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Article #660

Re: HOW EINSTEIN USED MAXWELL

#660
From: Rock Brentwood
Date: Thu, 01 Apr 2010 23:31
77 lines
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On Apr 1, 10:12 am, Uncle Al <Uncle...@hate.spam.net> wrote:
> Pentcho Valev wrote:
> > According to Maxwell's theory, (1) the speed of light is independent
> > of the speed of the emitter but (2) it does depend on the speed of the
> > observer:
> In 100,00 words or fewer, tell us how to measure absolute speed.

According to *Maxwell's* theory (i.e. the one actually published in
the treatise and the preceding papers, as far as the singular "theory"
can be used at all), the speed of light was independent of both the
emitter and observer and was characterized as 1/sqrt(mu kappa),
relative to the (presumed) unique frame of reference in which the
constitutive law between (D,H) and (B,E) becomes isotropic; where mu
and kappa are the coefficients characterizing the relation between
(D,H) and (B,E) in that frame.

The reference frame for light speed is any in which the constitutive
law becomes isotropic. The distinguishing feature of a Lorentzian
vacuum is that the reference frame is no longer unique. But it remains
unique even in relativistic theory, if one is not in a vacuum but an
isotropic medium.

(And one is never in a vacuum, since there is no such thing as a
perfect vacuum.)

In that frame, D = kappa E, B = mu H, where kappa is the dielectric
coefficient (kappa chosen because Maxwell wanted to draw an analogy to
the spring coefficient), while mu is the permeability (mu chosen to
draw an analogy to mass).

Applying the Galilean transform by a velocity G,
   r -> r - Gt, t -> t, del -> del, d_t -> d_t + G.del
where ().() denotes scalar product; and
   D -> D, E -> E + G x B, B -> B, H -> H - G x D,
   A -> A, phi -> phi - G.A, J -> J - G rho, rho -> rho.

This leaves both the equations:
   div D = rho, curl H - d_t D = J
   div B = 0, curl E + d_t B = 0
   E = -grad phi - d_t A, B = curl A,
   d_t rho + div J = 0
invariant, as well as the constitutive law generalized to non-
stationary frames:
   D = kappa (E + G x B), B = mu (H - G x D).

None of this changes with relativity. The same relations hold, with
the minor fix:
   D + 1/c^2 G x H = kappa (E + G x B)
   B - 1/c^2 G x E = mu (H - G x D).
So, the observations above about determining which frame is the
"reference" frame apply even in a relativistic context. It's whatever
frame the constitutive laws become isotropic.

Cosmologically, this is the co-moving frame; and G is the velocity
determined by the dipole moment of the cosmic microwave background
radiation.

So, that defines what you're essentially asking for.

Maxwell put the G term in with the E vs. A relation, so he wrote
   E = -grad phi - d_t A + G x B
instead, and wrote D = kappa E for the constitutive law. But in modern
terms it's D = kappa (E + G x B).

The other relation was missed by Maxwell because he failed to fully
distinguish B and H and (thus) failed to note that they transform
differently -- despite the fact that he wrote B as a 2-form
   B^x dy dz + B^y dz dx + B^z dx dy
and H as a 1-form
   H_x dx + H_y dy + H_z dz
-- which transforms differently than a 2-form. Thomas noted the -G x D
correction after Maxwell died and it was confirmed in experiments done
by Wilson and Wilson at the turn of the previous century.

The -GxD term remains in the relativistic form of the constitutive
law, so it's not an artifact of "non-relativistic theory" but is
paradigm-universal.
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From: Rock Brentwood <markwh04@yahoo.com>
Newsgroups: sci.physics.relativity,sci.physics,fr.sci.physique,fr.sci.astrophysique
Subject: Re: HOW EINSTEIN USED MAXWELL
Date: Thu, 1 Apr 2010 16:31:02 -0700 (PDT)
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