Friar Bellows

This article reports a new upper limit to the mass of a photon. An experiment by Chinese researchers found that the mass of a photon can be no more than about 10E-54 kg, or 7x10E-19 eV. I know that physicists expect it to have a mass of zero, but it's still cool to hear of experiments which check to see if it might have a mass. But a section from this article confuses me:

Why would a nonzero photon mass "make trouble for special relativity"?

I was under the impression that special relativity had nothing specific to say about the photon, other than predicting that if it does have a zero mass, then it should travel at a maximum possible speed which is invariant. Electrodynamics or QED has something specific to say about the photon, but not special relativity. It's true that one of the postulates of special relativity mentions the "speed of light", but that's based on the assumption that light has zero mass -- if it didn't have zero mass, then instead of using the term, "speed of light", you would use the term, "speed of a massless particle". I can't see how a nonzero photon mass would spell trouble for special relativity.

Where's the flaw in my thinking?

alek0

Look at the relativistic mass equation:

http://www.glenbrook.k12.il.us/gbssc...ity/U7l3a.html

You can see that the object moving at the speed of light would have infinite mass, unless m0 is zero.

wade-w

From a purely mathematical standpoint, if m0 = 0, and v = c, then we have m = 0/0, which is indeterminate. If m0 is nonzero, then m = m0/0, which is undefined.

If you look at mass as a function of velocity, and take the limit as v -> c of the given equation, the mass becomes arbitrarily large if m0 is nonzero, and is of course 0 if m0 is zero.

But the limit argument works only when we are considering an object whose velocity is approaching c. A photon is already there. So what is the mass of a photon? I don't know, but I don't think the mass transformation can tell us.

On the other hand, the length contraction gives us a length of zero for an object travelling at c, and therefore a zero volume. Does that imply it has zero mass? I think so, but I'm not a physicist.

Undercurrent

Agreed that the notion of the "rest mass of a photon" is more of an artifact than anything else.

No, it doesn't, because (1) photons and other elementary particles aren't normally considered to be extended objects anyway -- even at rest they hav no "volume" to speak of; And (2) for an extended object the preceived mass increases with velocity, even as the perceived volume decreases due to FL contraction.

Jesse

I'm not sure it's right to say the limit is zero if m0 is zero--it depends on how you set up your limit. If you do 0/x as x approaches infinity, I guess since the answer is 0 for every finite x, the limit would also be zero. But if you do y/x and have y approach zero and x approach infinity, then the limit will be undefined because you will get different answers depending on how x and y approach their respective limits.

edit: my bad, both x and y should be approaching zero.

wade-w

Thanks for the clarification. I hadn't considered that.

My understanding of the equation is that m0 is a constant. In the scenario you present, the limit of a quotient is the quotient of the limits, so regardless of how one or the other approaches its respective limit, the overall limit will still be zero.

wade-w

I don't have time to go into this right now, since I have to go to work, but Jesse's comment made me think a bit harder about the case of the limit as v->c and m0 = 0.

The case we are looking at is not x -> 0 and y -> inf, but x -> 0 and y -> 0. Thus we have to apply L'Hospital's rule. I'll look at it again when I get off work tonight.

echidna

Friar, here's another bearing on it asking if neutrinos can be massed, and travelling at c, although still not from a physicist, only undergraduate ...

http://www.physlink.com/Education/AskExperts/ae476.cfm

Answerer

Hi guys, first of all, is the article talking about the rest mass of the photons or just effective mass?

ex-xian

Special relativity predicts that: (momentum = p; u = velocity; m = mass; gamma = sqrt[1-(u/c)^2]; E = energy]
p = gamma * u * mass (1)
E = gamma *mc^2 (2)

It can be shown that

u/c = pc/E (3)

If m=0, then E = pc, which would imply from (3) that u=c.

So a massless particle travels at c. The converse is also true, if a particle travels at c, then it is massless. Since the photon travels at c, it is assumed to be massless.