Lecture 3Radiative and Convective Energy TransportMechanisms of Energy Transport Radiative Transfer Equation Grey Atmospheres Convective Energy Transport Mixing Length Theory

The Sun

I. Energy TransportIn the absence of sinks and sources of energy in the stellar atmosphere, all the energy produced in the stellar interior is transported through the atmosphere into outer space. At any radius, r, in the atmosphere:4pr2F(r) = constant = LSuch an energy transport is sustained by the temperature gradient. The steepness of this gradient is in turn dependent on the effectiveness of the energy transport through the different layers

I. Mechanisms of Energy Transport Radiation Frad (most important) Convection: Fconv (important in cool stars like the Sun) Heat production: e.g. in the transition between the solar chromosphere and corona Radial flow of matter: coronae and stellar winds Sound waves: chromosphere and coronaeSince we are dealing with the photosphere we are mostly concerned with 1) and 2)

I. Interaction between photons and matterAbsorption of radiation:InIn + dIn sdIn = knr In dxkn : mass absorption coefficient[kn ] = cm2 gm1Optical depth (dimensionless)Convention: tn = 0 at the outer edge of the atmosphere, increasing inwards

Optical DepthIn0In(s)dIn = In dtIn(s)=In0etnThe intensity decreases exponentially with path lengthOptically thick: t > 1Optically thin: t < 1If t = 1 In = 0.37 In0We can see through the atmosphere until tn ~1

Optical DepthThe quantity t = 1 has a geometrical interpretation in terms of the mean free path of a photon:t = 1 = krds = krss (kr)1s is the distance a photon will travel before it gets absorbed. In the stellar atmosphere the abosrbed will get re-emitted and thus will undergo a random walk. For a random walk the distance traveled in 1-D is sN where N is the number of encounters. In 3-D the number of steps to go a distance R is 3R2/s2. At half the solar radius kr 2.5 thus s 0.4cm so it takes 30.000 years for a photon do diffuse outward from the core of the sun.

Emission of RadiationInIn + dIndxjndIn = jnr In dxjn is the emission coefficient/unit mass [ ] = erg/(s rad2 Hz gm)jn comes from real emission (photon created) or from scattering of photons into the direction considered.

II. The Radiative Transfer EquationConsider radiation traveling in a direction s. The change in the specific intensity, In, over an increment of the path length, ds, is just the sum of the losses (kn) and the gains (jn) of photons:dIn = knr In + jnr In dsDividing by knrds which is just dtn

The Radiative Transfer Equationtn appears alone in the previous equation therefore try solutions of the form In(tn) = febtn. Differentiating this function:Substituting into the radiative transfer equation:

The Radiative Transfer EquationThe first two terms on each side are equal if we set b = 1 and equating the second term: etnor= Snt is a dummy variableSet tn = 0 c0 = In(0)

bringing etn inside the integral:

This equation is the basic intergral form of the radiative transfer equation. To perform the integration, Sn(tn), must be a known function. In some situations this is a complicated function, other times it is simple. In the case of thermodynamic equilibrium (LTE),Sn(T) = Bn(T), the Planck function. Knowing T as a function of x or tn amounts to a solution of the transfer equation.

Radiative Transfer Equation for Spherical GeometryAfter all, stars are spheres!zxyTo observerrqdInknrdz=In + Sn

Radiative Transfer Equation for Spherical GeometryAssume In has no f dependence and dr = cos q dzr dq = sin qInrcos qknrdrInqsin qknrr= In + SnThis form of the equation is used in stellar interiors and the calculation of very thick stellar atmospheres such as supergiants. In many stars (sun) the photosphere is thin thus we can use the plane parallel approximation

Plane Parallel ApproximationTo observerqdsTo center of starq does not depend on z so there is no second termCustom to adopt a new depth variable x defined by dx = dr. Writing dtn for knrdx:The increment of path length along the line of sight is ds = dx sec q

The optical depth is measured along x and not along the line of sight which is at some angle q. Need to replace tn by tnsec q. The negative sign arises from choosing dx = drIn the first case we start at the boundary where tn=0 and work inwards. So when In = Inin, c=0.In the second case we consider radiation at the depth tn and deeper until no more radiation can be seen coming out. When In=Inout, c=

Therefore the full intensity at the position tn on the line of sight through the photosphere is:In(tn) = Inout(tn) + Inin(tn) = Note that one must require that Snetn goes to zero as tn goes to infinity. Stars obviously can do this!

An important case of this equation occurs at the stellar surface:Inin(0) = 0Inout(0) = Sn(tn)etnsec q sec q dtn0Assumption: Ignore radiation from the rest of the universe (other stars, galaxies, etc.)This is what you need to compute a spectrum. For the sun which is resolved, intensity measurements can be made as a function of q. For stars we must integrate In over the disk since we observe the flux.

The Flux Integral Assuming no azimuthal (f) dependence

The Flux Integral Using previous definitions of Inin and InoutFn = 2p p/20tn Sne(tntn)sec q sin q dtn dq 2p ptn0 Sne(tntn)sec q sin q dtn dqp/2

The Flux Integral If Sn is isotropicLet w = sec q and x = tn tn

The Flux Integral Exponential IntegralsEn(x) =In the second integral w = sec q and x = tn tn . The limit as q goes from p/2 to p is approached with negative values of cos q so w goes to not

The Flux Integral The theoretical spectrum is Fn at tn = 0: Fn(0) = 2p Sn(tn) E2(tn)dtnFn is defined per unit areaIn deriving this it was assumed that Sn is isotropic. It most instances this is a reasonable assumption. However, in stars there are Doppler shifts due to photospheric velocities, stellar rotation, etc. Isotropy no longer holds so you need to do an explicit disk integration over the stellar surface, i.e. treat Fn locally and add up all contributions.

The Mean Intensity and K Integrals

The Exponential Integrals Exponential IntegralsEn(x) =1wndwEn(0) ==1n1

The Exponential Integrals Recurrence formulaEn+1(x) = ex xEn(x)

The Exponential Integrals For computer calculations:E1(x) = ex xEn(x)E1(x) = ln x 0.57721566 + 0.99999193x 0.24991055x2 + 0.05519968 x3 0.00976004x4 + 0.00107857x5 forx 1E1(x) = x4 + a3x3 + a2x2 + a1x + a0x4 + b3x3 + b2x2 + b1x + b01xexa3 = 8.5733287401a2= 18.0590169730a1= 8.6347608925a0= 0.2677737343b3= 9.5733223454b2= 25.6329561486b1= 21.0996530827b0= 3.9584969228En(x) =1xex[1 nx+ [Assymptotic Limit:x >1From Abramowitz and Stegun (1964)Polynomials fit E1 to an error less than 2 x 107

Radiative Equilibrium Radiative equilibrium is an expression of conservation of energy In computing theoretical models it must be enforced Conservation of energy applies to the flow of energy through the atmosphere. If there are no sources or sinks of energy in the atmosphere the energy generated in the core flows to the outer boundary No sources or sinks in the atmosphere implies that the divergence of the flux is zero everywhere in the photosphere.In plane parallel geometry:ddxF(x)= 0or F(x) = F0A constantF(x) = F0 (tn)dn =F0 for flux carried by radiation 0

dn Sn E2(tn tn)dtn tn0 Sn E2(tn tn)dtnRadiative Equilibrium[[ =F02pThis is Milnes second equationIt says that in the case of radiative equilibrium the solution of the radiative transfer equation is found when Sn is known that satisfies this equation.

Radiative EquilibriumOther two radiative equilibrium conditions come from the transfer equation :Integrate over solid angleddxIn cos q dw= knr In dw knr Sn dwSubstitue the definitions of flux and mean intensity in the first and second integrals= 4pknrJn 4pknrSn

Radiative EquilibriumIntegrating over frequencyBut in radiative equilibrium the left side is zero!ddxFn dn= 4pr knJn dn 4pr knSn dnknJn dn =knSn dnNote: the value of the flux constant does not appear

Radiative EquilibriumUsing expression for Jnkn[= 0 Sn E1(tn tn)dtn Sn E1(tn tn)dtn+ [dn First Milne equation

If you multiply radiative equation by cos q you get the K-integraldIncos2 qdx knrIn cos q knrSn cos qdn dKn= dtnF0 1st moment2nd moment= dwdwRadiative EquilibriumThird Milne EquationAnd integrate over frequency

Radiative EquilibriumThe Milne equations are not independent. Sn that is a solution for one is a solution for all threeThe flux constant F0 is often expressed in terms of an effective temperature, F0 = sT4. The effective temperature is a fundamental parameter characterizing the model.In the theory of stellar atmospheres much of the technical effort goes into iterative schemes using Milnes equations of radiative equilibrium to find the source function, Sn(tn)

III. The Grey AtmosphereThe simplest solution to the radiative transfer equation is to assume that kn is independent of frequency, hence the name grey. It occupies an historic place and is the starting point in many iterative calculations. Electron scattering is the only opacity source relevant to stellar atmospheres that is independent of frequency.Integrate the basic transfer equation over frequency and denote:Where dt = krdx, the grey absorption coefficient

The Grey AtmosphereThis grey case simplifies the radiative equilibrium and Milnes equations:F(x) = F0 J = S

The Grey AtmosphereS(t) = J(t) = 3K(t)But since the mean intensity equals the source function in this case

The Grey AtmosphereWe can now integrate the equation for K:Where the constant is evaluated at t = 0. Since S = 3K we get Eddingtons solution for the grey case:

III. The Grey AtmosphereUsing the frequency integrated form of Plancks law: S(t) = (s/p)T4(t) and F0 = sT4 the previous equation becomesT(t) =(t + )At t = the temperature is equal to the effective temperature and T(t) scales in proportion to the effective temperature.TeffNote that F(0) = pS(), i.e. the surface flux is p times the source function at an optical depth of .

III. The Grey AtmosphereChandrasekhar (1957) gave a complete and rigorous solution of the grey case which is slightly different:q(t) is a slowly varying function ranging from 0.577 at t = 0 and 0.710 at t =

IV. ConvectionIn a star the heat flux must be sufficiently great to transport all the energy that is liberated. This requires a temperature gradient. The higher the energy flux, the larger the temperature gradient. But the temperature gradient cannot increase without limits. At some point instability sets in and you get convetionIn hot stars (O,B,A) radiative transport is more efficient in the atmosphere, but core is convective.In cool stars (F and later) convective transport dominates in atmosphere. Stars have an outer convection zone.

IV. ConvectionConsider a parcel of gas that is perturbed upwards. Before the perturbation r1* = r1 and P1* = P1For adiabatic expansion:

IV. ConvectionStability Criterion:Stable: r2* > r2 The parcel is denser than its surroundings and gravity will move it back down.Unstable: r2* < r2 the parcel is less dense than its surroundings and the buoyancy force will cause it to rise higher.

IV. ConvectionStability Criterion:

IV. ConvectionThe left hand side is the absolute amount of the temperature gradient of the star. Both gradients are negative, so this is the algebraic condition for stability. The right hand side is the adiabatic temperature gradient. If the actual temperature gradient exceeds the adiabatic temperature gradient the layer is unstable and convection sets in:The difference in these two gradients is often referred to as the Super Adiabatic Temperature Gradient

Brunt-Visl FrequencyThe frequency at which a bubble of gas may oscillate vertically with gravity the restoring force:The Convection Criterion is related to gravity mode oscillations: is the ratio of specific heats = Cv/Cpg is the gravityWhere does this come from?

The Brunt-Visl Frequency

Dr = rADrFB = grAVDrIn our case x = Dr, k = grAV, m ~ rV

V. Mixing Length TheoryConvection is a difficult problem for which we still have no good theory. So how do most atmospheric models handle this complicated problem? By reducing it to a single free parameter whose value is left to guess work. This is mixing length theory. Physically it is a piece of junk, but at this point there is no alternative although progress has been made in hydrodynamic modeling.Simple approach:Suppose the atmosphere becomes unstable at r = r0 mass element rises for a characteristic distance L (mixing length) to r + L Cell releases excess energy to the ambient mediumThe cell cools, sinks back, absorbs energy and rises againIn this process the temperature gradient becomes shallower than in the purely radiative case.

V. Mixing Length TheoryRecall the pressure scale height:H = kT/mg

V. Mixing Length Theoryrv2 = gDrL = gDraH

Scale height H = kT/mgThe Sun: H~ 200 kmA red giant: H ~ 108 kmHow large are the convection cells? Use the scale height to estimate this:

Schwarzschild (1976)Schwarzschild argued that the size of the convection cell was roughly the width of the superadiabatic temperature gradient like seen in the sun.

In red giants this has a width of 107 km

Convective Cells on the Sun and Betelgeuse Betelgeuse = a OriFrom computer hydrodynamic simulationsThe Sun1000 km~107 km

Numerical Simulations of Convection in Stars:http://www.astro.uu.se/~bf/movie/movie.html

http://www.aip.de/groups/sternphysik/stp/box_simulation.htmlWith no rotation:With rotation:

cells small large total number on surface motions and temperature differences relatively small Time scales short (minutes) Relatively small effect on integrated light and velocity measurements cells large small total number on surface motions and temperature differences relatively large Time scales long (years) large effcect on integrated light and velocity measurements

RV Measurements from McDonaldAVVSO Light Curve