Re: massless or massive photon?

Daryl McCullough wrote on Sat, 12 Jul 2008 06:00:47 -0700:

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.

I wait just waiting how many time another crackpot would wait before
going in help of Eric and Dono crackpots.

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.

According to you a crackpot is one who writes the correct Hamiltonians
and Lagrangians but Eric, who is submitting *nonsense* for days, is not
named a crackpot.

To be precise, you don't call crackpot to someones (Eric) who said

(\blockquote
 No. The special relativistic Hamiltonian is
H = L = -mc^2 * [1 - v^2/ c^2 ].
)

Interestingly you also make several obvious mistakes below. I will add
this post to guidelines also.

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,

Do you mean the correct relativistic Lagrangian in my blog? The correct
Lagrangian giving the correct Hamiltonian in my original post?

The relationship between a Lagrangian and a Hamiltonian is this:

H = -L + p v

where p is @L/@v

That transformation was in the original poster. Nothing new but you do
not notice that. Is that an example you call "good faith arguments" Daryl?

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)

This may be more correctly called the energy E. Goldstein uses symbol h
because it is not the Hamiltonian H.

You lack rigor Daryl, as usual... but this kind of mistake already was
noticed before in the thread.

which, when expressed in terms of p becomes

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

Oh! I can see just the Hamiltonian (1) in the original poster!

That Hamiltonian that both crackpots, Eric and Dono, said was wrong. But
once again you don't notice that fact that Hamiltonian is already in the
original poster by me.

Is that another example you call "good faith" Daryl?

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

Oops, you were going right before this... but it may be too difficult to
you, Daryl, to write an entire post without crackpot-like mistakes.

You above Lagrangian is not using four velocities (v^u) but four proper
velocities (U^u). You confound both concepts.

The SR Lagrangian using 4-velocities is

L = -mc (\sqrt (v^u v_u) )

Moreover if you want use proper velocities then the Lagrangian is

L = T = - 1/2 m U_u U^u

as noticed in my blog article cited several days ago in this thread

http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-and-
limitations.html

See Daryl? See also the minus sign before the 1/2 factor?

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

already known and cited before in this thread

http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-and-
limitations.html

Is that another example you call "good faith" Daryl?

It looks a lot like the non-relativistic case, except that (1) you use
4-momentum and 4-velocity,

No, you use four proper velocity and four proper momentum. You confound
concepts Daryl.

--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org
.

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