### Need to Review?

<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle {#1}\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\hf{\frac12} \def\^{\hat } \def\.{\dot } \def\ddtau#1Template:D

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

\subsection*{ Bolometric Radiative Equilibrium}

%\iffalse %Here we wish to calculate the radiative transfer equation in the context of plane parallel atmospheres. In order to do that, we will be measuring from the surface up, where $\mu %= cos \theta$ is defined relative to the vertical upward direction. Therefore, we begin with the equation of radiative transfer, %\begin{align} %\mu \frac{dI_{\nu}}{d \tau} = I_{\nu} - S_{\nu} %\end{align} In this approximation we assume a grey atmosphere, meaning that opacity is frequency independent. Therefore, we integrate the Radiative Transfer Equation over frequency, \begin{align} \mu \frac{dI}{d \tau} = I - S \end{align} Now we use the Eddington Approximation to solve. First, we integrate the equation of radiative transfer over all angles (4$\pi$ steradian), \begin{align} \frac{d}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} I \mu d \phi d \mu = \int_{-1}^{1} \int_{0}^{2 \pi} I d \phi d \mu - \int_{-1}^{1} \int_{0}^{2 \pi} S d \phi d \mu \end{align} The integral on the left is by definition the second moment of intensity, the flux. Additionally the first integral on the right hand side is the intensity integrated over all solid angles, the mean intensity. Additionally, the source function, S, is angle independent. Therefore, \begin{align} \frac{dF}{d \tau} = 4 \pi J - 4 \pi S \end{align} In this approximation we assume the flux is constant through the atmosphere, yielding, \begin{align} 0 = 4 \pi J - 4 \pi S \end{align} \begin{align} J = S \end{align} To find the next important relation, we multiply the radiative transfer equation by $\mu$ and once again integrate over all solid angles. \begin{align} \frac{d}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} I \mu^2 d \phi d \mu = \int_{-1}^{1} \int_{0}^{2 \pi} I \mu d \phi d \mu - \int_{-1}^{1} \int_{0}^{2 \pi} S \mu d \phi d \mu \end{align} For the first integral on the right hand side, we know by definition that this is the flux. For the second integral on the right hand side, because S is independent of angle, \begin{align}

\int_{-1}^{1} \int_{0}^{2 \pi} S \mu d \phi d \mu = 0

\end{align} For the integral on the left hand side, we use the diffusion approximation that the intensity within the atmosphere is about isotropic and can therefore be taken out of the integral as the mean intensity. \begin{align} \frac{d}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} I \mu^2 d \phi d \mu = \frac{dJ}{d \tau} \int_{-1}^{1} \int_{0}^{2 \pi} \mu^2 d \phi d \mu = \frac{4 \pi}{3} \frac{dJ}{d \tau} \end{align} Therefore, plugging these results back into Equation 7, \begin{align} \frac{4 \pi}{3} \frac{dJ}{d \tau} = F \end{align} Using the prior result that $J = S$, \begin{align} F = \frac{4 \pi}{3} \frac{dS}{d \tau} \end{align} We can solve this simple equation, \begin{align}

\int dS= \int\frac{3}{4 \pi} F d \tau

\end{align} \begin{align} S = \frac{3}{4 \pi} F \tau + K \end{align} where K is a constant.

At the upper boundary of the atmosphere where $\tau = 0$, we assume that there is no incident radiation from beyond the boundary. Therefore, at this location, J = I/2. From our previous result that J = S, we know that $S= \frac{I}{2}$. We also know $F = I/ \pi$, so at the outer boundary, \begin{align} S = \frac{F}{2 \pi} \end{align} Plugging in for $\tau = 0$ to Equation 13, \begin{align} S = 0 + K= \frac{F}{2 \pi} \end{align} \begin{align} K = \frac{F}{2 \pi} \end{align}

Plugging in to our equation for S, \begin{align} S = \frac{3}{4 \pi} F \tau + \frac{F}{2 \pi} \end{align} Therefore, \begin{align} S = \frac{F}{\pi} \left( \frac{3}{4}\tau + \frac{1}{2} \right) \end{align} Here, if we assume that we are in Local Thermodynamic Equilibrium (LTE), we can assume the source function is a blackbody, \begin{align} S = \int_{0}^{\infty} B_{\nu} d\nu = \frac{\sigma T^4}{\pi} \end{align} Additionally, under the assumption that all energy is carried in radiation, we know the integrated flux is, \begin{align} F = \sigma T_{e}^4 \end{align} where $T_e$ is the effective temperature.

Therefore, plugging into our equation for S, \begin{align} \frac{\sigma T^4}{\pi} = \frac{\sigma T_e^4}{\pi} ( \frac{3}{4}\tau + \frac{1}{2} ) \end{align}

\begin{align}

T^4 = T_e^4 ( \frac{3}{4}\tau + \frac{1}{2} )

\end{align} %\fi \iffalse 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: $$\boxed{\sigma T^4=\sigma T_e^4\left(1\over2+{3\over4}\tau\right)}$$

Saying $F=2\sigma T_0^4\equiv\sigma T_e^4$, where $T_e$ is an effective temperature, then: $$T_0=\inv{\sqrt[4]{2}}T_e$$ \fi 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_e^4\left({1\over2}+{3\over4}\tau\right)\\ T&=T_e\tau^{1\over4}\\ \end{aligned} This is an expression of the greenhouse effect.

Recall that we had, using Bolometric Radiative Equilibrium, an equation which described the greenhouse effect: $$\sigma T^4=\sigma T_e^4\left[{1\over2}+{3\over4}\tau\right]$$ 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).

\section{Radiative Transfer Equation with Scattering}
To start, we will adjust the Radiative Transfer Equation with a few approximations and substitutions for scattering. To begin we have,

\begin{align} \frac{dI}{d \tau} = -I + S \end{align} where S is the source function. Here, because we have an absorption cross sections ($\sigma_{abs}$) and a scattering cross section ($\sigma_{scat}$), we can define the Optical Depth \begin{align} d\tau = n(\sigma_{abs} + \sigma_{scat}) ds \end{align} where n is the number density of particles in the medium. Plugging in Equation 2 to Equation 1, and simplifying, \begin{align} \frac{dI}{ds} = Sn\sigma_{scat} + Sn \sigma_{abs} -In(\sigma_{abs} + \sigma_{scat}) \end{align} Assuming isotropic scattering, the source function for scattering will be the mean intensity (averaged over all solid angles). Therefore, \begin{align} \frac{dI}{ds} = Jn\sigma_{scat} + Sn \sigma_{abs} -In(\sigma_{abs} + \sigma_{scat}) \end{align} Now, we wish to define the albedo, a coefficient of reflection, as \begin{align} a = \frac{\sigma_{scat}}{\sigma_{scat} + \sigma_{abs}} \end{align} Dividing Equation 4 by $n(\sigma_{scat} + \sigma_{abs})$ on both sides and substituting in a, \begin{align} \frac{dI}{d \tau} = Ja + (1-a)S - I \end{align} Now, we will move onto the two-stream approximation. Here we wish to approximate a plane parallel atmosphere where the radiation field propagates in only two directions. In this approximation z and $\tau$ are measured down from the top. For this set-up, \begin{align} ds = -\frac{dz}{cos \theta} \end{align} Here, $cos \theta = \mu$. Therefore, we can write the radiative transfer equation, \begin{align} \mu \frac{dI}{d\tau} = -Ja - (1-a)S + I \end{align} Here we are examining the common case where the source function from emission, S, is a blackbody, B, \begin{align} \mu \frac{dI}{d\tau} = -Ja - (1-a)B + I \end{align}

\section{Two-Stream Approximation, Only Scattering}
The Two Stream Approximation (part of Eddington's Approximations) is when the radiative transfer equation is approximated so that radiation only propagates in two directions. In this approximation we assume isotropic scattering.
We will do this derivation in two different ways- first with the approximation that only scattering is occuring, and second, we will include scattering and emission processes. The scattering only approximation simplifies the derivation and allows for a good first understanding of the math and approximations involved.

Two Stream Approximation
%twostream2.png

Because we are considering only scattering processes, we can simplify the radiative transfer equation derived above, \begin{align} \frac{dI}{d \tau} = I - aJ \end{align} Therefore, since we are approximating the scattering to be in only two directions ($\mu = \pm 1$), we can write two different radiative transfer equations that each must be satisfied (one for each direction), the upward intensity, \begin{align} \frac{dI_+}{d \tau} = I_+ - aJ \end{align} and the downward intensity, \begin{align} \frac{dI_-}{d \tau} = -I_- + aJ \end{align} We also define the mean intensity as, \begin{align} J = \frac{1}{2}(I_{+} + I_{-}) \end{align} and the flux, \begin{align} H =\frac{1}{2 }(I_{+} - I_{-}) \end{align}

We begin by summing the Radiative Transfer Equation,
\begin{align}
\frac{dI_+}{d \tau} + \frac{dI_-}{d \tau} = 2 \frac{dJ}{d \tau} = 2H
\end{align}
Thus,
\begin{align}
H = \frac{dJ}{d \tau}
\end{align}
Next, we subtract the equations for $I_-$ and $I_+$,
\begin{align}
\frac{dI_+}{d \tau} - \frac{dI_-}{d \tau} \Rightarrow 2 \frac{dH}{d \tau} = 2J - 2aJ
\end{align}
Thus,
\begin{align}
\frac{dH}{d \tau} = J(1-a)
\end{align}
We can find get an equation only in terms of J,
\begin{align}
\frac{d^2J}{d \tau^2} = \frac{dH}{d \tau} = J(1-a)
\end{align}
This is a simple differential equation with solution,
\begin{align}
J = K_1 e^{\sqrt{1-a}\tau} + K_2 \ e^{-\sqrt{1-a}\tau}
\end{align}
where $K_1$ and $K_2$ are constants that we can solve using boundary conditions. First, we know that the mean intensity is concentrated in the atmosphere, so $K_1$ = 0.
For the other boundary condition, we know that at the top surface of the atmosphere, $\tau$ = 0 and the only intensity will be from the downward incident light.
\begin{align}
I_-(0) = I_0 = J(0) - H(0)
\end{align}
First, we can find H,
\begin{align}
H = - K_2\sqrt{1-a}\ e^{-\sqrt{1-a}\tau}
\end{align}
Thus, solving the above boundary condition,
\begin{align}
I_0 = K_2 + \sqrt{1-a} K_2
\end{align}
Therefore,
\begin{align}
K_2 = \frac{I_0}{1 + \sqrt{1-a}}
\end{align}
Thus,
\begin{align}
J = \frac{I_0}{1 + \sqrt{1-a}} e^{-\sqrt{1-a}\tau}
\end{align}
And we can find H,
\begin{align}
H = \frac{dJ}{d \tau} = \frac{-I_0 \sqrt{1-a}}{1 + \sqrt{1-a}} e^{-\sqrt{1-a}\tau}
\end{align}
Additionally, we can find each component of the intensity $I_-$ and $I_+$,
\begin{align}
I_+ = J + H = I_0 \frac{1 - \sqrt{1-a}}{1 + \sqrt{1-a}}e^{-\sqrt{1-a}\tau}
\end{align}
\begin{align}
I_- = J - H = I_0e^{-\sqrt{1-a}\tau}
\end{align}
Now we have found an expression for each component of the intensity. Additionally, because the flux, H, is negative, we have a net flux downwards into the atmosphere.

\section{Two-Stream Approximation, Scattering and Emission/Absorption Processes}

Now we can undergo a similar derivation, this time including absorption and emission processes. The angles that the intensity propagate need not be $\mu = \pm 1$. Different texts chose different angles. Now I will assume that the radiation field is approximated by two angles, $\mu = 1/\sqrt{3}, -1/ \sqrt{3}$, the former being the outward intensity ($I_{+}$) and the latter being the inward intensity($I_{-}$). These are the angles chosen in Rybicki and Lightman. Therefore, we define \begin{align} J = \frac{1}{2}(I_{+} + I_{-}) \end{align} and a flux, H, as \begin{align} H = \frac{1}{2 \sqrt{3}}(I_{+} - I_{-}) \end{align}

Each stream must independently satisfy the Radiative Transfer Equation. Therefore, \begin{align} \frac{dI_{+}}{d\tau} = -\sqrt{3}aJ - \sqrt{3}(1-a)B + \sqrt{3}I_{+} \end{align} \begin{align} \frac{dI_{-}}{d\tau} = \sqrt{3}aJ + \sqrt{3}(1-a)B - \sqrt{3}I_{-} \end{align} Adding the equations, \begin{align} \frac{dI_{+}}{d\tau} + \frac{dI_{-}}{d\tau} = \sqrt{3}(I_{+} - I_{-}) \end{align} The left hand side of this equation equals $2 \frac{dJ}{d \tau}$ while the right hand side equals 6H. Therefore, \begin{align} 2 \frac{dJ}{d \tau} = 6H \end{align} \begin{align} H = \frac{1}{3} \frac{dJ}{d \tau} \end{align} Now, we will subtract equations 12 and 13 CHECK, \begin{align} \frac{dI_{+}}{d\tau} - \frac{dI_{-}}{d\tau} = \sqrt{3}(I_{+} +I_{-}) - 2 \sqrt{3}(1-a)B - 2 \sqrt{3} aJ \end{align} Subbing in J, \begin{align} \frac{dI_{+}}{d\tau} - \frac{dI_{-}}{d\tau} = 2 \sqrt{3}J - 2 \sqrt{3}(1-a)B - 2 \sqrt{3} aJ \end{align} The left hand side equals $2 \sqrt{3} \frac{dH}{d \tau}$, \begin{align}

2 \sqrt{3} \frac{dH}{d \tau}= 2 \sqrt{3}J - 2 \sqrt{3}(1-a)B - 2 \sqrt{3} aJ

\end{align} \begin{align}

\frac{dH}{d \tau}= J - (1-a)B -  aJ

\end{align} \begin{align}

\frac{dH}{d \tau}= (1-a)J - (1-a)B

\end{align} Now, substituting in equation 16, \begin{align} \frac{d^2J}{d \tau^2} = 3(1-a)J - 3(1-a)B \end{align} If we take B to be constant, the solution to this differential equation is, \begin{align} J = K_1 e^{\sqrt{3(1-a)} \tau} + K_2 e^{-\sqrt{3(1-a)}\tau} +B \end{align} We want the solution to be well enclosed in the atmosphere, therefore we must get rid of the positive exponential solution by setting $K_1 = 0$. \begin{align} J =K_2 e^{-\sqrt{3(1-a)}\tau} +B \end{align} In order to solve for $K_2$, \begin{align} \frac{dJ}{d \tau} = -K_2 \sqrt{3(1-a)}e^{-\sqrt{3(1-a)}\tau} = 3H \end{align} Solving for H, \begin{align} H = -\frac{K_2}{3} \sqrt{3(1-a)}e^{-\sqrt{3(1-a)}\tau} \end{align} Now we impose the boundary condition that at the surface of the atmosphere, where $\tau = 0$, $I_{-} = 0$ because there are no incident rays. From equations 10 and 11, we can solve for the intensity in the downward direction, \begin{align} I_{-}(0) = J(0) - \sqrt{3} H(0) = 0 \end{align} \begin{align} J(0) = \sqrt{3}H(0) \end{align} \begin{align} K_2 + B = -\frac{\sqrt{3}K_2}{3} \sqrt{3(1-a)} \end{align} \begin{align} K_2 + B = -K_2 \sqrt{(1-a)} \end{align} \begin{align} K_{2} = \frac{-B}{1 + \sqrt{1-a}} \end{align} Therefore, plugging this into the mean intensity, \begin{align} J = B - \frac{B}{1 + \sqrt{1-a}}e^{-\sqrt{3(1-a)}\tau} \end{align} To find H, \begin{align} H = \frac{1}{3}\frac{dJ}{d \tau} = \frac{\sqrt{3(1-a)}B}{3(1 + \sqrt{1-a})} e^{-\sqrt{3(1-a)}\tau} \end{align} Now we wish to calculate the intensity in the outward direction, $I_{+}$, \begin{align} I_+ = J + \sqrt{3}H= B - Be^{-\sqrt{3(1-a)}\tau}\frac{1-\sqrt{1-a}}{1 + \sqrt{1-a}} \end{align} \begin{align} I_- = J - \sqrt{3}H = B - Be^{-\sqrt{3(1-a)}\tau} \end{align} From this result, we can see that, far into the atmosphere at a high optical depth, J approaches B, the source function. Additionally, at a high optical depth, H = 0. This is due to the fact that we assumed B is constant, meaning there can be no net flux.

\end{document} <\latex>