# Are energy-momentum relation equations still valid with speed of light?

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?

## 3 Answers

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 $$E=pc$$.

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

• 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. Commented Jan 22 at 18:50
• $E = h\nu = \hbar\omega$ where $\nu$ is frequency and $\omega$ is angular frequency. $\hbar=\frac{h}{2\pi}$. Commented Jan 22 at 21:12
• @PM2Ring Thanks for the correction. I see I should have used h and not $\hbar$.
– KDP
Commented Jan 22 at 21:19
• I use f for frequency instead of v because v is confusing in relativity where we also use v for velocity.
– KDP
Commented Jan 22 at 21:22

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.