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\begin{document}


\begin{center}
{\large{\bf Astro 201 -- Radiative Processes -- Problem Set 6}}

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

\end{center}

\vspace{0.1in}
\ni
Readings: Hand-outs from Osterbrock; Rybicki \& Lightman 9.5; however much you like of Mihalas 108--114, 119--127, 128--137 (even skimming Mihalas can prove enlightening); however much you like of the on-line supplemental article by Purcell \& Field (1956).

\vspace{0.1in}
\ni
{\large{\bf Problem 1.}} Maxwellian is not quite LTE

\ni
A certain atom suffers collisions with another species (perhaps merely
its own). These collisions populate and de-populate a certain level
in the atom that lies above another level in that atom by energy $E$.
Prove that if the relative velocity distribution, $f(v)$, between
the atom and surrounding colliders
obeys a Maxwellian at temperature $T$ ($\int_0^{\infty} f(v) dv = 1$),
then the excitation rate coefficient,

\begin{equation}
q_{12} \equiv \int_{0}^{\infty} \sigma_{12} v f(v) dv \, ,
\end{equation}

\noindent
is related to the de-excitation rate coefficient,

\begin{equation}
q_{21} \equiv \int_{0}^{\infty} \sigma_{21} v f(v) dv \, ,
\end{equation}

\noindent
by

\begin{equation}
q_{12} = q_{21} \frac{g_2}{g_1} \exp (-E/kT) \, .
\end{equation}

\noindent where $g_i$ is the statistical weight of level $i$,
$\sigma_{ij}$ is the velocity-dependent collisional cross-section
for making a transition from level $i$ to level $j$, and $v$
is the relative velocity between the atom and the colliding species.
You should find this problem straightforward given the Einstein analogue
presented in lecture.

\noindent NOTE: Nowhere in this discussion have we assumed
that the level populations are distributed in a Boltzmann fashion
at temperature $T$. That is, none of the relations above
assume LTE (local thermodynamic equilibrium); LTE is a more
restrictive condition than merely assuming that the relative velocity
distribution is Maxwellian.


\vspace{0.1in}
\ni
{\large{\bf Problem 2.}} 21 (Flavors of) Temperatures for 21 cm Radiation

\noindent
This problem revisits the famous
hyperfine transition in neutral hydrogen.
Here we try to understand what sets the excitation (spin) temperature,
$T_{ex}$. (In a previous problem set,
we merely contented ourselves with
the statement that $T_{ex} \gg T_{\ast}$.) While we're at it, we learn
more astronomer's jargon.

\noindent In general, the excitation temperature of a transition is influenced
by two factors: (1) the radiation field, and (2) collisions with
surrounding species.

\noindent
The radiation field can either excite
the atom through photon absorption, or de-excite through
stimulated emission. Measure the strength of the ambient radiation
field at 21 cm by $\bar{J_{\nu}}$, the mean (i.e., angle-averaged)
intensity integrated over the hyperfine
line profile (recall Rybicki \& Lightman chapter 1). 

\noindent
As for collisions, consider here exciting and de-exciting collisions
with fellow neutral hydrogen atoms;
Purcell and Field (1957, hereafter PF) conclude that collisions
between a given electronic-ground-state H atom and other
electronic-ground-state H atoms are most important in the
predominantly neutral HI clouds of the ISM. (Electrons
are $\sim$42 times faster and tend to dominate the excitation dynamics
in other situations,
but assume here that there are too few of them in these cold clouds.)
Denote the collisional excitation rate coefficient by $q_{12}$,
and the collision de-excitation rate coefficient by $q_{21}$ (see problem 1).
Assume for this problem that both hyperfine-excited and hyperfine-ground
atoms can excite or de-excite the hyperfine level in an
atom.\footnote{In truth,
excitations and de-excitations proceed, as PF describe, by
``spin-exchange'' collisions, in which an electron with a certain
spin in one H atom swaps places with an electron having a different
spin coming from the colliding H atom. A given atom
can swap its way up to the hyperfine excited state, or swap its way down
to the hyperfine ground state. Here we follow PF and
place the relative probabilities of undergoing a swap-up versus
a swap-down into $q_{12}$ and $q_{21}$.}

\noindent
(a) Write down the equation of global (not detailed!)
balance for this transition.
That is, write down the statement that the rate of excitations (from all
possible channels) per
volume per time equals
the rate of de-excitations (from all possible channels) per volume per time.

\noindent Use only the following variables:
$n_1$ and $n_2$ are the number densities
of atoms in the ground and excited states, respectively,
$n = n_1 + n_2$,
$T_K$ is the kinetic temperature of the atoms that move according
to a Maxwellian, $q_{12}$, any Einstein coefficients you want,
$\bar{J_{\nu}}$, and the statistical weights $g_1$ and $g_2$
of the ground and excited states, respectively.

\noindent What you have written down is an equation for the excitation
temperature ($T_{ex} \leftrightarrow n_1/n_2$) in terms
of the radiation field and the rate of collisions. Regard
the latter two as given throughout this problem.

\noindent
(b) DEFINE a "radiation temperature," $T_R$, from
$\bar{J_{\nu}}$ as

\begin{equation}
\bar{J_{\nu}} \equiv B_{\nu}(T_R)
\end{equation}

\noindent Note that we are NOT saying the ambient radiation field
is Planckian. We are merely DEFINING a number $T_R$ by
using Planck's function, $B_{\nu}$, where $\nu = 1420$ MHz,
the frequency of the 21 cm line.

\noindent Re-write your equation in (a) to solve
for $T_{ex}$ in terms of the following variables:
$T_K$, $T_R$,
$T_{\ast} \equiv h\nu/k$ (recall last problem set),
and the dimensionless variable

\begin{equation}
z \equiv \frac{g_1 n q_{12} T_{\ast}}{g_2 A_{21} T_K}
\end{equation}

\noindent where $A_{21} = 2.85 \times 10^{-15} \s^{-1}$
is the Einstein decay coefficient.
Use the very likely condition that $T_K$, $T_R \gg T_{\ast}$
to rid your equation of all exponentials.

\noindent Verify that if $z \gg 1$, $T_{ex}\approx T_K$ (collisions
beat radiation; the transition is in LTE at $T_K$), but that
if $z \ll 1$, $T_{ex} \approx T_R$ (radiation beats collisions;
the transition is not in LTE at $T_K$).

\noindent (c) To order of magnitude (actually much better
than that),
what fraction of HI is in the excited hyperfine state?
Recall that $g_1 = 1$ and $g_2 = 3$
and use the very likely condition that $T_K$, $T_R \gg T_{\ast}$.

\noindent (d) Estimate the value for $z$ for an HI cloud
at $T_K = 100\K$, $n = 1 \cm^{-3}$. Use Table 1 and
equation (9) of PF; note that PF's collision frequency
$\nu = n\langle \sigma v\rangle$ is not
the same as our line frequency $\nu$; call PF's $\nu = \nu_{PF}$;
then $n q_{12} = 3 \nu_{PF} / 8$. (For those interested, the 3/8
can be understood easily; skim the first 3 pages of PF and use
your answer for part (c).)

\noindent Based on your answer, would you expect collisions
or radiation to be more important in determining the degree of excitation?

\noindent (e) "Critical densities," $n_{crit}$, for exciting the line
by collisions are defined by setting the rate of spontaneous decays
equal to the rate of collisional de-excitations.
Show that such a procedure gives

\begin{equation}
n_{crit} = \frac{A_{21}}{q_{21}}
\end{equation}

\noindent and solve for its value for this line
at $T_K = 100\K$.

\noindent One can define $n_{crit}$ for any line transition
at any temperature; it is a crude gauge of the density of colliders
required for collisions to be important in setting the level populations.

\noindent (f) Suppose radio observations are made that
spatially resolve emission from a uniform HI cloud that is optically
{\it thick} to its own 21 cm line radiation. Prove
that the observed specific intensity, $I_{\nu}$, equals

\begin{equation}
I_{\nu} = \frac{2kT_{ex}}{\lambda^2} \, .
\end{equation}

\noindent (The following comments are not important to solving
this problem. Radio
astronomers like to use "brightness temperature," $T_B$, as a measure
of specific intensity. DEFINE $T_B \equiv I_{\nu}\lambda^2/2k$;
as with $T_R$, we are NOT saying that $I_{\nu}$ looks like
a Rayleigh-Jeans tail of a blackbody; we are only USING
the Rayleigh-Jeans tail of a blackbody to DEFINE $T_B$.
This is just one of these astronomer's habits that one
must adopt to stay in the conversation; as if life weren't
already complicated with all the other temperatures!
Part (f) therefore asks you to show that the "brightness
temperature in the line equals the excitation temperature of the line
for an optically thick cloud.")


\vspace{0.1in}
\ni
{\large{\bf Problem 3.}} Photoionized Quasar Winds

\noindent
Quasars are luminous X-ray sources sitting in the cores of ancient
galaxies. They are supermassive black holes that accrete surrounding
gas; the gravitational potential energy of gas spiralling down the potential
well of the black hole is converted into radiation.
The luminosity of a typical quasar is $L \sim 10^{46} \erg \s^{-1}$,
mostly in the Lyman continuum, with a substantial fraction in a power-law
X-ray tail. The flux density in the X-ray tail obeys
$F_{\nu} \propto \nu^{-\beta}$,
with $1 < \beta < 2$.
($F_{\nu}$ has units of energy per time per frequency per area).

\noindent
This radiation streams out from regions closest to the black hole
and may illuminate more distant but still circumnuclear gas.
In so doing, it photo-ionizes
the more distant gas and threatens to turn it into a complete and
utter plasma, with {\it all} electrons stripped from parent
nuclei. This problem estimates the varying
degrees to which this threat is made good.
%It is inspired by work done by Norm Murray, a former visiting Miller
%professor in the astronomy department.

\noindent (a) Define the "ionization parameter," $\xi$, as the
number density of Lyman continuum photons, $\eta$, divided by the
total number density of hydrogen, $n_H = n_{H^+} + n_{H^0}$,
where $n_{H^+}$ is the number density of ionized hydrogen,
and $n_{H^0}$ is the number density of neutral hydrogen. Show that
in photo-ionization equilibrium (rate of photo-ionizations per volume per time
equals the rate of radiative recombinations per volume per time):

\begin{equation}
\frac{n_{H^0}}{n_{H^+}} \approx \frac{10^{-6}}{\xi}
\end{equation}

\noindent valid for $n_{H^0}/n_{H^+} \ll 1$.
This is a rule of thumb worth remembering. Just consider
photons near the ionization edge! Assume
a 1-dimensional geometry for the problem; consider only a
semi-infinite slab, upon which is incident a radiation
flux.  Further assume a kinetic temperature of $\sim$$10^4\K$.
In your derivation, you will see that the dimensionless factor
of $10^{-6}$ can be expressed in terms of fundamental constants.

\noindent (b) Prove for hydrogenic ions of species X and nuclear charge $Z$
(atoms that are holding on desperately to their last electron) that

\begin{equation}
\frac{n_{X^0}}{n_{X^+}} \approx \frac{10^{-6}}{\xi} Z^{2\beta + 4}
\end{equation}

\noindent where $n_{X^0}$ is the number density of hydrogenic
ions of species X that each still retain their last electron, and
$n_{X^+}$ is the number density of fully stripped ions of species X.
Assume that all of the hydrogenic metal ions are recombining with
electrons provided by hydrogen, and that nearly all of the hydrogen
is ionized. We check this last statement in part (d).

\noindent Steps you can take:

\noindent Use the Bohr model to understand how ionization energy
scales with $Z$.

\noindent Use our quick-and-dirty derivation of the photo-ionization
cross-section,

\begin{equation}
\sigma \sim \frac{\lambda^2}{8\pi} \frac{A_{21}}{\nu} \, ,
\end{equation}

\noindent to argue that $\sigma$ scales as the radius of the hydrogenic
ion squared. Then use the Bohr model to see how this radius scales with $Z$.

\noindent Finally, once you have determined how $\sigma$ scales with $Z$,
use the Milne relation to see how the radiative recombination coefficient
scales with $Z$.

\noindent (c) Notice that no matter how large $\xi$ is, there
are always a few electrons bound to nuclei at any given
moment (if there were none, there would be nothing for
photons to ionize, and photo-ionization
equilibrium would be violated). This tiny neutral/hydrogenic population
attenuates the UV-to-X-ray radiation as it tries to propagate
through gas. The gas will have an ionization gradient: at its
unshielded face, naked before the radiation,
nearly all the ions will be stripped, but as
we move further away from the face, the ionizing
radiation weakens due to increasing absorption
by the tiny neutral/hydrogenic population,
and the neutral/hydrogenic fraction grows.

\noindent Show that an element of fractional number
abundance $f_X = n_X / n_H$ relative to hydrogen
($n_X = n_{X^0} + n_{X^+}$)
goes from being fully stripped to predominantly
hydrogenic over a column density of {\it hydrogen} of order

\begin{equation}
N_H (Z) \sim 10^{23} \xi Z^{-2\beta - 2} f_X^{-1} \cm^{-2} \, .
\end{equation}

\noindent Assume that such columns are achieved over
regions sufficiently geometrically thin that
we can neglect the dilution of the radiation flux
by the inverse square law. Again, consider only ionizations
near the ionization edges of various species.

\noindent (d) If $\beta > 1.5$, show that the layers of fully
stripped carbon, nitrogen, oxygen, and neon are each thinner than
the layer of fully stripped hydrogen. In other words, soft X-rays
are stopped by the metals before the Lyman continuum photons
are stopped by hydrogen. This is a fact of relevance in understanding
how gas can be radiatively accelerated to form quasar winds.


\end{document}


