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I know that the mass of a photon is zero. However, others say no it is just the rest mass which is zero, the relativistic mass isn't, but I say that no it is also zero because the rest mass $m$ is related to the relativistic mass $M$ through the relation $M = \gamma m$ and if $m = 0$ then $M = 0$ accordingly.

But, they say, if for a photon the momentum is $p = mc$ and the energy is $E = mc^2$, you can obtain the right formula for the photon momentum, namely $p = E/c$. So even though I know I can't even use $E = mc^2$ and $p = mc$ for a photon, I don't understand why they give, if we assume that they are right, the right answer for the photon momentum $p = E/c$.

Qmechanic
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Jack
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    The relation $M = \gamma m$ doesn't apply to a photon, for which $\gamma$ is infinite. See: https://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html – 10GeV Aug 12 '22 at 21:35
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    If the mass is 0 and the gammafactor is infinity then 0 times infinity equals indeterminate – Yukterez Aug 12 '22 at 23:41
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    Relativistic mass is a badly named synonym for energy. – Connor Behan Aug 12 '22 at 23:42
  • Although in real life "two wrongs don't make a right", sometimes in mathematical manipulations several errors can bring us back to the true outcome. For example, two sign errors... :) So there's probably no deeper meaning to such episodes. – paul garrett Aug 12 '22 at 23:43
  • Possible duplicates: https://physics.stackexchange.com/q/3541/2451 , https://physics.stackexchange.com/q/30764/2451 and links therein. – Qmechanic Aug 13 '22 at 07:20

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It's a good question and one that has puzzled generations of students. It seems puzzling because you are coming at it from a Newtonian viewpoint but a photon is an inherently relativistic particle so Newtonian mechanics doesn't apply to it.

Our starting point is the expression for the total energy that we get from special relativity:

$$ E^2 = p^2 c^2 + m^2 c^4 \tag{1} $$

This comes from the energy-momentum four vector that is used in special relativity, though you won't study the details until you get to university. For now all we need to know is that this applies to all particles, both massive and massless. For a photon we set $m=0$ and we get the result you cite:

$$ p = \frac E c $$

The momentum in equation (1) is the relativistic momentum, and for a massive particle this is given by:

$$ p = \gamma mv = \frac{mv}{\sqrt{1 - v^2/c^2}} $$

The problem with your approach is that for a photon $m=0$ and $v=c$, and making this substitution we get:

$$ p_{\text{photon}} = \frac00 $$

and this is undefined so we cannot define a photon's momentum this way.

John Rennie
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  • Thanks very much for your response, I get it now that I can't use M=γm in the case of photons. In addition, I know that since p=mc is the Newtonian version of momentum, i can't use that as well, and I that should use E²=m²c⁴+p²c² for relativistic relation between energy and momentum. However, using the wrong p=mc with E=mc² still gives the right answer for p, that is p=E/c. The question is why is that? – Jack Aug 13 '22 at 19:50