Re: EM field, Noether's Theorem and Conservation Theorems

From: David McAnally (D.McAnally_at_i'm_a_gnu.uq.net.au)
Date: 07/18/04 

Date: 18 Jul 2004 00:31:34 GMT

dubious@radioactivex.lebesque-al.net (Bilge) writes:

> David McAnally:
> >While investigating consequnces of Noether's Theorem for the Lagrangian
> >density for the electromagnetic field, I found that for any C^2 function
> >f of x, y, z, t, the following 4-current density is conserved:
> >
> > (D.grad f + \rho f, - D df/dt - H x grad f + J f),
> >
> >i.e. d(D.grad f + \rho f)/dt + div(- D df/dt - H x grad f + J f) = 0.
> >I was wondering if there was any name for the quantity for which
> >D.grad f + \rho f is the density and - D df/dt - H x grad f + J f is the
> >current density. Of course, when f = 1, then the quantity in question is
> >the charge.

>I'm not sure what you call it, but it reduces to:

> D.\nabla f + \rho f = \nabla . (fD)

> -H x \nabla f = \nabla x (fH) - f(\nabla x H)

> = \nabla x (fH) - f(J + dD/dt)

>and

> D df/dt = (d/dt) (fD) - f dD/dt

>So, the second term in your ()'s is,

> -(d/dt) (fD) + f dD/dt + \nabla x (fH) - f dD/dt - J f + Jf

> = -(d/dt) (fD) + (\nabla x fH)

>which gives,

> (\nabla.(fD), (-d/dt)(fD) + \nabla x (fH))

>which guarantees that,

> (d/dt)\nabla . (fD) + \nabla . (-d/dt)(fD) == 0

>leaving the term, \nabla . (\nabla x fH), which is zero on its own.

Thanks for those observations. These interpretations for the densities
are very informative.

David

        "But I'm always true to you, darlin', in my fashion,
         Yes, I'm always true to you, darlin', in my way."
                -- Lois Lane

-----

Relevant Pages