Radiation Lecture 21

From AstroBaki
Revision as of 19:15, 14 February 2010 by WikiSysop (talk | contribs) (Created page with '<latex> \documentclass[11pt]{article} \def\inv#1{{1 \over #1}} \def\ddt{{d \over dt}} \def\mean#1{\left\langle {#1}\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} …')
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigationJump to search

<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\~{\tilde } \def\ddtau#1Template:D

\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \subsection*{ Monochromatic Radiative Equilibrium}

Last time we had the integro-differential equation: $$\mu{dI_\nu\over d\tau_\nu}=I_\nu-B_\nu(T(z))$$ We got rid of the ``integro part, but then we did an integral over $d\Omega$ to get the angle-averaged $I_\nu$, so we brought back our ``integro when we said: \def\inpTemplate:I \nu^+ \def\inmTemplate:I \nu^- $${dF_\nu\over d\tau_\nu}=4\pi(J_\nu-B_\nu)$$ Now recall our definitions of $\inp, \inm$: $$\begin{aligned}\inp(\tau_\nu)&=I_\nu(\tau_\nu,0<\mu<1)\\ \inm(\tau_\nu)&=I_\nu(\tau_\nu,-1<\mu<0)\\ \end{aligned}$$ Recall that $\mu\equiv\cos\theta$, but keep in mind that $\theta$ measures the angle of the {\it incoming flux intensity}, which is generally the opposite direction from which we are looking. Thus, $\inp$ is flux intensity coming from {\it below}.\par Now the angle-averaged flux $J_\nu$ is given by: $$J_\nu=\int{I_\nu d\Omega}=\hf(\inp+\inm)$$ Likewise, our flux is: $$F_\nu=\int{I_\nu\cos\theta d\Omega} =2\pi\int_0^{\pi\over2}{\inp\cos\theta\sin\theta d\theta} + 2\pi\int_{\pi\over2}^{\pi}{\inm\cos\theta\sin\theta d\theta} =\pi(\inp-\inm)$$ Now let's calculate our second moment, $\int{equation\cdot\mu d\Omega}$: $$\begin{aligned}LHS&=\ddtau{}\int{\mu^2I_\nu d\Omega} =\ddtau{}\left[2\pi\int_{-1}^0{\mu^2\inp d\mu}+2\pi\int_0^1{\mu^2\inm d\mu} \right] ={2\pi\over3}\ddtau{}(\inp+\inm)\\ &={4\pi\over2}\ddtau{}J_\nu\\ \end{aligned}$$ The right-hand side of the equation is: $$RHS =\int{\mu I_\nu d\Omega}-\underbrace{\int{\mu B_\nu d\Omega}}_{0} =\int{\mu I_\nu d\Omega} =F_\nu$$ where the last step was taken with the aid of Rybicki \& Lightman Eq. 1.3b. So our full second moment equation is: $${4\pi\over3}\ddtau{}J_\nu=F_\nu$$ Recall that our first moment equation was: $$\ddtau{F_\nu}=4\pi(J_\nu-B_\nu)$$ We may rewrite this as: $$J_\nu=\inv{4\pi}\ddtau{F_\nu}+B_\nu$$ So our first and second moment equations together give us: $$\boxed{{4\pi\over3}\ddtau{} \left[\inv{4\pi}\ddtau{F_\nu}+B_\nu\right]=F_\nu}$$ Our third moment equation, because we are in {\bf Monochromatic Radiative Equilibrium} is: $$\ddtau{F_\nu}=0$$ Thus, $F_\nu$ is constant. Now $\ddtau{B_\nu}=F_\nu=constant$, so $B(T(z))$ scales linearly with $\tau$. Now $\tau=0$ corresponds the the top of the atmosphere, and we'd expect that the temperature there should be $0$, but this is incorrect. Likewise $\tau_{max}$ corresponds to the full atmospheric depth, i.e. the ground. We'd expect that the temperature there should just be the temperature of the ground $T_g$. This is also incorrect. This is because there is an interface of conductivity at the surface of the ground. We'll call $T_1$ the temperature of the air at the base of the atmosphere, and try to see if get $B_\nu(T_1)$ in terms of $B_\nu(T_g)$. First, we need to get an expression for $J_\nu$ in terms of and $F_\nu$. Using that $J_\nu=\hf(\inp+\inm)$ and $F_\nu=\pi(\inp-\inm)$ $$\boxed{J_\nu=\inm+{F_\nu\over2\pi}=\inp-{F_\nu\over2\pi}}$$ Now since $\ddtau{F_\nu}=0=4\pi(J_\nu-B_\nu)$, $J_\nu=B_\nu$ everywhere. Specifically, in the atmosphere right above the ground, $J_\nu=B_\nu(T_1)$, so: $$J_\nu=\inp\eval{surface}-{F_\nu\over2\pi}\eval{surface}$$ Now $\inp\eval{surface}=B_\nu(T_g)$, so: $$B_\nu(T_1)=B_\nu(T_g)-{F_\nu\over2\pi}$$ So we have shown why $B_\nu(T_1)\ne B_\nu(T_g)$. We should also discuss why, at the top of the atmosphere, $T_0\ne 0$. For this, we'll write: $$\begin{aligned}J_\nu=B_\nu(T_0) &=\underbrace{\inm\eval{altitude}}_{0}+{F_\nu\over2\pi}\\ B_\nu(T_0)&={F_\nu\over2\pi}\\ \end{aligned}$$ Thus, $T_0\ne0$. So we could solve for $F_\nu$ and find that it is a function of $\tau$, but solving it for the fairly contrived case of MRE (Monochromatic Radiative Equilibrium) doesn't give us anything very meaningful. Instead we'll solve it for BRE (Bolometric Radiative Equilibrium).

\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: $$4\pi R_\oplus^2F=(1-\tilde\omega_{eff}){L_\odot\over4\pi d^2}\pi R^2$$ $\~\omega_{eff}$ 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, $\~\omega\sim0.3$. 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.

\end{document} <\latex>