Homework 3 solutions

1. LTE

The equation of HSE is:

\[\frac{dP}{dr} = -\rho |g|\]

we can take \(g\) as a constant, using the value at the surface of the Sun:

\[g = \frac{GM_\odot}{R_\odot^2} = \frac{6.67\times 10^{-8}~\mathrm{dyn~cm^2~g^{-2}} \cdot 2\times 10^{33}~\mathrm{g}}{(7\times 10^{10}~\mathrm{cm})^2} = 2.7\times 10^4~\mathrm{cm~s^{-2}}\]

We use the ideal gas law:

\[P = \frac{\rho k T}{\mu m_u}\]

and substituting this into HSE, we have:

\[\frac{1}{\rho} \frac{d\rho}{dr} = \underbrace{\frac{\mu m_u g}{k T}}_{H^{-1}}\]

where \(H\) is the scale height. Upon integrating, we get:

\[\rho = \rho_0 e^{-r/H}\]

We can evaluate \(H\):

\[H = \frac{kT}{\mu m_u g} = \frac{1.38\times 10^{-16}~\mathrm{erg~K^{-1}} \cdot 6000~\mathrm{K}}{0.6 \cdot 1.66\times 10^{-24}~\mathrm{g} \cdot 2.7\times 10^4~\mathrm{cm~s^{-2}}} = 300~\mathrm{km}\]

Now, we know that the mean free path is:

\[\lambda_\gamma = \frac{1}{\kappa \rho}\]

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:

\[\lambda_\gamma = \frac{1}{0.264~\mathrm{cm^2~g^{-1}} \cdot 2.5\times 10^{-7}~\mathrm{g~cm^{-3}}} = 150~\mathrm{km}\]

Since \(\lambda_\gamma \sim H\), we cannot assume LTE in the photosphere.

2. Limb darkening

From class, we know that

\[\langle I \rangle = \frac{1}{2} \int_{-1}^1 I(\mu) d\mu\]

and

\[U = \frac{2\pi}{c} \int_{-1}^1 I(\mu) d\mu\]

together this tells us that

\[U = \frac{4\pi}{c} \langle I \rangle\]

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

\[P(\tau) = \frac{F}{c} \left (\tau + \frac{2}{3} \right ) = \frac{4\pi}{3c} S\]

which says that

\[S = \frac{3}{4\pi} F \left (\tau + \frac{2}{3} \right )\]

The outgoing solution to the rad transfer equation is:

\[I(\tau, \mu> 0) = \frac{1}{\mu} \int_\tau^\infty e^{-(t-\tau)/\mu} S(t) dt\]

(we took the reference to be \(\tau_0 \rightarrow \infty\)). Substituting in our \(S(\tau)\) and evaluating at \(\tau = 0\), we have:

\[I_0(\mu> 0) = \frac{1}{\mu} \int_0^\infty e^{-t/\mu} \left [ \frac{3F}{4\pi} \left ( t + \frac{2}{3} \right ) \right ] dt\]

making the substitution \(\xi = t / \mu\), we have:

\[I_0(\mu > 0) = \frac{3F}{3\pi} \int_0^\infty e^{-\xi} \left (\mu \xi + \frac{2}{3} \right ) d\xi = \frac{3F}{4\pi} \mu \Gamma(2) - \frac{1}{2} \frac{F}{\pi} \left . e^{-\xi} \right |_0^\infty\]

or

\[I_0(\mu > 0) = \frac{3F}{4\pi} \left (\mu + \frac{2}{3} \right )\]

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:

\[\frac{I_0(\mu)}{I_0(\mu = 1)} = \frac{3}{5} \left (\cos\theta + \frac{2}{3} \right)\]

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 \(\tau = 0\):

\[I_0 = e^{-\tau_0} I(\tau_0) - \int_{\tau_0}^0 e^{-t} S(t) dt\]

Finally, taking \(S\) as constant, we can do the integral:

\[I_0 = e^{-\tau_0} I(\tau_0) - S \int_{\tau_0}^0 e^{-t} dt = e^{-\tau_0} I(\tau_0) + S \left (1 - e^{-\tau_0} \right )\]

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:

\[I_0 = I(\tau_0) - \tau_0 [I(\tau_0) - S] \]

• 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.