> > This is not true. Maxwell’s equations predict a medium for light to
> > propagate through at a constant speed relative to the stationary
> > background of this medium. In doing so, what is observed as empty
> > vacuum exists this illusive and still undetected medium that gives the
> > properties of the permeability and the permittivity in free space.
> > <shrug>
>
> In geometry if you claim precise dimension you must have precise reference
> points.
> And just how does Maxwell manages to find the reference points of such
> medium?
Not reference points, but reference velocity. Early on, he introduced
a reference velocity for the "stationary frame". Eventually, by the
time we get to the treatise, it got its own name, G; and generally
appears as the G vector (occasionally written just as v).
Since isotropy is not a boost-invariant property, then in an isotropic
frame one can generally define a unique "stationary" frame as the one
where the constitutive relations are isotropic. There is only one
occasion where the frame of isotropy is not unique -- only one
exception to the rule. That was the case found by Einstein. More on
that below.
The point of the title "on the electrodynamics of *moving bodies*" was
in fact to resolve the question of why every frame of references
appeared to be "stationary" (i.e. one where G = 0). The abstract of
Einstein's 1905 SR paper also makes mention of Maxwell's vector,
though not by name.
This is why the alphabet soup of letters, that Maxwell coined, now has
a gap where G used to be. The other Maxwell letters A, B, C, D, E, F,
G, I and J are still used (though C rarely so -- C was "total current"
= J + D_t; I somewhat rarely, we now use M for magnetization instead
of I).
Maxwell's constitutive law read D = KE. But his equation for E was E =
A_t + grad phi + G x B. Our E, nowadays, is defined by E = A_t + grad
phi. So, what Maxwell did -- when expressed in modern notation -- was
write the constitutive law D = K (E + G x B). His E is our E + G x B.
Maxwell's treatment of B and H was never fully resolved, and was
inconsistent (meaning both that (a) he changed his story from year to
year, from paper to paper and even from section to section in the
treatise and (b) he did the analysis of the transformation properties
wrong). On (b), Thomson corrected Maxwell's mistake on (b) and pointed
out that since H has to transform as H - (Delta v) x D under a change
in frame of reference by velocity Delta v, then the correct
constitutive law would NOT be anything like B = mu H, but B = mu (H -
G x D).
Hence, the Maxwell-Thomson account of the field would be that couched
in the constitutive laws
(1) D = epsilon (E + G x B)
(2) B = mu (H - G x D).
where K is rewritten in modern notation as epsilon.
The velocity G marks the motion of the "isotropic frame" -- the unique
frame of reference where the constitutive law becomes isotropic (D =
epsilon E, B = mu H).
Of necessity, in a Galilean-covariant theory, you have to write the
constitutive laws in this way in terms of a medium, since no in-vacuo
law is possible that produces a finite propagation speed (simply
because no finite speed is Galilei-invariant).
For that reason (and for an entirely separate, but much deeper reason
having to do with the self-force and self-energy problem), Maxwell
expressed doubt as to the existence of any such thing as a true honest-
to-goodness vacuum.
The passage over to Relativity does NOT eliminate G. You STILL have
the notion of an "isotropic frame" in any isotropic medium. Outer
space, it bears mentioning, is one such medium, since not even space
is a complete vacuum. (There is no such thing as an absolute vacuum).
The relativistic version of the Maxwell-Thomson relations are the
Einstein-Laub-Minkowski relations,
(3) D + a G x H = epsilon (E + G x B)
(4) B - a G x E = mu (H - G x D)
where a = (1/c)^2, and c is the invariant speed. In this setting, a
distinction is made between the in-medium speed of light V = 1/root(mu
epsilon) and the invariant speed that is often misnamed "the speed of
light".
You can solve these equations for (D, B) vs. (E, H) or for any of the
other modes; e.g. (E, H) in terms of (D, B); (B, E) in terms of (D, H)
or (D, H) in terms of (E, B).
One mode has a solution that remains regular as |G| -> c(!) ...
actually two of the modes do. One of the modes has a solution that
remains regular even as |G| goes ABOVE c (up to c/V^2). The other mode
is regular only up to |G| -> V.
If you go to a vacuum -- i.e. where mu epsilon = a -- then the vector
G will generally become "superfluous". It will generally be the case
that the equations above will be equivalent to their "stationary" (G =
0) forms:
(5) D = epsilon E
(6) B = mu H.
This can only occur if
* mu epsilon = a
* a > 0.
This is the one case where the frame of isotropy is no longer unique.
But it is also worth underscoring that these conditions are necessary
BUT NOT SUFFICIENT. That is, G does NOT necessarily become superfluous
-- not even for a relativistic vacuum. The one case where a vestige of
G remains is in the limit as |G| -> c and V -> c.
Ironically, that one exception is also the one case that provides the
answer to Einstein's question "what's it like to travel alongside a
light beam?"