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Q & A: photon gravity

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Most recent answer: 10/22/2007
Q:
I have a question about photons. Actually, I would like to know if my interpretation is correct. I’m aware that the photon is given as massless, yet this mass is a value when the object is at rest. So this mass would represent the inherent curvature of space-time that something has, independently of the effect of velocity (i.e. when velocity=0). But since velocity does lead to further space-time, the photon would have a mass in the sense that curves space-time but not independently of its motion. It would have a mass as suggested by E=mc^2, but only because it moves. Is this interpretation correct and could you please tell me the truth if I am (most probably) completely wrong?
- Ben Kemp (age 16)
UK
A:
You have made a mistake here- by guessing that your physics is probably wrong. The rest of the argument is fine.

Some people quibble about the use of the word 'mass' but the essential physical fact is that a box of photons will indeed cause as much gravity as a box of coal with the same total energy. I use a box so that the net momentum of all the photons will cancel and we can just consider their energy.

Mike W.

The version practitioners use instead of E=mc^2 is this one:

E^2 = (mc^2)^2 + (pc)^2

where "m" is what some people call "rest mass" and others the "invariant mass", and is zero for a photon. The variable p is the momentum. Nonrelativistically, p=mv, but this doesn't work relativistically, and the problem is very noticeable in the case of a photon. The formula above describes the relationship between energy, momentum, and mass, and is valid in all frames of reference. Unfortunately there is no frame which moves with a photon
(there is no frame in which a photon is at rest).

Mike's box of photons is a great example because in general there will be a frame in which the sum of all the momenta of the photons in the box is zero. It turns out the above relationship between E, p, and m works also for combinations of particles. Add up the energies, add up the momenta (as vectors -- they have x, y, and z components), and you can compute the invariant mass of the collection of particles. It's nice this works out because most objects we work with are made up of component parts. It's nice that we can think of an atom or a block moving relativistically without having to worry about each subatomic particle inside. We don't even know if the particles we think of as fundamental aren't made up of pieces themselves.

Tom

(published on 10/22/2007)

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