Energy TransportEnergy Transport

• Formal solution of the transfer equation

• Radiative equilibrium• The gray atmosphere• Limb darkening

The Transfer EquationThe Transfer Equation

Recall: for radiation passing through a gas, the change in I is equal to:

dI = intensity emitted – intensity absorbed

dI = jdx – Idx or

dI/d = -I + S

The Integral FormThe Integral Form

• A solution usually takes the form • Where • One must know the source function to

solve the transfer equation

• For LTE, S() is just the Planck function B(T)

• The solution is then just T() or T(x)

)())((

0)0()()(

eIdteSI t

bfeI )(

00cdteSf t

Toss in GeometryToss in Geometry

• In real life, we are interested in I from an arbitrary direction, not just looking radially into the star

• In plane parallel geometry we have azimuthal symmetry, so that

SId

dIcos

SId

dIcos

dtetSI t

csec)()( sec)(

Radiative EquilibriumRadiative Equilibrium• To satisfy conservation of energy,

the total flux must be constant at all depths of the photosphere

• Two other radiative equibrium equations are obtained by integrating the transfer equation over solid angle and over frequency

dFFxF

00)(

IntegratingIntegrating Over Over Solid AngleSolid Angle

• Assume and Sare independent of direction, and substitute the definitions of flux and mean intensity:

• becomes:

• Then integrate over frequency:

SIdx

dIcos

dSdIdIdx

d cos

SJdxdF 44/

(integrating over frequency…)(integrating over frequency…)

• LHS is zero in radiative equilibrium, so

• The third radiative equilibrium condition is also obtained by integrating over solid angle and frequency, but first multiply through by cos Then

dSdJdFdx

d

00044

dSdJ

00

44

40

0

TeffF

d

dK

3 Conditions of Radiative 3 Conditions of Radiative Equilibrium:Equilibrium:

• In real stars, energy is created or lost from the radiation field through convection, magnetic fields, and/or acoustic waves, so the energy constraints are more complicated

dFF

0

dSdJ

00

44

40

0

TeffFd

d

dK

Solving the Transfer Equation Solving the Transfer Equation in Practicein Practice

• Generally, one starts with a first guess at T() and then iterates to obtain a T() relation that satisfies the transfer equation

• The first guess is often given by the “gray atmosphere” approximation: opacity is independent of wavelength

Solving the Gray AtmosphereSolving the Gray Atmosphere• Integrating the transfer equation over frequency:

• gives or

• The radiative equilibrium equations give us:

F=F0, J=S, and dK/d = F0/4

dSdIdIdx

d

000cos

SIdx

dI cos SId

dI

cos

Eddington’s Solution (1926)Eddington’s Solution (1926)

• Using the Eddington Approximation, one gets

• Chandrasekhar didn’t provide a rigorous solution until 1957

• Note: One doesn’t need to know since this is a T() relation

TeffT 41

))3/2(4

3()(

0)3

2(

4

3)()( FSJ

Class ProblemClass Problem

• The opacity, effective temperature, and gravity of a pure hydrogen gray atmosphere are = 0.4 cm2 gm-1, 104K, and g=2GMSun/RSun

2. Use the Eddington approximation to determine T and at optical depths = 0, ½, 2/3, 1, and 2. Note that density equals 0 at = 0.

Limb DarkeningLimb Darkening

This white-light image of the Sun is from the NOAO Image Gallery. Note the darkening of the specific intensity near the limb.

Limb Darkening in a Gray Limb Darkening in a Gray AtmosphereAtmosphere

• Recall that

• so that as increases the optical depth along the line of sight increases (i.e. to smaller and smaller depth and cooler temperature)

• In the case of the gray atmosphere, recall that we got:

deSI sec)0( )sec(

0

)0()3

2(

4

3)( FS

Limb darkening in a gray atmospehreLimb darkening in a gray atmospehre

• so that I(0) is of the form

I(0) = a + bcos

One can derive that

and

cos32),0( I

5

cos32

)0,0(

),0(

I

I

Comparing the Gray Comparing the Gray Atmosphere to the Real SunAtmosphere to the Real Sun

0

0.2

0.4

0.6

0.8

1

1.2

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

Cos theta

I(th

eta)

/I(0

)

Eddington Approx.

Observed Intensity