Re: massless or massive photon?
- From: "Juan R." González-Álvarez <juanREMOVE@xxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 10 Jul 2008 15:50:44 +0200 (CEST)
Eric Gisse wrote on Wed, 09 Jul 2008 18:10:55 -0700:
On Jul 8, 5:11 am, "Juan R." González-Álvarez
<juanREM...@xxxxxxxxxxxxxxxxxxxx> wrote:
The usual argument for massless photons uses the Hamiltonian
No, the usual argument for the massless photon is the following, in this
order:
a) Experiments and observation don't support it.
Don't true. Experiments only provide a upper limit for m_ph.
b) Proca's equations
with m--> 0 recovers Maxwell's equations and the expected vacuum
dispersion relation.
And Maxwell equation with massless photons is a special case of a more
general theory with massive photons. Your comment is irrelevant.
c) SR + Quantum field theory ---> photons travel along null paths --->
photons must be massless.
And QED with massless photons is a special case of a more general quantum
theory with massive photons. Your comment is also irrelevant.
H = (\sqrt (m^2c^4 + p^2c^2) ) (1)
No. The special relativistic Hamiltonian is H = L = -mc^2 * [1 - v^2/
c^2 ].
Nonsense.
There is no reference to electromagnetism regardless, so any
argument about electromagnetism that _neglects_ the electromagnetic
contributions to the Lagrangian is destined to be rather silly.
Misreading or off-topic or what else...
and set (m = 0) to yield
H = pc
However, the original Hamiltonian (1) was derived using the Legendre
transformation
No, you derived the original Hamiltonian is by assuming it is the energy
in the magnitude of four-momentum.
No.
H = pv - L = (\sqrt (m^2c^4 + p^2c^2) )
If alternatively we start assuming (v = c) in the transformation, the
result is
H = pc - L = pc (2)
How does pc - (\sqrt (m^2c^4 + p^2c^2) ) = pc in your world? You can
only obtain that by _assuming_ a massless particle.
No. Read.
where no assumption was taken about the mass.
By equating v = c you have DEFINED the particle to be massless.
No, a well-known counterexample is the electron in Dirac theory. It is
massive and its |v| = +-c.
Either
that or you have defined a contradiction by allowing a massive particle
to travel on a null path.
It is to be proved that (2) and (1) represent the same particle.
(...)
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org
.
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