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E/c = m x c; so momentum requires a mass. without mass does momentum exists?

2006-07-25 04:34:18 · 5 answers · asked by goring 6 in Science & Mathematics Physics

5 answers

no, a massless particle would have no momentum by definition

2006-07-25 04:37:33 · answer #1 · answered by Firstd1mension 5 · 1 1

An object without any sort of mass has no momentum. However, some of an object's mass comes from its relative motion. An object that moves rapidly (in my reference frame) gains mass (again, in my reference frame). This is because the object has kinetic energy, and energy is mass by e=mc^2.

Usually, when we talk about an object's mass, we really mean "rest mass" and not "total mass". The "rest mass" of an object is what we measure in the reference frame of the object. That is, in the reference frame where the object is stationary, or at rest.

There *are* objects with zero rest mass, but which still have momentum (and energy, and "total mass"). Photons, or particles of light, fit the bill. If we could "stop" a photon and weigh it, it would have a mass of zero. Yet photons carry momentum, contain energy, and are affected by gravity just like they had mass.

NOTE: As far as we know there is no way to "stop" a massless particle. Massless particles *always* travel at the speed of light. That is, they move at the speed of light in the reference frame of any object *with* mass. So, the idea of "rest mass" is a little funny when applied to massless particles. Nevertheless, its mathematically valid and a useful way to do things.

2006-07-25 04:56:08 · answer #2 · answered by Aaron 3 · 0 0

1) Momentum is mass * velocity = mv. E / v = (mv^2 /2) / v = mv /2
2) I'm really not familiar with any 'massless particle'. Sounds like an oxymoron...........

2006-07-25 04:44:22 · answer #3 · answered by Steve 7 · 0 0

if by massless particles you mean photons, then yes they do. look up the photoelectric effect.

2006-07-25 05:23:09 · answer #4 · answered by frickinaj 1 · 0 0

Aaron is precisely correct.

2006-07-25 06:42:30 · answer #5 · answered by Tom H 4 · 0 0

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