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Classical non-relativistic newtonian mechanics can be derived from special relativity by letting the speed of light tend to infinity $c \rightarrow \infty $. But doing this for Einsteins famous formula $$ E = mc^2 $$ would yield $$ E = \infty $$ This would mean, that in classical mechanics the rest energy is infinite? Or how does one interpret this fact? Or should one rather look at $m =\frac{E}{c^2}$ and conclude that in classical mechanics the rest mass of a particle must be zero?

Qmechanic
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asmaier
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    there is some controversy among the laity (the physics priesthood seems to have little disagreement) about the use of terms "relativistic" and "rest" as modifiers to the term "mass". turns out that photons, if they really do move at the speed of $c$, can have no rest mass (a.k.a. "invariant mass") because they would have infinite energy if they did. but they do have relativistic mass of $\frac{E}{c^2} = \frac{h \nu}{c^2}$ (for those to do not deprecate that term) and everyone agrees that they have momentum of $p=\frac{h \nu}{c}$. – robert bristow-johnson Nov 06 '16 at 17:19

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In relativity, the absolute value of the energy of an object is important and measurable, because it contributes to the object's inertia by $E = mc^2$, as you wrote. But outside of relativity, energy and mass have no relation whatsoever. So it doesn't matter what the rest mass contribution to energy is, even if it's formally infinite -- we can just subtract it out, and nothing changes.

However, we do need to make sure that changes in energy agree. To do this, we need to use the full relativistic energy formula $$E = \gamma mc^2 = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$ which applies to moving bodies as well as stationary ones. Performing a Taylor expansion in $v/c$, $$E = mc^2 + \frac12 mc^2 \frac{v^2}{c^2} + O(v^4/c^4)$$ which approaches $mc^2 + mv^2/2$ in the nonrelativistic limit. Subtracting out the constant rest energy contribution, we find $E = mv^2/2$, as we should.

knzhou
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  • But isn't the non-relativistic limit is $\infty + mv^2/2$? This substraction of a formally infinite, but constant quantity reminds me a bit of https://en.wikipedia.org/wiki/Renormalization. Is this the same idea here? – asmaier Nov 06 '16 at 18:05
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    @asmaier I would say it's somewhat different. Renormalization is usually applied to physically measurable quantities. But the absolute value of the energy is never measurable in non-relativistic mechanics in the first place, so there is no 'physical infinity' to deal with. – knzhou Nov 06 '16 at 23:51