2
$\begingroup$

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 $v<c$ 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.

$\endgroup$
2
  • $\begingroup$ What experiment do you propose to test your hypothesis? $\endgroup$
    – John Doty
    May 22, 2023 at 22:13
  • $\begingroup$ 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. $\endgroup$
    – EEH
    May 23, 2023 at 1:06

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ There are conserved quantities besides energy and momentum, though. Intrinsic angular momentum or any charges could also be measured. That said, if such a particle exists, we would probably never have come up with the concept of a conserved quantity ... $\endgroup$
    – tobi_s
    May 23, 2023 at 6:40
  • 1
    $\begingroup$ @tobi_s yes, this is a good point. I have read that a massless particle cannot have charge, but I don’t remember the justification. I wonder if there is a justification why a massless energyless particle cannot have spin. $\endgroup$
    – Dale
    May 23, 2023 at 11:28
  • $\begingroup$ Massless particles can have charge: gluons carry color charge. Admittedly, that may miss the point of what you read as the gluons live in a different representation of the color group than the quarks. So a better example may be the neutrinos in the standard model which are massless and carry weak hypercharge. That's a consistent theory, but unfortunately experiment indicates that neutrinos are a bit more complicated. $\endgroup$
    – tobi_s
    May 26, 2023 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.