Homework 5 Due Thursday Nov 17 by class time
1) A blackbody surface is illuminated by visible light, and comes to radiative equilibrium at a given T characteristic of infrared temperature.
a) Imagine the surface is covered by a screen that is fully transparent in the visible, but reflects half the infrared that impinges on it (transmitting the other half), and do an infinite sum of the infrared light emitted by the blackbody as it reflects back to the blackbody surface, to determine the new temperature of radiative equilibrium.
answer: The sum is, 1/2 + 1/4 + 1/8 + ... = 1, so every erg of energy initially radiated by the blackbody is re-radiated an average of 1 more time due to the partially reflecting screen. Thus the blackbody must radiate twice as much energy as without the screen, so T^4 is raised by a factor of 2, so T is raised by a factor of 2^(1/4).

b) Now imagine embedding small white spheres into the screen, that do not interact with infrared light but reflect some of the visible light that impinges on the screen. What fraction of the visible light would need to be reflected in order to return the blackbody to its original T?
answer: Now the same thing happens in the visible as in the infrared, so there can be no effect on the T of the screen because if there were, it would violate the zeroth law of thermodynamics (if the blackbody had the same T as the light source, energy could not flow into or out of it). You can also just do the calculation that says 1/2 the visible light gets in, and from (a) it is doubled when radiated from the surface, so the outgoing flux from the surface is the same as the incident flux from the source, so same T.

2) The Earth has a surface T of about 290 K and an escape speed of 11 km/s.
Using a Maxwell-Boltzmann velocity distribution at that T, determine the fraction of each of the following particles that has a speed faster than the escape speed: O2, N2, H2. Do these results help understand why the Earth is not a gas giant but does have an atmosphere?
answer: The Maxwell-Boltzmann distribution says that the fraction of particles with speed above a given v is the same as the fraction with kinetic energy above a given E = 1/2 m v^2, and the fraction above E is the Boltzmann factor, e^(-E/kT), times the density of states in energy space, dN/dE = dp/dE * dN/dp, where dp/dE comes from p(E) = (2mE)^(1/2) and dN/dp comes from the density of states in phase space, which is proportional to p^2 which is proportional to E. (We only need proportionalities here because we are looking for a fraction of the particles, so we will divide by the integral over all velocities and so the constants will cancel out.) So dN/dE is proportional to E^(-1/2)*E = E^(1/2). So the fraction of particles above a given v is the fraction with energy above 1/2 m v^2, so the numverator is the integral over dE of E^(1/2)*e^(-E/kT) from 1/2 m v^2 to infinity, and the denominator is the same integral from 0 to infinity. The integral function is pi^(1/2) / 2 *erfc[y^(1/2)] + y^(1/2)*e^(-y) where y = 1/2 m v^2 / kT. Thus if we plug in T=290 K and the masses of O2, N2, and H2, for v_esc = 11 km/s, we find y= 808, 707, and 50.5 respectively. Then the fractions above v_esc are 0, 106^-306, and 9.5x10^-22, respectively. So both O2 and N2 are held very tightly, and even H2 seems to be held tightly enough that the Earth should have retained it over the roughly 10^17 second age of the solar system (since a particle moving at 11 km/s takes about a second to cross the atmospheric scale height, so if only 10^-21 particles of H2 did that every second, the Earth could only lose about 10^-4 of its H2 in its lifetime). Apparently there is more to the question-- for one thing, H2 can sometimes be dissociated into H, and H would have y=25, so the fraction goes up to about 10^-10, a huge increase that would greatly assist escape. There are other factors as well-- H can be ionized to H+, and then electromagnetic forces can assist escape. Planetary escape is a complicated topic that we largely breeze through with oversimplifications, as is often the case.

3) The primary cause of global warming is the rising CO2 in our atmosphere.
a) Given that Earth has hundreds of thousands of times less CO2 in its atmosphere than Venus, how is it that CO2 plays any role at all in warming the Earth?
answer: The infrared opacity is not constant across all wavelengths, so most of the heat escapes through "holes" in the opacity where the optical depth is extremely small. So even on Venus, those holes are barely filled, on Earth they are not filled at all. But when the CO2 increases, it closes in the "walls" of the holes, making them less wide rather than less deep. This creates a much weaker dependence on the CO2 number, which is lucky because we have doubled the CO2 in our atmosphere so far, and it has only risen the T by about 0.3%. Still, even a 1 C rise in T has had effects, and those effects will increase significantly more in the coming decades. How much more depends on our efforts to prevent this.

b) Why is CO2 being released by human activity?
answer: Primarily because C-O bonds are very strong, much stronger than the C-H bonds that have been built by photosynthesis. Hence the ability to replace C-H bonds with C-O bonds represents the release of a lot of energy, the energy stored by photosynthesis. So creation of CO2 is how we essentially tap into solar energy over a huge fraction of the planet surface (by growing food) and over a huge amount of time (by digging up petrochemicals). Avoiding the creation of CO2 requires rethinking our ways of accessing solar energy, or using nuclear energy.

c) What do you think would be the best way to improve the situation, going forward?
answer: no "right answer" here, just interested in the responses.

4) The coriolis force looks a lot like the magnetic Lorentz force.
a) Use this analogy to find the effective qB/m (in cgs units) that would produce the same coriolis effect as Earth's rotation.
answer: The Lorentz force (I'll break down and use SI units here because that's what is easiest to look up) is qvB which we equate to 2*m*omega*v to yield qB/m = 2*omega = 1.5x10^-4 s^-1. (If done in cgs, this comes out 44 m/s^2.)

b) Put the mass and charge of a proton into your expression to solve for the B field (along the Earth's axis) that would produce the same coriolis effect on a proton as Earth's rotation does.
answer: B = 0.00015 s^-1 * m/e = 1.5 pT (a pT is a Tesla times 10^-12). For comparison, the Earth's magnetic field is about 50 million times larger than that, so charged particles really don't care about the Earth's rotation!).
c) Calculate the rotation period that the coriolis effect would produce if it were the only effect on moving gas in the Earth's atmosphere.
answer: If the only acceleration was 2*omega*v, that would produce circular paths at speed v that have a frequency of 2*omega, which is a rotation period half of a day, or 12 hours, since omega corresponds to the rotation of the Earth.

d) There should be something that seems odd at first glance with your result for (c). Say what other rotational effect corrects the problem, and show how it does it (in a particular case if you like).
answer: It should seem odd that the period is only half a day, rather than a whole day. This is because a particle moving in a circle (say, around the north pole, for simplicity) experiences not only the coriolis force, but also the centrifugal force. The centrifugal force has half the magnitude of the coriolis force and is opposite in direction (for a circle around the north pole), so this reduces the frequency by 1/2, and doubles the period to a full day (which of course is what we would witness for a particle with no real forces on it, or for a Foucault pendulum at the north pole).