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The total energy $E$ of a particle is given by $$\tag{1}E=\sqrt{(pc)^{2}+({mc^{2}})^{2}}.$$ We also have

$$\tag{2}E=\gamma mc^{2}.$$

For photons, if we plug $m=0$ in the first equation, we get the correct formula describing its energy: $E=pc$. Whereas if we do the same in the second equation, we get an indeterminate form (since for photons $\gamma =\infty$ and $m=0$, we get $E=0\times \infty$).

Why do these two equations disagree for the photon even though they are algebraically equivalent?

I don't really have a sophisticated background in math, so an explanation that doesn't involve four-vectors would be greatly appreciated.

N.B. I've already went through this post, but all the answers relied on advanced math (four-vectors, Minkowski metric...).

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    $\begingroup$ What do you mean by the equations being "algebraically equvalent"? Try writing out your argument for why you think they are equivalent and pay special attention to divisions by zero. $\endgroup$
    – ACuriousMind ♦
    Commented Jan 2, 2020 at 17:28
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    $\begingroup$ Related: physics.stackexchange.com/q/3541/2451 , physics.stackexchange.com/q/2229/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jan 2, 2020 at 17:32
  • $\begingroup$ @ACuriousMind I meant that we can derive $(1)$ from $(2)$ in the same way that we can derive $(2)$ from $(1)$. See for example equation $(2.1.30)$ and $(2.1.33)$ of this article. $\endgroup$
    – Hilbert
    Commented Jan 2, 2020 at 17:36
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    $\begingroup$ @hilbert try going through the derivation line by line and checking if any step requires $m\neq 0$ (hint: one step does!) $\endgroup$
    – Jahan Claes
    Commented Jan 2, 2020 at 17:55

2 Answers 2

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It is not really correct to say that those two equations disagree. An indeterminate form does not give a specific value, it could have any value. So it doesn’t disagree with any specific answer.

When you see an indeterminate form it is an indication that you need to use a different formula. Often, if you look at the derivation for the equation you will find that one of the assumptions involved in the derivation is violated for the case where it generates the indeterminate form.

In this case, the derivation for equation 2 is based on a massive particle (one which can be at rest in some inertial frame). So it does not apply to a massless particle (which cannot be at rest in any inertial frame).

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Your equation (2) is an expression for a particle with mass; it does not apply to photons.

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  • $\begingroup$ So even though either equation can be derived from the other, one is more fundamental? $\endgroup$
    – Hilbert
    Commented Jan 2, 2020 at 17:47
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    $\begingroup$ @Hilbert, as you "derive" one from the other be sure you are not doing something illegal in the case where m = 0. There are famous math problems that "prove" 1 = 2 due to an unseen divide by 0 in the steps. You may be doing this and not realizing it. Also, I'd say that the first equation is more fundamental as it can be derived from first principles using the norm of a 4-vector in Minkowski space. $\endgroup$
    – user196418
    Commented Jan 2, 2020 at 20:04

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