Mass and Speed
Most recent answer: 07/06/2011
1. Yes. The basic equation relating energy (E), momentum (p), rest mass (m0) and the speed of light (c) is E2=m02c4+p2c2. When m0=0, you just get E=pc. Now we also have p*sqrt(1-v2/c2)= m0v, where v is the velocity. So that means that when m0=0 p*sqrt(1-v2/c2) = 0. When p isn't zero, we must have v=c.
2. It's not that light is taking a longer route. To the extent that there are gaps between the wave-functions of the atoms (partially the case), you can think of the speed in the gaps as being just plain c. The slowing down is, as you say, a larger-scale average effect that shows up when you include the parts of the wave that are scattered by the material.
3. To the best of my knowledge, the mechanism by which mass couples to spacetime geometry is not yet understood. That's what the string theorists are trying to do.
4. This one will require some help from somebody who understands the Higgs field. We'll update at some point.
5. There are probably more direct ways, but here's an old example. The angular momentum of the electrons in say a piece of iron can be measured by taking a magnetized block and heating it up. When the magnetism melts, the spin angular momentum transfers to the ordinary mechanical angular momentum of the block. Einstein was in on the first version of that experiment. The angular momenta of electromagnetic waves can be then measured by letting them be absorbed by electrons in a spin resonance experiment, flipping the direction of a known number of electron spins. Translating from the total wave momentum to the photon pieces just requires knowing how many photons there are, obtained from simply accounting for the energy. The angular momentum is always either plus or minus in the direction of the wave propagation.
(published on 07/06/2011)
Follow-Up #1: photon spin rules
That's a very profound question, for which I will provide the beginning of an answer. If I can find a more knowledgeable colleague, we can then deepen it.
First, let's discuss other particles, e.g. electrons. It would be impossible for them to have a rule like "spin is always aligned along the direction of motion". The reason is that you can look at the same particle from any legitimate reference frame, moving but not rotated with respect to your first one and choose to see it as traveling at any speed up to c in any direction. The angular momentum doesn't transform under these changes of coordinates in the right way to follow the direction of motion. You put your finger on a key step in seeing this. Pick a reference frame in which the particle isn't moving. Nothing breaks rotational symmetry for it so it can spin any direction. Now look at it from a frame where it's slightly moving. Its spin is hardly transformed. So any direction of spin can go with any direction of motion. This argument applies to any particle with rest mass, since we can go through the same range of velocities with any such particle.
That argument does not apply to particles with zero rest mass, such as photons. They always travel at c. We can't pick a frame where they are at rest, so that argument is out altogether. It turns out the direction of the angular momentum for these spin-1 particles transforms the same way that the direction of travel transforms as we change coordinate frames. So the rule that the angular momentum lies along the propagation direction is at least logically consistent with the symmetry of Special Relativity. I'll see if a colleague can explain why the rule is not only allowed but actual. There's a discussion of the issue from the classical point of view here: http://en.wikipedia.org/wiki/Photon_polarization.
(published on 07/16/2011)
Follow-Up #2: spin of massless bosons
I can answer much of this, and will see if someone else can answer the remainder. First, let's introduce a helpful vocabulary word, "helicity", for the the spin along the direction of motion.
Massive particles are indeed always in states of at least a tiny bit of mixed helicity, since the distribution of velocities has smooth tails extending at least a tiny bit on both sides of zero along any axis. Since the distribution of velocities always extends through zero, I suppose you can say "all the massive particles are (in the mixed states of??) at rest and in motion at once". However "rest" is just one point out of a continuous distribution, so the probability of being found at rest is zero.
That photon state are combinations of helicity +1 and helicity -1 with no helicity 0 component is confirmed to good accuracy by many experiments. The same applies to gluons, also massless spin-1 particles. Those are the only other ones I know of.
The set of frames described by special relativity does not include the boundary, i.e. the point of view of the photons. So I can't say anything about that.
You ask several versions of the question about why this helicity rule (no zeros) applies to photons and gluons. That's a truly excellent question. I don't know the answer, beyond some words about it being a requirement for gauge bosons. If Lee understands it or a colleague can teach us to understand it, you'll get an update.
(published on 07/18/2011)
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