on the line's wavelength (thelevel's energy). Different lines of the sameatoms have different shifts. This is in contradiction with the observational data inwhich blue- and redshifts of different linesare alike.

This property is more typical of a gravitational redshift, in which a photon givenoff by an atom in a gravitational field hasto do work to overcome the gravity and losethe "kinetic energy" defined by Planck's formula. Since photon's veloci ty cannot bereduced, its frequency and wavelength arechanged. Although extremely small andhardly detectable for ordinary stars, theshift can reach several tens of nanornetresfor relativistic stars.

An outside observer will see a gravitational shift in which the photon's energy isalways reduced, i.e. an increase in wavelength, or a redshift. So how do we accountfor the blueshifts in the spectrum of SS433?

The Doppler effect is the only mechanismleft for the shift. This means that V 1343Aquilae must have three separate radiatingregions. The first is practically stationaryand gives us the lines that are not shifted.


The other two regions move in opposite directions at gigantic velocities, and theirradial velocity components must be equalin magnitude and opposite in direction.

I have already mentioned an interestingfeature of the radial velocity curves: whenthe "moving" lines meet, they are stillshifted in relation to the stationary component. That this couple "seeks privacy"seems to run counter to the Doppler mechanism of line shifting. If the radial velocities are in antiphase, they can meet onlywhen their radial velocities are zero. Besides their position should coincide withthat of the stationary line, whose radial velocity is always zero. In contrast, the shifted lines show evidence of a great positiveradial velocity of about 20000 km/s whenthey meet. This paradox is resolved by special relativity. Special relativity must beinvoked because the observed velocitiesare significant fractions of the velocity oflight.

In special relativity, the static observeron Earth measures the time in a moving ref-erence frame a factor 1/V 1 - V2/C2 slower(v is the total veloci ty). Hence, the electromagnetic wave's frequency, or the wavelength, will be shifted toward the red endof the spectrum. This effect still occurs evenif the radial velocity is zero, i.e. the sourceis moving across the line of sight. Therefore it is known as a transverse Doppler


(20)

effect. The correct formula for the Dopplereffect is

Arec 1+ vrad/C

Asource y 1- (v/C)2 •

In this formula, Vrad is, as before, the radial veloci ty, and v is the total veloci ty ofthe source.

Where the moving lines meet, the radialvelocity is zero, and the received radiation'swavelength is determined by the total velocity v. Now we can find the total velocityof the source by measuring the line's wavelength at the meeting point. (In practice,we don't even have to wait for the lines tomeet. I t is enough to find the ari thmeticmean of the wavelengths of the red and bluecomponents, which will be the wavelengthat the meeting point.)

The total velocity turned out to be80000 krn/s, i.e. about one third of the veloci ty of ligh t.

A Kinematic Model

The radial velocity curve indicates that theradiating regions producing the shifted spectral lines move in opposite directions whilethe velocity's radial component changesstrictly periodically.

We might surmise that SS 433 is a triplestar system in which one star is much moremassive than its partners. The less massive


stars revolve with a period of 164 days(Fig. 83) and the shifted line componentsare formed by the less massive companions

•.•....:.~..~.~.: ....::..~~/

IA

Blue StationaryCOn7ponpnt line

Red(:ompollt'llt

Fig. 83. A triple star system as described by thekinematically incorrect model or SS 433

which move in antiphase along the SHine

orbit while the unshifted Jines are generated by the central star.

This arrangement, however, cannot bejustified quantitatively. Given the orbitalvelocity (80000 km/s) and the orbital period, we can use Kepler's third law (see

CII. 7. Cosmic Miracle 265

formula (2)) to find the central star's mass,I t is equal to 1010AI0, i.e, only an orderof magni tude less than the mass of theGalaxy. Clearly, this quantity is far toolarge, but eV{\H this is not the only argumerit against the model.

Let us estimate how much time it takes]ight to traveI across the tri ple system'sorbit along the diameter. The stars cornplete an orbit at 80000 km/s in 164 days.A diameter is a factor n smaller than thecircnmference. In other words, light cantravel across the system in (164/a.14 X80000/300000) == 14 days. The radial veloci ty of one of the stars would thus he zero14 days earl ier than of its companion,which is in con trad iction wi t11 ohserva tionaldata. Thus, this arrangement is incorrectkinematically.

I t is obvious that the only way we canget a less massive object is to give up theerroneous assumption of a 164-day orbitalperiod. Otherwise we shall never get belowthe 101u1\10 mark. One model is a gaseousring (or a disc) revolving about some unusual body at a velocity of 80000 km/s(Fig. 84). Tho central body radiates twohigh-energy beams that cross the ring intwo small areas and make these areas hotand luminous, I t is easy to see that oneof these spots will generate redshifted linesand the other blueshifted lines. Now imagine that this system rigidly rotates about a


certain axis with a period of 164 days. Thiswill result in the periodic movement of thelines in the spectrum.

Yet, in this model, too, the central body'smass is unreasonably large. Luminosity inthe shifted lines exceeds that of the Sun

Fig. 84. Another erroneous model of SS 433: agaseous ring illuminated by a beam of rad iation

hundredfold, while the temperature, judging by the spectrum, is close to 10000kelvins, that is only one and a half timesmore than the temperature of the Sun'sphotosphere. The radiating regions thuscannot be much smaller than 10 radii of theSun. The ring's radius must be at least10 times larger, i.e. approximately 100 radii of the Sun. This is close to the size ofEarth's orbit. The Earth revolves aboutthe Sun at a velocity of 30 km/s, while

ChI 7. Cosmic Miracle 267

the gas in the ring of the same radius at avelocity of 80000 km/s. In this model, thecentral body's mass must be (80000/30)2 ==7000000 times larger than the Sun's.

To reduce the mass still more, we haveto admit that the orbital velocity is not

Fig. 85. Matter is ejected into space in two oppositedirections

80000 km/s. Thus we arrive at a correct model. Imagine a body ejecting gas jets inopposite directions at 80000 km/s, Onceejected, the gas expands, cools down, andrecombines. This is where the shifted linescome from. The ejected jets revolve aboutsome axis (Fig. 85) with a period of 164days. This changes the radial component of


the jets' velocity. The jets are slow enoughfor the gas to give off its energy beforethe jets travel very far.

In the mechanics of rigid bodies, there isa phenomenon called precession: the axisof a rotating body subject to an appliedtorque changes periodically i is direction andbegins to sweep out a cone. This periodicchange in direction is called a forced precession. Free precession occurs when thebody's rotation axis does not pass throughits centre of mass.

In our case, the 164-day rotation of thelines of the ejected jets was in advancecalled precession, and the 164-day periodthe precession period. Today nobody doubtsthis kinematic model, but the origin of therotation of the directed jets still causesargument.

We are sure that SS 433 really ejectsthe gas jets at subrelativistic veloci tiesbecause we see them. True, they are notvisible in optical light but in the radioand X-ray ranges (Fig. 86) the picture isclear. The visible light is generated comparatively near the central body which iswhy we cannot see the jets even throughthe best telescope under optimal weatherconditions. The radio waves are generatedfurther away, and the capabili ties of a radiotelescope are much better. It gives us adistinct image of separate blobs of matterleaving the centre, the direction of the

eh. 7. Cosmic Miracle 269

ejected jets changing with a period of 164days. From our knowledge of the ejectedmatter's velocity and its angular displacement, we can find how far away SS 433 is.

Fig. 86. An X-ray image of S5 433 made from aboardthe Einstein orbital observatory

I t is 10000 Iight-years away, i.e. SS 433is in our Galaxy. (OUf Galaxy is 100000light-years across.) Nobody has ever seenanything like that in our Galaxy. To get abetter idea of what is happening, we neededa clue or something familiar in its behaviour.


Something Familial

There are no poor telescopes, there may beunskilled observers looking through them.This was proved by the Soviet astrophysicist Anatoly Cherepashchuk, who made avery important discovery shedding light onthe nature of S8 433 using a small (bymodern standards) telescope.

Within the first months of the "goldrush" around S8 433, American astronomersfound that the central unshifted linestrembled slightly with a semiamplitude of70 km/s and a period of ahout 13 days.The light was later reported to fluctuatewith the same period. These data, however,were not reliable and were not much of aclue. What astronomers needed was longterm observations.

This was done with a small reflecting telescope with a 600-mm mirror. Unofficially, atelescope wi th a mirror of more than twometres in diameter is large, one with a mirror between one and two metres in diameteris medium, and all the rest are small.

The advantage of small telescopes is thatthere are many of them and they are nottoo busy. To make sure the light fluctuateswith a period of 13 days, you need severalmonths of observations. Not 13 days!

The star's periodic light fluctuations werecombined with chaotic changes in light intensity. These may be due to atmospheric


scintillations or the intrinsic (sometimesknown as physical) variability of a star.Averaging random quantities takes time. Ifthe periodic and chaotic fluctuations havethe same amplitude, the light curve mayonly be plotted after you have observedmany periods.

Having spent several months observingSS 433, Cherepashchuk plotted a reliable

~ f3~5 I. je...~ I'· I • · • • •~ 14.0 .,. II.:::'.· .. I I II. I~ ·1·· I: - ••.~ 14.5 • - • •~

0.0 0.5 toOroitol phose

Fig. 87. The light curve of SS 433 obtained byCherepashchuk

light curve. Its shape resembled Algol'scurve and this established that SS 433 isan eclipsing binary system (Fig. 87).

It was the first time that we saw somefamiliar traits in this mysterious object. Thenext step was to see what kind of binary itis. We can see no absorption lines in thespectrum of SS 433. At least, no one hasever seen any. The star's continuous spectrum has a maximum in the red range,which indicates a low temperature. As isknown, cold stars are usually low mass stars

272 In tho World of Binary Stars

and so SS 4;i3 was initially believed to be alow mass binary.

The mistake was corrected when astronomers took into account the Iight absorptionby interstellar dust. The constellation Aquila is located near the Galaxy's plane wherethere is a great deal of dust. Given the distance to SS 4aa, we can determine howmuch the dust dims tho light, ill Iact, itreduces the light by approximately eightmngni tudes, or 1500 times, Relatively moreblue light is absorbed by the dust which, bythe way, is why the Sun at sunset looks fed.

We may thus conclude that V 1~14a Aquilac is not red And cold but blne and hot. Ananalysis of the light curves has shown thatthe both objects (I choose to call the binary's components this way so far) have ahigh temperature (more than 20000 kelvins) and nearly equal luminosity in visiblelight (the depths of tho eclipses do not differmuch).

Even if one of the components is an ordinary star, it is massive, The star's size iscomparable wi th the size of the binary,which makes one wonder whether the normal star tills its Roche lobe and, like pLyrae, is flowing into its neighbour.

This, however, seems to contradict thoapparent spectrum. The width of the lines'unshifted components corresponds to severalthousands kilometres per second. It follows that they occur in the optically thin


matter outflowing from the binary. Thislooks very much like a stellar wind from ablue supergiant. The similarity, however,is misleading. The rate of outflow in SS 433is hundreds of times that of ordinary supergiants. It amounts to approximately 10-4M0

per year. Such is the theoretically estimatedflow of mass from a massive star which hasfilled its Roche lobe to its less massive partner.

Now let us see whether the mass inV 1343 Aquilae is outflowing from one starto the other or is being ejected into space.

Supererltieal Accretion

Before we start choosing between hypotheses, we must know for sure the differencebetween what we see and what we imagine.I t is certain that SS 433, otherwise knownas V 1343 Aquilae, is a binary with a period of about 13 days. This period is nearlythe same as that of ~ Lyrae. The systemejects material at a velocity of several thousands of kilometres per second. The outflowvelocity is usual for hot O-B stars but theoutflow rate is unexpectedly high. If thiswere the only strange thing about SS 433,we might consider it a binary, though withsome reservations.

But apart from the other things, the binary ejects two jets of gas at an extremelyhigh velocity of 80000 km/s for some un-

18-0667

274 ln the World of Binary Stars

known reason. The direction of the jets precesses with a period of 164 days. The jetsradiate in the radio and X-ray ranges. Thebinary is surrounded by a nebula whichlooks very much like what remains after asupernova explosion. I'd like to add thatthe star itself only weakly radiates in theX-ray range. This seems to be all we knowfor sure about it.

I t is possible that one of the binary'scomponents is a blue star of spectral class Bwith a mass of no less than 10M0. Thenature of the second star is little known.The high velocity of the jets and the remnants of a supernova indicate that the companion may be a relativistic star (a neutron star or a black hole).

Why does a relativistic star behave sostrangely? Unlike other massive binaries,the relativistic star in SS 433 is not a brightsource of X-radiation. Judging by the observational data, the binary's second companion looks very much like an ordinaryhot star, and it would be listed as such ifit were not for the relativistic jets.

Have a look at Table 2 which lists thephysical properties of X-ray pulsars. Notone has a luminosity in excess of the Eddington limit (formula (19»* and there must be an

• The mass of an X-ray pulsar is several timesthat of the Sun; therefore, its Eddington limit is50-100 thousand L0 (L0 is the luminosity of theSuo).


explanation for this. All X-ray pulsars"work" in the su bcri tical accretion 1110de or"slightly exceeding the Eddington limit.A relativistic star with a supcrcritical accret.ion ]ooks q1Ii te d i rreren t.

Supercri tical accretion must occur ill aclose binary d uri ng the second ITIaSS exchangethrough the inner Lagrangian point. Inthis case the ou tflow ra te exceeds the cri tical value equal to the Eddington limit bytens of thousands of times. If all the matterwere to fallon the relativistic star, the radiation pressure would be as many timesmore than the gravity. This means that allthe matter cannot fall onto the star.

Is it possible that the matter falls anddoes not fall at the same time? Yes, it is.This can be easily arranged in the case ofdisc accretion. In 1973, Shakura and Syunyaev presented a quali tative scenario of howthis might happen. They considered supercritical accretion onto a black hole (Fig. 88).Far from the black hole, the accretiondisc's structure is hardly different fromthat of an ordinary subcritical disc, andthe energy produced in the disc is muchless than the Eddington limit, the lightpressure being insignificant. At a certaindistance, however, the radiation pressurebecomes comparable to or exceeds the gravity. Some of the matter is, therefore,"blown off" away from the disc. The remaining matter continues to move toward the

18*

2i6 In the \Vorld or Binary Stars

black hole. In order for the accretion to besteady-state, i.e. not to vary with time,the mass of the matter falling onto theblack hole must be such that the totalluminosity of the disc exceeds but little theEddington limit. This limit was calculated

F i..~. 88. Supercritical disc accretion

for a static star radiating isotropically. Thesituation is different for a disc. Therefore, aminor excess over the Eddington limit willnot destroy the disc.

Only a small fraction of the matter initially captured reaches the black hole inthis "semiaccretion" mode. Almost all thematter is "blown off" in the form of a qua-


sispherical stream resembling a stellar wind.There is so much material that any hardX-radiation occurring in the vicinity of theblack hole is almost wholly absorbed by it.Consequently, a relativistic star undergoingsupercritical accretion cannot be a brightsource of hard X-radiation. An outside observer would see only a "coat" with properties like those of an ordinary star.

An accretion disc has a preferred direction,or an axis. The gas jets are ejected from thecentral regions of the accretion disc at arelativistic velocity along this axis. Thisscenario was developed six years before thediscovery of the unique properties of theSS 433. Today it might be regarded as aforecast if investigators had not found relativistic jets in galaxies and quasars longbefore.

A classical example of jet discharge is inthe galaxy Virgo A (Fig. 89), which wasfirst observed very long ago. Now we knowof dozens of similar examples. True, thejets in the nuclei of galaxies move at relativistic velocities, not subrelativistic velocities. Jets are not ordinary matter and looklike blobs .of relativistic particles whichhave "become entangled" in the magneticfields and been ejected together with themfrom the nucleus of the galaxy.

The model of jets moving along the axisof a supercritical disc was intended to explain this phenomenon.


If we place a neu tron star in the centre ofa supercri tical disc instead of a black hole,much will remain unchanged. The interiorwi ll still be covered by an opaque "coat",and an ou tside observer would see an ord inary star. The strong magnetic field of aneutron star destroys the supercritical disc

Fig. 89. A jet from the nucleus of the galaxy Virgo A

over distances of several tens or hundreds oftimes the neutron star's radius. All matterreachi ng t he Inagnetosphcre falls on to theneutron star's surface. This is accompaniedby an outburst of energy which is tens orhundreds of times more t han the Eddingtonl imi t. A fraction of matter affected by theradiation pressure may be ejected along themagnetic axis of the neutron star at velocities of 100000 km/s, i.o. the escape velocityfor the neutron star. If the neutron star'srotation ax is coi ncides \vi til i ts magnetic


axis, an outside observer will see jets ofmatter ejected from behind the "coat" in

Fig. 90. A binary in a supercritical accretion mode

opposite directions at relativistic velocities(Fig. 90).

280 Jn the World of Binary Stars

SS 433 may be a binary undergoing thesecond mass exchange accompanied by theformation of a supercritical disc around therelativistic star. This would solve the apparent paradox that, on the one hand, theunshifted lines indicate the presence' of astellar wind and, on the .other hand, thevalue of the mass-flow rate indicates mass exchange between the components. This iswhat we expect from supercritical accretion.The matter, in the form of a "gravity wind",flows from the normal star to the relativistic star and then is almost completelyejected in the form of a stellar wind fromthe relativistic satellite.

How long can such a process last? Nobodyknows exactly. I t may be tens of thousandsof years, or thousands of years. If we bearit in mind that it takes light 10000 years totravel from SS 433 to the Earth, we mayconclude that now the mass exchange inSS 433 is already over. Today we see whatwas happening 10000 years ago.

At present, we don't know for sure what is(or was) underway in the V 1343 Aquilaesystem. I t is not yet clear why the jet direction "moves" with a period of 164 days.There are many hypotheses but, probably,the phenomenon is similar to the 35-daycycle of Hercules X-i. If the normal star'srotation axis does not coincide with that ofthe binary, the normal star starts precessing because of the tidal force acting from


the relativistic star. Both the direction ofthe matter jets and the plane of the accretion disc will rock together with the star.

Such a "slaving" disc in the HerculesX-1 periodically obscures the X-ray pulsarfrom us and changes the direction of jets in58 433. This attractive hypothesis is ratherpopular. Yet I believe it has many drawbacks, and probably here we deal with anabsolutely different phenomenon.

These questions still await final answers,but every year makes us more certain thatS8 433 is the first massive system in whichwe have come across the second mass exchangeaccompanied by the formation of a supercritical disc. Let us see what happens tothe binary next.

The "Nightmare"of the Second Exchange

The second exchange in a massive binary isalways accompanied by a loss of mass. Therelativistic star is physically unable to accept all matter that leaves the normal star,and radiation pressure ejects it from thesystem. Besides the mass, the system alsoloses its angular momentum which causesthe binary's components to move closer together. This reduces the Roche lobe, butthe normal star does not respond quicklyto this reduction and loses still more mass,which brings the stars still closer together.


The process continues ever more quickly until the relativistic star finds itself insidethe supergiant; the supergiant "swallows"the tiny neutron star.

Thirty years ago this idea would haveseemed fantastic. Now attitudes are different. Many hundreds of hours of computertime have been spent trying to understandwhat happens to a relativistic star "swallowed" by a supergiant. The situation seemssimple in that the normal star's structureand its density distribution are known. Therelativistic star's motion will be similar tothat of a satellite reentering the Earth'satmosphere. The satellite is decelerated bythe atmospheric drag and approaches theEarth along a steep spiral, its speed increasing all the time. The reason for this is thesame as the reason for a star's negative heatcapacity.

The satellite's kinetic energy grows dueto the work done by gravity, but a relativistic star differs from a satellite in onevery important aspect (which complicatesthe situation enormously): the satellite entering the Earth's atmosphere does notchange its physical structure. The satelliteburns up producing energy which is insignificant as compared to the energy binding theatmosphere to the Earth. Things are different with stars. A relativistic star insideany ordinary star is dangerous ecologically.The energy produced by the compact


star's deceleration is so great that the normal star starts losing its matter. This altersthe star's structure and the decelerationforce. A rigorous solution of such a dynamic problem is only possible by a computerwhich can operate at billions of operationsper second. There are not so many suchcomputers and the calculations so far doneare rough approximations and only be considered qualitative research.

The relativistic star entrains the supergiant's material with its gravity. One layerafter another is accelerated and pulled awayfrom the star for good. Like a sharp knifecutting lemon peel, the compact star slicesaway the supergiant's upper layers. Eventually the "swallowed" star arrives at thecentre.

Do you remember that by the time thesecond exchange begins, there is a heliumnucleus at the centre of a massive star? Thereason is that the exchange begins after thestar has expanded to the size of the Rochelobe or, in other words, has left the mainsequence. Let us see what will happen whenthe relativistic star reaches the nucleus.

I-Iere the following alternative is possible.Either the binary containing a helium anda relativistic stars occurs, or the relativisticstar is "swallowed" by the helium nucleusonce and for all and "settles down" at thesupergiant's centre.

An interesting resul t of stellar "cannibal-


ism" was considered by the American astrophysicist Kip Thorne and his colleagues.They calculated what the structure of a supergiant which "swallowed" a neutron starwould be. The source of energy in such a"cannibal" would mainly be matter accretiononto the neutron star. By the way, Landaufirst suggested in 1937 that accretion by aneutron star inside an ordinary star mightbe the reason for a whole star's energy generation.

How long does a "cannibal" star live?These "freaks" will live for a long time ifthe accretion is slow. The matter must support the neutron star's radiation by settlingon its surface slowly.

Suppose "cannibalism" is a rare exception. A binary remains a binary. Then comesthe next stage in a close binary's evolutionwhen a helium star is orbiting with a relativistic star. The possibility that such starsmay exist was stressed by the Soviet astrophysicists Tutukov and Yungelson. Heliumstars (Wolf-Rayet stars) intensely lose theirmatter in the form of a stellar wind. Itseems that after the second exchange therelativistic star must again become a brightX-ray source. Yet no hard X-radiation hasso far been discovered from a Wolf-Rayetstar. So perhaps stellar "cannibalism" maybe the rule rather than the exception.

Chapter 8

Wanderers

We have traced a binary's evolution via ~

Lyrae and Algol through X-ray binaries likeCentaurus X-3 and Cygnus X-1 and havecome to the critical point, the second massexchange. This is when the binary's life isat stake. The second exchange may only becompared to the explosion of one of thebinary's components. The second exchangeis accompanied by supercritical accretiononto the relativistic star and, probably, byit being "swallowed" by the normal starsupergiant.

VVhether a binary can survive can onlybe determined by observations. If it can,the result will be a binary of a helium anda relativistic star. What we need to do isto look for them in the sky. The heliumstars must be Wolf-Rayet stars and thereare many of them. I t would be easier ifthey differ in some way from other binaries. Fortunately, systems with relativisticstars do have distinguishing features,namely, birthmarks which they obtained during earlier stages of their evolution. In the


Halo ~-z--

.\

late 1970s astronomers used these featuresto restrict their search and they had anastonishing success.

Runaways

We live in a spiral galaxy. Most of its starsmake up a plane figure, which is shownedgewise in Fig. 91. The flat disc is surround-

I

~~.~~...... ~

1Caseous

disc

I... tooooo lilJht-years ..I

Fig. 91. The structure of the Galaxy

ed by a less massive spherical subsystemcalled a halo. In the last few billion years,stars have only been born in the Galaxy'sdisc. The disc is only of the order of threethousand light-years in thickness and onehundred thousand light-years in diameter.

ChI 8. \Vanderers 287

In order to describe intragalactic phenomena, astronomers use a coordinate systemin which the z axis is directed along the axisof the disc. Then it is usually said that thethickness of the star disc along the z axis isthree thousand light-years. New stars, however: are born in a much thinner layerthat is several hundred light-years thick,i.e, 5-10 times thinner. The reason is thatthe gas which is a kind of material you needto form a star is concentrated in a thin discseveral hundred light-years thick.

The disc is rotating. This means that eachstar revolves in a gravity field created bythe other stars of the Galaxy. Alongside thegeneral revolution, each star moves at alow random (chaotic) velocity. But forthis, the star disc would be infinitely thin.Random star motions smear the disc to itsapparent size. Young stars move randomly10 times more slowly than the revolutionvelocity which is close to 200 km/s nearlyin the whole Galaxy. The greater the star'srandom velocity, the further along the zaxis it can travel.

The average random velocity of massivestars in the Galaxy does not exceed 510 km/s, A star moving at such a velocitycannot move more than several hundredlight-years. Indeed, most massive stars areconcentrated in the vicinity of the plane ofthe Galaxy. Yet there are some exceptions.

There are stars in the Galaxy which move


tens of times faster. They are called runawaystars. At such velocities they can climb toaltitudes of thousands of light-years abovethe plane of the Galaxy. They do not leavethe Galaxy, but they do not remain in the"herd".

Blue runaway stars of spectral classes 0and B were carefully studied by the Dutchastronomer Adrian Blaauw in the 1950s.Although there are only a few of them,they consti Lute several percent of the normal stars of the same spectral class.

Could this be due to the cumulative effectin Pokrovsky's experiment, or when coinsjump onto the floor in the underground, orwhich turns multiple stars into binaries(see Chap. i)?

Can anything like this happen in the Galaxy? Although each star's momentum issmall, when stars collide Of, to be moreexact, when they approach each other, theycan transfer momentum from one to another and some may attain velocities of several hundreds of kilometres per second and soleave the Galaxy's plane.

This simple explanation for runawaystars is wrong quantitatively. Star collisions are so rare that they cannot explainthe observed number of runaway stars. Thiswas understood by Blaauw, who found another explanation.

He suggested that the runaway stars usedto be components of close binaries. At that

Cit. 8. Wanderers 289

time, the idea that a change of roles, whichexplained the Algol paradox, could happened had not been generally accepted. Blaauw suggested that the binary's more massive component evolves faster and thatthere is no mass exchange. Then the 1110remassive star explodes first, Most. of thebinary's mass is ejected, and the systemdisintegrates (see Chap. 4, Section "TheStory Will Be Continued"). The bond binding the system's components (gravity) isbroken due to the radically decreased gravity caused by the tenfold loss of 1118SS by oneof the stars. The stars or, to be more exact,the normal star and the collapsed remnantflyaway like a stone from a sling. Eachstar's velocity is about its orbital velocity,which in close binaries is of the order ofhundreds of kilometres per second. This isBlaauw's explanation for runaway stars.

But we now know about the Algol paradox and role changing, and that the Iirstexplosion in a binary does not lead to itsdisintegration. Thus there is role changingbut no Blaauw mechanism, If so, there areno runaway stars either.

In fact, the Hlaauw mechanism as suchdoes not seem to be correct but role changing may in itself be used to explain runaway stars. After the mass has been ejected,the binary acquires additional momentumbut does not disintegrate (Fig. 92). Theejection must occur quickly because the

19-0667


Fig. 92. The effect of "recoil" caused by the ejection of matter by one of the stars in a binary.Before the explosion, the envelope's matter hadbeen moving with the star along its orbit and thushad a certain momentum. If the matter is ejectedrapidly, the envelope takes this momentum, andthe remaining binary starts moving in the oppositedirection according to the law of momentum conservation. As a result, the centre of mass of theenvelope-and-binary system as a whole remainsstatic. Note that the envelope is ejected symmetrically about the exploded star

Ch. 8. Wanderers 291

ejected mass takes the orbital momentumwith it. According to the law of momentumconservation, the binary must acquire a recoil momentum in the opposite direction.If it does, it will develop additional velocity of several hundred kilometres per second. On average, binaries with relativisticcomponents must have greater velocitiesand z coordinates.

"Single" Wolf-Rayet Stars

Now it is clear where to look for binarieswith relativistic and helium stars. They areamong the Wolf-Rayet stars moving rapidly and having large z coordinates.

These, however, are not their only "birthmarks". During the second mass exchange,the binary ejects a great amount of matterinto the surrounding space. This mattermust be observable for at least an astronomically short period of time. Some "wanderers" must be surrounded by matter thatappears as nebulae.

The "single" Wolf-Rayet stars fit thispattern. They are said to be single becausetheir spectra show the lines of only onestar. Either the star has no component atall, or the component is so weak that itslines are invisible against the backgroundof the bright emission lines of a WolfRayet star. These stars were for a long timeconsidered singles because no trace of an

19*


eclipse or radial velocity variation had everbeen recorded. III 1978, astronomers setont in search of binaries among the "single"

Fig. 93. A photograph of the ring nebula NGC 6888surrounding the "single" Wolf-Hayet sturH D 192163 (shown wi th the arrow). The photograph was made by Tatyana Lozinskaya (a t theCrimean station of the Sternberg State Astrouonucal Institute)

W olf-Rayet stars. Some of them are surrounded by ring-shaped nebulae (Fig. 93).The nebulae could be the Blatter ejected bybinaries during the second exchange. These


stars were selected as the most prospectivebinary candidates.

Very soon a binary was discovered. Thestar numbered 50896 in the HD (HenryDraper) catalogue and surrounded by a

~+40"';::;

~0~

....... V)

,~~ -40~~~

"Single"wR star

I', '~ I~ ;--,

• I ~ I \ •• , \....!- ..'. \,,:.,' '.-:..+ \.. . ..... . ~ '-...:.

o 0.5 1.0Orbi tal phose

Fig. 94. Light and radial velocity curves of the"single" Wolf'-Hayet star HD 50896

planetary nebula ReW 11 was changing itsbrigh tness and rad ial veloci ty wi th a periodof 3.8 days (Fig. 94). 1t is enough just tolook at the radial veloci ty curve to seehow difficul tit was to uncover the binarynature of this "single" star. The widths ofthe lines in the spectrum correspond tothousands of kilometres per second and thesemiamplitude of the radial velocity variations is only :35 krn/s. You must be very


skilled and very optimistic to notice andtrust the trembling of the lines whichamounts to only several percent of its width.I t is like trying to find the wid th of a razoredge with a school ruler.

Note that the time of the secondary minimum in the light curve (see Fig. 94) coincides with the passage through a gammavelocity in the radial velocity curve. Thisis excellent evidence for the eclipsing natureof the minimum. Astronomers in variouscountries have found about ten "single"Wolf-Rayet stars with periodic variationsin their radial veloci ties and brightness(Table 5).

The most amazing thing concerns the massfunction obtained from the measurementsof radial velocities. For all stars, the massfunction is much smaller than the mass ofthe SUD, which indicates that the secondcomponent has a small mass.

Now let us do some arithmetic. In a closebinary containing a massive star and aneutron star, the relativistic star's orbi talvelocity can be as high as 300 km/s. Sincethe orbital momenta of the binary's components are equal, the orbital velocity ofthe normal star, which is tens of times moremassive, must be the same tens of timeslower. It follows that the massive star'sradial velocity amplitude must, therefore,be several tens of kilometres per second,which is exactly what we see in Table 5.

Table 5. "Single" Wolf-Bayet stars with possible relativistic companions o~

Amplitude ?JApparent Period. z axis, Mass Presence <Star function. of ringmagnitude days light, velocity. light..years Mev ~

magnitude km/s nebula ='Q..('t)...,~...,en

HD 50896 6.9 3.8 0.08 35 -900 10-2 yes

HD 192163 7.7 4.5 0.03 20 200 10-2 yes

HD 191765 8.3 7.4 0.04 30 100 7X10-a yes

HD 197406 10.5 4.3 0.06 90 3300 0.3 no

HD 96548 7.8 4.8 0.04 to 1000 3X1o-' yes

HD 164270 9.0 1.8 0.05 20 700 10-3 nor,,:,~<:J1


If here we deal with relativistic stars,it is not clear why none of them radiates inthe X-range. Wolf-Rayet stars supply theirrelativistic neighbours with vast amount ofmatter, and their stellar winds are tens oftimes stronger than those of O-B supergiants, near which bright X-ray sources canbe found.

This discrepancy makes us doubt thatWolf-Rayet stars have relativistic companions. Is it possible that astronomers havefound something they were not looking for?This happens fairly often, indeed, radiopulsars were discovered this way. I nvestigators were looking for scintillations andthey discovered neutron stars.

However, in this case there are too manypoints of contact between theory and observation. So it was asked whether there mustbe any X-radiation at all? The astronomers'reasoning was as follows. Given that thestellar wind from a Wolf-Rayet star is veryintense and dense, could it be too dense?Then it won't let X-radiation pass through,just as the Earth's atmosphere blocks offX-radiation. This explanation became widely accepted because it allowed us to seethings clearly. A relativistic star capturingthe stellar wind matter radiates in the Xrange with a luminosity hundreds of timesthat of the Sun. All the X-radiation, however, is absorbed by the stellar wind and isreemitted in, say, the optical range. The


relativistic star's eclipse causes the systemto fade several percent (several hundredth ofmagnitude) in the visible range, which is inconformity with the variations in the apparent light intensity of the "single" WolfRayet stars.

I t was after the binary nature of the"single" stars had been discovered that thehypothesis of the X-radiation self-absorption began to crack. The observational datagave us the components' period and velocity, which allows us to find the distancebetween them. The result was 10-15 timesthe radius of the Sun, while theoreticallyWolf-Rayet stars should be 2-:3 radii of theSun in size. So the relativistic star is notthat deep in the envelope outflowing fromthe helium star. However, at such a distance, the stellar wind is already transparent to hard X-radiation, and the relativistic star should be visible in this wavelengthrange.

Some investigators sought an answer inanother direction. Suppose all the "single"binaries so far discovered contain neutronstars. We know that in order for matter tofall onto a neutron star's surface, at least twoconditions must be met: first, the materialmust be available to fall and, second, theneutron star must not rotate too quickly asotherwise the magnetic field would cast thematerial in different directions due to thepropeller effect. In systems with Wolf-Rayet


stars, the first condition is satisfied but thesecond one is not. Recall that during thesecond exchange, i.e. at the stage immediately preceding the formation of a binarywith a helium star and a relativistic star,the system undergoes supercritical disc accretion. As a result, the neutron star receivesboth mass and angular momentum. Thesecond exchange gives the neutron star ahigh-rate axial rotation. The propeller effectprevents the star from being a bright X-raysource. So you see that the binary's evolution itself is such that no bright X-raysource appears after the second exchange.If it does, it happens very seldom.

The research on the single Wolf-Rayetstars still continues, and astronomers mayget a clearer picture in years to come.

The Second Explosion

A helium star's evolution inevitably resultsin the exhaustion of its nuclear energy andcollapse. This is what happens during thefirst collapse. The central iron core turnsinto a neutron star, and 90% of the matteris spewed out into space. The difference isthat now the more massive component ofthe binary explodes. The total energy ofthe remaining stars becomes positive, whichcauses the binary to disintegrate. The newneutron star becomes a young radio pulsar.


I t will produce radio pulses for several millions of years, but this is not all.

The most amazing thing is that the second explosion may bring about two radiopulsars, not one. This result of a binary'sevolution was first described by the Sovietastrophysicists Gennady Bisnovaty-Koganand Boris Kornberg in 1974. They started from the observational data. Some ofthe X-ray pulsars in binaries have periodsof less than four seconds and are still accelerating. On the other hand, radio pulsarsalso have periods less than four seconds.This means that a neutron star may becomea radio pulsar if the optical component is"taken away" from a Hercules X-1 binarywith an X-ray pulsar.

This is done by nature. The optical component is "eliminated" by itself after thesecond explosion.

Hence, massive binaries are good producers of radio pulsars. Note that their efficiency exceeds 100 %.

Another general conclusion is very tempting. Suppose that all (or almost all) radio pulsars come from binaries. Strange asit might seem, there are no serious objections to this because it is statistically true.

If every massive binary gives birth toone or two radio pulsars, their birth-ratesmust he approximately equal. Let us seehow we can estimate the birth-rate ofpulsars. We already know how to count


stars and so we divide the number of radiopulsars in the Galaxy by their average lifetime. Unfortunately, it is difficult to estimate the total number of pulsars in theGalaxy because we only see the nearest onesand not all at that. Pulsars radiate in narrow beams and we do not see even nearbypulsars if their beams do not strike theEarth. We could take this into account byfinding the probability that the Earth ishi t but it will not be an accura te calculation because we lack information about theshape of the pulsar beam.

These difficul ties make us trea t estimates of pulsar birth-rates with suspicion.Even so most experts are of the opinionthat one pulsar is born every 15-20 yearsin the Galaxy, which is very close to thebirth-rate of massive binaries determinedby the Salpeter function (see formula (14)).The Salpeter function gives us the following birth-rate for binaries with a massmore than 10M0: one system every 3040 years. Of course, these are only estimates.

In general, there are many coincidencesin astronomy. The well-known Soviet astrophysicist Ernest Dibai used to joke tha t ifwe took two arbitrarily chosen numbers andmultiplied them together whi le preservingtheir dimensions, we would get a thirdquantity which is what astronomers observe.This joke has a grain of bitter truth. You


wi] l hardly count all ideas ruined by suchcoincidences. This "law of large numbers"may easily be explained. Many quantities

~ 1tJkmjs'* First explosion

~l- 100km/s

1ft'" Second explosion

+•• 500km/s

IFig. 95. As a result of t\VO explosions, a radio pulsarmay reach a velocity of several hundreds of kilometres per second

in astronomy are known to an accuracy ofthe order of magnitude and are extremelylarge. So how do we know whether 10~1±1

is the same as 10 2 0±1?!Yet if m assi ve systems are the ancestors

of radio pulsars, we would get a much clearer picture. For example, we know fromobservations tha t there are many radio pul-


sars whose intrinsic velocities are severalhundreds of kilometres per second. Thiswould mean that radio pulsars are also"runaways", but we know where they comefrom. A radio pulsar may attain a highvelocity in two ways, viz. during the firstexplosion (together with a binary) or during the second explosion (disintegration)(Fig. 95). A lucky pulsar could be accelerated to a velocity of 500 km/s, which is exactly the velocity of the fastest pulsars.

Binary Radio Pulsars

None of the radio pulsars found in the firstfive years following their discovery werecomponents of binaries. It seemed thatsome inexorable law caused radio pulsars toavoid binaries.

This strange "law" was violated in July1974 when the American radio astronomersR.A .. Hulse and Joseph Taylor discoveredthe pulsar PSR 1913+16.* I ts average periodwas 0.059 s, and the period's instantaneousvalue varied, in turn, with a period of7.75 hours.

In 1974, this was not a riddle. Thestudy of X-ray pulsars showed that thesevariations result from the Doppler effect

• PSR stands for a "pulsar", and the numbers areits right ascension ex = 19h13m and declination6 = +16°.


and the pulsar's orbital motion. Figure 96shows the radial velocity curve right afterthe discovery of the pulsar. I t differs froma sinusoid, which means that the pulsar'sorbit is very elliptical. The system's mass

w-•.•.,....,....."""""'-,.~.

: "'.fOO

~ ....~"~ 0 t--'\~--+---..........--#---~ \

:~ -fDa·~

~ -2(JO.......~~-~oo

~ o as 1.0Orbitol phose

Fig. 96. The radial velocity curve of the radiopulsar PSR 1913 + 16

function is O.13M0. As soon as it becameclear that it was a binary, investigators began looking carefully for a companion ineach of the electromagnetic wavelengthrange, but in vain.

When discussing the Algol paradox, I saidthat in order to collect all informationabout a binary both components must be observed. The mass function, which is obtainedfrom the apparent radial velocity of oneof the stars, is a function of three unknowns,namely, the masses of the components M 1and M 2 , and the angle of the system's in-


clination i, You cannot find the mass ofei ther star from the radial veloci ty curve,and yet this curve is the only source of information about a pulsar and its invisiblecompanion available to the researcher.

Yet the pulsar's mass has been measured.The mass of its invisible companion hasalso been measured, and the accuracy ofthese measurements was unprecedented. 111deed, these are the only stars in the wholeUniverse whose masses are known to withina few percen t.

The reason for this is that the pulsar'sintrinsic radiation is extraordinarily steadystate. If instead of considering the periodwe speak abou t i ts frequency, we may saythat the pulsar behaves like a star thatgives absolutely thin spectra] lines. Basically, the width of a star's spectral lines isdetermined by the thermal motions of theatoms in its atmosphere, Thermal motionusually has a veloci ty of several ki lometresper second and so the line is "smeared out"by hA/A ==' vTlc == 111000 around the central wavelength. "Smearing" a pulsar's rotation period caused by the slowing downof its rotation occurs in the course of 200 million years. The width of the pulsar's "spectral line" is the reciprocal of the "smearing" time; so the relative width is about10-17!

Since this line is so thin, we may measure the pulsar's radial velocity with a very


high accuracy. All we need is a long periodof observations or, in other words, a largenumber of received pulses. The accuracyachieved in the first year of observationswas such that effects described :by specialand general relativity could be measured.Measuring new effects is like adding newequations for unknowns. The mass functiondetermined by the classical Doppler effectgives us a relationship between three quantities. What we need is another two equations.

Here the transverse Doppler effect comesto rescue. The pulsar's apparent period depends on both the radial velocity and thepulsar's total velocity. If a pulsar movedalong a circle and its total velocity wereconstant, the transverse Doppler effectwould be useless. For we do not know the pulsar's "true" period! But a pulsar's orbit islargely eccentric. This means the totalorbital velocity changes, and so we must observe a periodic change in the pulsar's period: at the periastron (when the stars areclosest together) the pulsation period is longer than at the apastron,

Time dilation in a gravi tational field isanother relativistic effect; we covered itwhen we were discussing the gravity-induced redshift. Then we spoke about it interms of energy (8 quantum reddens becauseit does work against gravity). In terms ofgeneral relativity, quantum reddening oc-:!u-UliU7

306 In the World of Binary stars

curs because time in a strong gravitationalfield flows more slowly than it does for adistant observer.

So the measurement of the relativisticDoppler effect and the gravity-induced redshift gives us another equation. But in reality the classical Doppler effect changes thepulsar's apparent period with the same period (the binary's period) as the relativisticeffect does. They are inseparable. If we onlycould have a look at the binary "from theside"! Then the classical Doppler effect willchange while the relativistic effects willremain the same. We would be able to divide these effects and obtain an additionalequation.

Strange as it might seem, this is possible.There is another quite observable relativistic effect which makes a "space voyage"unnecessary. According to general relativity, Newton's law of universal gravitationis not applicable in strong fields and is notaccurate in weak fields. A binary's gravityat the distance of semimajor axis is notstrong because the orbital velocity is athousand times less than the veloci ty oflight.

In fact, gravity is not exactly inverselyproportional to the square of distance. Under such conditions, the binary's orbit cannot be closed (Fig. 97). I t looks as if theorbit's ellipse were turning slowly in thebinary's orbital plane. This is the same as


to fly around the binary and see it from allsides.

This means that we can both differentiatethe classical and the relativistic Dopplereffect and have a third equation to find allthree parameters, i.e. both masses and theinclination angle.

The pulsar's elliptical orbit makesone revolution in 86 years. I hope you

Fig. 97. The effects of general relativity result inthe star's orbit being open

understand that we do not have to waitthat long. ~easurement accuracy is suchthat only a few years of observation can beused to find the masses of the binary'scomponents. They happen to be similar andequal to (1.42±O.10) M0. The obtainedmasses are in impressing conformity withthe theory of stellar evolution and structure.

20·


In recent years, investigators have discovered two more pulsars which are com ponents of binaries (Table 6). Characteristically, the companions of these stars do notreveal themselves either.

Table 6. Binary radio pulsars

PulsarOrbitalperiod,days

Pulsarpe

riod,s

Mass Mass ofEccen tri- tunc- second

city tton, star,M/M0 M/A10

PSR1913+16 0.32 0.059 0.617 0.1322 1.4PSR0655+64 1.03 0.196 0.000 0.0712 1±O .03PSR0820+02 1.232 0.865 0.012 0.0030 0.2-0.4PSR1953+29 appro 120 0.006 < 0.05 0.0027 0.2-0.4

A binary radiates gravitational waves,General relativity predicts a certain intensity for this radiation. The binary losesboth energy and angular momentum. Thestars "slide" toward each other, with anaccompanying reduction in the system's period.

The radio pulsar's stability helped measure a binary's period and its changes veryaccurately. The period does indeed decrease, and this is happening in compliancewith Einstein's theory.


Options

Only a few of the 400 known radio pulsarsare components of close binaries. All of thebinary radio pulsars hide their partners. I tis probable that these invisible componentsare stars which have come to the end oftheir evolution such as white dwarfs, neutron stars, or black holes. This means thatoccasionally binaries do not disintegrateafter the second explosion.

I hope you remember that a system disintegrates if the explosion happens veryquickly and more than a half of the system'smass is lost. The second condition will besatisfied if the second explosion is precededby an evolutionary loss of a great deal ofmass by the normal star. This may happenduring the second mass exchange.

We know that the second exchange usually ends with a binary consisting of a heliumstar and a neutron star. The helium star'smass is no less than (B-10)M0' Clearly, theexplosion of such a star will destroy thesystem. Given that a would-be supernova'smass is of the order of (2.-3)M0' the starassociation will survive.

Do you still remember the scenario suggested by van den Heuvel for the pulsarHercules X-1? If the star masses had differed much initially, the system would havelost much of its matter already during thefirst exchange (see Fig. 67). Thus a massive


binary would have become a system withmedium mass.

Imagine an explosion in the HZ Herculis-Hercules X-1 system. The system's total mass is (3.5-4) M0' of which the neutronstar accounts for 1.5M0. If the ~ collapseturns the second star into a neutron starwith a mass of 1.5M0. this means the system, if explodes, will lose (0.5-1) M0' i.e.less than a half of the binary's mass. Thusthe system will continue to be binary.

This scenario has one peculiarity, namely,the resulting system of two degeneratestars always has an elliptical orbit. I t ismade elliptical as a result of the rapid lossof mass. True, if the system lost half ofits mass, its total energy would be zeroand the orbit's eccentricity (formal) wouldbe unity. If less is lost, the circular orbitbecomes elliptical.

The system remembers this eccentrici ty.The orbits of the pulsars PSR 1913 + 16and PSR 0820 -t- 02 are truly elliptical (seeTable 6), while the orbit of the radio pulsarPSR 0655 + 64 is circular. I t looks as ifthe binary in the second collapse either lostno mass or it lost it slowly. A third possibility is that there was no collapse at all.Shklovsky called the first possibili ty a"quiet collapse" and it must end up withthe formation of a black hole. The secondpossibility was supported by the Italianastrophysicist Franco Pacini and it may

cu. 8. Wanderers 3ft

occur when the collapsing star rotates veryquickly. Then centrifugal forces stop thecollapse, and the star continues to shrinkgradually only as it is losing its angular momentum. When I say "gradually", I meanslowly enough for the collapse and the resulting loss of mass to go on longer than onerevolution of the 'system. Then the effect ofthe mass loss is averaged and the orbit remains circular.

Finally, the third variant is also possible. If the normal star's mass before itsnuclear sources are exhausted is less thanthe Chandrasekhar limit, there is no collapse. The star shrinks slowly into a whitedwarf.

You see that there are various alternatives. The later the stage we look at in a binary's evolution, the more difficult it is tosee how a binary could have so evolved.

Anyway, it is small wonder that thereare so few binaries consisting of a radiopulsar and a compact star. This is accountedfor by the following. First, binaries composed of stars that differ greatly in mass arerare. Second, a binary's evolution is longand full of hardship, and the binary runsthe risk of disintegrating during the firstmass exchange.

Generally speaking, a binary's later evolutionary stages are still poorly understood.Many ideas do not make up for few calculations. Let us see how all this may end.


The inner (nuclear) evolution of the binary's components has ended. Our currentunderstanding is that the inner evolution isthe slow exhaustion of fuel in the depths ofthe star as it burns out. The next stage ofthe evolution may only be a gradual approach of stars to each other caused by gravitational wave radiation. A moment comeswhen the stars collide. The coalescence occurs almost instantaneously, during thedegenerate star's hydrodynamic timescale.For a neutron star and a black hole it equalsabout 10-°-10-4 s. Their coalescence is followed by a strong gravi tational pulse whichtakes almost all the energy of the binary.What remains is, most probably, a blackhole which, like a radio pulsar, becomes awanderer.

The coalescence of binaries containingwhite dwarfs may be more varied. Theirmaterial still has a nuclear structure andtheir coalescence may be accompanied bythe emission of neutrinos and the ejectionof matter into space. In principle, this phenomenon may be called a supernova explosion.

Glaring Peaks

Our story is drawing to a close. We havetraced the evolution of a binary star fromthe first stages when the components stilllive "separately" to the final stages whenthey can't live apart.

Their binary nature is enriching. This iseven more so at the final stages of the evolution. Their evolutionary paths branch,and the range of possibili ties looks verymuch like a tree with a beautiful crown.Many branches are poorly studied. Someof the branches are doomed to "wither and21-0667

314

~~..:---=::.-.._----....

Evolution scenario


fall". For this reason, when discussing abinary's evolution, I have sought to keepto the major paths. Now let us review mat-

Glaring Peaks 315

terse What else might be seen on this evolution tree? Have we overlooked anythingimportant?

Summing Up

A binary's evolution is like a film shown atan increasing speed (Fig. 98). The stars gothrough the first evolution stages on thenuclear timescale. This takes massive starstens and hundreds of millions of years, andless massive stars billions of years. A moremassive star will fill its Roche lobe first andwill begin the first mass exchange. We considered this when we were discussing the pLyrae system.

There are various outcomes of the firstexchange. In systems with a large ratiobetween the masses of components, the binary loses a great deal of matter (a scenario resulting in systems like Hercules X-1).As a rule, the system's total mass is not reduced radically. The exchange ends upwith systems with a pronounced Algol paradox: the less massive star seems olderthan its partner. Its evolution has gone sofar that its partner can't catch up withit even if it has increased its mass. Depending on the binary's mass, the evolutionof the initially more massive star ends in awhite dwarf, a neutron star, or a blackhole; white dwarf ancestors have massesless than 10M0.21*


~u ving aporttogether

~ 1Jlyro/'

~Al90l,WR+ 0-6

~Ru.nawaystors

Cenfllurus X- 3,CY;RlIS 'X-f

G4Hercules X-I,U6eminorum,55 43~, 1Jurstl'/J

~IISln9/p ,i

WR stars

I \SRfond1'zploslon,or gradual cooling

\ 00 tfin!ll' gnd binary·• • radio p,,/sorJ

"Fig. 98. The evolution of close binaries. Someobservational data are given on the right

Glaring Peaks 317

The formation of the first relativistic star,even though accompanied by an explosionand the ejection of matter into space, doesnot cause the system to fall apart becauseit is the less massive star that explodes.This lucky circumstance connected withrole changing brings about some qualitatively new phenomena in a binary.

My story now continues. Next comes theX-ray stage with binaries like CentaurusX-3 and Cygnus X-i. Strong sources arebrighter than a thousand suns while capturing a small fraction of the stellar windfrom the neighbouring star. The first explosion, however" will leave "birthmarks": thebinary will acquire additional velocity andwill begin wandering throughout the Galaxy.

As a result of its nuclear evolution, thenormal component leaves the main sequenceand inevitably fills its Roche lobe. Thesecond mass exchange begins. The normalstar returns the matter previously takenfrom i ts neighbour.

In low mass binaries with neutron stars,the second exchange resul ts in the forma tionof a bright X-ray source (an X-ray system like Hercules X-1 or an X-ray burster).In systems with white dwarfs, the secondexchange leads to the formation of cataclysmic variable stars, recurrent novae, novae, etc. In dwarf binaries, the second fillingof the Roche lobe does not usually arise


due to nuclear evolution (it is too slow) butdue to the stars' "rapprochement", whichis caused by the binary's loss of angularmomentum. The reason lies in gravitationalwaves and the effect of a magnetic propeller.

In massive binaries, the second exchangeis so rapid that it brings about supercriticalaccretion onto the relativistic star. Thelatter is wrapped in a dense "coat" of outflowing matter and is invisible to an outsideobserver. I t looks like an ordinary star.At the same time, supercritical accretionmay cause relativistic jets. This is not un.likely that we came across this very phenomenon in SS 433.

As a result of the second exchange, therelativistic star is "swallowed" by its companion. This is where a branch forks out.The relativistic star may be totally "swallowed", and the binary becomes a single star,or, having ejected a part of its mass, thebinary may turn into a system composed ofa helium core and a relativistic star. Themost promising candidates for such a binary are "single" Wolf-Rayet stars.

Several hundred thousand years later, thehelium star runs out of its nuclear fueland explodes to give birth to two relativistic stars. The usual result is the formation of, on average, more than one radiopulsar. When the binary disintegrates, theneutron star attains additional velocity. Forthe first million years, the wanderers con-

Glaring Peaks 3i9

tinue to send radio signals into space untilthey become dead stars for ever.

On rare occasions, a binary evolves intoa system of two degenerate stars. Here nature gives us a pure experiment of generalrelativity. Thanks to such systems, astronomers have managed no t only to "weigh"a neutron star with an unprecedented accuracy but also prove that gravitational wavesreally exist.

Much of what I have been telling you inthis book has been discovered and studiedin the recent 10-15 years. New and quiteunexpected phenomena take place in systems with relativistic stars, The discoveries made in this field have caused a "revolution in astronomy", and yet these arebut the first steps. Today we know enoughto realize that there are still many thingsyet to be discovered about the world ofbinaries.

Evolution Squared

Before telling you about the relativisticstages in the evolution of binaries, I madesome calculations of their possible number.My arguments are extremely simple andtherefore quite reliable. The lifetime of abinary after the formation of a relativisticstar in it is close to the nuclear time of thenormal star on whose surface the materialhas flown. The normal star's nuclear time,


however, is not very different from thatof the initially more massive star. I t followsthat there must be the same order of magnitude of X-ray binaries in the Galaxy asthere are ordinary massive binaries, or, inother words, tens and hundreds of thousands.

Compared to this, the number of discovered X-ray binaries (several hundreds orthousands) is ridiculous.

We may get a far more sensitive telescopebut the difference between what we see andwhat we expect to see will not disappear.Today we already see all the galactic sourcesof hard X-radiation with a luminosityseveral hundreds and ~ore the luminosityof the Sun.

We seem to have missed something reallyimportant. A naive approach to the binaries' relativistic stages is ineffective. Theremust be some reason why a relativisticstar in a binary does not necessarily becomea source of X-radiation. The latter is ratherthe exception than the rule. I t is not anordinary star that is to blame. All starshave wind. and all stars may supply fuel foran accreting relativistic star. This meansthat the relativistic star is not that simple.There must be some other condition or, maybe, a number of conditions required toswitch on the accretion machine.

Recent research has helped us understand that compact stars evolve too. I t has

Glaring Peaks 32i

a great impact on the apparent properties ofbinaries. The overwhelming majority ofrelativistic stars in massive binaries areneutron stars. Although the same, eachneutron star may behave differently. Before X-ray pulsars had been discovered,

Shvartsman showed that a neutron star in abinary undergoes at least three stages. Initially it rotates very rapidly ejecting electromagnetic waves and relativistic particles into space like a radio pulsar. True,the radio radiation itself is absorbed bythe stellar wind from the normal companion. This is one of the main reasons whytoday we do not observe radio pulsars pairedwith a normal star.

During its ejection stage, the neutronstar loses energy and slows down. At a


certain moment, the ejection ends. Yet thematerial can't yet fallon the neutron star'ssurface because it scatters the falling plasma by its magnetic field like a gigantic propeller. The star goes on decelerating, and,finally, gravity wins. This is when accretion begins, and a bright X-ray source flaresup.

Each of these stages lasts for tens to hundreds of thousands of years. This is a realevolution. We shall add to this the evolution of the normal component. Now we seethe life of a binary with a neutron star in anew light. This is more than just one oreven two evolutions, it is, rather, evolutionsquared because the states of the normalstar and the neutron star are not directlylinked. We must multiply the four evolutionary stages of a binary with a relativisticcomponent (a star in the main sequence, asupergiant, a star at a stage of exchange,and a helium star) by the number of statesof a neutron star. True, the state of a neutron star is connected (though not rigidly)with the state of its normal component.Therefore, the number of possible states abinary may have is half this number. This,however, does not change the" essence.

In fact, the neutron star may be in atleast eight, not three, states. Only one ofthese states, namely, accretion, is accompanied by strong thermal X-radiation. Inother cases which, I must confess, have

Glaring Peaks 323

been less studied, the neutron star is notthat noticeable. This is why we first cameacross the X-ray stages in the Galaxy. They

"Non-X-ray stages in a binary's evolution"

are simply eye-catching. I t is a glaringpeak, but the peak of an iceberg.

Weare now making our first steps to findnon-X-ray relativistic binaries, and we already have had some success.

We have already counted the relativisticsystems and have seen that they are only alittle less numerous than ordinary systems.


Every second or third massive star whichseems to be alone has, in fact, a relativisticcompanion. In the early 1980s, astronomersset out to look for such binaries, and theirsearch was successful. This work is extremely time- and effort-consuming. First theydecided to check O-B runaway stars whichhad been always considered singles.

These stars are accelerated to great velocities during the binary's first explosion.If there had been a collapse, there must bea neutron star. Being attracted by the neutron star's gravity, the massive star mustmove along its orbit at several tens of kilometres per second. Traces of such motionshave recently been discovered in a numberof O-B stars.

This, however, is a subject for a bookwhich is yet to be written.

To the reader

Mir Publishers would be grateful for your comments on the content, translation and design ofthis book. We would also be pleased to receive anyother suggestions you may wish to make.Our address is:Mir Publishers2 Pervy Rizhsky PereulokI-tiO, GSP, Moscow, 129820USSR

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