Difference between revisions of "Radiative Diffusion"

Reference Materials

<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

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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.

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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\section{The Eddington Approximation and Bolometric Radiative Equilibrum Approximation } 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}(\frac{3}{4}\tau + \frac{1}{2}) \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}

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


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