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Re: [Phys-l] Absolute four-momentum of massless particles




Now this thread has got me wondering about a fact that I think is supposed to obvious, but isn't (to me). Massless particles must travel at the speed of light. Everybody assumes it, but what is/are the strongest argument/s for this assumption?

One that comes to mind, given the arguments put forward in this thread, is that if a massless particle didn't travel at the speed of light, then its four momentum would have to be zero by virtue of the (now well-defined formula) P = mU, and conservation of four momentum would fail (I think). But this is a pretty weak argument because I can always declare that the formula P = mU only holds for massive particles (which is actually what we end up doing, in fact). There must be something really silly (i.e. inconsistent) about the very idea of a sub-luminal massless particle that I'm just not seeing. Which usually means I'm looking at things the wrong way...

And that leads me to yet another issue. On what basis do we declare a particle to have zero mass anyway? The popular argument seems to be that assigning any positive mass leads to a contradiction, but that only convinces me that the mass is either zero for that particle OR meaningless. In the latter case we might assign the particle a mass of zero on the basis that this leads to a logically coherent system of thought, but that is different from the idea of actually measuring the mass and finding it to be zero. So I'm curious as to which of these approaches lies behind a statement of the form 'photons have zero mass'. (I mean 'rest' mass, of course - but it's tiring to say that all the time.)

Derek

P.S. John - I read that Berry article. Very interesting, thanks. It's something I've thought about, much less coherently than Berry, many times before. And possibly connected quite closely to the above question, since limits are probably not the right way to assign meaning to the zero mass case. But I'm not sure how reasonably direct measurement could work here either.

Date: Thu, 30 Sep 2010 09:47:25 -0400
From: jsd@av8n.com
To: phys-l@carnot.physics.buffalo.edu
Subject: Re: [Phys-l] Absolute four-momentum of massless particles

On 09/30/2010 08:30 AM, Derek McKenzie wrote:

2. It can be helpful to think of the four velocity vector as a
'proper time vector' so that we can see intuitively what goes wrong
in the massless case (loosely - time stands still). I'm tempted to
say the proper time vector of a photon is the zero vector,

Be careful there. 4-velocity is related to proper time,
but it's not the same thing. Indeed it is _inversely_
related. U goes like d/dτ, with τ in the denominator.

Let's see how that works, focusing on U rather than τ.
We consider the case of a particle with a very small but
nonzero mass. I wish to _contrast_ the infinitesimal-mass
particle with the zero-mass particle -- not equate them!

Suppose a highly relativistic particle whizzes past the
laboratory. We know the mass is very small but we don't
know how small. We can measure the 4-momentum and we
find it is very close to P = [1, 1], well within the
uncertainty of measurement. If we are sure the mass is
nonzero we can write U = P/m = [1/m, 1/m] approximately,
subject to the constraint that U.U remains equal to -1,
which is no problem if we write U = [cosh ρ, sinh ρ] and
consider very large values of the rapidity ρ. We see
that the components of U are approaching infinity, not
approaching zero.

For the small-mass particle, there will always be SOME
frame in which it is at rest. For the zero-mass particle,
not so. The limit as m goes to zero is a singular limit,
so intuition can be expected to fail, and you have to be
very careful about the limiting cases.

Saying a function has a singular limit means that the
value of f(0) is not equal to the limit of f(x) as x goes
to zero. For more on this, including the unforgettable
"worm in the apple" example, I highly recommend the
article by Berry (yes, THAT Berry) at
http://www.phy.bris.ac.uk/people/berry_mv/the_papers/Berry341.pdf
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