# Massless Particle SR

I was just reading through the excellent answer to Special relativity and massless particles by @user4552 which says:

For the reasons given in the comment above, I think the argument from the $$m\rightarrow 0$$ limit is valid. But if one doesn't like that, then here is an alternative. Suppose that a massless particle had $$v in the frame of some observer A. Then some other observer B could be at rest relative to the particle. In that observer's frame of reference, the particle's three-momentum $$\mathbf{p}$$ is zero by symmetry, since there is no preferred direction for it to point. Then $$E^2=p^2+m^2$$ is zero as well, so the particle's entire energy-momentum four-vector is zero. But a four-vector that vanishes in one frame also vanishes in every other frame. That means we're talking about a particle that can't undergo scattering, emission, or absorption, and is therefore undetectable by any experiment.

I don’t quite understand the logic in the last sentence. I think understand why a energyless particle can’t scatter, but I don’t understand why it couldn’t absorb. For instance, why couldn't absorption take place coinciding with an increase in speed to c.

• What experiment do you propose to test your hypothesis? May 22, 2023 at 22:13
• Absolutely no idea! But that’s not really the point. I don’t understand the logic in the last line of the answer and I’m hoping someone can elaborate on it.
– EEH
May 23, 2023 at 1:06

Suppose that you have a particle with four-momentum $$(E,\vec p)$$ in natural units. So if it absorbs a massless energyless particle then the conservation of four momentum would give that the final momentum is $$(E, \vec p) + (0,0)=(E,\vec p)$$ In other words the particle that has absorbed the vanishing particle looks the same as one that hasn't absorbed the vanishing particle. There is no way to detect this event and distinguish it from not having absorbed it.