# Radiative Diffusion

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### Reference Materials

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\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document}

\subsection*{ Bolometric Radiative Equilibrium}

Let's start again with our equation:
$$\mu\ddtau{I_\nu}=I_\nu-B_\nu$$
We can rewrite this in terms of atmospheric height $z$ (recall that as
$z\to\infty$, $\tau\to0$, so:
$$\mu{dI_\nu\over\rho\kappa_\nu dz}=-I_\nu+B_\nu$$
Then the first moment of this equation is:
$${\mu\over\rho}{d\over dz}\int{{I_\nu\over\kappa_\nu}d\nu}=-I+B$$
where $I\equiv\int{I_\nu d\nu}$ and $B\equiv\int{B_\nu d\nu}$. Now we'll make
the (dubious) approximation that $\kappa_\nu=\kappa$ does not depend on
frequency (``Grey atmosphere*). This gives us:*
$${\mu\over\rho\kappa}{dI\over dz}=-I+B=-{dI\over d\tau}$$
Now we make the assumption of Bolometric Radiative Equilibrium, so that:
$${4\pi\over3}{dB\over d\tau}=F=constant$$
Recall that $B=\int{B_\nu d\nu}={\sigma T^4\over\pi}$, so BRE is telling us:
$${4\pi\over3}\ddtau{}\left({\sigma T^4\over\pi}\right)=F$$
Or rewriting this:
$$\boxed{\sigma T^4=\sigma T_b^4+{3F\over 4}\tau}$$
At infinite altitude, $J\equiv\int{J_\nu d\nu}=B$, so using that:
$$\begin{aligned}J&=I^-+{F\over2\pi}\\
&=I^+-{F\over2\pi}\\ \end{aligned}$$
we have:
$$J=B=\underbrace{I^-\eval{altitude}}_{0}+{F\over2\pi}$$
Therefore:
$$F=2\pi B\eval{altitude}=2\pi{\sigma T_0^4\over\pi}=2\sigma T_0^4$$
Thus, our BRE equation gives us:
$$\boxed{\sigma T^4=\sigma T_0^4\left(1+{3\over2}\tau\right)}$$
Saying $F=2\sigma T_0^4\equiv\sigma T_e^4$, where $T_e$ is an ``effective
temperature*, then:*
$$T_0=\inv{\sqrt[4]{2}}T_e$$
This is apparently a classical result. Let's do an example by calculating
the effective temperature of the Earth. $F\eval{earth}$ is given by:
$$4\pi R_\oplus^2F=(1-\tilde\omega_{eff}){L_\odot\over4\pi d^2}\pi R^2$$
$\tilde\omega_{eff}$ is a measure of how much of the sun's energy we get. We'll
say that, since it's cloudy about a third of the time, $\tilde\omega_{eff}\sim0.3$.
Plugging in the numbers, we find that $T_e=258K$, so $T_0=217K$. This is
indeed about the mid-latitude temperature of air in the troposphere. Now
temperature scales with optical depth by:
$$\begin{aligned}T^4(\tau)&=T_0^4\left(1+{3\over2}\tau\right)\\
T&=T_0\tau^{1\over4}\\ \end{aligned}$$
This is an expression of the greenhouse effect.

\subsection*{ Radiative Diffusion}

Recall that we had, using Bolometric Radiative Equilibrium, an equation which described the greenhouse effect: $$\sigma T^4=\sigma T_0^4\left[1+{3\over2}\tau\right]$$ Now we want to talk about the effects of the diffusion of photons. For this, we have the general diffusion equation: $$F=-D\nabla n$$ For photons, $F$ is the energy flux, $D$ is $\lambda_{mfp}\cdot c$, and $n\sim{\sigma\over c}T^4$ is the number density of photons. Then: $$F\sim\underbrace{\lambda_{mfp}\over L}_{1\over\tau}c{\sigma\over c}T^4 \sim{\sigma T^4\over\tau}$$ Recall that $F\equiv\sigma T_e^4$, so: $$T^4\sim T_e^4\tau$$ This says that as we go deeper into the atmosphere, the temperature increases, but slowly (as the fourth root).

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