Homework 4 due by class time, Thurs Nov 10
1) The 21 cm line stems from a spin interaction between the electron
and proton in an isolated hydrogen atom.
a) Describe the nature of the spins of these particles in the true ground state.
answer: the true ground state has opposite spin between the electron
and proton, but it is a superposition (with a minus sign) of up-down
and down-up, such that the actual spin of either particle is indeterminate.
b) Find the characteristic temperature T sufficient for exciting
this transition. What does this tell you about the probability that
a typical H atom will be found in its true ground state?
answer: hc / k*lambda = 0.069 K. That is such a low T that any
normal T, and even the blackness of deep space (2.7 K), is much larger
than it. So all four n=1 states are equally likely, and none of the
other states are likely to be highly populated, so about 1/4 of all H
atoms should be found in the true ground state when their energies
are measured.
c) The observed brightness of this line depends on the
column depth of neutral H, times the product of the Planck function
at 21 cm (for the assumed temperature T) times the difference between
the population in the true ground state and the population in any of
the triplet excited states (which will be the same, they're degenerate).
The subtraction in that last term stems from stimulated emission, which
is normally neglected because excited state populations are normally
much smaller. Why is this not the case for the 21 cm line, and how
does the presence of the stimulated emission correction lead to a
brightness that depends only on column depth, and not on T? (Hint:
assume the populations are in thermodynamic equilibrium at T, and
consider the form of the Planck function.)
answer: The stimulated emission correction looks like 1 - x_u/x_l, where
x_u is the upper-level population and x_l is lower-level. Normally this
is much less than 1 and is neglected, but the 21 cm line is such a low
energy line that x_u and x_l are nearly the same, so the stimulated emission
correction is quite important. Basically, putting HI in front of a light
beam is just as likely to add light to it (stimulated emission from the
upper level) as it is to remove light from it (excitation of the lower
level). Thus it is often optically thin, so when seen against a
black background, it just looks like the Planck function times the
effective optical depth. The Planck function in the limit of low energy
transitions is proportional to T (a common approximation in radio
astronomy). Meanwhile, the effective optical depth that would normally
be proportional to the lower level population x_l must be corrected
for stimulated emission by subtracting the upper level population x_u
(the latter is like negative opacity because it adds to the beam of light
instead of subtracting from it), so that amounts to multiplying the opacity
by a correction factor 1 - x_u/x_l, and in thermodynamic equilibrium
for low energy lines we know that 1 - x_u/x_l is near 1 - Exp[-h nu / k T],
and that is proportional to 1/T. So the Planck function is proportional
to T, but the effective optical depth is proportional to 1/T, so T cancels
out in the brightness that we see. All that remains is the column depth,
because more stuff produces more emission.
d) Say why it is relatively easy to observe the 21 cm line from the
surface of Earth, by mentioning atmospheric transmission and potential
background sources.
answer: 21 cm is in the radio, which passes easily through the
Earth's atmosphere because the wavelength is much larger than the size
of any particle it encounters, allow it to diffract right around (or
put differently, it is in the Rayleigh regime, where opacity drops
to large wavelength like wavelength to the -4 power). Also, there are
weak background sources near 21 cm, because the galactic radio background
is seen mostly at longer wavelengths, and the atmospheric infrared is
seen mostly at shorter wavelengths (and also the extragalactic microwave
background peaks at about 0.2 cm).
2) a) Why do optical photons penetrate the atmosphere better than UV photons?
answer: This is an example of Rayleigh scattering, where the optical
depth is proportional to frequency to the 4th power because slower
oscillations produce weaker scattering of light when the oscillation
has a frequency-independent amplitude (because it is so slow that it
reaches the amplitude of force balance). This is also the reason
the sky is blue, blue light interacts more (and UV even more).
b) What processes do EUV and X-ray photons allow that makes the
atmosphere very opaque to them?
answer: EUV and X-ray photons have enough energy to excite and ionize
electrons in molecules,
further increasing the opacity.
c) Given the above,
why do optical photons not penetrate flesh as well as X-ray photons?
answer: because flesh is not like isolated gas molecules, it is more
like solid matter which supports communal excitations of all kinds,
due to forces between the constituents. This allows almost any energy
photon to resonate with something, but higher energy photons penetrate
deeper before their energy is absorbed. This allows X-rays to reach,
and scatter off, the bones before they are absorbed in flesh.
3) The fossil record shows that Earth's oceans have been liquid for
at least 4 billion years, yet well-tested stellar evolution theory
establishes that the Sun's luminosity has risen about 30 percent since
core fusion initiated.
a) If the average surface T of Earth today is around 290 K, what would
it have been at the start of the main sequence, assuming nothing other
than the solar luminosity was different then?
Why is that a problem?
answer: A 30 percent increase in L corresponds to 1/4 of that, or about
8 percent, increase in Earth's T (in order to emit the same energy flux
as is absorbed). An 8 percent drop would drop more than 20 C, which would
freeze the oceans.
b) In order to keep the surface T of Earth the same over this time via
a growing greenhouse effect, what would have needed to happen to the
infrared optical depth of the Earth's atmosphere? (Assume it averages
about an optical depth of 2 today.)
answer: 2 is actually a little high, it's more like 1.7, but we can just
round up. The flux reaching the surface from all sources scales with
the solar flux, so if the solar flux is down by 30 percent, the factor
1+tau that multiplies that flux due to the greenhouse effect (where tau
is the excess infrared optical depth in the greenhouse gases) would need
to go up by 30 percent. So if it is about 3 now, it would need to have
been more like 4 then, for an optical depth in the infrared that was like 3.
These are approximate, a full radiative transfer calculation is beyond
our current scope.
4) The most commonly outgassed volatiles from terrestrial planets are
H2O, CO2, and N2, in that order. That happens to also be the order of
their freezing temperatures, with the highest freezing temperature first.
a) How does this explain why moons of Jupiter and Saturn (when they have
any atmosphere at all) tend to have N2 atmospheres, whereas closer planets
like Mars and Venus have CO2 atmospheres?
answer: CO2 freezes out at distances like Jupiter and Saturn, but N2
is much harder to freeze, so remains a gas.
b) The Earth also has a predominantly N2 atmosphere, quite similar to
Titan actually, but Earth is much warmer
than the reasoning you invoked in part (a).
What explains this aberrant behavior, and what role does the
anthropic principle play in understanding why the Earth doesn't fit
the expected trend?
answer: Earth should have a CO2 atmosphere, but because its conditions
are right for liquid water (and has been for a very long time), it
sustains liquid oceans on its surface. These are effective at absorbing
CO2 and locking it up into rocks, leaving behind the N2.
It may be very unlikely for that to happen, since it might be extremely
hard to sustain liquid oceans, but any planet that has not
managed somehow to do that cannot have
intelligent life there to wonder how it got that way (hence the
anthropic principle).