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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?


marked as duplicate by ACuriousMind, Kyle Kanos, HDE 226868, Danu, user10851 Jul 30 '15 at 20:54

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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.


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