I

_{l}_{ }(0,q) = ò_{o}^{z }B_{l}(t) exp [(-t_{l}_{ }/ cos q)] dt/ cos q
where B

_{l}(t) is the Planck function, and the integral is taken from z = 0 to the point z in the interior.
Next, since we’re dealing with the passage of radiation

*out of the star*, we need the relevant equation of transfer. This is best dealt with using integral with dx, and hence recasting the emergent intensity in the form:
I

_{l}_{ }(0,q) = ò_{o}^{¥}^{ }B(x) e^{(-x / cos }^{q}^{)}dx/ cos q
Then the appropriate equation of transfer would be:

dI = j dx - s I (q) dx = (j - s I (q)) dx

or: dI/dx = j - s I (q))

After some further manipulation, and replacing x with t:

(cos q) dI/ dt = I

_{ }(q) – j/ s
This is the important equation, in terms of emergent intensity, that embodies the conservation of radiant energy (i.e. no more radiation can flow out of a star’s surface than can be generated within it and which approaches that surface). Next, we want to be able to obtain an even more improved basis for our calculations and this entails getting the moments of the intensity. These are defined as follows:

J = 1/4p ò

_{4}_{p}I (q) dw (*Mean Intensity*)
Where dw is an element of solid angle - defined as (A/r

^{2}) for a sphere, for example. Thus a sphere with surface area A = 4p r^{2}has solid angle (4p r^{2}/ r^{2}) = 4p steradians. If we are only dealing with a sliver of emergent beam of area 0.01p r^{2 }then the element of solid angle is:
dw = (0.01p r

^{2}/ r^{2}) = 0.01p sr
Or 1/ 400 the volume of a sphere.

The next moment is:

H = 1/4p ò

_{4}_{p}I (q) cos^{ }q dw
This is defined as the “

*net flux*” or the net energy breaching the stellar surface in units of net energy per second per unit area of that surface. Finally, we come to the last moment of intensity:
K = 1/4p ò

_{4 }_{p}I (q) cos^{2 }q dwThis is the

*energy density*.

At this stage, we are in a position to use each of the above moments to further manipulate the equation 0f transfer. We start by multiplying the original equation of transfer: (cos q) dI/ dt = I

_{ }(q) – j/ s through by 1/4p ò

_{4}

_{p}I (q) dw to get:

dH/ dt = 1/4p ò

_{4}_{p}I (q) dw - 1/4p ò_{4}_{p}j/ s dw
This makes use of the definition: dH/ dt = J - j/ s

Next, we multiply the equation of transfer through by H, or 1/4p ò

_{4}_{p}I (q) cos^{ }q dw to get: dK/ dt = H or more simply: 4p H = const. So, dH/ dt = J - j/ s = 0, and the equation of transfer now becomes:
(cos q) dI/ dt = -I

_{ }(q) + 1/4p ò_{4}_{p}I (q) dw = - I + JFrom here a number of specific assumptions are made in order to not have to evaluate the integral. The main one is the Eddington approximation which will apply to the quantities J, H and K. We will also discriminate the radiation intensity I into two components: I

_{1}(in the forward direction) and I

_{2 }(in the backward direction). We can then write as follows:

1) J = ½(I

_{1}+ I_{2})
2) H = ¼( I

_{1}- I_{2})
3) K = J/3

Now, we focus on the boundary in Fig. of instalment (4) and note that here the optical depth t= 0, and we must have I

_{2 }= 0 also. Since I_{2 }= 0 then: H = ¼ I_{1}and J = ½I_{1}so clearly: J = 2H.Further, K = J/3 or (½I

_{1})/3 so K = Ht + const. This follows, since we had: dK/ dC = H or dK = H dt and we know t= 0, hence Ht + const. on integration. From this it follows that:

J = H(2 + 3 t) and K = J/3 = 2H/3

At the boundary everywhere. And since H = ¼( I1 - I2) = const.

I

_{1}(t) = H(4 + 3t) and I_{1}(t) = 3Ht
A special case occurs if the mean intensity J = B, the Planck function, then (since B » sT

^{4}/ p):
J = H(2 + 3 t) = sT

^{4}/ p
Therefore, the boundary temperature (T

_{o}) approaches the value of the effective (or surface) temperature when t = 0. So we have the basic relationship:
sT

_{o }^{4}/ p = 2H
And: sT

^{4 }= sT_{o }^{4}/ 2 [a + 3t]Problems for the budding astrophysicist:

1) Find the mean intensity if I (q=p/4) is taken over all space.

2) Estimate the net flux, H, passing through the Sun’s surface.

3) Consider the solar half-sphere and the energy going into it each second. We know on average photons are absorbed after traveling a distance with optical depth t =1 in the propagation direction. Averaged over all directions this corresponds to a vertical optical depth of t = 2/3. Based on this find:

a) The energy going into the half sphere each second.

b) The change in (a) over each absorption and re-emission over vertical optical depth.

c) The total absorption and total emission and the relationship between then over all space.

(All content extracted from my book: 'Astronomy & Astrophysics: Notes, Problems and Solutions', Chapters 15, 17)

## No comments:

Post a Comment