\documentstyle[11pt]{article}
\input answermac.tex
\input units.tex

\addtolength{\oddsidemargin}{-.4in}
\addtolength{\evensidemargin}{-.4in}
\addtolength{\topmargin}{0in}
\addtolength{\textwidth}{0.8in}
\addtolength{\textheight}{1.00in}
\setlength{\parskip}{0.15in}
% This last command gives the possibility of variable line spacing.
%\renewcommand{\baselinestretch}{1.5}

\def\ni{\noindent}
\def\J{{\rm\,J}}
\def\W{{\rm\,W}}
\def\s{{\rm\,s}}
\def\erg{{\rm\,erg}}
\def\cm{{\rm\,cm}}
\def\m{{\rm\,m}}
\def\km{{\rm\,km}}
\def\kg{{\rm\,kg}}
\def\mm{{\rm\,mm}}
\def\gm{{\rm\,g}}
\def\g{{\rm\,g}}
\def\d{\rm\,d}
\def\h{\rm\,h}
\def\mum{\,\mu{\rm m}}
\def\K{{\rm\,K}}
\def\yr{{\rm\,yr}}
\def\Hz{{\rm\,Hz}}
\def\days{{\rm\,days}}
\def\cals{{\rm\,cals}}
\def\mole{{\rm\,mole}}
\def\gal{{\rm\,gal}}
\def\calo{\,{\rm calorie}}
\def\dyne{\,{\rm dyne}}
\def\s{\,{\rm s}}
\def\at{\, {\rm atmosphere}}
\def\J{\,{\rm joule}}
\def\pomega{\tilde{\omega}}

\begin{document}


\begin{center}
{\large{\bf Astro 201 -- Radiative Processes -- Problem Set 1}}

{\large{\bf Due in class.}}

\end{center}

\vspace{0.1in}
\ni
Readings: Rybicki \& Lightman Chapter 1 except the last section 1.8.

\vspace{0.1in}
\ni
%{\large{\bf Problem 1.}} Practice with Solid Angles
%
%\ni The solid angle subtended by an object is merely the square area
%that it presents to an observer. It has dimensions of angular
%area---e.g., square degrees, or square radians, or square arcseconds, etc.
%One steradian equals one square radian. Radians and steradians
%are the units of choice for physicists.

%\ni (a) Take a cube of edge $a$, and place it a distance $r \gg a$ away
%from the observer. If the observer views the cube face-on,
%write down {\it approximately} the perceived angular arclength of
%the cube's edge. Express in radians.
%Write down approximately the solid angle subtended by the cube.
%Express in steradians.

%\ni (b) Calculate, approximately,
%the solid angle subtended by the Sun to an observer on the
%Earth. Express in steradians, and in square degrees.

%\ni (c) Imagine an observer at the center of a sphere of radius
%$r$. (When $r$ goes to infinity, we refer to this sphere as the
%``celestial sphere.'') A circular cap is painted on the inside
%of the sphere. The vector from the observer to the
%center of the cap defines the polar axis,
%and the edge of the cap lies at a polar angle of $\theta$
%away from the polar axis. Calculate exactly the solid angle
%presented by the cap to the observer, for arbitrary $\theta$
%(express in steradians).

%\ni (d) If the entire inside of the sphere were painted,
%what is the solid angle presented by the inside of the sphere
%to the observer (in steradians)?

%\ni (e) Calculate exactly the solid angle presented by an
%open umbrella to the person
%holding it. Assume the person is viewing the umbrella from the tip
%of the handle, and that the umbrella is of the
%type used for tropical cocktail drinks---that is, when folded,
%the edge of the cloth just touches
%the tip of the handle, and the umbrella has no joints.
%Express in steradians.

\vspace{0.1in}
\ni
{\large{\bf Problem 1.}} Blackbody Flux

\ni Derive the blackbody flux formula

\begin{equation}
F = \sigma T^4\,,
\end{equation}

\ni from the
Planck function for specific intensity,

\begin{equation}
B_{\nu} = \frac{2h\nu^3}{c^2 (e^{h\nu/kT} - 1)} \, .
\end{equation}

\ni You may use the integral $\int_{0}^{\infty} \frac{dx}{x^5 (e^{1/x}-1)} = \pi^4/15$.

\ni Those who know what they have to do can stop reading here.
Those who need a bit of help should read on.

\ni Imagine a perfectly flat, blackbody patch at the center of a sphere
of radius $R$.
The patch has tiny surface area $dA$ and is lying flat at $r = 0$
in the $\theta = \pi/2$
plane (in spherical coordinates, where $r,\theta, \phi$
are the radius, polar angle, and azimuth). Only one side
of the patch is warm; the other side is at absolute zero.
Assume that the warm side radiates as a perfect thermal blackbody.

\ni A thermal blackbody emits radiation
{\it isotropically}---i.e., with no preference for direction.
In other words, imagine each point on the patch beaming out
a tiny, uniform, hemispherical dome of rays (like an umbrella).
It is a hemispherical dome and not a full sphere because only one
side of the patch is warm. Note that the Planck blackbody function has no
variable inside it that specifies direction of emission; the
direction of emission doesn't matter for purely thermal emission.
The only variables that need to be specified for the Planck
function are temperature and frequency (wavelength).

\ni Despite the isotropic nature of the emission, because
the patch is geometrically flat, has definite area, and emits only on one side
of itself, a detector glued to the inside of the sphere at $r=R$
detects different amounts of radiation depending on where it is placed. 
For example, if the detector is glued on the half of the sphere
that can't see the bright side of the patch, the detector detects
nothing. If the detector is glued to the sphere directly above
the patch, the detector picks up the most number of photons,
because it sees the full face-on area of the patch (namely, $dA$).
If the detector is glued at an angle to the pole, then it
sees less than area $dA$. If it is glued in the plane of the patch,
it sees nothing but a infinitesimally thin line segment.

\ni This problem asks you to place detectors everywhere on the inside
of the sphere and sum up the radiation collected;
this operation is equivalent to
``integrating the Planck function over all solid angles into
which the radiation is beamed.''

\ni Calculate the total {\it luminosity} (in units of energy/time)
emitted by the patch of area $dA$. You must integrate
the Planck function over the entire emitting
area (which is maximally $dA$ but in general will be less,
depending on the viewing geometry), over all solid angles into
which the radiation is beamed, and over all
frequencies.
Think about placing tiny detectors all over the inside
of the sphere and asking how much energy/time each
of the detectors receives.

\ni Then divide the luminosity
by $dA$ to calculate the total {\it flux} (in units
of energy/time/area) emitted by the patch. You should
recover the usual blackbody flux formula, $\sigma T^4$.
By definition, $\sigma T^4$ {\it
is the total amount of energy radiated per time per unit area of a blackbody
surface, radiated into all solid angles and over all frequencies.}

\vspace{0.1in}
\ni
{\large{\bf Problem 2.}} Flat Disks

\ni This problem forms the foundation for understanding the spectral
energy distributions of circumstellar disks, e.g., those surrounding
pre-main-sequence stars and AGN.\footnote{The problem has the further distinction that for some reason, whenever I teach this course, it is the one problem that drives some students crazy and they never forget it.}

\ni Consider a perfectly flat, blackbody disk encircling
a blackbody star. The star has radius $R_{\ast}$ and effective
temperature $T_{\ast}$. The disk begins at a stellocentric
radius far from the star, i.e. the inner disk radius $r_i \gg R_{\ast}$.
%where $\epsilon \ll 1$ is a
%small but finite constant.
The disk extends to infinity.

\ni Calculate the temperature of the disk, $T(r)$, as a function
of radius, $r$. Make whatever approximations you deem necessary
in light of the fact that $r_i \gg R_{\ast}$.
%$\epsilon \ll 1$.
Be sure you get the {\it scaling} of $T$ with $r$ correctly.
It is less important that you get the numerical coefficient correctly.

\ni This disk is PERFECTLY flat. Do not consider individual particles in the disk. It is a blackbody SHEET.

\ni Also, neglect radial transport of energy. Each annulus is independent
of neighboring annuli. This is a fine approximation for thin disks since they transport more radiation vertically than radially (the temperature gradient is steeper vertically than radially).

\vspace{0.1in}
\ni
{\large{\bf OPTIONAL Problem 3.}} Albedos

\ni Optical astronomers detect rocky bodies in the solar system
by virtue of their reflected sunlight. 
The received flux is proportional to $aA$, where $a$ is the albedo
(reflectivity) of the object in some optical passband,
and $A$ is the area of the object. Typically, albedos
of minor bodies in planetary systems range from $\sim$0.01 to $\sim$0.7.

\ni The proportionality of the flux to $aA$
presumes that all of the optical
light received from an object is from {\it reflected} sunlight. Of course,
that is not quite true; there is some contamination
at optical frequencies from the object's {\it thermal emission}
(the fact that it is warm).

\ni Give a quantitative explanation as to why this contamination
is not worth troubling over. If you wish, you can adopt as your ``case study''
a Kuiper belt object of optical albedo $0.07$ and heliocentric
distance $40$ AU, observed in the visual (V) passband.
Flesh out the problem as you deem appropriate
(``make it up as you go along.'') 
You may find tables of the
integrated Planck function, found in the old edition of Allen's
Astrophysical Quantities, useful. Such tables might also
be found in the new Allen, but I have not checked.

\ni Assume that the efficiency of emission at the (infrared)
wavelengths at which most of the object's thermal power
is radiated is one. Of course, its efficiency of emitting
at optical wavelengths is assumed to be less than one (by Kirchoff's law).

\vspace{0.1in}
\ni
{\large{\bf Problem 4.}} Make like a tree and ...

\ni If all the leaves of a tree fall to the ground,
how thick is the layer of leaves on the ground?
Give your answer in leaf thicknesses, to order-of-magnitude.

\ni What is the order-of-magnitude of the
optical depth presented by the leafy tree
to someone lying below the tree? Why does this answer make sense
from the perspective of the vegetation? (Be the tree.)

\ni The above two paragraphs are flip sides of the same question.
You can try the second paragraph before the first, or vice versa.

\end{document}



