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Feb. 2, 2011 Rosseland Mean Absorption Poynting Vector Plane EM Waves The Radiation Spectrum: Fourier TransformsPowerPoint Presentation

Feb. 2, 2011 Rosseland Mean Absorption Poynting Vector Plane EM Waves The Radiation Spectrum: Fourier Transforms

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### Feb. 2, 2011Rosseland Mean AbsorptionPoynting VectorPlane EM WavesThe Radiation Spectrum: Fourier Transforms

Rosseland Mean Opacity

Recall that for large optical depth

In a star,

is large, but there is a temperature gradient

How does Fn(r) relate to T(r)?

z

dz

ds

so

Equation of radiative transfer:

emission

scattering

absorption

(1)

“Zeroth” order approximation:

Independent of m

1st order approximation:

Equation (1)

Depends on m

Define Rosseland mean absorption coefficient:

Combining,

Equation of radiative diffusion

In the Rosseland Approximation

- Flux flows in the direction opposite the temperature gradient
- Energy flux depends on the Rosseland mean absorption coefficient,
- which is the weighted mean of
- Transparent regions dominate the mean

Poynting Vector

One of the most important properties of EM waves is that they transport

energy—

e.g. light from the Sun has traveled 93 million miles and still has enough

energy to do work on the electrons in your eye!

Poynting vector, S: energy/sec/area crossing a surface whose normal

is parallel to S

Poynting’s theorem: relates mechanical energy performed by the E, B

fields to S and the field energy density, U

so...

where U(mech) = mechanical energy / volume

= field energy / volume

Now

So (1) says

rate of change

of field energy

per volume

rate of change of

mechanical energy

per volume

+

Maxwell’s Equations in a vacuum:

(3)

(1)

(4)

(2)

These equations predict the existence of WAVES

for E and B which carry energy

Curl of (3)

use (4)

operates on each component of

so these 2 vector wave equations are actually 6scalar equations

So, for example, one of the equations is

Similarly for Ey, Ez, Bx, By and Bz

What are the solutions to the wave equations?

First, consider the simple 1-D case --

Wave equation

A solution is

A = constant (amplitude)

[kx] = radians [kvt] = radians

So the wave equation is satisfied

Ψ is periodic in space (x) and time (t)

The wavelength λ corresponds to a change in the argument of

the sine by 2π

frequency

angular frequency

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