Chronology Current Month Current Thread Current Date [Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

[Phys-l] Why is the photon massless?

Here I want to approach the question about massless photon raised in the
previous discussion, from 2 other perspectives different from those discussed
before (someone on this List has said that it is always good to have more than
First of all I would slightly reformulate the question, introducing the term
"invariant speed" instead of speed of light or a graviton. Let c stand for the
invariant speed. Then we could ask: "If there exists a particle moving with
invariant speed, what is its mass m?" (For brevity, "m" here stands for the
invariant or rest mass).
For any non-zero m, by definition of the rest mass, we can find a frame
comoving with the particle in question. In this frame, the particle's speed is
zero. Clearly, the invariant speed cannot be zero. Therefore the particle in
question cannot have a non-zero m.
Interestingly enough, it directly involves QM, since now we will use the wave
aspect of matter. If the particle in question is in a state with definite
4-momentum P = (E/c, p), it is described by the monochromatic de Broglie's wave
Aexp (kx - wt), with the phase velocity u = w/k. In another reference frame
(RF), w and k would have other numerical values, but, by definition, their ratio
must remain the same (otherwise, it would be possible to "cook up" wave packets
with different group velocities, which would contradict the condition that the
particle has the invariant speed). Thus, we must have u = w/k = const = c. It
follows that the dispersion equation for this particle must be (w/c)^2 - k^2 =
0.
Now, when we switch to another RF, E and p undergo exactly the same Lorentz
transformation as w and k (this similarity was one of the factors that inspired
de Broglie to introduce his famous equations.) Therefore, there must be also u =
E/p = const = c, and accordingly, (E/c)^2 - p^2 = 0.
This is the dispersion equation for a massless particle, so there must be m = 0.

Moses Fayngold,
NJIT

________________________________
From: Derek McKenzie <derek_s_mckenzie@hotmail.com>
To: phys-l@carnot.physics.buffalo.edu
Sent: Fri, October 1, 2010 8:02:35 PM
Subject: Re: [Phys-l] Absolute four-momentum of massless particles

Ok - I've finally got on top of the arguments in this thread (or at least those
relating to my questions about massless particles)...

On 09/30/2010 08:23 PM, Derek McKenzie wrote:
On what basis do we declare a particle to have zero mass anyway?

What means "declare"? Strictly speaking, I reckon you
can declare anything you want.

At the other extreme, if you want to /prove/ that the
photon mass is zero, then you're probably out of luck.
According to Popper's view, which is nowadays a rather
conventional view in the scientific community, no theory
ever gets proved right. Roughly speaking: wrong theories
can be disproved, but valid theories always remain open
to further testing.

It was precisely because you can't 'prove' things in physics that I used the
word 'declare'. The physics community (more or less unanimously) concur that a
photon has zero mass, so they declare it to be the case (albeit possibly
temporarily). Nothing deep was meant by the word. Perhaps I should have asked
'What makes physicists 'believe' a given particle has zero mass?'.

Your description of the photon case was very enlightening, but my question is a
different one (sorry I didn't express it properly). Firstly, I do realize that
relativity does not require photons to travel at the speed c (c being defined
here as the geometrical constant in the theory of relativity). But the argument
accepted in the community is that IF a particle (who cares whether it's a photon
or a neutrino) travels at speed c (again, I mean here the geometrical c), THEN
it has to be massless (and conversely). This assumption is not usually justified
in texts - perhaps because it is supposed to be totally obvious.

Is the idea that IF a particle has a timelike worldline, THEN it has a well
defined four velocity and therefore a well-defined P = mU, and therefore the
latter must be zero? In other words, a time-like particle of zero mass would
have no momentum or energy in any frame. Is that all there is to it?

Derek

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.

We can do much better than that. A massless photon
means that electromagnetism is an infinite-range
interaction. A positive mass would dictate the range
of the interaction via the Yukawa formula. To say
the same thing in slightly different words, you would
see a departure from the 1/r^2 law.

This has been checked experimentally on the laboratory
length-scale and on cosmological length scales. In
all cases that have been checked, the exponent n in
the 1/r^n law is 2 within the precision of measurement,
and some of the measurements are good to one part in
10^12. Reference: Jackson; also
http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass

There is no 11th commandment that requires the photom
mass to be zero. Maybe tomorrow somebody will discover
that the photon has a mass ... but if so, it must be
reeeally small.

Note that neutrinos were for many years believed to be
massless, but are now believed to have a very small but
nonzero mass. This demonstrates that it is possible to
reclassify a particle from massless to non-massless.
Such a reclassification is not hard to carry out; it's
not like we've never seen a massive particle before.
On the other hand, such a reclassification is very
unusual and newsworthy.
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l

_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l