The relativistic momentum formula is [itex]p = \frac{mv}{\sqrt{1-v/c^{2}}}[/itex]. Since photons are massless and travel the speed of light, this formula doesn't apply to them directly, but it seems logical that we could find the momentum of a photon by calculating [itex]lim_{(m,v) \rightarrow (0^{+},c^{-})} \frac{mv}{\sqrt{1-v/c^{2}}}[/itex]. Unfortunately, this double limit does not exist, but we can still calculate the limit along specific "path" in the "m-v plane". If we choose a "curve of constant energy", namely [itex]\frac{mc^{2}}{\sqrt{1-v/c^{2}}} = E[/itex], then we find the that the limit along this path is [itex]\frac{E}{c}[/itex], as expected.(adsbygoogle = window.adsbygoogle || []).push({});

But what a priori reason do we have for choosing such a path, as opposed to any other path? There doesn't seem to be any physical rationale; is there any situation where a physical object gets lighter and lighter as it approaches the speed of light, all the while keeping its total energy constant?

Any help would be greatly appreciated.

Thank You in Advance.

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# Can calculus find the momentum of light?

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