Re: massless or massive photon?

Eric Gisse wrote on Sat, 12 Jul 2008 06:10:41 -0700:

On Jul 12, 5:00 am, stevendaryl3...@xxxxxxxxx (Daryl McCullough) wrote:
Eric Gisse says...

I didn't bother writing the Hamiltonian in the correct set of
generalized coordinates, instead I left it be because IT IS SO FUCKING
OBVIOUS and it was irrelevant to my point.

But since you are pitching a shitfit over it, here: v = p/m --> H = -
mc^2 [1 - p^2 / (mc)^2 ]^1/2

Eric, for one thing, I don't think that anything good can come out of
arguing with Juan. I don't believe that his arguments are in good
faith.

Agreed. The entertainment value of this is rapidly converging to zero -
it was only extended by Pete/Juan freaking out over something trivial.

Yes so trivial you got it completely wrong :-)


Peter, on the other hand, sometimes makes a good point. Unlike Koobee
and Juan and Sue, he's not a crackpot, there are just some
(unfortunately, permanent) gaps in his understanding of physics.

Anyway, getting to the point: You're wrong on this point. You were
right, that you can use - m c^2 square-root(1-v^2/c^2) as a
relativistic Lagrangian, but you *can't* use it as relativistic
Hamiltonian.

The relationship between a Lagrangian and a Hamiltonian is this:

H = -L + p v

where p is @L/@v

H and L are only equal if (1) there is no potential *and* (2) the
Lagrangian is quadratic in the velocity: 1/2 m v^2. In that case, p =
mv, and H = -1/2 m v^2 + mv^2 = +1/2 mv^2.

I didn't know about the quadratic condition. I know the relation between
the Hamiltonian and Lagrangian in addition to the covariant formalism,
but I assumed that the equivalence absent potentials remained true in
relativity.

The correction is appreciated.

[...]

See this is what happen when you think you know physics :-)

Poor Eric.

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