# Maxwell's Equations in a Medium

Let us consider Maxwell's equations in a homogeneous isotropic medium. We may look for a set of transformations for which the form of the equations remain unchanged[in accordance with the first postulate of Relativity].Of course we get the same Lorentz transformations but with a different value of "c".Here $$c^{'}{=}{\frac{1}{{\sqrt{\mu\epsilon}}}$$
and $$c^{'}{<}{c}$$

Let us re-examine the second postulate of Special Relativity in matter.If a moving source emits light, the speed of light before it strikes the molecules/particles ,is the vacuum speed c=3*10^8 m/s.After interaction with the particles it takes on an average value $$c^{'}$$ and this value[defined to be the average value] should again be independent of the motion of the source.We may develop the Lorentz transformations with $$c^{'}{<}{c}$$

The value $$c^{'}$$ should accommodate fluctuations up to the value of c[and these fluctuations occur in vacuum] but the mean value[of particle velocity] should not exceed $$c^{'}$$. But in Cerenkov effect the mean value of the particle velocity definitely exceeds the value $$c^{'}$$. How does this happen? Should it affect causality in any manner?

Staff Emeritus
should again be independent of the motion of the source.

No, because there is a preferred frame here - the frame where the medium is at rest.

Let us consider a hypothetical experiment. We have two inertial frames S and S' sliding along the x-x' axes.The medium is at rest in the S' frame.We have two torches A and A' which are fixed in the frames A and A' respectively.This means that A is in relative motion wrt to the medium.
We flash the torch A' first. The light ray hits the medium with a speed c and the speed gets reduced to the value v=c/n,where n is the refractive index of the medium.
Now we flash the torch A after putting off the torch A'.Light again moves at the rate c/n wrt to the medium, which is in motion wrt to the unprimed frame. Observer in the unprimed frame should see light moving in the medium at a rate $${u}{\neq}{c/n}$$

Therefore,

Refractive index of the medium as calculated the second observer=c/u

But ,$${RI}{=}{\sqrt{\frac{\mu\epsilon}{\mu_{0}\epsilon_{0}}}}$$

Does this quantity change with the speed of the medium? I am not sure.

Dale
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The speed of light in a moving medium follows the results predicted by relativity. The seminal experiment on this topic is the Fizeau experiment confirming Frensel drag. This is probably the first experiment really showing the way to relativity, several decades prior to Einstein.

This clearly shows that the product relative permeability*relative permittivity should change with the speed of the medium.

For non relativistic motion we have:

$${v}{=}{c}{/}{n}{\pm}{v_{m}}{[}{1}{-}{1}{/}{n^{2}}{]}$$

By using the relations,

$${v}{=}{c}{\sqrt{\frac{{{\mu}_{0}}{{\epsilon}_{0}}}{{{\mu}_{v}}{{\epsilon}_{v}}}}$$

And,

$${n}{=}{\sqrt{\frac{\mu\epsilon}{{{\mu}_{0}}{{\epsilon}_{0}}}}}$$

We may find a relationship between the product relative permeability*relative permittivity at different speeds.

In fact by keeping a chunk of a dielectric on a table and then by running backwards or forwards we could change the electrical properties[pertaining to relative permittivity and rel. permeability] of the medium.If somebody could manage to attain relativistic speed a piece of glass would look like a piece of diamond.Of course the above formulas have to be modified.

Let us have another look at Maxwell's equations in a medium[homogeneous and isotropic].These equations do not retain their form if we apply the Lorentz Transformations using the vacuum speed of light.This could mean that by loading a piece of bakelite into the carriage of a train we could upset Gauss Law or Ampere's Circuital Law

That would be interesting.

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Observations

Maxwell's Equations in a medium are not Lorentz Covariant. One may try out the transformations[for a homogenous linear medium] using the Lorentz transformations (and the Field Transformations) using the vacuum speed of light. The divergence E and curl B equations change their form while the other two retain their form.As a whole the equations are not Lorentz covariant.The transformations do not indicate the dependence of relative permittivity*relative permeability on the relative speed between frames.

If one uses the Lorentz transformations and the field transformations using the reduced value of c[$${{=}\frac{1}{\sqrt{{\mu\epsilon}}}}$$] the equations do not change their form.But having one set of transformations in vacuum and another for a medium is quite peculiar.Moreover there is no indication of any dependence of the product relative permittivity*relative permeability on relative speed between two frames.

Dale
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Maxwell's Equations in a medium are not Lorentz Covariant.
Obviously not, the medium establishes a prefered frame.

Given all that[I mean the assertions of thread#7],if I keep a chunk of a dielectric on a table and start running backwards or forwards Gauss Law or Curl B law should change.

Dale
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It also depends if you are doing your equations using the total charge and current or the free charge and current. It also depends if your medium is isotropic and the direction of any anisotropy relative to the motion. When you introduce a medium there are all sorts of things that you have to take care of.

So what? This is all reasonably well known, but it is mathematically cumbersome so it is not used in practice very often.

The Principle of Relativity:

The laws of physics are the same in all INERTIAL frames of reference.No preferred inertial system exists

Special principle of relativity: If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K.

– Albert Einstein: The Foundation of the General Theory of Relativity, Part A, §1

If Maxwell's equations in a medium are treated to be a set of laws they should have identical form in all inertial frames of references[bearing in our minds that there is no preferred frame among the inertial systems,according to the principle of relativity.

Dale
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You can derive the general form for Maxwell's equations in a moving medium simply by giving an arbitrary boost to the standard equations for a medium at rest. Be warned, is a little ugly and it is not the textbook form you are used to. However, this form is Lorentz covariant and would be considered the fully relativistic "law of physics" governing classical EM in a medium. For a medium at rest or for vacuum it reduces to the usual form.

Well, then how do you incorporate the dependence of $$\sqrt{\frac{1}{\epsilon\mu}}$$ on the relative speed between a pair of inertial frames?An arbitrary boost will not give you such a relation.

In fact if the form of equations change there is a lot of doubt as to whether you will get the wave equation at all! If the "simplest form"[of Maxwell's equations in matter] remains unchanged you will always get the same wave equation in each inertial frame

Dale
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Well, then how do you incorporate the dependence of $$\sqrt{\frac{1}{\epsilon\mu}}$$ on the relative speed between a pair of inertial frames?An arbitrary boost will not give you such a relation.

In fact if the form of equations change there is a lot of doubt as to whether you will get the wave equation at all! If the "simplest form"[of Maxwell's equations in matter] remains unchanged you will always get the same wave equation in each inertial frame
I seriously doubt both of those claims. Can you derive either one of them?

We consider a dielectric at rest in an inertial frame of reference and write Maxwell's equations in the absence of free charges and free currents.

$${div} {E}{=}{0}$$
$${div}{B}{=}{0}$$
$${curl}{E}{=}{-}\frac{{\partial }{E}}{\partial t}$$
$${curl}{B}{=}{\epsilon\mu}\frac{\partial E}{\partial t}$$

A primed frame is assumed to move in the x=x' direction with a speed v wrt to the umprimed frame where the dielectric is at rest.
Lorentz Transformations

$${x^{'}}{=}{\gamma}{[}{x}{-}{vt}{]}$$
$${y^{'}}{=}{y}$$
$${z^{'}}{=}{z}$$
$${t^{'}}{=}{\gamma}{[}{t}{-}{vx}{/}{c^{2}}{]}$$

We have

$$\frac{\partial}{\partial x}{=}{\gamma}{[}{\frac{\partial}{\partial x^{'}}}{-}{v}{/}{c^{2}}{\frac{\partial}{\partial t^{'}}}{]}$$
$${\frac{\partial}{{\partial}{y}}}{=}{\frac{\partial}{{\partial}{y}^{'}}}$$
$${\frac{\partial}{{\partial}{y}}}{=}{\frac{\partial}{{\partial}{y}^{'}}}$$
$$\frac{\partial}{\partial t}{=}{\gamma}{[}{\frac{\partial}{\partial t^{'}}}{-}{v}{\frac{\partial}{\partial x^{'}}}{]}$$

Field transformations:

$${E_{x}}{=}{{E^{'}}_{x}}$$
$${E_{y}}{=}{\gamma}{[}{{E^{'}}_{y}}{+}{v}{{B^{'}}_{z}}{]}$$
$${E_{z}}{=}{\gamma}{[}{{E^{'}}_{z}}{-}{v}{{B^{'}}_{y}}{]}$$

$${B_{x}}{=}{{B^{'}}_{x}}$$
$${B_{y}}{=}{\gamma}{[}{{B^{'}}_{y}}{-}{{v}{/}{c^{2}}}{{E^{'}}_{z}}{]}$$
$${B_{z}}{=}{\gamma}{[}{{B^{'}}_{z}}{+}{{v}{/}{c^{2}}{{E^{'}}_{y}}{]}$$

c=Vacuum speed of light
$${\gamma}{=}{\sqrt{\frac{1}{{[}{1}{-}{{v^{2}}{/}{c^{2}}{]}}}}$$

Using the above relations one may show that the first three equations remain invariant[if free charge density is not zero the first equation also changes]
The fourth equation transforms to:

$${\frac{{\partial}{B^{'}}_{z}}{\partial {{y}^{'}}}}{-}{\frac{{\partial}{B^{'}}_{y}}{{\partial }{z^{'}}}{=}{\epsilon\mu}{\frac{\partial{E^{'}}_{x}}{{\partial}{t_{'}}}{-}{v}{\mu\epsilon}{\frac{\partial{E^{'}}_{x}}{{\partial}{x_{'}}}}{-}{{v}{/}{c^{2}}}{\frac{\partial{E^{'}}_{y}}{{\partial}{y_{'}}}{-}{{v}{/}{c^{2}}{\frac{\partial{E^{'}}_{z}}{{\partial}{z_{'}}}$$

$${\frac{{\partial}{B^{'}}_{x}}{\partial {{z}^{'}}}}{-}{{\gamma}^{2}}{[}{1}{-}{{v^{2}}{\epsilon\mu}{]}{\frac{{\partial}{B^{'}}_{z}}{{\partial }{x^{'}}}{=}{{\gamma}^{2}}{[}{\mu\epsilon}{-}{{v^{2}}{/}{c^{4}}}{]}{\frac{\partial{E^{'}}_{y}}{{\partial}{t_{'}}}{-}{{\gamma}^{2}}{v}{[}{\epsilon\mu}{-}{{1}{/}{c^{2}}}{]}{\frac{\partial{B^{'}}_{z}}{{\partial}{t_{'}}}}{-}{{\gamma}^{2}}{v}{[}{\epsilon\mu}{-}{{1}{/}{c^{2}}}{]}{\frac{\partial{E^{'}}_{y}}{{\partial}{x_{'}}}$$

$${{\gamma}^{2}}{[}{1}{-}{{v^{2}}{\epsilon\mu}{]}{\frac{{\partial}{B^{'}}_{y}}{\partial {{x}^{'}}}}{-}{\frac{{\partial}{B^{'}}_{x}}{{\partial }{y^{'}}}{=}{{\gamma}^{2}}{[}{\mu\epsilon}{-}{{v^{2}}{/}{c^{4}}}{]}{\frac{\partial{E^{'}}_{z}}{{\partial}{t_{'}}}{-}{{\gamma}^{2}}{v}{[}{\epsilon\mu}{-}{{1}{/}{c^{2}}}{]}{\frac{\partial{B^{'}}_{y}}{{\partial}{t_{'}}}}{-}{{\gamma}^{2}}{v}{[}{\epsilon\mu}{-}{{1}{/}{c^{2}}}{]}{\frac{\partial{E^{'}}_{z}}{{\partial}{x_{'}}}$$

If one replaces $${\mu\epsilon}$$ by $${{\mu}_{0}}{{\epsilon}_{0}}{=}{{1}{/}{c^{2}}$$ he gets the invariant curl B equation[vacuum equation]. This could be treated as a check.

From the above equations I could not get the equation:$${{\nabla}^{2}}{E}{=}{\epsilon\mu}{\frac{{{\partial}^{2}}{E}}{{\partial}{t^{2}}}$$ or the equation

$${{\nabla}^{2}}{B}{=}{\epsilon\mu}{\frac{{{\partial}^{2}}{B}}{{\partial}{t^{2}}}$$

The speed dependence of $${\epsilon\mu}$$ is not discernible.

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Staff Emeritus
None of that is correct.

You can't just plug in mu's and epsilons. A moving magnitization gives rise to a polarization and vice versa.

Let me repeat: in this problem, there is a preferred frame: the one where the medium is at rest.

If the First Postulate of Special Relativity is correct then there should be no preferred frame among the INERTIAL FRAMES.One may refer to thread #10

If a chunk of a material is at rest in some fame it simply has different speeds in different inertial frames. The value of E or B could be different in different frames of reference. That has nothing to do with the preference of one frame over another.If a LAW has a DIFFERENT FORM in some frame of reference against the others it becomes a preferred frame.Individual variables can have different values in different frames,inertial or non inertial.

Incidentally there is no preferred frame among the INERTIAL FRAMES. This happens to be a well known principle

Staff Emeritus
It sounds like you are trying to argue SR is not correct. Is that the case?

I have no reason to say that SR is incorrect. I have simply tried to emphasize the fact that Vanadiun 50 and DaleSpam are incorrect when they say that a chunk of a material [at rest]considered in an inertial frame of reference makes it a a preferred frame[Threads #2 and #7].This statement clearly defies Special Relativity[the first Postulate]

I have clarified my stand in thread #17.

Dale
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2020 Award
See sections 2 and 3 of
http://arxiv.org/PS_cache/arxiv/pdf/1006/1006.3118v1.pdf [Broken]

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atyy
But having one set of transformations in vacuum and another for a medium is quite peculiar.

Only the vacuum form is fundamental. And only Maxwell's equations in vacuum are fundamental (in the sense of having Lorentz covariance). Maxwell's equations in a medium (which are not Lorentz covariant, since they may include things like a frequency dependnet speed of light in a medium) are an approximation to a case where Maxwell's equations in vacuum interact with other fundamental fields (like the electrons, protons etc.).

Regarding the Paper in Thread #20

We consider the relation[equation (9) of the paper]:
$${G}{=}{\chi}{(}{*}{F}{)}$$
When we go over to a different frame
G transforms to $${G^{'}}$$
$${\chi}$$ should transform to $${{\chi}^{'}}$$
Indeed $${{(}{*}{F}{)}$$ transforms to $${{(}{*}{F^{'}}{)}$$
We should have,
$${G^{'}}{=}{{\chi}^{/}}{(}{*}{F^{'}}{)}$$

$${G^{'}}{=}{\chi}{(}{*}{F^{'}}{)}$$
As indicated by equation equation (16) Section III of the paper.
I find this difficult to comprehend and I am requesting the audience to comment on the issue.
[NB: All quantities in equation (9) of the paper are individually tensors]

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DrDu
I have my doubts on the validity of the approach of the article cited by DaleSpam.
It is highly formal and, more importantly, it assumes that the relation between the tensors G and F is local (i.e. the epsilon-mu approach you are also using).
That this is not sufficient is shown e.g. in Landau Lifshetz "Electrodynamics of continuous media" or here: http://ufn.ru/en/articles/2006/10/c/references.html
The non-locality will also be important in the problem you are considering, as, even if in the rest frame of the medium, the response is local,i.e. excitation and de-excitation both take place at x, this will not be so in another frame: An atom that gets excited at point x will travel to point x+vt before it emits the radiation again.

DrDu