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# Re: [Phys-l] Why is the photon massless?

Derek McKenzie wrote on Mon, October 4, 2010 10:49:31 AM

If mass is defined in terms of a rest frame, and you require definitions to be
empirically meaningful (your requirement below), then surely a photon does >not
have a mass of zero, but rather an UNDEFINED mass.

"Empirically meaningful" means experimentally executable in principle. It does
not (and cannot!) require for a measurement result to be absolutely exact
since no experiment is ideal even classically (let alone QM indeterminacy). All
our theories are mathematical models of the real world, which are based on
idealizations, and we are talking about theory here. If we insist that
"empirically meaningful" must mean exactly measurable, then ALL physical
parameters are UNDEFINED including the invariant speed.

A limiting procedure is surely unreliable in 'singular-limit' cases like this -
as discussed >recently in another thread.

True, but this is not the case in the discussed situations.
As has been stated recently in another thread,
"... a singular limit means that the value of f(0) is not equal to the limit
of f(x) as x goes to zero."
In other words, there must be a discontinuous jump at the limiting point, be it
at x=0 or at any other value of x. What we are discussing, does involve a limit,
it could be called an "improper limit" but it is NOT singular. If you plot the
graph of the Lorentz factor gamma (v) vs. v, it diverge at v --> c, but it does
it continuously. There are no discontinuous jumps there, at least for v not
exceeding c - within the domain we are restricted to. We have gamma (c) = Inf at
v = c, AND the limit of gamma(v) is Inf as v --> c. So the value gamma(c) =
limit gamma (v) as v --> c. It is Not a singular limit!

Moses Fayngold,
NJIT