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

One reference frame that is useful to have in the calculation tool box is the Instantaneously co-moving reference frame (some authors say momentarily co-moving reference frame MCRF.) These have lots of calculational uses, one being that a sequence of these reference frames can be used to handle acceleration calculations within the framework of SR. Even as a conceptual idea they are a nice handy thing for use when you hear people who suggest that SR can not handle kinematics of accelerating particles and then claim that you need GR for such objects.


Joel Rauber, Ph.D 
Professor and Head of Physics
Department of Physics
South Dakota State University
Brookings, SD 57007
605.688.5428 (w)
605.688.5878 (fax)

-----Original Message-----
From: [mailto:phys-l-] On Behalf Of John Denker
Sent: Thursday, September 30, 2010 11:49 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Absolute four-momentum of massless particles

On 09/30/2010 08:23 PM, Derek McKenzie wrote:
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...

You can prove that the massless particle (moving at
less than the speed of light) has zero energy and
zero momentum in all frames. You can prove this in
about two lines, without assuming P = m U.

Hint: What are the components of P in the rest frame
of the particle?

Take-home message: There is a style of doing relativity
calculations that goes like this: Calculate something
in a frame where it is easy to calculate, then express
it in a frame-independent way, whereupon it can be used
in other frames. A serious calculation may switch back
and forth many times, switching from frame-based to frame-
independent representations, using several different frames
along the way, including the lab frame, the CM frame, et

Knowing how to express things in a frame-independent way
is important, but it is not the only tool in the toolbox.
Forum for Physics Educators