# Radiation Lecture 23

<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\ppt#1Template:\partial \def\ddtau#1Template:D \def\wtTemplate:\tilde\omega \def\aiaio{\left[a_1I(\mu_1)+a_2I(\mu_2)\right]} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} Recall that we were solving the following equation for $I$: $$\mu\ddtau{I(\mu,\tau)}=I(\mu,\tau)-{\wt\over2}\left[I_1(\mu_1,\tau)+ I_2(\mu_2,\tau)\right]-{\wt\over4}\mathfrak{F}e^{-{\tau\over\mu_0}}$$ We then broke $I$ into two components measuring the upward and downward directed components of the specific intensity: $$\begin{aligned}\mu_1\ddtau{I(\mu_1)}&=I(\mu_1)-{\wt\over2}\aiaio-{\wt\over4}\mathfrak{F} e^{-\tau\over\mu_0}\\ \mu_2\ddtau{I(\mu_2)}&=I(\mu_2)-{\wt\over2}\aiaio-{\wt\over4}\mathfrak{F} e^{-\tau\over\mu_0}\\ \end{aligned}$$ The final solution is a sum of exponentials: $$\begin{aligned}I(\mu_1,\tau)&\equiv I_1=I(\mu_1=\inv{\sqrt{3}},\tau)\\ &=I_{1,hmg}+I_{1,part}\\ I(\mu_2,\tau)&\equiv I_2=I(\mu_2=1\inv{\sqrt{3}},\tau)\\ &=I_{2,hmg}+I_{2,part}\\ \end{aligned}$$ and last time we solved for the homogeneous component. All that remains to be done at this point is to fit the boundary conditions.

\subsection*{ Fitting Boundary Conditions }

The first condition is that $I_2$, which measures the intensity directed toward the ground, should vanish at $\tau=0=$the surface of the atmosphere (we are only considering reflected light here): $$I_2(\mu_2,\tau=0)=0$$ The second condition is that at the ground the intensity directed upward out of the ground should be equal to whatever is reflected from the downward directed intensity. Thus: $$I_1(\tau=\tau_{max},\mu_1>0)=I_2(\tau-\tau_{max},\mu_2<0)\Lambda$$ where $\Lambda$ is the ground albedo. There is another component we should examine. First, let's define: $$I_*(\tau,\mu)=\pi\mathfrak{F}\delta(\mu+\mu_0)\delta(\phi-\phi_0)e^{\tau\over\mu}$$ Then we'll examine the total flux (specific intensity integrated over solid angle, weighted by $\mu$) directed into the ground: \def\taum{{\tau_{max}}} $$\begin{aligned}F_{in,*} &=\int{\pi\mathfrak{F}\delta(\mu+\mu_0)\delta(\phi-\phi_0)e^{\taum\over\mu}\mu \,\underbrace{d\mu d\phi}_{d\Omega}}\\ &=-\pi\mathfrak{F}e^{-{\taum\over\mu}}\mu_0\\ \end{aligned}$$ Since the flux out of the ground is just the reflected portion of the incoming flux: $$\begin{aligned}F_{out,*}&=\Lambda F_{in,*}\\ F_{out,*}&\equiv\int{I_{1,*}\mu d\Omega}\\ &=I_{1,*}\int{\mu d\Omega}=I_{1,*}\pi\\ \end{aligned}$$ $$\boxed{I_{1,*}=\lambda\mathfrak{F}\mu_0e^{-{\taum\over\mu_0}}}$$ Therefore, the full second boundary condition reads: $$\boxed{I_1(\tau=\taum,\mu_1>0)=\Lambda I_2(\tau=\taum,\mu_2<0) +\Lambda\mathfrak{F}e^{-{\taum\over\mu_0}}\mu_0}$$ These two boundary conditions together give us answers to the two right-hand components of our original equation: $$\mu\ddtau{I(\mu,\tau)}=I(\mu,\tau)-{\wt\over2}\left[I_1(\mu_1,\tau)+ I_2(\mu_2,\tau)\right]-{\wt\over4}\mathfrak{F}e^{-{\tau\over\mu_0}}$$ So the equation had just become the equation of radiative transfer: $$\mu\ddtau{I}=I-S$$ and now we know the source function $S$.

\subsection*{ Effective Albedo of the Atmosphere }

Using 4.1.29 in the handout, we have that an atmosphere of optical depth
$\tau=\taum$, with ground albedo $\Lambda$ is identical to an infinite
atmosphere ($\tau\to\infty$) if $\xi=0$. In order for this to be the case:
$$\xi={\sqrt{1-\wt}(1+\Lambda)-(1-\Lambda)
\over\sqrt{1-\wt}(1+\Lambda)+(1-\Lambda)}=0$$
$$\boxed{\Lambda={1-\sqrt{1-\wt}\over1+\sqrt{1-\wt}}}$$
where $\Lambda$ here is the effective, macroscopic albedo of a semi-infinite
atmosphere.\par
We can get this same result by solving the diffusion equation, this time with
an absorption modifier:
$$\ppt{n}=\underbrace{D\nabla^2n}_{scattering\atop term}
-\underbrace{nn_d\sigma_ac}_{absorption\atop modifier}$$
where $n$ is the \# density of photons (which don't change in frequency),
$\sigma_a$ is the cross-section for
absorption, $n_d$ is the \# density of the ``dust* (any absorbing/scattering*
particle), and $D$ is the diffusivity, $D=\lambda_{mfp}c$. We have to be
careful that $\lambda_{mfp}$ contain both aspects of absorption and scattering:
$$\lambda_{mfp}=\inv{n_d(\sigma_a+\sigma_s)}$$
We have a steady-state solution to this diffusion equation. In 1-D, this
equation looks like:
$$\ppt{n}=0=D{\partial^2n\over\partial z^2}-n(n_d\sigma_ac)$$
We'll guess that $n=Ae^{z\over z_0}$ is a solution. Doing some algebra, we
can show that:
$$z_0=\inv{n_d[\sigma_a(\sigma_a+\sigma_s)]^\hf}$$
where we are assuming $z=0$ is the top of the atmosphere, where $\tau=0$, and
that $z<0$ as we move down through the atmosphere. Now we'll break our
solution into two components, evaluated at the surface of the atmosphere:
$$\begin{aligned}n&=Ae^{z\over z_0}\\
n(0)&=(n_i+n_e)e^{0\over z_0}\\ \end{aligned}$$
where $n_i$ are the incident photons from space, and $n_e$ are the escaping
photons from below. Then the diffusion equation (this time written in terms
of flux) tells us:
$$\begin{aligned}F&=-D\nabla n=c(n_e-n_i)\\
{-DAe^{z\over z_0}\over z_0}\eval{z\sim0}&=c(n_e-n_i)\\ \end{aligned}$$
We are saying $z\sim0$, and not $z=0$, because we have to be careful about
using the diffusion equation at low optical depth, which we have at the surface
of the atmosphere. Ignoring the subtleties inherent in the assumption, we'll
say that $e^{z\over z_0}\sim1$, giving us:
$$\boxed{{-D\over z_0}(n_i+n_e)}$$
Now we use the definition $\Lambda\equiv{n_e\over n_i}\eval{z=0}$ to get our
previous equation. First, note that:
$$n_e={\left(\sqrt{\sigma_a+\sigma_s\over\sigma_a}-1\right)\over
\left(\sqrt{\sigma_a+\sigma_s\over\sigma_a}+1\right)}n_i
={\left(\sqrt{\inv{1-\wt}}-1\right)\over\left(\sqrt{\inv{1-\wt}}+1\right)}n_i$$
where we used that $1-\wt={\sigma_a\over\sigma_a+\sigma+s}$. Using this
result, we can solve for $\Lambda$:
$$\Lambda={n_e\over n_i}={1-\sqrt{1-\wt}\over1+\sqrt{1-\wt}}$$

\subsection*{ Quick Reality Check }

We can use the fact that $$F_{out}\eval{surface}=\int{I_1\eval{surface}\mu\,d\Omega}$$ to independently find the macroscopic albedo: $$\Lambda={F_{out}\eval{surface}\over F_{in}\eval{surface}}$$ Doing this by hand, using the test case of looking at reflected flux directed straight up ($\mu_0=1$), from perfectly reflective particles $\wt=1$, with $\mu_1=\hf$ (an approximation of a quadrature gaussian integral), we get: $$\begin{aligned}F_{out}\eval{surface}&={3\pi\over2}\ln2\mathfrak{F}\\ F_{in}\eval{surface}&=\pi\mathfrak{F}\\ \end{aligned}$$ Numerically, we find $\Lambda=1.03$. We were, of course, shooting for an answer of $1$, so this hand-calculation gives us an indication that all of the various assumptions we made have introduced a couple \% error in our final calculations.

\end{document} <\latex>