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Given a particle with mass $m$ moving at velocity $v$, total energy is:

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

Note I am not using the relativistic - rest mass convention, as I was taught to think in terms of rest - total energy instead. "Relativistic mass" would be represented as $\gamma m$ , where the mass of the particle is being changed by a factor of $\gamma$ depending on it's relative velocity $v$, where gamma is equal to,

$$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$$

To obtain momentum of Particle, we expand the original equation to account for relativistic momentum:

$$E^2 = (\gamma mvc)^2 + (mc^2)^2$$

If the above equation is an accurate representation of the energy-momentum relationship, how does my professor use this equation to derive $E=pc$ for a photon? He says for a massless particle (photon),

$$E^2 = (pc)^2 + ((0)c^2)^2$$

therefore,

$$E = pc$$

But he appears to me to neglect the fact that you can expand the $p$ variable into relativistic momentum, which is a function of mass, gamma, and velocity. Was his move acceptable? If so , why? How can you make one mass 0 but not the other?

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    $\begingroup$ Why is your math in ALL CAPS? $\endgroup$ – G. Smith Oct 9 at 20:07
  • $\begingroup$ better question is, why not? $\endgroup$ – Jdog1998 Oct 9 at 21:25
  • $\begingroup$ Because it is rude (it comes across as SHOUTING) and nonstandard. It’s usually crank physicists who write $C$ when they mean $c$. And it is possible that it will aggravate readers enough that they downvote your question. $\endgroup$ – G. Smith Oct 9 at 21:47
  • $\begingroup$ Ah, well you know I am a grumpy guy. I will continue just to spite them. $\endgroup$ – Jdog1998 Oct 10 at 0:02
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Your equation involving $\gamma m$ is useless for photons because $\gamma$ is infinite and $m$ is zero. That product is indeterminate.

Your professor is correct.

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  • $\begingroup$ I am confused by your answer. First, you say the equation is useless for photons. But then you say my professor is correct in what he did in utilizing the equation. So is the energy-momentum relationship equation applicable to photons or not? If it is, how can you neglect the fact P is a function of mass, gamma, and velocity? $\endgroup$ – Jdog1998 Oct 9 at 21:30
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    $\begingroup$ $p=\gamma mv$ is meaningless for photons. Your first equation is valid for all particles. $\endgroup$ – G. Smith Oct 9 at 21:50
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    $\begingroup$ For a photon, momentum has nothing to do with mass and speed. How could it? Photons have one mass, zero, and one speed, $c$, but can have an infinite range of momenta. You have to give up your Newtonian understanding of momentum when dealing with massless particles. $\endgroup$ – G. Smith Oct 9 at 21:55
  • $\begingroup$ It is coming to me now. Last question: if my Newtonian conception of momentum is too narrow and wrong, what is the proper universal conception? What I mean is, people say Energy is "the capacity to do work," so what is momentum the capacity to do? $\endgroup$ – Jdog1998 Oct 10 at 0:08
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    $\begingroup$ The modern conception of a system’s energy and momentum is that they are components of a single Lorentz four-vector which happens to be conserved due to the invariance of the laws of physics under translations in spacetime. See Noether’s Theorem. The modern conception of mass is that it is the Lorentz-invariant length of this four-vector, which is what $E^2-\vec{p}^2=m^2$ (in units where $c=1$) is saying. $\endgroup$ – G. Smith Oct 10 at 0:26

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