Maxwell's Equations in a Medium

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SUMMARY

This discussion centers on the implications of Maxwell's equations in a homogeneous isotropic medium, particularly regarding Lorentz transformations and the speed of light in such media. The modified speed of light, denoted as c', is defined as c' = 1/sqrt(με), where c' < c. The conversation highlights that Maxwell's equations do not retain their form under Lorentz transformations when using the vacuum speed of light, indicating that these equations are not Lorentz covariant in a medium. The discussion also emphasizes the existence of a preferred frame, specifically the frame where the medium is at rest, which contradicts the principle of relativity that asserts no preferred inertial frame exists.

PREREQUISITES
  • Understanding of Maxwell's equations in electromagnetism
  • Familiarity with Lorentz transformations and their implications
  • Knowledge of the concepts of refractive index and electromagnetic properties of materials
  • Basic principles of special relativity and its postulates
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  • Research the implications of the Cerenkov effect in relation to light speed in media
  • Study the Fizeau experiment and its significance in the context of relativity
  • Explore the mathematical derivation of Maxwell's equations in moving media
  • Investigate the relationship between relative permittivity, relative permeability, and their dependence on the speed of the medium
USEFUL FOR

Physicists, electrical engineers, and students of electromagnetism seeking to deepen their understanding of the behavior of electromagnetic fields in various media and the implications for special relativity.

  • #31
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) ]
 
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  • #32
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.
 
  • #33
Anamitra said:
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".
 
  • #34
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.
 
  • #35
Anamitra said:
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.
 
  • #36
DaleSpam said:
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.
 
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  • #37
Anamitra said:
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.

Anamitra said:
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.
 
  • #38
Anamitra said:
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.
 
  • #39
Anamitra said:
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.

Anamitra said:
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.
 
  • #40
Tantalos said:
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?

Tantalos said:
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.
 
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  • #41
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]
 
  • #42
Anamitra said:
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.
 
  • #43
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?
 
  • #44
Anamitra said:
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.
 
  • #45
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.
 
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  • #46
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]
 
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  • #47
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?
 
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  • #48
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.
 
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  • #49
Anamitra said:
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.
 
  • #50
Anamitra said:
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.


Anamitra said:
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.
 
  • #51
Is it at all possible to frame boundary conditions wrt a moving dielectric[moving wrt the observer] that conform to the fact that the speed of light in the moving dielectric is different directions is different in accordance with the rules of addition in SR. If the answer is in the affirmative I would request the concerned person/persons to present the exact calculations.[ One may assume that Maxwell's equations in a medium remain invariant or otherwise]
 
  • #52
Anamitra said:
Is it at all possible to frame boundary conditions wrt a moving dielectric[moving wrt the observer] that conform to the fact that the speed of light in the moving dielectric is different directions is different in accordance with the rules of addition in SR. If the answer is in the affirmative I would request the concerned person/persons to present the exact calculations.[ One may assume that Maxwell's equations in a medium remain invariant or otherwise]
Why don't you attempt it on your own first and post the details?
 
  • #53
Persons thinking in the "affirmative direction" do have the responsibility of proving their assertions.Assuming the existence of such persons I have made a simple "request"
 
  • #54
It just seems lazy to me. You have a question but rather than working it out yourself you request others to do it for you. I don't think that anyone else has a responsibility of proving anything to you, particularly since nobody here is your employee and you are not paying anyone for their time and effort.

If you want to learn then do the work. Especially for something as cumbersome as what you are requesting. You shouldn't think that other people have an obligation to spend several hours to answer your questions in precisely the way you request simply because you request it.
 
  • #55
Persons thinking in the affirmative direction may respond.Others don't need to worry.
 
  • #56
I think in the affirmative direction. That does not obligate me to spend hours preparing a problem for your perusal.
 
  • #57
It is not obligatory at all ,on your part, to answer my question--that is to work out the thing.Thank you very much.
 
  • #58
Good, then I will say "yes" and leave the remainder as an exercise for you or someone else.
 
  • #59
DaleSpam said:
It just seems lazy to me.You have a question but rather than working it out yourself you request others to do it for you. I don't think that anyone else has a responsibility of proving anything to you, particularly since nobody here is your employee and you are not paying anyone for their time and effort.

DaleSpam said:
If you want to learn then do the work. Especially for something as cumbersome as what you are requesting. You shouldn't think that other people have an obligation to spend several hours to answer your questions in precisely the way you request simply because you request it.

I had placed a request before the audience to show certain calculations. DaleSpam volunteered to make some assertions/statements which are visible in the interaction from threads #52 through #58

1)It is quite clear that the meaning of the word "REQUEST" is not known to this person.
2) Persons making assertions usually try to prove them without being paid.The statement--"particularly since nobody here is your employee and you are not paying anyone for their time ad effort." is quite interesting. It is clear picture of frustration if not impoliteness.
3) He has accused a member of the forum by saying---"It just seems lazy to me..."Equally true, "unlazy" persons are unwilling to spend "several hours" on long calculations.But they are ready to make unproved assertions-----Thread #56
4) Decorum/politeness is an important issue in a forum--it applies to the case of the Science Adviser concerned
 
  • #60
Anamitra said:
3) He has accused a member of the forum by saying---"It just seems lazy to me..."Equally true, "unlazy" persons are unwilling to spend "several hours" on long calculations.
:smile: Ha ha, you do have me there. I am indeed being just as lazy as you.

But in the end it is your question and you are the one with the motivation to spend the effort here. It is not a simple calculation.
 

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