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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...).

Hilbert
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2 Answers2

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

dmckee --- ex-moderator kitten
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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).

Dale
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