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I'm confused about the theory regarding the energy- momentum relation when I consider the speed of an object being $v = c$. Using for example the relations $$E^2 = (pc)^2 + (m_0c^2)^2$$ $$E = \gamma_{(v)}m_0c^2$$ $$\textbf{p} = \gamma_{(v)}m_0\textbf{v}$$ to find that $m_0 = 0$. I want to ask how why these equations valid in this case? I mean all the theory come from the definition of the four-momentum $p^\mu$ $$p^\mu = m \frac{\mathrm{d}x^\mu} {\mathrm{d}\tau} = m \gamma_{(v)} \frac{\mathrm{d}x^\mu} {\mathrm{d}t}$$ but i used the relation $t = \gamma{(v)} \tau$ that come from Lorentz transformation between two inertial frames, but an object that move with speed $v = c$ is not inertial, doesn't that mean I can't use it to infer anything about light?

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These equations are not valid in the case $v=c$ and you can indeed not properly use them to infer properties about light. It is actually meaningless to use $\gamma$ for massless particles, as this quantity diverges. At the fundamental level, the expression you are using for the linear momentum $p^\mu$ comes from the Lagrangian density of a free relativistic massive particle. This is where you should start (see here https://en.wikipedia.org/wiki/Relativistic_Lagrangian_mechanics) as this is the most general way to define the linear momentum $p^\mu$, as the quantity conserved under space-time translations.

When taking a proper generalization of the Lagrangian working also for massless particles, the only one remaining valid is $E=pc$.

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A photon does not have rest mass because it is never at rest in any inertial reference frame, but it does have momentum as can be observed by photons impinging on a light sail or in the photoelectric effect. The momentum of a photon (p) is $h f/c$ where $h$ is the Lorentz constant and f is the frequency. Using the equation

$E^2 = (pc)^2 + (m_0c)^2)$

the calculation is $E = \sqrt{( hf )^2 + (m_0c)^2} = \sqrt{( hf )^2 + 0} = hf $.

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  • $\begingroup$ Note that this derivation mixes special relativity and basic quantum mechanics. However, even within special relativity alone, light and massless particles have a linear momentum $E=pc$, without even having to consider the notion of photon. $\endgroup$ Commented Jan 22 at 18:50
  • $\begingroup$ $E = h\nu = \hbar\omega$ where $\nu$ is frequency and $\omega$ is angular frequency. $\hbar=\frac{h}{2\pi}$. $\endgroup$
    – PM 2Ring
    Commented Jan 22 at 21:12
  • $\begingroup$ @PM2Ring Thanks for the correction. I see I should have used h and not $\hbar$. $\endgroup$
    – KDP
    Commented Jan 22 at 21:19
  • $\begingroup$ I use f for frequency instead of v because v is confusing in relativity where we also use v for velocity. $\endgroup$
    – KDP
    Commented Jan 22 at 21:22
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You cant use it for objects moving at $c$ since $d\tau = 0$ and the momentum becomes $\infty$ for any massless object which cannot be true.

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