1
$\begingroup$

If supposedly someone is to argue that photon might be carrying a mass which is much smaller than our current detector can measure, and it could decay into something that moves at the speed of true massless object which is faster than speed of light in vacuum. So is there any experiment to probe whether photon can decay? btw is the lightest neutrino lighter than a photon and could photon decays into these neutrino?

p.s. I'm not saying generally neutrino is moving at or faster than c but I'm not sure about the lightest neutrino if they could exist.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

If supposedly someone is to argue that photon might be carrying a mass which is much smaller than our current detector can measure,

This would effectively be zero mass. If we can't detect a mass, it's zero for all intents and purposes.

The only test we could make would be on velocity. If it traveled more slowly than $c$ then we'd know it was a non-zero mass. It's actually how we define the limits on the mass of the photon - by measuring it's velocity and using the limits of accuracy to determine the limits for mass.

What about faster than the speed of causality (the speed of light) ?

It's worth remembering that in general relativity light has to travel at the same speed for all observers. There is no reference frame where light does not travel at that speed. So any move away from that would mean a big problem for all the theories that rely on this (and at this point that's all the mainstream theories !).

and it could decay into something that moves at the speed of true massless object which is faster than speed of light in vacuum.

This is wrong.

What is commonly referred to as "the speed of light" is not defined in that way any more. It's better referred to as "the speed of causality". It's a limit imposed by the structure of spacetime. Here's a link to a PBS Spacetime video on YouTube explaining why "The Speed of Light is NOT about Light".

That limit is (as far as we know) the absolute limit for the speed of any object, massive or not. Nothing faster is expected under current mainstream theories. Finding something faster would require a massive rethink of physics as we understand it.

So is there any experiment to probe whether photon can decay?

We watch the little devils closely for any signs of trouble. :-)

Photons can change into massive particles. You can have a sufficiently energetic photon transform into pair of particles ( see Pair Production ). I'm not sure if you'd call that a decay, as such, but that might be semantics. The problem with a decay is that the net energy and momentum has to balance and for low energy photons (and we think they can get to very low energy levels) the numbers don't provide us with any particles they can transform into.

btw is the lightest neutrino lighter than a photon and could photon decays into these neutrino?

Neutrinos are not massless. However the problem with such a decay is that even if it's possible how would be detect it ? Neutrinos are very hard to detect and if we had our detectors running what would be detect ?

Well we can't detect the photon except by the interaction it makes with something else. And in this case the something else would be a neutrino and we won't detect a neutrino, so we see nothing going in and nothing coming out.

So we'll probably never be able to detect such a thing even if we thought it could happen.

p.s. I'm not saying generally neutrino is moving at or faster than c but I'm not sure about the lightest neutrino if they could exist.

They can't move faster than the speed of causality regardless of whether they're massless or not. It just happens that we think that coincides with the speed of light because we think massless photons travel at that speed.

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ A free photon in a vacuum can't decay into an electron and a positron, there's no way for that to happen that conserves both energy and momentum. But if it has enough energy it can decay in the presence of something that can participate in conserving momentum, like a nucleus. $\endgroup$
    – PM 2Ring
    Commented Apr 20, 2018 at 4:51
  • 2
    $\begingroup$ I believe that's explained in the Wikipedia page I linked to on pair production. $\endgroup$
    – StephenG - Help Ukraine
    Commented Apr 20, 2018 at 4:53
  • 1
    $\begingroup$ Fair enough. I just wanted to mention it directly here, for the convenience of future readers. $\endgroup$
    – PM 2Ring
    Commented Apr 20, 2018 at 4:56
0
$\begingroup$

Of course there are many theories on this as you will find if you do some research on the subject. The generally accepted view is that photons have zero rest mass, but some have proposed photons with a non-zero rest mass which would imply they could decay.

Now much of this rests on determining whether the speed of light can be exceeded. Of course some people hypothesize that Tachyons could move faster than the speed of light, and there are models where this works as long as no particle can move with both $v < c$ and $v > c$. So in this scenario, I suppose it would be possible to consider the photon decaying into a Tachyon.

In current models though, we consider the photon to have zero rest mass and this works best for the theory that is available. In short, there is no definite solution to this as it depends on your theoretical framework.

Improve this answer
$\endgroup$
2
  • 3
    $\begingroup$ Hmmm ... I think this answer fails rather miserably to convey an important fact: strong experimental limits on the possible mass of photons imply that even if they possess non-zero mass there is nothing lighter for them to decay into. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 20, 2018 at 4:29
  • 1
    $\begingroup$ If photons aren't perfectly massless then they travel at less than $c$. Sure, we originally defined $c$ as the speed if light in a vacuum, but that's kind of a historical accident. It's better to think of $c$ as the space:time scale factor, i.e., 1 second of time has the same magnitude as 1 light-second of distance. $\endgroup$
    – PM 2Ring
    Commented Apr 20, 2018 at 4:45

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.