4 ms·

I'd suggest looking at the Invariant mass page. What you are talking about isn't rest mass of a photon, rather the center of mass invariant mass[1]. This aris

by hpcjoe 5y ago

I'd suggest looking at the Invariant mass page. What you are talking about isn't rest mass of a photon, rather the center of mass invariant mass[1]. This arises from the equivalence principle, the relation between mass and energy, not because a photon (a massless particle) has a rest mass (it doesn't). It might be a pedantic argument on my part, but still, it is worth being clear on this point. [1] https://en.wikipedia.org/wiki/Invariant_mass#Collider_experiments https://en.wikipedia.org/wiki/Invariant_mass#Collider_experi...

im3w1l 5y ago
(Center of mass) invariant mass is synonymous with rest mass.
- AnimalMuppet 5y ago
  Is it? It sounds like you are explicitly disagreeing with hpcjoe's position, but just a bare "no, I'm right" isn't informative. Can you give some reason or argument that supports your position over hpcjoe's?
  - im3w1l 5y ago
    > In physics, the energy–momentum relation, or relativistic dispersion relation, is the relativistic equation relating total energy (which is also called relativistic energy) to invariant mass (which is also called rest mass) and momentum. It is the extension of mass–energy equivalence for bodies or systems with non-zero momentum Those are the first few sentences in the article I already posted. > The invariant mass, rest mass, intrinsic mass, proper mass, or in the case of bound systems simply mass, is the portion of the total mass of an object or system of objects that is independent of the overall motion of the system. More precisely, it is a characteristic of the system's total energy and momentum that is the same in all frames of reference related by Lorentz transformations.[1] If a center-of-momentum frame exists for the system, then the invariant mass of a system is equal to its total mass in that "rest frame". Those are the first few sentences in the article he already posted.
    - AnimalMuppet 5y ago
      In the last sentence of the article he posted, it's unclear to me how that applies to a system with two photons in it. If I have two photons heading off in opposite directions, I'm in the center-of-momentum frame for that system. Then the invariant mass is equal to the system's "total mass" in that frame. For two photons, is that 0, or E/mc^2? I know what your position is. I'm just not sure that the quote from his article supports your position. I will grant you that if you take the energy-momentum equation and apply it to such a system as a whole, and plug in the total energy (non-zero) and the total momentum (zero), then out comes a non-zero mass. I just am unclear on whether that's a valid thing to do. (I also don't know that it's not valid.) Can you make a convincing case that it's reasonable to apply that equation in that way?
      - im3w1l 5y ago
        Well there are a couple of things to address here. The first is that the invariant mass is invariant - it's the same in any reference frame. So the invariant mass of a system will either equal the sum of the masses of the particles or it won't, irrespective of frame. And equality will only hold if there is no internal movement. > Can you make a convincing case that it's reasonable to apply that equation in that way? Even hpcjoe did not object to that, only to the name "rest mass" for the result. We saw that the number you get out was useful upthread, because it allowed predicting if pair production can occur or not.

by hpcjoe 5y ago

im3w1l 5y ago

(Center of mass) invariant mass is synonymous with rest mass.

AnimalMuppet 5y ago
Is it? It sounds like you are explicitly disagreeing with hpcjoe's position, but just a bare "no, I'm right" isn't informative. Can you give some reason or argument that supports your position over hpcjoe's?
- im3w1l 5y ago
  > In physics, the energy–momentum relation, or relativistic dispersion relation, is the relativistic equation relating total energy (which is also called relativistic energy) to invariant mass (which is also called rest mass) and momentum. It is the extension of mass–energy equivalence for bodies or systems with non-zero momentum Those are the first few sentences in the article I already posted. > The invariant mass, rest mass, intrinsic mass, proper mass, or in the case of bound systems simply mass, is the portion of the total mass of an object or system of objects that is independent of the overall motion of the system. More precisely, it is a characteristic of the system's total energy and momentum that is the same in all frames of reference related by Lorentz transformations.[1] If a center-of-momentum frame exists for the system, then the invariant mass of a system is equal to its total mass in that "rest frame". Those are the first few sentences in the article he already posted.
  - AnimalMuppet 5y ago
    In the last sentence of the article he posted, it's unclear to me how that applies to a system with two photons in it. If I have two photons heading off in opposite directions, I'm in the center-of-momentum frame for that system. Then the invariant mass is equal to the system's "total mass" in that frame. For two photons, is that 0, or E/mc^2? I know what your position is. I'm just not sure that the quote from his article supports your position. I will grant you that if you take the energy-momentum equation and apply it to such a system as a whole, and plug in the total energy (non-zero) and the total momentum (zero), then out comes a non-zero mass. I just am unclear on whether that's a valid thing to do. (I also don't know that it's not valid.) Can you make a convincing case that it's reasonable to apply that equation in that way?
    - im3w1l 5y ago
      Well there are a couple of things to address here. The first is that the invariant mass is invariant - it's the same in any reference frame. So the invariant mass of a system will either equal the sum of the masses of the particles or it won't, irrespective of frame. And equality will only hold if there is no internal movement. > Can you make a convincing case that it's reasonable to apply that equation in that way? Even hpcjoe did not object to that, only to the name "rest mass" for the result. We saw that the number you get out was useful upthread, because it allowed predicting if pair production can occur or not.