Re: massless or massive photon?

Daryl McCullough wrote on Sun, 13 Jul 2008 05:50:00 -0700:

Dono says...

How about if you showed how the relativistic Lagrangian is DERIVED

In Jackson's Electrodynamics, the following heuristic argument is used:

We want the action A = Integral L dt to be a scalar, independent of
which coordinate system is used. However, the parameter t is
coordinate-dependent. So convert it to tau, the proper time:

dt = gamma d tau

Then the action becomes

Integral gamma L d tau

For this to be independent of coordinate systems, the quantity (gamma L)
must be a Lorentz scalar. The simplest choice is for it to be a
constant, K. With this heuristic,

gamma L = K
L = K/gamma = K square-root(1-(v/c)^2)

Working out the corresponding momentum gives:

p = @L/@v = -Kv/(c^2 square-root(1-(v/c)^2))

To give the usual expression for momentum, we let K = -mc^2.

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

Alternatively, you can just start with the equation

p = @L/@v = mv/square-root(1-(v/c)^2)

and integrate with respect to v:

L = m Integral of v/square-root(1-(v/c)^2) dv
= -mc^2 square-root(1-(v/c)^2)

A less convoluted approach starts from arguments about worldlines for
free particles with invariant action

A = \Int k ds

where k is a constant which is found to be k = -mc

From the above action one can find the coordinate frame Lagrangian

A = \Int k ds = \Int L dt

in one much more easy and direct way.

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