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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).
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: $$T_0=\inv{\sqrt{2}}T_e$$ 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.