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).