Energy Transport Formal solution of the transfer equation Radiative equilibrium The gray atmosphere Limb darkening

The Transfer Equation Recall: for radiation passing through a gas, the change in I is equal to: dI = intensity emitted – intensity absorbed dI = j  dx –   I dx or dI /d  = -I + S

The 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)

Toss 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

Radiative 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

IntegratingOver Solid Angle Assume   and S  are independent of direction, and substitute the definitions of flux and mean intensity: becomes: Then integrate 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

3 Conditions of Radiative 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

Solving the Transfer Equation in 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 Atmosphere Integrating the transfer equation over frequency: givesor The radiative equilibrium equations give us: F=F 0, J=S, and dK/d  = F 0 /4 

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

Class Problem The opacity, effective temperature, and gravity of a pure hydrogen gray atmosphere are  = 0.4 cm 2 gm -1, 10 4 K, and g=2GM Sun /R Sun 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 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 Atmosphere 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:

Limb darkening in a gray atmospehre so that I (0) is of the form I (0) = a + bcos  One can derive that and

Comparing the Gray Atmosphere to the Real Sun