Re: massless or massive photon?

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.

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.

For the relativistic Lagrangian, the hamiltonian is not
quadratic in the velocity, and so it's not equal to the
Lagrangian. Instead, we have:

H = mc^2/square-root(1-v^2/c^2)

which, when expressed in terms of p becomes

H = square-root(p^2 c^2 + m^2 c^4)

An alternative way to do Lagrangian mechanics in relativity
is to use 4-velocities and 4-momenta. Then you can write
the Lagrangian as:

L = 1/2 m U_u U^u

where U^u is the 4-velocity dX^u/ds, where s is proper time.

This Lagrangian is quadratic in the 4-velocity, so the
Hamiltonian is equal to the lagrangian. The canonical
momentum is just

P^u = m U^u

The lagrangian equations of motion are just

d/ds (m U^u) = 0

It looks a lot like the non-relativistic case, except
that (1) you use 4-momentum and 4-velocity, and (2)
you use proper time instead of time.

--
Daryl McCullough
Ithaca, NY

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