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In A.P. French's Special Relativity, page $20$, the author wrote,

For photons we have $$\tag{1}E=pc$$ and $$\tag{2}m=\frac{E}{c^2}$$ (the first experimental, the second based on Einstein's box). Combining these we have $$\tag{3}m=\frac{p}{c}$$

He didn't give any justification for combining equation $1$ (valid only for photons) and equation $2$ (which can't be applied to photons since they are massless) into equation $\textbf{3}$.

To which particles does equation $\textbf{3}$ apply to: massless or non-massless?

I couldn't find a satisfactory answer in this post.

Hilbert
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    (2) does not hold for massless particles if $m$ is the invariant mass, which is the only kind of mass you should ever use. – G. Smith Dec 28 '19 at 11:32
  • All you have to remember is $E^2-(pc)^2=(mc^2)^2$. It holds for both massive and massless particles. – G. Smith Dec 28 '19 at 11:34
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    Relativistic mass has only one use: confusing students. – G. Smith Dec 28 '19 at 11:39
  • On relativistic mass: https://physics.stackexchange.com/questions/133376/why-is-there-a-controversy-on-whether-mass-increases-with-speed/133395 – PM 2Ring Dec 28 '19 at 13:50
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    If the mass in (2) is the invariant mass then (2) shouldn't hold for photons as $E=mc^2$ only holds in the rest frame of the object – bemjanim Dec 28 '19 at 15:18

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Unfortunately, you’re trying to learn from a 50-year-old book that presents the concepts in what we now consider to be a very confusing way.

Now, we (generally) use “mass” m to denote the rest mass of a particle: the mass that’s characteristic of any electron, for example. Then in the modern convention:

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

French uses “mass” to denote a form of total relativistic energy. (I'm describing the following, not defending it; this is not a good idea) He's using a concept that we now call "relativistic mass", a "mass" that increases with the gamma factor: a moving body "has more mass". Mathematically, this is like saying that $mc^2 = E$ always, so as the body gets more energy $E$ with the addition of momentum $p$, the mass goes up. (I agree this is confusing; that's my point) All of this goes completely sideways with photons, where the rest mass is zero, the gamma factor is infinite, and their product has to miraculously work.

Learning from this book is going to be difficult because you’ll be learning a terminology and form of calculation that will cause confusion when you ask questions, and probably isn’t consistent with what you already know.

Bob Jacobsen
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  • "and “mass” to denote a form of total relativistic energy." Could you clarify this passage? – Hilbert Dec 28 '19 at 21:38
  • If understood the other comments, the mass used in equation $(3)$ is $\gamma m_{0}$, where $m_{0}$ is the rest mass, correct? – Hilbert Dec 28 '19 at 21:45
  • This is a quote from the same book, "the variable mass $m(v)$ is just a convenient construct which, for example, allows us to preserve the form of the Newtonian statement that momentum is mass times velocity. Many physicists prefer to reserve the word mass to describe the rest mass $m_{0}$, a uniquely defined property of a given particle, But this is essentially a matter of taste." – Hilbert Dec 28 '19 at 22:10
  • It may have been a matter of taste back then. Today, one convention is immensely more useful than the other. – Bob Jacobsen Dec 28 '19 at 22:18