Mass and Schwarzschild Radius Do free massless particles have a Schwarzschild radius?  I'm curious about the mass in the equation for the Schwarzschild radius.  I know that you can calculate a Schwarzschild radius for any massive object, but does, for instance, a single (free) photon have a corresponding Schwarzschild radius as a result of its momentum, or does something need to actually have mass?  If mass is required, what is the difference between mass and energy in this case?
 A: The curvature of spacetime is determined by the stress-energy tensor, and in the stress-energy tensor we do not distinguish between matter and energy. The two are treated as equivalent and interconverted using Einstein's famous equation $E = mc^2$.
The Schwarzschild radius of a mass is conventionally written as:
$$ r_s = \frac{2Gm}{c^2} \tag{1} $$
where $m$ is the mass. We write the equation this way because it's convenient - when we're talking about astronomical black holes we usually know the mass from observing orbits. But there is nothing to stop us from writing:
$$ r_s = \frac{2GE}{c^4} \tag{2} $$
where $E$ is the energy of the black hole. This is exactly the same equation and gives the same value of $r_s$.
So if we take a light beam and focus it down narrowly enough we could create a black hole. We just need to get the focus tight enough and the energy high enough that it satifies equation (2) and a black hole will form. The only problem is that this task is outside our current capabilities and will remain so for the foreseeable future.
So you could make a black hole from a light beam, but this isn't the same as making a black hole from a photon. The trouble is that while it's tempting to think of a photon as a particle like an electron, this is a very poor way to describe light in most circumstances. In most situations where we have propagating light the photons are delocalised over a wide area and don't have a position in the usual sense of the word. Generally speaking the only time we can pin down photons is when the light is interacting with something else e.g. with a photomultiplier or a photographic film.
And given photons rarely have any precise position it's therefore meaningless to ask if a high enough energy photon would form a black hole. The photon energy won't be concentrated in a precisely defined spot and can't form a black at that spot.
