Maxwell's Equations in a Medium

[Anamitra]

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?

 

[Vanadium 50]

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.

 

[Anamitra]

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]

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.

 

[Anamitra]

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 LawThat would be interesting.

 

[Anamitra]

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]

Maxwell's Equations in a medium are not Lorentz Covariant.

Obviously not, the medium establishes a preferred frame.

 

[Anamitra]

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]

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.

 

[Anamitra]

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]

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.

 

[Anamitra]

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]

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?

 

[Anamitra]

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.

 

[Vanadium 50]

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.

 

[Anamitra]

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

 

[Anamitra]

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

 

[Vanadium 50]

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

 

[Anamitra]

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]

See sections 2 and 3 of
http://arxiv.org/PS_cache/arxiv/pdf/1006/1006.3118v1.pdf

 

[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.).

 

[Anamitra]

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^{'}}{)}

Instead of
{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]

 

[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]

Also, it should be clear that you can no longer assume epsilon and mu to be scalar quantities but tensors if the substance is moving, even if the substance as such is isotropic.

 

[Dale]

I have my doubts on the validity of the approach of the article cited by DaleSpam.

I wasn't terribly happy with it either, but it was the best that I could find in a freely accessible format. I am sure there are much better textbooks etc.

 

[stevenb]

I wasn't terribly happy with it either, but it was the best that I could find in a freely accessible format. I am sure there are much better textbooks etc.

I recommend the following text which considers fields in media and curved space, using differential geometery.

"Foundations of Classical Electromagnetics", by Hehl and Obukhov

http://www.google.com/products/cata...og_result&ct=result&resnum=1&ved=0CEAQ8wIwAA#

 

[DrDu]

I think the easiest way is to proceed as follows:
Instead of working with the fields and polarizations or the like, the properties of the matter can be taken care of considering a relation of the 4-current vector of the matter as a function of the vector potential in k space:
J_\nu(k_\mu)=Q_{\nu}^\rho(k_\mu) A_\rho (k_\mu)
The transformation of J, A and k transform standard under Lorentz trafos, so the transformation of Q can also be derived. If wanted, the relation of P and F can be derived from their relation to J and A, respectively.

 

[grav-universe]

I have worked on this recently. We can imagine two separate sets of a "medium contained within a box", with the boxes moving through the vacuum of space at a relative speed to each other. Each box also contains an observer that is stationary to the medium. To each observer, then, the other entire box contracts by 1/γ as with ordinary SR, and all observations are made the same as with SR in this case. Then if we figure that a particle, even a photon, travels at less than c through one of the boxes according to the observer in that box, c/n, it's speed relative to the other observer will be determined using relativistic addition or subtraction of speeds as usual, the same as if the particle were moving through the vacuum of space at less than c. It is all mathematically consistent, so another observer moving through one of the boxes would also use the relativistic addition of speeds to find an anisotropic speed of the medium. The only thing that could indicate otherwise would be another actual M-M type experiment performed in fast running water or other medium, but this has already been done and is the Fizeau experiments, which were found to be consistent with SR, even before SR was born as Dalespam mentioned earlier.

 

[DrDu]

I tried to work out that relation:
I use the B, P, E approach as in Landau Lifshetz. That is \mathbf{B}=\mathbf{H}
and \mathbf{P}(\omega,\mathbf{k})=(\hat{\epsilon}(\omega, \mathbf{k})-\hat{1})\mathbf{E}(\omega,\mathbf{k}).
Bold face signifies 3 vectors with roman indices ranging from 1 to 3 and a hat signifies a 3x3 matrix. I use Einstein sum convention
Using this convention, \rho=-i \mathbf{k}\cdot \mathbf{P}, \mathbf{j}=-i\omega \mathbf{P} and
\mathbf{E}=-i \mathbf{k} \phi +i\omega \mathbf{A}.
Using all this it is easy to work out
<br /> j^\nu(k^\mu)=Q^{\nu}_\rho(k^\mu) A^\rho (k^\mu) <br />
j^\mu=(\rho, \mathbf{j}^T)^T
A^\mu=(\phi, \mathbf{A}^T)^T
k^\mu=(\omega,\mathbf{k}^T)^T
I find
Q^\mu_\nu=-\left( \begin{array}{l}\mathbf{k}^T\\ \omega \mathbf{1} \end{array} \right) (\hat{\epsilon}-\hat{1}) \left(\mathbf{k}, -\omega \hat{1}\right)
In the case of a homogeneous, nonchiral, isotropic material,
the dielectric tensor epsilon can be split into a longitudinal part
\mathbf{k}^T \cdot \hat{\epsilon}\mathbf{k}=\epsilon_l
and a transversal part \epsilon_t.
Then we get
<br /> Q^\mu_\nu=\left(\begin{array}{ll} k^2 (\epsilon_l-1) &amp; \mathbf{0}^T\\<br /> \mathbf{0} &amp; -\omega^2 (\hat{\epsilon}-\hat{1}) \end{array} \right)<br />
Inserting expression for the current four vector into the inhomogeneous Maxwell equation, yields a covariant expression for the propagation of light. .

 

[Anamitra]

Special Relativity requires that the form of a law should remain unchanged when it is transformed from one inertial frame to another. Now a law is usually expressed in the form of a differential equation involving certain variables. These variables I believe, are the exact values—say functions of space and time.
How would the situation be like if we were to consider the statistical averages instead of the exact values? We investigate the situation with reference to Maxwell’s equations in vacuum . In two inertial frames K and K’ they have identical forms. We may derive Maxwell’s equations in matter in each frame from the identical Maxwell’s equation in vacuum in each frame. We should get the same results unless the result of averaging process is influenced by the relative motion of the inertial frames.

We consider some variable,say E, in a general way:
{&lt;}{E}{&gt;}{=}{\frac{1}{{\Delta}{V}} {\int E}{d}{V}
Now,
{&lt;}{E}{&gt;} is a set function and not a point function. We may forcefully convert it to a point function by considering a sphere of small radius at each point.

We have
{\frac {{\partial }{{&lt;}{E}{&gt;}}}{\partial x}}{=}{&lt;}{\frac{\partial E}{\partial x}}{&gt;}-------- (2)
[One may refer to the proof in Igor Tamm’s "Fundamentals of Electricity",Chapter 2[Dielectrics],Section 2.6[Micro and Macroscopic Values of Physical Quantities]
[Relation 2, helps in preserving the form of the law in the same frame of reference[when we pass to the statistical form,from the exact value form] if it is expressed as a linear differential equation. Thankfully the fundamental laws have a linear form when referred to the inertial frames. It is relevant to note that {{&lt;}{a}{&gt;}}^{2}{\neq}{&lt;}{a}^{2}{&gt;}]
Is the same relation ie relation (2) going to be valid if we consider the averaging process from a moving frame?We may ,for example,consider a chunk of dielectric moving wrt the observer's frame.The spheres would get compressed along the direction of motion and the Lorentz factor would make its appearance in a significant manner. Any failure of relation (2) would imply the fact that the statistical form of the law may change when we pass from one inertial frame to another, though the same law remains unchanged so far as the exact values are concerned.

Interesting to note that that in experimental evaluations we get only the average values of a quantity and not the exact ones.
If we try to make things too precise we land into the quantum mechanical area, where an exact value is usually treated with a certain amount suspicion.

 

[Anamitra]

The Logical Alternatives:

1)The averaging process does not produce different results in different reference frames.The problem with this alternative is that it should produce Maxwell's equations in matter in the same form in all frames[inertial].That would defy facts like the speed of light in a medium should depend on the state of motion of the medium.

2)If the averaging process produces different results in different frames[inertial], the First postulate of Special Relativity may not hold for average values-----possibly this alternative has a poorer acceptability since the gadgets measure the average values and not the exact values in experiments.

I would request the audience to comment on the alternatives.

[The success of the relation below [for moving frames]
{\frac {\partial{&lt;}{E}{&gt;}}{\partial x}}{=}{&lt;}{\frac {\partial E}{\partial x}}{&gt;}
would favor alternative (1) ]

 

[Anamitra]

Let us have a look at the relations:

{&lt;}{a^{2}}{&gt;}{\neq}{{&lt;}{a}{&gt;}}^{2}

{&lt;}{(}{\frac {\partial E}{\partial x}}{)}^{2}{&gt;}{\neq}{{&lt;}{\frac {\partial E}{\partial x}}{&gt;}}^{2}

{&lt;}{(}{\frac {\partial E}{\partial x}}{)}^{2}{&gt;}{\neq}{\frac {\partial {&lt;}{E}{&gt;}}{\partial x}}^{2}

For a nonlinear equation the exact value form and the average value form would have different appearances in the same frame.Fortunately the fundamental equations are linear at least when we consider therm in the inertial systems.The exact value form and the average value forms have the same appearance.

But if the average value form changes on transformation the same trouble should reappear in other frames.

 

[Dale]

The problem with this alternative is that it should produce Maxwell's equations in matter in the same form in all frames[inertial].That would defy facts like the speed of light in a medium should depend on the state of motion of the medium.

How do you arrive at that conclusion? As long as the laws are covariant (guaranteed by casting them in terms of tensors) then nothing is "defied".

 

[Anamitra]

If a light ray is flashed in different directions on a moving dielectric[having uniform motion wrt to some inertial frame,K] the speed of light inside the dielectric should be different in different directions,for observations made from K.The relativistic law of addition ensures such an observation[see thread #28] Now if Maxwell's equations[in matter] have the same form as for a stationary dielectric[reference frame containing a stationary one, to be specific],such anisotropy cannot be predicted.
Important to note that the covariance of a law reduces to invariance when we think in terms of the inertial frames.The subtle difference between covariance and invariance should be taken care of.
One might object to a direct transformation of Maxwell's equations in matter from one inertial frame to another. The best way to over-rule such an objection would be to consider the vacuum equations from the two frames and start the averaging process to get the matter equations.
An important issue to consider is the question ---"What is a law?" or"What is a fundamental law?"---when one thinks of Lorentz covariance[wrt inertial frames].The issue came up in thread #21.
I have tried to analyze the situation in terms of the exact value laws and the statistical laws.

 

[Dale]

Now if Maxwell's equations[in matter] have the same form as for a stationary dielectric[reference frame containing a stationary one, to be specific],such anisotropy cannot be predicted.

As long as you write the equations in terms of tensors you are guaranteed that they will have the same form in all reference frames, including any required anisotropies due to motion.

 

[Anamitra]

As long as you write the equations in terms of tensors you are guaranteed that they will have the same form in all reference frames, including any required anisotropies due to motion.

We consider Maxwell’s equations in covariant form:
{{F}^{\mu\nu}}_{;{\mu}}{=}{-}{J}^{\nu} -------------------- (1)
{F}_{{\mu\nu}{;}{\lambda}}{+}{F}_{{\nu\lambda}{;}{\mu}}{+}{F}_{{\lambda\mu}{;}{\nu}}{=}{0}
---------------------------- (2)
Using the relations:
{F}_{\lambda\kappa}{=}{g}_{\lambda\mu}{g}_{\kappa\nu}{F}^{\mu\nu}
{{A}^{\mu\nu}}_{{;}{\mu}}{=}{\frac{1}{\sqrt g}}{\frac{\partial {(}{\sqrt g}{A}^{\mu\nu}{)}}{\partial x^{\mu}}}
We may write,
{\frac{\partial {(}{\sqrt g}{F}^{\mu\nu}{)}}{\partial x^{\mu}}}{=}{-}{\sqrt g}{J^{\nu}} ---- (3)
{\frac{\partial {F}_{\mu\nu}}{\partial x^{\lambda}}}{+}{\frac{\partial {F}_{\nu\lambda}}{\partial x^{\mu}}}{+}{\frac{\partial {F}_{\lambda\mu}}{\partial x^{\nu}}}{=}{0} --------------- (4)
[Reference:Steven Weinberg, Gravitation and Cosmology, John Wiley and Sons(Asia) Pte Ltd,2004,page 124-125]
Equations (1) and (2) have the same appearance in all frames of reference [not necessarily inertial]
The same is true for (3) and (4)
Now we consider two coordinate systems where g has different functional forms. For arguments sake let us assume that in one frame {g}{=}{{(}x^{0}{)}}^{2}{+}{3}{x^{1}}{x^{2}}{+}{5}{x^{2}}{x^{3}} while in the other {g}{=}{{(}x^{0}{)}}^{2}{+}{7}{{(}x^{2}{)}}^{2}{+}{8}{x^{1}}{x^{3}}
When we make the above substitutions into equation (3) the form of equation changes . Covariance does not guarantee invariance, except for inertial frames. An arbitrary accelerating frame may be described by suitable metric coefficients[similar to those characterizing a gravitational field]

One may consider the following link to interpret the equivalence between an accelerating frame and a gravitational field.
https://www.physicsforums.com/showpost.php?p=2845093&postcount=8
Here we are working in flat spacetime but if one wishes to consider an accelerating frame an equivalent curved spacetime situation is created.

 

[Dale]

Equations (1) and (2) have the same appearance in all frames of reference [not necessarily inertial]
The same is true for (3) and (4)

Exactly.

When we make the above substitutions into equation (3) the form of equation changes . Covariance does not guarantee invariance, except for inertial frames.

All measurements are scalars, and covariance of the laws does guarantee invariance of all scalars determined by the laws.

 

[DrDu]

Special Relativity requires that the form of a law should remain unchanged when it is transformed from one inertial frame to another. Now a law is usually expressed in the form of a differential equation involving certain variables. These variables I believe, are the exact values—say functions of space and time.
How would the situation be like if we were to consider the statistical averages instead of the exact values? We investigate the situation with reference to Maxwell’s equations in vacuum . In two inertial frames K and K’ they have identical forms. We may derive Maxwell’s equations in matter in each frame from the identical Maxwell’s equation in vacuum in each frame. We should get the same results unless the result of averaging process is influenced by the relative motion of the inertial frames.

We consider some variable,say E, in a general way:
{&lt;}{E}{&gt;}{=}{\frac{1}{{\Delta}{V}} {\int E}{d}{V}
Now,
{&lt;}{E}{&gt;} is a set function and not a point function. We may forcefully convert it to a point function by considering a sphere of small radius at each point.

First you write you want to consider statistical averages (which trivially commutes with spatial derivation), but then you use averaging over some space region. I do not quite understand why.

 

[Tantalos]

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.

The first equation does not imply that n has changed. You cannot use the speed v to calculate a new refraction index.

If a light ray is flashed in different directions on a moving dielectric[having uniform motion wrt to some inertial frame,K] the speed of light inside the dielectric should be different in different directions,for observations made from K.The relativistic law of addition ensures such an observation[see thread #28] Now if Maxwell's equations[in matter] have the same form as for a stationary dielectric[reference frame containing a stationary one, to be specific],such anisotropy cannot be predicted.

If the dielectric is moving, then the solution to the Maxwell equations will be diffrent than it is in the stationary case, because boundary conditions are changed. It has nothing to do with relativity.

 

[Anamitra]

The first equation does not imply that n has changed. You cannot use the speed v to calculate a new refraction index.

The value of "n" observed in the first equation is the refractive index measured by a person at rest wrt to the dielectric. It should not change. What about a person moving wrt to it?

If the dielectric is moving, then the solution to the Maxwell equations will be diffrent than it is in the stationary case, because boundary conditions are changed. It has nothing to do with relativity.

One can always think of changed boundary conditions when the dielectric is moving
The observer [wrt whom the dielectric is in motion] should always be able to predict the speed of light in the medium[and of course in different directions] using the knowledge of Special Relativity[Addition rule]. The result he gets from this process should tally with the solution of Maxwell's equations in the moving dielectric,using the changed boundary conditions.
What are these boundary conditions that should match with the Special Relativity considerations[addition of velocities]?Anisotropy should have a place in the solution.

 

[Anamitra]

On Scalars and Covariance:
We consider an equation:
f(t,x,y,z)=0
The right hand side is a scalar[zero].So the left hand side is also a scalar and it should not change with some transformation. But the form of the function f(y,x,y,z) can always change.

[It is important to remember that the relation f(t,x,y,z) is an equation and not an identity]

 

[Dale]

We consider an equation:
f(t,x,y,z)=0
The right hand side is a scalar[zero].So the left hand side is also a scalar and it should not change with some transformation. But the form of the function f(y,x,y,z) can always change.

No, if f is a scalar then it transforms as a scalar so f=0 in all coordinate systems.

 

[Anamitra]

Invariance of form and the invariance of value have different connotations.

We consider the equation: f(t,x,y,z)=0
Where,
f(t,x,y,z)=Atx+Byz A,B are constants

We carry out the following transformations:
{x}{=}{x}^{&#039;}{+}{y}^{&#039;}
{t}{=}{t}^{&#039;}
{y}{=}{z}^{&#039;}
{z}{=}{z}^{&#039;}

Original equation:
Atx+Byz=0

Transformed equation:
At'[x'+y']+By'z'=0

The value of the function does not change. But what about its form?

 

[Dale]

Invariance of form and the invariance of value have different connotations.

We consider the equation: f(t,x,y,z)=0
Where,
f(t,x,y,z)=Atx+Byz A,B are constants

We carry out the following transformations:
{x}{=}{x}^{&#039;}{+}{y}^{&#039;}
{t}{=}{t}^{&#039;}
{y}{=}{z}^{&#039;}
{z}{=}{z}^{&#039;}

Original equation:
Atx+Byz=0

Transformed equation:
At'[x'+y']+By'z'=0

The value of the function does not change. But what about its form?

It is a tensor equation, so its form is the same in all coordinate systems: f=0.

A tensor is not the same as its representation in some specific coordinate basis. As you change the basis you clearly change the components of the tensor, but not the tensor itself. The fact that the components are different in different bases is trivially true. I think you are "missing the forest for the trees", or more directly, mistaking the tensor for its components or its algebraic expression in some coordinate system.

 

[Anamitra]

A tensor equation[or the component equations to be specific] has the same form in all frames of reference--- absolutely true.But if we cast the equation[I mean, the component equations] in
terms of coordinates,derivatives related to coordinates or other physical quantities[like E and B] do we get the same differential equation or the same set of differential equations[by same ,I mean the same form]?One may refer to the geodesic equation itself.The tensor equation continues in its own elegant form while the differential equations related to the coordinates,metrics etc seem to have different appearances in different frames.In my previous thread[#44] I simply wanted to highlight such an instance[hypothetical]One may also consider the link:https://www.physicsforums.com/showpost.php?p=2979702&postcount=1

In contrast Maxwell's equations in the inertial frames have the same set of differential equations in terms of coordinates ,derivatives,physical quantities like E and B[so far as the form is considered]and not only the same tensor equations.

 

[Anamitra]

I would request the audience to insert comments/postings related to covariance/invariance into the thread "On Covariance and Invariance"---- https://www.physicsforums.com/showthread.php?t=446835.
We would be referring to those ideas here whenever required.
[One may of course post such ideas here directly if he considers it essential]

 

[Dale]

In some coordinate bases it may not even make sense to speak of E and B. For instance, if you are using a null basis, then which components are E and which are B. You need to stop thinking in terms of the components and think in terms of the underlying geometry. That is the purpose of tensors.

Once you have expressed a law in terms of tensors you are done. The law is automatically Lorentz covariant (and generally covariant). You are free to express the equation in terms of any weird coordinates you like, but that expression is not the law, it is the components of the law (tensor) in the coordinate basis. Do you understand the distinction?

 

[Anamitra]

Let us come down to the form of Maxwell's equations in flat spacetime. The differential equations have the same form in all inertial frames.But in curved space time the differential equations change notwithstanding the fact that the tensor equations continue in full force of elegance.One is not supposed to apply the same traditional form of Maxwell's equations when working in curved spacetime.

If we are to solve a boundary value problem [say in curved spacetime]we have to consider coordinates ,differential equations etc.----the so called weird things[referred to in thread #47].Can we solve these problems directly from the tensor equation[expressing the law independent of frames]?

The laws of physics have an elegant[and unchanging] form when tensor equations are used.There is no reason to deny this issue at all.But when we think in terms of application we must come down to the individual equations/component equations.In such an event we do not get invariant forms of differential equations despite the elegant covariant form of the law.

 

[Tantalos]

The value of "n" observed in the first equation is the refractive index measured by a person at rest wrt to the dielectric. It should not change. What about a person moving wrt to it?

Then the rule of addition of velcities applies.

One can always think of changed boundary conditions when the dielectric is moving
The observer [wrt whom the dielectric is in motion] should always be able to predict the speed of light in the medium[and of course in different directions] using the knowledge of Special Relativity[Addition rule]. The result he gets from this process should tally with the solution of Maxwell's equations in the moving dielectric,using the changed boundary conditions.
What are these boundary conditions that should match with the Special Relativity considerations[addition of velocities]?Anisotropy should have a place in the solution.

The boundary condition is that the B and E field functions must be continuous at the boundary of the medium. In stationary case we need only to consider the time dependent part of the functions outside the medium, but when the medium is moving both time- and position-dependent parts need to be considered.
In relativity the observer can't tell whether the medium is moving and he is at rest or the opposite is happening. The solution to the Maxwell equations will be the same in both cases.

 

[Dale]

The laws of physics have an elegant[and unchanging] form when tensor equations are used.There is no reason to deny this issue at all.

Exactly. Which is why the tensor form is the one considered to be the law of physics.

But when we think in terms of application we must come down to the individual equations/component equations.In such an event we do not get invariant forms of differential equations despite the elegant covariant form of the law.

Yes, and proper choice of coordinates can mean the difference between being able so dolve those differential equations or not.