Homework 3 solutions#
1. LTE#
The equation of HSE is:
we can take \(g\) as a constant, using the value at the surface of the Sun:
We use the ideal gas law:
and substituting this into HSE, we have:
where \(H\) is the scale height. Upon integrating, we get:
We can evaluate \(H\):
Now, we know that the mean free path is:
If we take \(\kappa = 0.264~\mathrm{cm^2~g^{-1}}\) and \(\rho = 2.5\times 10^{-7}~\mathrm{g~cm^{-3}}\) (both typical of the photosphere), then we find:
Since \(\lambda_\gamma \sim H\), we cannot assume LTE in the photosphere.
2. Limb darkening#
From class, we know that
and
together this tells us that
Now, if we:
Use the closure \(P = (1/3) U\)
Assume we are isotropic and in radiative equilibrium, so \(\langle I \rangle = S\)
then we have
which says that
The outgoing solution to the rad transfer equation is:
(we took the reference to be \(\tau_0 \rightarrow \infty\)). Substituting in our \(S(\tau)\) and evaluating at \(\tau = 0\), we have:
making the substitution \(\xi = t / \mu\), we have:
or
What does this mean?
\(\theta = 0 \rightarrow \mu = 1\) means the center of the disk, where the intensity is the greatest
\(\theta = \pi/2 \rightarrow \mu = 0\) means the limb. But notice that the intensity is not zero there!
Normalizing by the intensity at the center of the Sun, we have:
3. Atmospheres#
Starting with the solution to the radiation transfer equation, with our referenece optical depth \(\tau_0\):
we now assume:
We are looking from above, so we care about the \(\mu = 1\) direction
We are at the top of the slab, so we can take \(\tau = 0\):
Finally, taking \(S\) as constant, we can do the integral:
Now we can consider different cases:
a. If \(\tau_0 \gg 1\), then \(e^{-\tau_0} \rightarrow 0\), and $I_0 = S = B–we see blackbody radiation.
b. If \(\tau_0 \ll 1\), then \(e^{-\tau_0} \sim 1 - \tau_0\), and we have:
If \(I(\tau_0) > S\), then we see absorption
If \(I(\tau_0) < S\), then we see emission lines
These are known as Kirchhoff’s laws.