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SOLUTIONS
2.2-The flux at the film plane is
F,= Z,cos8dQ~Ir,cost?AQ, I since A!d< 1. Now we have
AA cos 8 r2
A&! =
where hA = n(d/2)’ and r = L/cost?. Thus
where f i=L/d.
1.2-We first find the energy absorbed in volume dV and time dt due to radiation in solid angle dQ and frequency range dv. Let dA be the cross-sectional area of the volume normal to the radiation. Then the
313
RADIATIVE PROCESSE S IN ASTROPHYSICS GEORGE B. RYBICKI, ALAN P. LIGHTMAN
Copyright 0 2004 W Y - V C H Verlag GmbH L Co. KCaA
314 solvtim
energy absorbed is - dI,dA dt dvdQ, where dI, = - q,Z,dl by the absorp- tion law (1.20) and where dl is the thickness of the volume along the direction of the radiation. Since d V = dA dl, the energy absorbed is %I, dVdtd9dv . Integrating over solid angle gives the energy absorbed per unit volume per unit time per unit frequency range to be 47rq,J,, which is also equal to cq, u, by (1.7). Noting that each photoionization requires an amount hv of energy and that 4. = nao, we obtain the number of photoioni- zations per unit volume per unit time:
00 u, u,dv ~ 477n0~0w dyJy dv = ena
hv 1, hv '
1.3
a. The transfer equation with no absorption is
dl . r - =J= - cis 477.
Here I is defined in terms of photon numbers rather than energy (photons-cm-2-s- l-ster- '). Integrating along a line through the center gives
b. The observed average intensity is equal to the total flux divided by the solid angle accepted by the detector, AQ2,, = a(A0,J2, and AOBt is the detector half angle. Since the total luminosity L of the source (photons-s-') is simply equal to (477/3)R3r, the flux is
where d is the distance to the source. Thus
- R 'r I =
3 ~d 2( A OmJ2
Noting that AO,= R / d is the angular size of the source, we can write
For a completely unresolved source AO,<AO,,,, so that I<<I,.
soiutiom 315
1.4
a. Assume that the luminous object has spherical symmetry, so that the flux F at distance r is just L/(4nr2). From Eq. (1.34) the outward radiation force per unit mass on the cloud is
KF KL f =-=- c 4ncr2’
rad
The inward gravitational force per unit mass due to the object is
The condition of ejection is that fgrav<frad, which can be written
M K -<- L 4nGc
b. The cloud experiences an inward force per unit mass GeffM/r2, where the “‘effective” gravitational constant is given by
K L 4nMc
Geff=G--.
Thus the effective potential per unit mass is - GeffM/r. Note that Geff is negative under conditions for ejection. Setting up the conservation of energy connecting the state at r = R and r = m, we obtain
-- G M eff - R ; u2,
which can be written
G M ) .
c. The minimum luminosity occurs when the inequality in part (a) becomes an equality. Substitution of K = a,/m, then gves the stated result.
I . 5
a. The brightness is I, = Fv/A!2, where A 0 = n(A0)’. Here A 0 = 8/2 = 2.15 arc min= 6.25 X radian. Thus
I,= 1 . 3 ~ lopi3 ergcrn-’s-’Hz-’ster-’
C 2 Tb= - Ip=4 .2X 10’ K . 2v2k
316 SoIutions
Since hv<<kT,, the use of the Rayleigl-Jeans approximation is ap- propriate.
b. TbaZva(All)-*. If the true All is smaller, the true Tb will be larger than stated above.
c. From Eq. (1.56b) we find vm,,=2.5 X 10l8 Hz.
d. The best that can be said is T > Tb. This follows from Eq. (1.62) with Tb(o)=O. In general, the maximum emission of any thermal emitter at given temperature T will occur when the source is optically thick (see Problem 1.8 d) .
1.6-Since u( T ) = aT4, Eqs. (1.40) can be written
(g) =4aVT2, ( $ ) T = ?aT3 4 V
It follows that S = (4 /3 ) aVT3 +constant. The constant must be chosen to be zero, so that S+O as T+O (third law of thermodynamics).
1.7
a. The equation of statistical equilibrium [Eq. (1.69)] with B,, = O be- comes
n,B,,J= n,A,,.
With the Boltzmann law (1.70) this implies
This cannot equal the Planck function with A * , / B , , independent of temperature. However, the choice (A2,/ B12)( g2/gI) = 2 h v i / c 2 does yield J= By, in the Wien limit, hv,>>kT, [cf. Eq. (1.54)].
b. The main difference between the interactions of neutrinos and photons with the atom is that the former particles are fermions, whereas the latter are bosons. Stimulated emission in a fermion field would place two particles in the same state and thus violate the Pauli exclusion principle. This process is replaced by inhibited emission, in which an atom in the excited state is prevented from emitting a fermion when one is already present. The analysis using the Einstein coefficient is
Sdutiom 317
identical to the photon case, except BZl+ - Bzl . One then has, at equilibrium, for atoms of temperature T,
n , B , J = n,A,, - n,B2,J
and one obtains the same relationship as before for the Einstein coefficients. These coefficients are properties of the atom alone and clearly must be the same, regardless of the external interactions used to derive them.
I . 8
a. Note: that j , = P,/477 and that, effectively, a, =0, since the cloud is optically thin. Then, using Eq. ( I .24),
b. The total power emitted by the cloud is L = ( 4 / 3 ) r R 3 P , where P = P,, dv. Then
L = 4nR ,aT:,,
by definition of Teff, so that
PR T e f f = ( X ) .
c. Let d be the distance from the spherical cloud to the earth. Energy conservation gives a relation between F,, the flux at the earth, and P,,:
4 3
4ad2F,, = - VR 'P,,
R 3
3d F,,= P . 2 .
318 Solutions
d. From Eq. (1.30), with S,=B,(T) , I,(O)=O, ~,<<1,
- Z, = B,( T ) ( 1 - e ‘~)=T,B,( T)<<B,( T ) .
With the definition of Tb from Eq. (1.59),
Bv(Tb)<<Bv(T) ,
and the monotoncity of B,( T) with T, we have Tb< T
e. For the optically thick case the results are:
a’. From Eq. (1.30) with T,>> 1 and with S, = B,( T) we have Z, = B,( T)
b. Since Z,=B,, the flux at the surface is the blackbody flux, so
c’. The monochromatic flux at the surface is TB,( T) [cf. Eq. (1.14)], so
independent of 6 .
T,,, = T.
using the inverse square law gives
d. From (a’) and Eq. (1.59) we have B,(T,)= B,(T), which implies Tb = T.
Z.9-Ray A starts on the central object with intensity B,(T,), and this is essentially the observed intensity at Y = Y,, where the absorption in the shell is negligible. The observed intensity at v = v, depends on whether the source function in the shell, namely, B,(T,), is greater or smaller than the incident intensity B,( T,). (See Eq. (1.30) and subsequent discussion.) When T, < T, we have B,( T,) <B,( Tc), and the intensity is reduced by passing through the shell, so that Zz is larger than Z;. When T, > T,, Z( will be larger than Z;f .
Ray 3 starts with zero intensity, which is the observed intensity at Y = Y, : Z: = 0. At v = v,, the observed intensity will be somewhere between zero and the maximum B,(T,), depending on the optical depth. In any case, Z: >Z: always. These cases are illustrated in Fig. S. 1.
Z.ZO-The radiative diffusion equation is of the form [cf. Eq. (1.119b)l
Solutions 319
‘I! f ( ‘ I )
I&! t
vg y : y o y 1
Figure S.1 Intensity from a bhckbody surrounded by a thermal absorbing shell (a) along ray A when T, < T,, (6) along ray A when T, > T,, (r) along ruy B when 1;, < T,, (d) along my B when T, > T,.
where
a” a + a, €=-
is the probability per interaction that the photon will be absorbed. The general solution to Eq. (1) is
J, - B, = C,eT* + C2e-*=, (3)
where
and C , and C, are independent of r and to be determined by boundary conditions. The proper boundary conditions for a semi-infinite half-space are that J , remain finite as 7-60 and that there be no incident intensity at r S O : Z,(r =O,p<O)=O. The first boundary condition requires C , =0, that
320 Solutim
is,
J, (T) - B, = C,e-'*. ( 5 )
Note that J, approaches B, after an effective optical depth 7* [answer to
Now, using the two-stream approximation to the boundary condition at Part (b)l*
7=0, Eq. (1.124a)
a J - ~ a t ? = & v3 a7
we obtain
c*= - B , ( l + v i ) - ' ,
In the Eddington approximation the flux, F,(T), satisfies [cf. Eqs. (1.1 13b) and (1.118)]
Thus, we have the result for the emergent flux:
For small E this differs from Eq. (1.103) only by a factor I /* . (Note F,(O)= & / A ) . Note, further, that from J,,(T), one may compute the source function
S,( T ) = ( I - E) J, + EB,
= B, [ 1 - ( I - G ) e -'*I and thus the intensity at any optical depth. In the two-stream approxima- tion the intensity in the outward direction ( p = 1/@ ) at 7=0 is
2 B , G 1,+(0)= ___
l + f i (9)
2.1-Writing
B( t) = Re 9?1 e
and noting the results (1) = 1 and (e - c2 iw ' ) = 0, we obtain
= Re '?!?I * e iw',
(AB) =+Re@*% = +Re@% *.
2.2
a. Maxwell's equations (2.6), with j - aE, are
V *E=477p V*B=O,
We seek plane-wave solutions, so we assume solutions of the form (2.18a, b) and p = poexp[i(k*r - or)]. This gives
ik 08, Eo = 477p0 ik*12Bo= 0
iw i k x l , E o = - i 2 B o
C
where mz = p( 1 + 477ui/a). Dotting the vector k into the last equation, we find k*i ,=O, which implies po=O from the first equation. Thus these equations have the same form as Eqs. (2.19), except for the additional m2 factor. The solution proceeds analogously, leading to the dispersion relation
w2m2
C2
k 2 = - .
b. Take k along the z-axis, with E and B along the x- and y-axes, respectively. Then
E = g~~~ - Im(m)zw/c , i [Re(m)zw/c -wf ]
B = $ B , ~ - I m ( m ) a w / c e i [ R e ( m ) z w / c - w f ]
9
7
and the Poynting vector is
where the absorption coefficient a, is given by
2w a, = - Im( m).
C
2.3
a. Substituting FLorentl = Q E and F,,,, = - flv into the force equation gives v=QE/fl. The direction of the velocity rotates uniformly in a plane normal to the propagation direction with period 2n/w. Thus the radius is found from
to be r = Q E / P w .
b. The power dissipated is P = - v . Fvisc = flu2 = Q 2E '/ f l . Since the orbit of the charge is constant in time, this is the power transmitted to the fluid.
c. The magnetic force is in the direction of propagation and has magni- tude Fmag= QBu/c = Q E u / c = Q Z E 2 / P c . Here we have used [El = IBI for a free wave.
d. Using the center of the charge's motion as an origin we find the magnitude of the torque to be 7 = IFLorentz x rI = Q2E ' / f lu. For a left-hand circularly polarized wave the E-vector, and thus the charge, rotates counterclockwise as viewed facing the wave. This imparts a torque along the direction of propagation. The opposite holds for right-hand polarization. Thus 7 = & Q 2 E 2 / P w .
e. The absorption cross section can be found from P = a S where the Poynting flux is S = cE '/4n. Thus u = 4mQ 2/ f lc .
f. P, Fmag, and 7 are the rates of energy, momentum, and angular momentum, respectively, given to the fluid. From the results above we have the ratios Fmag/ P = (momentum)/(energy) = I / c and r / P = (angular momentum)/(energy) = ? 1 /a. Assuming the quantum rela- tion E=hw for a single photon then implies the relations p = E / c = h / X = h k and J = ? E / w = ? A . Since these refer to properties of pho- tons, they are applicable in general, not just to the limited problem considered here.
g. The case of linear polarization leads to the (primed) results:
solutionr 323
a'. Again v = QE/P. Since E oscillates harmonically along one axis (Ex = E,coswt), so does the particle. Taking an appropriate origin, we find for the displacement: x ( t ) = ( QE0/q8) sin wt. The maxi- mum displacement from the origin is QE,,/wP.
b. The average power dissipated is (P ) = - ( v - F,,,) = P ( u 2 ) = (Q2EOZ/P)(~os2wt> = Q2Ei/2P.
c'. The average magnetic force is along the direction of propagation arid has the magnitude (Ems,) = ( Q u B / c ) = (Q/c)(v - E) =
( Q 2E: / fic>(cos2 wf ) = Q 2E,2/2fic.
d . There is no torque on the fluid, since F,,,, always acts along a line through the origin.
e'. The absorption cross section can be found from the relation (P) =
o ( S ) . The average Poynting vector is cE,2/87rr, by Eq. (2.24b). Thus u = 47rQ '/Pc, as before.
f. The power and magnetic force are the same as before, with E2+ E,:/2. Their ratio is the same, and we conclude that p = E/c. The angular momentum is zero, however. Quantum mechanically this comes about because a linearly polarized photon is a superposition of two circularly polarized photons of opposite helicity.
2.4-Suppose that V x H=4ac- 'j. Taking the divergence of both sides yields V.j=O. But the equation of charge conservation is V * j = - + / a t . Therefore, this form of the field equation applies only to the special case
Furthermore, omitting the displacement current form the derivation b=O.
leading to (2.17) gives: 4rup aE
c 2 a t
47rup aB c2 at .
V2E= - -
V2B= - -
That is, the equations for E and B become (parabolic) diffusion equations rather than (hyperbolic) wave equations.
3.1
a. By analogy with the Larmor formula for electric dipole radiation, the power radiated by a magnetic dipole is
324 soluriovu
To evaluate this expression, we first note that the component of m along the rotation axis, Imlcosa, is a constant. Thus
ImI = w21ml sina, (2)
since m, = m sin a sin wt and q, = m sin a cos a t . Next, we wish to ex- press Iml in terms of B,, the magnetic field strength at the star surface. In electrostatics, the electric field due to a dipole of strength d Eel is obtained in the following way:
E=V+,
At the “pole,” 0 = 0, the electric field has a magnitude E = 2d/r3. By analogy, the magnetic field B, at the magnetic pole has the value
2m B,= -
R 3 *
Substituting Eqs. (2) and (3) into (l), we obtain
w4R 6Bi sin2 (Y P =
6c3
(3)
(4)
b. Assuming the neutron star is a homogeneous body, its rotational energy , ErOt, satisfies
E,,, = M R 2u2. ( 5 )
Now, using P = - E,, = - 2 / 5 M R ’wrj and substituting from Eq. (4), we obtain
c. To obtain quick quantitative estimates of functions as their arguments assume particular values, astrophysicists frequently write equations with all the quantities normalized to some standard values. Thus, for
this problem, we can express Eqs. (4) and (6) in the form
Thus for a = lo4 s-I, Id s-I, Id s-I, P/3.1 x l v 3 erg s - ' has values 1, lo-', and 7/42 yr has values 1, Id, and lo4.
3.2-Since vI<<c we can compute the radiation field Erad from Eq. (3.15a):
e
c r Erad= ,nx(nxu).
Also, since the system is axially symmetric, and we will eventually average the motion over time, no generality is lost by taking n in they-t plane (see Fig. S.2). The plane normal to n contains the unit vectors ii, = - f and 8,,
X
F'igure S.2 Geometry for pdrrriration decompositiion of radiation emined by a eireulating charge.
326 SoIutim
which lies in they-z plane. Letting w = v l / a , we have for the particle
r=a(fcoswt+jrsinot)
u = uL ( - S sin wt + y coswr)
u = - v,w(ii coswt + 9 sin w t ) .
The components of n and are given by
n = 9 sine + icose,
I,= -ycos8+isin8.
We now have
n xu = - D ,w(a, cos B sin wr - a, coswt),
n x (n xu) = - D lw(8, coswt + 8, cosBsinwt), ev,w
rc * Erad= - - (a, coswt +a2cos#sinwr).
a. The power per solid angle is found from
2 2 2
~ = - IErad12r2 = ___ dP c e v I w ( cos2 wt + cos2 0 sin2 wt).
di2 47r 4ac3
Averagng this over time gives
2 2 2 dP e v l w (-)=- ( 1 + COSZB).
di2 87rc3
b. Comparing the formula for Qad with Eq. (2.37) (taking the i and 9 directions of that equation to be now 8, and a,, respectively), we find
eu,w ev,w
rc ’ rc G , = - - , E,=-- cos e
+, -0, +2 = n / 2 .
The Stokes parameters are, therefore, from Eqs. (2.40),
where A =(euLo/rc2)’ . The radiation is 1w0 elliptically polarized ( I 2 = U z + Q 2 + V 2 ) , with 8, and a2 being principal axes. Special cases are
8 = 0 : left-hand circular polarization.
8 = n/2 : linear polarization along H,
8=a : right-hand circular polarization
c. Sinct: Qad contains only sinot and cosof terms, the radiation is monochromatic at frequency w. (See, however, Problem 3.7 when radiation of higher order than dipole is included.)
= ewrB/c, d. Setting the magnetic force equal to the centripetal force gives w:mw’r
eB a= - mc
Using the result of part (a) gives (P):
2 2 2 2 e v l o =-- 3 c3 ’
Using ro= e 2 / m c 2 and PI = u I / c , this becomes
( P > = ;r;cp:B’.
e. The Lorentz force law for an incident electric field E gives maw2 = eE or t‘ = e E / m o . Thus
dP &E2
d 0 477 (->=- ( 1 +c0s2fl).
Now use the results
c E 2 477
( S ) = -
328 Solutiom
to obtain
To obtain the cross section for unpolarized radiation we should average this cross section with one for circular polarization of the opposite helicity. But since these cross sections do not depend on helicity the unpolarized results are the same:
This is the same result obtained previously in Eq. (3.40). The total cross section is just equal to the Thomson cross section, independent of the polarization:
3.3
a. Use Eq. (3.15a) with qti= - a2dcos~f for each dipole, noting that the retarded times for each differ by A t = (L/c)sinB (see Fig. S.3). Then
W L
rc IE,,J = - 7 [ d,coswt + d, cosw(t -At)] sin0
W L
rc2 =- - [(d,+d,cosS)coswt+ d2sin6sinwr]sin8,
To observer
Figure S.3 distance L.
Geometry for emission fmm two dipole radiators sepmted by
where S = wAt = W L sin 8 / c . Squaring and averaging over time, we find
w4 sin2 8 (IEr,dl2> = [ (d, + d2cosI?)2+(d2sin8)2]
w4 sin2 8 =--- (df+2d,d2cosS+d~) .
2 ~ 4
We have finally,
u4sin2B
8 T C 3 =- (d: + 2d,d2cosS + dj).
b. When L<<X, we have 6 r2~LsinB/?,<< 1, and
dP ~ ~ s i n ' 8
d!2 8nc3 (-)=- (dl + dd2.
which is the radiation from an oscillating charge with dipole moment d, + d2.
3.4
If the cloud is unresolved, then by symmetry there can be no net polarization. Physically, the polarization from different regions of the cloud cancel.
Figure S.4a shows a typical scattering event in the scattering plane. Radiation from the object can be decomposed into two linearly polarized beams of equal magnitude, one in the plane of scattering and one normal to it. The first produces scattered radiation with polariza- tion direction in the plane of scattering, the second having direction normal to it. These are not of equal magnitude, being in the ratio cos28 : 1 respectively [cf. discussion leading to Eq. (3.41)) The normal component thus dominates for each value of 8. Integration along the line of sight then gives an observed intensity with dominant compo- nent normal to the scattering plane, or, to the observer viewing the plane of the sky, normal to the radial line connecting the object and the point of observation. The plane of the sky with its observed polarization directions is illustrated in Fig. S.4b.
That the central object can be clearly seen implies that T~ = n,a ,R5 1, and thus neS(Ra , ) - '=5X 1 6 cmp3.
330 ~olutions
To observer
( ' I )
Figure S.4a Scattering event from a spherical cloud
I / > )
Figure S.4b Obseroed po&uization directions in the prclne of the sky.
3.5-Since the field inside the sphere is independent of position, it acts like a single dipole with moment:
where the incident field is E,coswt%. From Eq. (3.23b) with k = w / c we obtain the time averaged power:
( P ) = ( 4 7 ~ a / 3 ) ~ ( 1 + 4 ~ a / 3 ) - ~ k ~ u % E i / 3 .
From u = ( P ) / ( S ) with ( S ) =cEi/8n, we get the required result.
Solutiom 331
3.6
a. The Fourier transform can be performed explicitly by changing vari- ables in each term of the sum over i:
N I k(o) = 2 g Eo( u ) e '""e iwr, du
r = l --oo
b. Explicitly, we have
Now, since ti and 5 are randomly distributed, the second term averages to zero.
c. Equation (2.33) gives the spectrum:
d. In this case we may take each ~ , x O , because all the pulses have the same arrival time, to order (size of region)/(wavelength). Thus
and
- = N 2 ( - ) dW dW . dA do dA do single pulse
332 ~o~utions
3.7-Let the charge move in the x-y plane, and its position be denoted by ro(t). Then
ro( t ) = ro( cos w,t j i + sin wot 9 )
ro( t ) = wore( - sinw,t 5 + cos wot 9) .
The current is
j(r,t)=ero(t)6(r-ro(t)). (1)
Since j(r,t) is clearly periodic, we write it as a Fourier series:
to
j(r,t)=fjA(r)+ C [jl,(r)cosnwOt+j2,(r)sinnwot] (2) n - 1
with
From Eq. (3.31), we have the Fourier terms for the I-pole contribution to A;, A;:
where i = 1,2 for the coefficient of the cosine or sine term, respectively, in the series. Now, substituting Eq. (3) into Eq. (4) and performing the d3r‘ integral first, we have for the dipole contribution (I = 1):
[ A!,(r)] I a / 2 T ( - sinwot f + cosw,tf) cosnwold(wot)a an, I 9 , (5a) 0
[ A , Z ( ~ ) ] 1 a an , 11, (5b)
where we have used the orthogonality property of sines and cosines and 6 is the Kronecker delta. Thus the dipole contribution to the vector potential is nonzero only at n = 1 (w = w,) and the cosine and sine coefficients are vectors along the 9 and f directions, respectively.
solutiom 333
For the quadrupole contribution we obtain, using Eqs. (l), (3), and (4) and performing the r’ integration,
[ A!,(r)I2a: J2n( - sinw,t 22 + cosw,t 9) 0
where nx and ny are the x and y components of the unit vector n. Now, using standard trigonometric identities, we can write this as
Analogously, we have the result,
Thus the quadrupole contribution is nonzero only at n = 2 (0=2w,) . It should be clear that, in the general case, the I-pole contribution is solely at the harmonic o = Lo.
4.1-The key idea in this problem is that, because of relativistic beaming of radiation, portions of the surface that are at sufficiently large angles from the observer’s line of sight never communicate with him. Because the sphere is optically thick, only surface elements can be observed. Referring to Fig. S.5 we see that the cones of emission at any surface point (half-angle of order y -’) include the observer’s direction only for a limited region of the sphere, between points B and B‘. Emission from points such as C will not reach the observer.
The observed duration of any pulse has as a lower bound the time delay between the observed radiation from A and B:
But 0, = y-’ from the geometry, so R 5 2 y 2 c A t .
334 Solutions
Fiem S.5 Geometry of erniksion cones fmm p i n t s on the surface of a rapidlv expanding shelL
4.2-Suppose that K and K‘ count the same set of M stars. Since M is a scalar, they must agree on the count:
P(8,+)d0 = P(8’, +’)dM,
P(@’,+’)= W,+)-&7. d 0
Now, if we take 8 as the angle of the incident light ray with the velocity axis (ie., 8=?r corresponds to the forward direction), we may use Eq. (4.95) for dft/dQ‘ to obtain
Note that P(O’,+’)=P(O,@) if p=O, and that
I P ( e t , + r ) m = N
= N .
Finally, since P(f?’,(p’) has a maximum at 8’= T , the stars bunch up in the forward direction.
Sdutions 335
4.3
a. Use Eqs. (4.2) and (4.5) to compute changes in times and velocities measured in different frames:
dux = y-2u-2du:,
Hence
A simdar result holds for a,.
Then a = 1, and from part (a), b. If the particle is at rest instantaneously in K ‘ , then u:=u;= ui=O.
a;, = y3a,,,
2 a; = y a, .
4.4
a. An inertial frame instantaneously at rest with respect to the rocket measures its acceleration as g . Transforming from this frame (the “primed frame”) to the earth frame with problem 4.3 gives
Note that the choice of which frame is “primed” is not arbitrary, because in only one is the rocket instantaneously at rest.
336 Solutiom
b. Since y = ( 1 - p 2, - ' I2, the variables are immediately separable:
Let /3 = sin u. Then the integral becomes
P - tanu = du
y q ' so that
' = s' +constant.
Since /3 -0 at t = 0, the constant vanishes. Inverting the last expression gives
g t l c . P ( t ) = d m
c. Set p = c - ' d x / d t and separate variables again:
Substituting u =( gt /c )2+ 1 to transform the integral,
x = gd?l2+l g +constant.
Since x = 0 at t = 0, we obtain
d. To find the proper time of the rocket we set
e- du
gr +constant,
sinh-' sf - +constant. ( c l - c
Since: t = 0 when r = 0,
C = sinh( F). e. By symmetry, the journey consists of four segments of equal distance
and time. The maximum distance away from earth is twice the result of part (a), which with part (d) can be written
With r = 10 yr, g =980 cm sK2, this yields d=2.8 X ld2 cm. (One could visit the center of our galaxy).
f. Four times the time given in part (d) is
Unless their friends have also been exploring, the answer is "no."
g. Changing g, to 2g gives T = 8.8 X lo8 yr and d=4.2 X ld6 cm. (Were it not for energetic and shielding considerations, round-trip intergalactic travel within one's lifetime would be possible.)
4.5-We demonstrate a simple counterexample: A " = B"=(1,0,0,0). Using the boost (4.20) in the transformation law (4.30), we obtain A'" = B'" = y ( l , - p , O , O ) . Now A"B"= 1, which does not equal A'"B'"=(1+p2)/ ( I - p 2 ) , unless P = O .
338 so~utions
- - 0 0 0
y’u’ y‘ 0 0 U 0
0 0 1 - u (1 - 0 y 2 0
4.6-Make an arbitrary boost in a direction that lies in the y-z plane. The photon now may have nonvanishing py and pz as well as px. A pure rotation lines up the coordinate frame again so that only p x is nonvanish- ing, but p x does not now have its original magnitude. So, make a final boost alongp” either to redshift it or to blueshift it to the original value. Since E2-p2=0, E also has its original value. You can easily convince yourself that the product of these transformations is not a pure rotation; there is in general a net boost left over. An example is:
where u’ is chosen to satisfy the equality yy’(1 + u’)= 1 , which 0’ = U’/ (U’ - 2).
gives
4.7
a. Suppose that the blob moves from points 1 to 2 in a time A t (see Fig. S.6). Because 2 is closer to the observer than 1, the apparent time difference between light received by him is
(c.f. discussion of Doppler effect, $4.4). The apparent velocity on the sky is
Solutiom 339
1
Figuw S.6 Geometry of emission fmm a mooing source.
b. Differentiation with respect to 6 and setting to zero yields the critical angle R, :
U cos@,=-=rp; C ~ i n @ ~ = j / q = y - ' .
The maximum apparent 0 is thus
This clearly exceeds c when y >> 1.
4.8-Let K be a frame in which the two velocities are v1 and v2. The four-velocities are
2 2
u, = Y ( U , ) ( L V , h u2= Y(U2)(1,V2).
Let K' be observer 1's rest frame. The four-velocities are:
2 3
u; = ( I , O), u,l = y( u)( 1 f v), 2 -2 - 2 - 3 2 3
and U,'. Ui = y(u), so t@t u2 = 1 - (U;. U;)-'. But U , . U, = U,'. ll; = scalar, so that u2 = 1 - (d,. U2)-2. Using the above expressions for U , and G, we obtain the desired result.
340 Solutim
4.9-We must find a tensor expression that reduces to j = aE in the fluid rest frame. From Eq. (4.59) for Fry, the “time components” Fok, contain the electric field. We also know that the fluid four-velocity, Up, has only time components in the fluid rest frame. Thus we try
(we use units where c = I.) This equation has the right space components in the rest frame
j k = aFk’Uu = aEk,
but the time component ( p = 0) in the rest frame gives p = 0, an unaccept- able constraint. Thus we want to subtract out of Eq. (1) its time compo- nent in the rest frame, that is, we need to project out only that part which is orthogonal to U :
j a - jaUp U“ = aFavU,,, (2)
where we have used Fa”Ua U,, = 0. Now, Eq. (2) is correct. It is manifestly a tensor equation; its space components give j = aE in the rest frame (where Uk=O) and its time component in the rest frame gives O = O , that is, no constraint on p.
4.10
a. If the radiation is isotropic in K‘, there is no preferred direction in that frame, so by symmetry the particle must remain at rest in K‘, that is, a’P = 0. Since a‘ is a four-vector that vanishes in K‘, it also vanishes in K.
b. In K‘:
4
PA = ( W‘,O),
in units where c = 1. Transforming to K , one obtains
giving a spatial momentum, P,,, = y W’p.
c. As shown in (a), the acceleration vanishes in both frames. Momentum is conserved because the particle loses mass, even though its speed in K is unchanged. The mass loss Am‘ = W’ in K’ is measured as Am = y W’
solutiom 341
by an observer in K . Since the particle has speed p, the associated momentum change is - P A , = -pyW‘ , which just balances the momentum of the radiation.
4.11
a. Let y ’ ~ ( 1 - u’’ ) -~/ ’ and P‘ and P” be the total four-momentum vectors before and after absorption. Then ( c = I),
P’ = ( m + hv,hv, o,o),
F” = ylml( l,v’,o,o).
Conservation of energy and momentum gives P = P ’, or m + hv = y’m’ and hv= y’u’m’. Thus
- 1
b. Suppose that in the lab frame K the particle now initially has velocity u. In the frame K’ in which the particle is at rest the photon has frequency
v’= yv( 1 - pcose).
In frame K‘ we now perform the same computation as in part (a), and we obtain
Beca.use m/m’ is a scalar quantity (the ratio of rest masses), this equation now holds in any frame, including K .
4.12
a. Since the photons carry no angular momentum with respect to the star (unpolarized radiation), none can be given to the particle.
b. By consideration of a Lorentz frame instantaneously at rest with respect to the particle, plus the result of problem 4.10(a), we have immediately that the particle’s u and direction cannot change at the
342 solutions
instant of emission. Then
Thus from Problem 4.1 I@) we have
I m 1, m' ,
where f? = 71/2 in the Doppler formula.
c. Expanding to first order in hu/mc2,
hu N
A1
10 mc2 ' _---
d. The net effect of absorbing and reemitting many photons is for the particle to slowly spiral in towards the sun (assuming AI/I,<<l per orbit), with no change in its mass; the entire effect, in the lab frame, comes from a nonradial redirection of the incident photons. Letting e be the sun's luminosity and be the rate of photon absorption,
Now, using the fact that Iccr1/2 for circular orbits,
1 dl - I 1 dr (y) Eja
- - - - - -= I dt 2 r dt per photon
Combining the above equations with part (c) we have an equation for d r l d t , whose solution is
Substituting C=4X ergs- ' , rnwlO-I ' g, u - I O - ~ cmP2, R,-7x 10" cm, 1 AU-1.5x l O I 3 cm, we find that the time for r to decrease from 1 AU to R, is
1-5 x lo4 yr
Sdutiom 343
4.13
a. From Eq. (4.1 10) we have
I” I”’! y 3 v13 . -= -
Substituting the Planck function I ,=B,(T) from Eq. (1.51) we obtain
using the Doppler formula (4.12b) with 8 + 8 + ~ , since radiation propagates in the direction opposite to the viewing angle. If we define
then lu’, = flu,( T’), and for each direction the observed radiation is blackbody.
b. Expanding for /3<<1, we obtain T‘xT(l+pcosO‘), so that TAm= T( I + p ) and Tkin= T( 1 - 8). Then
for hv/kT-0.18<<1. Thus
so that 05300 km s-’ .
4.2l-Let the primed frame be the instantaneous rest frame of the particle. Then
4,l - Y(&,h+PdE’) dP;, FIi 1
=-= FlI = - - dt y( dt’ + p/ c dx’) dt‘
344 Solutions
because the particle is instantaneously at rest in this frame. Similarly,
Then, from (4.92),
4.25-The two scalar invariants of the electromagnetic field strengths are E-B and E2 - B2. If we can show that W:m - IS12/c2 can be written solely in terms of these two invariants, then it must be an invariant. Since We, = ( 8 r ) - ’ ( E 2 + B2) and S = (c/4?r)Ex B, we have
=(E2+B2)’-41ExBI2
= E4 + 2 E ’B + B - 4E2B2sin20
= E4+2E2B2+ B’-4E2B2 ( 1- ___ ‘E;;!) = ( E - B 2)2 + 4(E* B)’.
4.16-The solution to this problem requires some tensor index manipu- lation plus use of Maxwell’s equations in tensor form.
Solutiom 345
Now,
relabeling indices and using antisymmetry of F. Thus
because the two quantities in parentheses vanish according to Maxwell's equations in free space. [See Eqs. (4.60) and (4.61).]
5. I
a. The optically thin luminosity is equal to the volume V=(4/3)77R3(t) times the power radiated per unit volume, Eq. (5.15b):
Clhln = 1 . 7 ~ 10-27n,npT~/2V,
where we have taken gB = 1.2. Now, n, = np = Mo/mp V, where mp = hydrogen mass. Thus
b. The optically thick luminosity is equal to the surface area 477R2(t) times the blackbody flux, Eq. (1.43):
c. The transition between thick and thin cases occurs roughly when tthm % Cthlck. Setting the above expressions equal for t = to we obtain
R( to) ~ 4 . 7 x 104M,2/5T[ '"O.
[An alternate solution follows by setting c ~ { ~ R ( t ~ ) a l , using Eq. (5.20). This yields a result of the same form, but with coefficient 2.0x 104.1
d. See Fig. S.7.
5.2-The knee in the spectrum gives T:
346 Solutim
L
' 0
Figure S. 7 Luminosity as a fwretion of time, from a collpping sphere.
From Eq. (5.15b) we obtain
(1.4X 1 0 - ~ ~ ~ ~ / % , @ g g ) . 1 4nR3 F = - -
4aL2 3
At T= lo9 K the gas is completely ionized. If we can assume it is pure hydrogen, n, = n,. (Including a typical helium abundance makes only a negligible difference.) Then
Taking Z = 1 and gB = 1.2 gives
Hydrostatic equilibrium gives another constraint on p and R. From the virial theorem we know that 2 X (lunetic energy/particle) =
- (gravitational energy/particle) or
GMm, 3 k T - P
R '
For T= lo9 K this implies
R = 5 x LO8( ") cm, M a
where M a =mass of s u n x 2 x g .
Combining Eqs. ( 1 ) and (2) gives
p w 4 ~ - ( Substituting in the measured values of F and L we obtain
pw 1.2 x 10-’g cmP3 - ( i i 3 - 3 / 2 * (3)
Now, to get an optical depth we must first determine the dominant opacity source, free-free (bremsstrahlung) K~ or scattering K,. Using Eq. (5.20) for the Rosseland mean of K# we have
K{ 0 . 7 ~ i d 3 p ~ - 7 / 2 810-15(x) - 3 / 2
- N N
0.4 M a Kes
Thus, for M / M a >> 10- lo, K{<
[cf. Eq. (1.97)], and the “effective” opacity coefficient is
The effective optical depth T* is
Thus, for M/M,>>10-5 , the source is effectively thin, 7*<<1, and the assumption of bremsstrahlung emission is justified. For complete con- sistency, however, one must also check to see whether inverse Compton cooling (Chapter 7) is important. [See Problem 7.2.1
6.1-By conservation of energy, d/dt(ymc2)= -power ralated. There- fore, using Eq. (6.5) and B, = Bsina we have
- P mc
y = - T = - A B 5 2 m - A y 2 ,
since 1. This equation is easily integrated to yield
- - y ‘=-Aftconstant.
348 Solutionr
The boundary condition implies constant = - y o I , so that
Yo
At t = t1,2 we have y = y0 /2 . Thus
- I f1 ,2=tAYO)-I=( -%A) 2e4
3m3c5
To correctly account for the radiation reaction (and decrease of y) in the particle equation of motion, the electric field of the self-radiation must be added to Eq. (6.1).
6.2
a. We assume that the magnetic field is frozen into the gas. [This is almost always a good approximation for problems on a cosmic scale. See, e.g., Alfven, H., Cosmical Electrodynamics (Clarendon, Oxford, 1963)J. The magnetic flux through a loop moving with the gas is then a constant, and since area scales as 12, the magnetic field is proportional to
b. The action integral for a particle in a periodic orbit is defined as
where p=momentum and where the integral is taken over one period. This quantity is an adiabatic invariant, that is, it is approximately constant for slow changes in the external parameters, such as the magnetic field [see, e.g., Landau and Lifshitz, Mechanics 3rd ed., (Pergamon, New York, 1976)l. The motion of the electron separates into uniform motion along the
field and circular motion around the field. We may apply the adiabatic invariant to the circular motion alone. From Eq. (6.3), we find
p,c= eBa,
where a is the radius of the projected motion on the normal plane. From the adiabatic invariant we have p,a=constant, and from part (a) we have B a l - 2 . Thusp ,a I - ' and a a l . Since the orbit contracts at the same rate as the contraction of the gas, the flu through it remains constant.
Sdutiom 349
To observer
\ \ \ \ \ \ \ \ \
Figure S.8 field with pitch an& a.
Geometry of Doppler shp for a parti& spimling in a magnetic
c. For relativistic particles p a y so that y a 1 - I . From Eq. (6.7b) and B a I --2 we have P a I -6. From Eq. (6.17) we have a, a I -4. From Problem 6.1 we have t , ,2 a I ’.
6.3-Suppose that the time for the particle to go from 1 to 2 and back from 2 to 1 is 7. For the second half of each cycle, of duration 7/2, no radiation from the particle reaches the observer, since the radiation beam, of halfwidth y-’<<l, is directed at an angle 7-2a or greater away from the observer. Now, from simple time delay, the apparent duration of the first half of the cycle is (Fig. S.8)
Thus the fraction of each cycle in which the particle appears to radiate is
1 ’ 2 p i n a.
6.4
a. When the source is optically thin the observed flux F u a j v = q S v a v-(P-’)/~; when it is optically thick F v a S u a v 5 / 2 . From the given spectrum, (p - 1)/2 = 1 /2 or p = 2. At the critical frequency v2, where
the synchrotron source becomes optically thick, we have two pieces of information:
J ols, d r ~ : a ; ” ~ R = 1,
and
nSV2R
d 2 Fo= Sv2Q= - .
Therefore,
and we have
b. The spectrum below v 1 seems to be dominated by bound-free absorp- tion, exhibiting the clear v2 dependence of an optically thick thermal emitter. The frequency v I is thus the frequency at which the hydrogen gas becomes optically thick, so that we have an additional relation for the distance to the source
or
Using the expressions for R, d, and Q = nR 2 / d 2 we have
soiutions 351
6.5
a. Integ,rating the two intensities of Eqs. (6.32) over the electron distribu- tion, we have for the linear polarization
We now write y in terms of x, ~ c c x - ” ~ , obtaining
Now., using Eq. (6.35), and the property of the r function, r(q + 1)= qr(q), we obtain
b. The polarization of the frequency-integrated ( 6 . 3 2 : ~
J G(x) dx n=
(2)
emission is, using Eqs.
where we have used the fact that w a x . Comparing this integral with that of Eq. ( 1 ) above, we see that they are equal for p = 3 . Thus substituting this value into Eq. ( 2 ) above, we obtain
= 75%. n=- 4
3 + ;
7.2-The energy transfer to a photon of energy c, in a single scattering, has the form [cf. Eq. (7.36)]
A € = E ( - - - ) . 4kT c
mc2 mc2
352 SOW^
Thus, for photons of energy ~<<4kT, the energy gain per scattering can be put into the approximate differential form
dc 4kT --e- dN mc2’
where dN is the differential number of scatterings. After N scatterings, the energy of a photon of initial energy q is thus
-u cN e(4kT/mc2)N for ~,<<4kT. ci
The exponential nature of the energy gain is apparent from the initial equation. In a medium of optical depth res>>l, the characteristic photon scatters -~,f, times before escaping, because of the random walk nature of the scattering process. Thus setting N = r:.? gives
a.
b. When c,-4kT, the initial equation shows that photons stop gaining energy from the electrons; the process has saturated. Thus, to obtain ‘crity
4kT - -eYc”l
cr
(4kT/rnc2)& = In
c. From part (a) the parameter is y =(4kT/mc2)r$
7.2-From the solution to Problem 5.2, the size R , density p , and temperature T, of the emitting region satisfy
M R = ~ X lo8 cm-
M ,
p = 1 . 2 ~ 10-’g cm-3 - ( MM,)-3’2 T = 109 K.
From Problem 7.1, inverse Compton is important if the “Comptonization
parameter” y = (4kT/mc2)7’ exceeds unity. Now,
so that 7es - Kes PR 2
Thus if A4>>400 M a , inverse Compton can be ignored, and the determina- tion of 7’, p, and R on the assumption of pure bremsstrahlung cooling is self-consistent. On the other hand, if M5400 M,, then the model is self-inconsistent, because inverse Compton cooling was ignored in de- termining the energy balance.
7.3
a. From Eq. (6.17~) for the characteristic synchrotron frequency, we have., in normalized units, (taking sina = 3-’12)
The ratio of the photon’s energy to the electron rest mass energy, in the electron rest frame, is then given approximately by
b. The energy associated with a temperature of 1 K is -0.86X eV. The blackbody spectrum peaks at -2.8 kT. Thus the characteristic photon in a blackbody spectrum of temperature T has an energy -2.4 x 10-4T eV. The ratio of a microwave photon energy to electron rest mass in the latter’s rest frame is, therefore,
mc2
Note that in both (a) and (b), for the second scattering, the relevant ratio is a factor y 2 higher and no longer less than unity for y-lo4!
7.4
a. Computation of A, Eq. (7.53): First we set c = 1 in our computations. Let the initial photon f y r - momentum be =ho(l,n), final photon four-momentum be P,, =
3
ha,( l,n,), initial electron fozr-momentum be P, = (E,p), and final electron four-momentum be P,, . Then, expanding out the expression
gives
EAw - hop* n = h2uu I ( 1 - n - n , ) + Awl E - Awl p n, . (1)
Here n, n,, and p are the initial and final photon directions and the initial electron momentum, respectively. From Eq. (1) we obtain
h(o, - w ) A = - xp* (n, - n) - x2kT( 1 - n n ,) - kT E-p*n,+xkT( l -n*nl)
where xrAw/kT. Now, since p/m is of order aE(kT/m)1/2, to lowest order in a we may replace the denominator of Eq. (2) by E = m, where m is the electron mass, and neglect the second term in the numerator, o(a2), in comparison with the first, thus obtaining (putting back factors of c )
b. Computation of I,, Eq. (7.54): Let x be the angle between the vector p and the vector (n, - n). Then, using Eq. (7.53) for A2, we obtain
I , - JJd% f,A2- do dS2 dS2
=( z)2 /d$p2cos2Xf , / ln , - n 1 2 x d Q . da mc (4)
Now, since du/da does not depend on p, to lowest order in u / c - a , the integral over p may be done independently of the integral over photon directions. Next, substitute in Eq. (7.49) for the Maxwellian electron distribution, f,, and let x be the polar angle for the d$ integration, that is, d$ =p2dpdcosXd@. The integration over d$ then gives
( 5 )
Solutions 355
Finally, let nI lie along the polar axis for the dS1=dcos8d+ integra- tion, so that In, - .I2 = 2( l - cos8). Substituting Eq. (7.lb) for du/d!d, we obtain the desired result,
c. Computation of &/at , Eqs. (7.55): To conserve the total number of photons, an/& must have the funciional form of Eq. (7.55a), since
that is, the change in total photon number arises only from a flux through the boundaries of energy space. Next, write Eq. (7.55a) in the form
and Eq. (7.52) in the form
an - at = C , ( x ) n ” + C2(n,x)n’+ C3(n ,x ) . (9)
Equations (8) and (9) must be functionally identical. Comparing the highest x derivatives in these two equations, we see that j must contain a term linear in n’, with coefficient independent of n and no terms in n”. Thusj must be of the functional form
1 = g(x)[ n’ + h( n , ~ ) ] . (10)
8.2-Consider two media, of refractive indices n, and ni. Let 8 and 8‘ be the angles of incidence and refraction of a beam of radiation incident on an area du of the surface of separation of the two media. Let I, and I: be the intensities of the incident and refracted beam, respectively. Then, assuming that no energy is lost by reflection at the interface, we have
Ivcos8dud!d= I,fcos8’dudQ2’, (1)
where dO = d cos B d+, dill = d cos B' d+'. Now, d+ = d+', and by Snell's law
n,sinB=n;sin8'. ( 2 )
Squaring and differentiating Eq. (2) leads to
n,? cos8d cos 8 = n; cos B'd cos6'. ( 3 )
Now, combining Eqs. ( I ) and (3) gives
In a medium in which the refractive index changes continuously and slowly on the scale of a wavelength, we can imagine that the photon path is made up of a number of short segments in regions of constant refractive index, to which the above result applies. Thus l v / n : is seen to be an invariant over general paths. Note that the assumption of no reflection loss at the interface between media becomes completely valid in the continuous limit.
8.2-The Fourier transform of $(r , t ) with respect to r is simply A (k)exp[ - iw(k) t ] from its definition. Therefore, from Parseval's theorem we have
W
1$/2dr=(271)-11m [ A exp( - i ~ ~ i ) / ~ d k = ( 2 n ) - ' l ~ IA(k)('dk, (1)
since w is real. Thus the normalization of the packet remains constant in time (no absorption). Now consider the result
l- W --to -03
00 i a ~( r , t ) = 1- W A (k)e-'"' 7 - elkr dk I ilk
where we have integrated by parts. This shows that the Fourier transform of r$ is ia/ak(Ae-'"'). Using the generalized Parseval's theorem [cf. Eq. (2.3 1 )I,
/ A *( r)B( r)dr = (27r) - ' 1 a* ( k ) l?( k)dk,
Sdutim 357
with A = 4) and B = r+, we obtain
=
which depends on time linearly. Therefore,
and dividing by Eq. ( l ) , which is independent of time, yields the desired result:
Suppose that the wave packet is localized in both space and wave number, within the restrictions of the uncertainty relation Ak A r z l , of course. If ao/ak changes slowly over the scale Ak about the central wave number k,, then the packet will move with the group velocity (ao/ak)k,k,. This is the usual statement of the group velocity property, but Eq. (3) also holds when the packet is spread arbitrarily in space and wave number.
8.3-Taking the derivative of Eq. (8.31) with respect to w and dividing the resulting equation by Eq. (8.20), we obtain
where (BI,) is measured in Gauss. Note the interesting result that the frequency dependence cancels out of the above expression. Substituting dAO/dw== 1 . 9 ~ lop4 s and d'p/dw= 1.1 x low5 s2, we obtain the result
<Bll) = I.0x
without needing to know the frequency at which the measurements were made!
3 s Solutions
9.1
a. The properly normalized antisymmetric wave function is:
where 1 and 2 include space and spin coordinates. The operator whose expectation value we want is
2 R = (rl -r2) = r: - 2r , . r , + r:.
Now, use Dirac notation for integrals,
1 u,*(rI,Sl)rlub(rI,Sl)d3rl ( a ( r l b ) ,
use the fact that rl only operates on functions of rl,
J u,*(r1)r2uu(r1)d3r1 = ‘2,
and use the orthogonality of orbitals
(aJb)=O for a f b ,
to obtain
b. Note that the dipole operator vanishes between states of the same
c. Separating space and spin parts,
parity (see 810.4)
%(rlJl) = uu(rl)l’l>9
we have
Sdutim 359
since r does not operate on Is). Since different spin states are orthogo- nal, for example, Is,) =(1,0), Isb) =(O, I )% electrons of different spins give (s,Isb) =O=(alrlb).
d. Since I ( ~ l r l b ) ( ~ is a positive definitive quantity, comparison of parts (a) and (c) shows that ( R 2 > is larger for same spin electrons by a term 21(alrlb)12.
9.2
a. 2s': These are equivalent electrons and must be in opposite spin states, so S - 0 , and only singlets occur. Each orbital has zero orbital angular momentum, so L=O, and only S states occur. The parity is even: (- I)'+'= 1. There is only one possible value for J = L + S, that is, zero. The one possible term and level is, therefore, IS,,.
These are nonequivalent electrons, so that all combinations of spins are allowed. Therefore both S=O and S = 1 are possible, and singlets and triplets occur. The orbital angular momenta of the orbitals are I = 1 and I = O , so that L = 1 and only P states can occur. The panty is odd: ( - I)'+' = - 1. The angular momentum of the singlet state (S=O, L= 1) can be only J = 1. The triplet state ( S = 1) can combine with L = 1 to yield J=O, 1,2. Therefore, the terms and levels are: ' P p and 3P-&0.
These are nonequivalent electrons, allowing S=O and 1, so that singlets and triplets occur. The values of L, found from combining I = 1 and 1 = 1, are L =0, 1 and 2, so that S, P , and D states can occur. The panty is even: (- l ) '+ '=l . There are six ways to pick L and S which lead to the following terms and levels: 3D3,2,1, P2,1,0, 3S1, ID2,
There are four equivalent 2p electrons and one 3p electron. In cases such as this, find the terms of the equivalent electrons first, then combine each in turn with the remaining nonequivalent electron. The terms of p 4 are the same as forp', which are IS, ID, 3P. (See $9.4.) The ' S term plus the 3p eiectron gives rise to zP30/2, The 'D term plus 3p
b. 2 ~ 3 s :
c. 3p4p:
3
'PI, Is,. d. 2p43p:
yields 2F&,2, 2 0 D5/2 .3 /2 , and 2P,0/2,1/2. The 3P term plus 3p yields 4 0 4 0 2 0 2 0
4Dp/2,5/2,3/2. 1/23 p 5 / 2 3 3 / 2 , 1 / 2 ~ s 3 / 2 ; and 20,0/2,3/2, ' 3 / 2 , 1 / 2 and '1/2*
9.3-Recall the statistical weights: 2(21+ 1) for each nonequivalent electron in a configuration; (2L+ 1) (2S+ 1) for a term; and ( 2 J + 1) for a level.
360 Soiutions
a. From I, = l2 -0, we have 2(21, + 1) 2(24 + 1)=4, but two states violate the Pauli principle, and the remaining two are indistinguishable. Thus Nconf = 1. From L = 0 and S = 0 we obtain N,,, = (2L + 1)(2S + 1) = 1. From J = 0 we obtain Nleve, = (2J + 1) = 1.
b. From I , =O, I , = 1 we have N,,,,=2(2-0+ 1) 2(2.1+ 1)- 12. The (L= 1,S=O) and ( L = l , S = l ) terms give Nte,=(2~1+1)(2~0+1)+(2~1+ 1)(2.1+ 1)= 3 + 9 = 12. The levels have J = 1,2,1,0 so that Nleve; = 3 + 5 + 3 + 1 = 12.
c. From Z,=l, 12=1 we have N,,=2(2.1+1) 2(2.1+1)=36.The terms 'D, 'P, 3S, I f f , 'P, 'S yield Ntem = 15 + 9 + 3 + 5 + 3 + 1 = 36. The levels have J=3,2,1,2,1,0,1,2,1,0, so that Nlevel=7+5+3+5+3+ 1+3+5 +3+ 1-36.
9.4-Using the definitions of A, Z, and y, the Saha equation can be written
since y is large compared to ln(2 U, + , / q). a. The transition from stagej t o j + 1 is defined by Nj=N,+I . From Eq.
b. From Eq. (1) we have
(1) we have kT-x,/y.
since (a T-'''. From part (a) we know x/kT=y>>3/2, so that
c. The ratio of excited to ground state populations in statej is gven by the Boltzmann law:
where g; and go are statistical weights and xiJ is the excitation
Sdutim 361
potential. Using part (a) we have
X i J X , J k T “’xi - N
Except for very low-lying states, x i J is of order xi, so that the exponential term is very small and NoJ>>Nij.
9.5
a. The Saha equation has the following form:
The statistical factor 2 V,/ U, is unity. Eliminate N, by writing it in terms of p and 6, using N, = Np (neglect H- and H2):
p = NHmH + Ne(m, + mp)%(NH + Ne)
= mHNe( 1 + 6 -’)
Pa m,(S + 1) ‘
N , =
Thus
- E ~ ( ~ , T ) = - h - ~ e x p -- , S 2 mH 6 + 1 ( 3 P
where h is the thermal de Broglie wavelength of Problem 9.4.
b. Solving the quadratic equation in S we obtain
6 = [ A + (A2+4A)’/’].
ZO.I--.The selection rules are:
1. Configuration changes by exactly one orbital, for which AI= 2 1.
2. AS-0.
3. AL=: ? l ,O.
4. A J = 2 1,0, except J = O to J = O .
362 Solutions
r 2 3s 1 r 0 3s 1 1
3 p { 1 4 L 0 I 2
0 2
1 1
3s 1 2 3s 1 3p I 0
(a 1 (bl IC) (4
Figuw S.9 Energy spacings for 'S and 'P lewls when 'P term is (a) normal, upper, (b) inoerterl, lower, (c) h r t e a ! , upper, (d) n o m l , lower.
Without knowing the configurations we can say nothing about (1) because, for example, ls2s 3S,-+ls2p is not. Rules (2) and (3) are satisfied, both terms being triplets and A L = 2 1. Finally, all the transitions
is allowed while 132s 3S+2p3d
'sl +3P0
3S,+3P,
3s, -+3P2
satisfy (4) and are therefore allowed. The four possible arrangements of the two terms with both normal and
inverted 3P term are illustrated in Fig. S.9. Let C be the energy difference between the closEst 3P levels (J=O and J = 1 levels). Then the energy difference between the other two adjacent levels (J= 1 and J = 2 ) is 2C, by the Lande interval rule. Also, let AE, be the energy difference between the 3S, level and the closest 'P level, corresponding to the least energetic spectral line.
The energies of the three spectral lines are given in terms of two cases:
1. Normal upper or inverted lower 'P term (Fig. S.9a and b):
'S,-'PO, A E = AEo,
3S1-3P,, A E = A E , + C ,
'Sl-'P2, A E = AEO + 3C.
Sdutiom 363
2. Normal lower or inverted upper 3 P term (Fig. S.9c and d ) :
'S1++'P2, AE = AEo,
'S , ++'PI,
'S, ++'Po,
A E = A E, + 2 C,
A E = A E, + 3 C.
10.2
a. 3s 's , / , c-)
b. 2p 2P1/2 t,
c. 3s3p 'P, t,
e. 2p2 3.P0 t,
d. 2p3p 3D, t-)
f. 3s2p 'PI t)
g. 2s3p 'Po t,
h. 1s' 'So t)
i. 2p3p ' S , t)
j. 2p' 'DSl2 t,
4s 2s,/2.
3d 2D,/2.
3p2 ID,.
3p4d 'F,.
2p3s 'Po.
2p3p 'P I .
3p4d ' P I .
2s2p IP,.
2P3 2Dl,,.
2p4d 'D2.
Not allowed. Parity does not change; jump- ing electron has A1 = 0.
Not allowed. A J = 2 .
Not allowed. A S = 1.
Allowed.
Not allowed. J = 0-J = 0.
Allowed.
Not allowed. Parity does not change; jump- ing electron has A1 = 2.
Not allowed. Two electrons jump.
Not allowed. AL = 2.
Not allowed. Parity does not change; con- figuration does not change.
Z0.3-Comparison of Eqs. (lo. IOc), (10.23) and (10.29b) shows that the oscillator strength may be written as
where g, is the statistical weight of the initial state, the sum is over degenerate levels of the initial and final states, q, is the frequency of the transition, and
The initial wave function is the (n, I, m) = ( 1, 0,O) ground state of hydro- gen, which has the form [cf. Eqs. (9.10) and (9.16)]
(3) 4 = - 112 - 3 4 - r /ao. 100 a0
364 sotutiw
The final wave function may be any of the three states (n,I,m) =(2,1, - l), (2,1,0), (2,1, l) , corresponding to wave functions
where
R =2-3/2a-5/23- 1/2r2e-r/2ao, 21 - 0 ( 5 )
and Y , are the spherical harmonics. Now, it is easily shown that the operator may be written as
where the latter “spherical operators” are conveniently expressible in terms of spherical harmonics:
x + & = r ( 8 ~ / 3 ) ’ / ~ Y , . , , ( 7 4
z = r ( 4 ~ / 3 ) ’ / ~ Y l o . (7b)
Using Eqs. (3) to (7) the position matrix element may then be written as
where
Now, by the orthogonality relations of the Y,,,,, only one term in contributes for each m, and t h s contribution is unity.
Finally, performing the sum indicated in Eq. (l), that is, summing over m=O, 5 1 and multiplying by 2 for the two possible spin states of the
sdutions 365
initial electron we obtain
Now, wi/ is the frequency of transition between the n = 2 and n = 1 levels:
Substituting Eq. (10) into (9), we obtain the desired result:
L g.f. = - . ' cf 3 9
IO.l--.The initial wave function $i for the 1s state of a hydrogen like ion is
The final wave function 4, for the continuum electron in the Born approximation is taken to be the free electron state
where q=p/h is the wave number of the electron. The normalization V - ' l 2 is consistent with the derivation of Eq. (10.52), the final electron being localized to a volume V.
In the nonrelativistic regime we can neglect the retardation factor exp (ik'r) in the matrix element (dipole approximation), so we must evaluate I(fll. Vli)I2. By the Hermitian property this is equivalent to evaluating \( i l l . Vlf)I2. Now
This latter integral can be evaluated by choosing a polar coordinate system with respect to the direction of q. Let 8 be the polar angle between r and q,
and let p=cosO. Then
the integrals being elementary. Using these results in the cross-section formula (10.52) we obtain
When the energy of the electron is large compared to the ionization energy, huwh2q2/2m~Z2e2/2ao, it follows that q2>>Z2/u& so that we may write
- dubJ 32ah z '( ieq)' dQ +=(;) 7
We now integrate this over solid angles to obtain the total cross section. It is convenient to use now polar coordinates with respect to the direction 1. Note that
Theref ore, we obtain for the integrated cross section,
Using the relation )iu&22q2/2m to eliminate q, we obtain finally Eq. (10.53) of the text.
~oiut iom 367
10.5-Since the source is optically thin the spectrum of the emitted radiation is proportional to the emission function (1.73), and thus has the same shape as the profile function r#~(v). For this case +(v) is given by Eq. (10.78), where r= y is the natural width of Eq. (10.73). In the limiting case of AvD<<y/4a the profile is essentially given by Eq. (10.73), and the observed. half-width is independent of temperature. In the limiting case of Avo >>y/477 the profile is essentially given by Eq. (10.68), and the observed half-width grows as the square root of temperature. The critical tempera- ture T, separating these two cases is found by setting AvD(TC)=y/4a . Using Eq. (10.69) we obtain
if we assume that hydrogen atoms, m, = rnH, are emitting. For Lyman-a we have y = A,, and using Eq. (10.34) we obtain
Now g, := 2, g , = 2(21+ 1) = 6 for the 2p state and gf= 214/39 from Problem 10.3. The value of wo is found from Eq. (10.42a), with n = 1 and n’=2,
With these values Eq. (1) becomes
22’
318 kT, = - a6mHc2,
where a - e2 /hc is the fine structure constant. Numerically,
T, = 8.5 x K.
For most cases of astrophysical interest T >> T,, and the Doppler broaden- ing dominates, at least near line center. It should be noted, however, that far from line center the Lorentz part of the broadening, which falls off as (v- v J 2 , will eventually dominate the Doppler part, which falls off extremely rapidly as exp[ - ( v - V ~ ) ~ / A V ~ ] [cf. Eqs. (10.77) to (10.79)].
368 Solutim
10.6-The dipole matrix elements of r = ( x , y , z ) can be conveniently expressed in terms of the matrix elements zg and ( x k &)w The matrix element of z = r cos B between states with (I, m ) and (I ', m') is proportional to
P;I.'( p) pP,"'( p) d p l i (m - m')+ d $J 0
where p = COSB and P,"'( p ) is the associated Legendre function. Since the second integral vanishes unless m' = m we need consider only
The recurrence relation
(21 + I)&"' = ( I - m + 1)PG + ( I + m)P,?
and the orthogonality relations for the P,"' (see, e.g., Arfken, G. 1970, Mathematical Methods for Physicists, Academic, New York) imply that
zg=O, unless m'= m and I f = I + 1 or l'= I - 1.
The matrix element of (x 2 9) = r sin 8ef i+ is proportional to
which vanishes unless m' = m + 1 or m' = m - 1. If m'= m + 1 we use the
recurrence relation ( 2 f + 1 ) Y g P,"-' = P,? I - PZ I and the orthogonal- ity relations to show that
(x + 0, unless m' = m + 1 and I' = I + 1 or I' = I - 1.
Similarly, we show that
(x-&),,=O, unlessm'=m-1 a n d f ' = f + I orI '=I-1.
Taken together, these results imply the electric dipole selection rules: Am=O,k 1 and A f = 2 1.
10.7-Let +c(t) be the collision-induced random phase at time t. Then the electric field will be
(1) E( t ) = Ae ''4 - u r n + i+Ar)
so~utiom 369
where A is a constant, wo is the fundamental frequency, and y is the rate of spontaneous decay. We wish to compute the averaged power spectrum
From Eq. ( l ) , we have
Now, the only random function in Il?(w)12 is c~ , ( t ) . Thus we obtain
Now, we can write
where Ag,(t, - t2) is the change of phase during the time interval t , - t,, and we wish to compute
Since changes in phase are random, this average vanishes if one or more collisions occurs during A t = It, - t,l and is unity if no collisions occur during this interval. We are given the mean rate of collisions is vCol, thus implying that the probability for no collisions to occur during At is e - ~ f ~ - f z ~ ” ~ ~ ~ (assuming a Poisson distribution for the collisions). Thus
and Eq. (4) becomes
Equation (6), using Eq. (3) for G(t,12), can be integrated in terms of elementary functions and yields Eq. (10.75).
370 ~o~ut iom
11.1
a. In order of magnitude the equilibrium separation is the Bohr radius of an atom, since the electron binding is what holds the two positive charges together. So
h2
me2 . ro- a, 3 -
b. Since the molecules will be electrically neutral in the temperature range considered, they can be treated approximately as hard spheres of size -ro-ao. The cross section is thus the simple geometrical form for the area
2 2 u- n-r, - m,.
c. For Doppler broadening, the line width, A v D , is
For collisional broadening, the line width, Avc, is the collision frequency uCol. From part (b), we estimate
For low p, the line width is dominated by AvD and is thus independent of p. At high p, vc0l>>AvD and the line width increases in proportion to p. The transition occurs at a po such that
or
or
h me p o - T = a- -5 x cuo a;
g ~ r n - ~
Sdutiom 371
11.2
Make the changes of variables R r R , -RB, y=lr-RAl/R, x-cost?, where 8 is the angle between (r - R,) and R. Then, after doing the trivial C+ integral to obtain a factor 277, one gets
Now, reverse the order of integration, doing the x integral first, and make the change of variables a2-y2+2yx + 1, dx = y - ’ a h , to obtain
This elementary double integral is easily done by (e.g., integration by parts) to yield Eq. (1 1.12).
II.3-Regard the H ( J ) term in the expression for and as a function of a,, and a,,,, written in the form
where the “coefficients” j , ( J ) and j 2 ( J ) are to be determined by equating the above expression to Eq. (1 1.42) for H ( J ) . For the P branch, the two resulting equations for j , ( J ) and j 2 ( J ) yield
j , ( J ) = -(J+ 1)+
j2 ( J ) = j 2 .
As J ranges over its allowed positive values, j ranges over negative integer values, j < - 1. For the R branch the two resulting equations for j , ( J ) and j 2 ( J ) yield
j , ( J ) = J =j,
j d J > =j2
and j ranges over positive integer values j > 1. Combining the two formulae yields
H ( J ) = j ( a,, + a,,.) + j 2 ( a,, - a,,.),
372 so~utions
where j ranges over negative and positive integers, excluding zero, the “band origin.”
12.4-Regarding j as a continuous variable in the result of Problem 11.3, we find the band head (i.e., the reversal of line spacing with increasingj) by setting the derivative of H ( J ) equal to 0, aH/ajljhead=0, giving
If jhcad < O (i.e., an.<an), the band head clearly falls in the P branch. If jhcad > 0, (i.e., a, <an.), the band head falls in the R branch. Generally, jhead, as deduced above, is not an integer, so that the true value of the band head corresponds to the nearest integer value to jhead. Fromjhead, the value of Jhead may be deduced from the solution to Problem 11.3, that is,
JheadNjhead forjhead > 0, ( R branch)
Jhead“ - (’ +jhead) forjhead < (P branch)
The frequency of the band head is found by substituting jhcad into the expression for H ( j ) :
1 (a, + a,J2
4 (an - . , , ) ’ %oJl head = O O - -
The band head frequency is below or above wo, depending on whether a, >a,, or a, <fin,.
22.5-Using the same arguments as in Problem 11.4, we see that the Q branch would have a band head at the J value satisfying
or, using Eq. (1 1.42b) for the Q branch form of H(J),
Jhead = - 1 /2.
Since J only has positive integer values, this band head is never actually realized. However, this value is sufficiently near J = I that the Q branch (at low resolution), at J = 1, resembles a band head.
21.6--.Rotational energy levels have energies
h2 2r Erot= - J ( J + I) ,
where Z - q r i - T a i , and vibrational and electronic transitions have energies
The probability of a given energy level being occupied is proportional to exp( - E / k T ) . Thus if k T 5 E r o t ( J = 2 ) , most molecules will be in the J = 1 rotational ground states, and few rotational transitions can occur. On the other hand, if k T z E E v i b r vibrational levels will be excited, and rotational- vibrational spectra will be produced. To have pure rotation spectra pro- duced, then, one requires
