Re: massless or massive photon?

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.


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.

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