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- Thread starter Anamitra
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- #51

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- #52

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Why don't you attempt it on your own first and post the details?

- #53

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- #54

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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

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Persons thinking in the *affirmative direction* may respond.Others don't need to worry.

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- #58

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Good, then I will say "yes" and leave the remainder as an exercise for you or someone else.

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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.

I had placed a request before the audience to show certain calculations. DaleSpam

1)It is quite clear that the meaning of the word "REQUEST" is not known to this person.

2) Persons making assertions

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

4) Decorum/politeness is an important issue in a forum--it applies to the case of the

- #60

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Ha ha, you do have me there. I am indeed being just as lazy as you.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 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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Why don't you attempt it on your own first and post the details?

"affirmative direction"do have the responsibility of proving their assertions.Assuming the existence of such persons I have made a simple"request"

Persons thinking in theaffirmative directionmay respond.Others don't need to worry.

Isotropy and homogeneity of space breaks down in this situation----we simply cannot apply the Lorentz Transformations[or the velocity addition rules of Special Relativity]

We should not apply Special Relativity if we have dielectrics around us!

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- #62

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I am not rehashing this old conversation with you. Go get a standard textbook on the subject.

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I am not rehashing this old conversation with you. Go get a standard textbook on the subject.

I simply don't understand what you mean to say.

Special Relativity is an APPROXIMATE science in the physical world JUST like classical physics

It[SR] puts up a better performance in regard to speeds close to/comparable to the the value"c"--that too with the idealized assumptions of homogeneity of space[and time]and isotropy of space. These conditions are too remote in view of physical considerations.

Assuming such conditions with reckless abandon may lead to erroneous conclusions from calculations/experiments. A dangerous thing to happen!

In this case[the present problem with the dielectric with boundary conditions etc] it falis miserably as expected----the conditions of homogeneity and isotropy of space are not present!

You may go through the following link:

https://www.physicsforums.com/showthread.php?t=545002

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- #64

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Of course special relativity is an approximation, and so is all the rest. But as you should expect, people are suspicious of any ideas that say that maxwell's equations, in sufficiently flat spacetime seem to be incompatable with special relativity. This seems to be your thesis, correct?

In any case, your maxwell's equations used in post #14 are not the right ones to use. A dielectric medium may have no free charge, howwever it is not in general free of charge nor current under changing fields. Therefore the divergence of E is nonzero in Gauss law should be nonzero and you must include the term for current in Ampere's law.

-------------------

If you really want to get down to brass tacks, the faraday tensor given in the post Dale came up with is somewhat in error. A two form in four space does not partition into vectors in three dimensional subspaces given as vectors E and B. This is crazy. I know; I have entire thesis that is a lot of very upsetting, and embarassing garbage. The partitions should be three dimensional two forms [itex]\epsilon_{ab}^c E_c[/itex] and [itex]\epsilon_{ab}^c B_c[/itex]. The errror just happens to be disguised in special relativity--if you don't look too closely. The units are wrong, and it fails CTP symmetry. You may get erronious results. If you use equations that work on a manifold locally isotropic to R4 and the Lorentz metric, you shouldn't go very far wrong. Maxwell's equations expressed in tensors (or tensor densities, depending upon where the squareroot of -g is attached) need only be composed of connection free antisymmetric tensors of lower indices.

In any case, your maxwell's equations used in post #14 are not the right ones to use. A dielectric medium may have no free charge, howwever it is not in general free of charge nor current under changing fields. Therefore the divergence of E is nonzero in Gauss law should be nonzero and you must include the term for current in Ampere's law.

-------------------

If you really want to get down to brass tacks, the faraday tensor given in the post Dale came up with is somewhat in error. A two form in four space does not partition into vectors in three dimensional subspaces given as vectors E and B. This is crazy. I know; I have entire thesis that is a lot of very upsetting, and embarassing garbage. The partitions should be three dimensional two forms [itex]\epsilon_{ab}^c E_c[/itex] and [itex]\epsilon_{ab}^c B_c[/itex]. The errror just happens to be disguised in special relativity--if you don't look too closely. The units are wrong, and it fails CTP symmetry. You may get erronious results. If you use equations that work on a manifold locally isotropic to R4 and the Lorentz metric, you shouldn't go very far wrong. Maxwell's equations expressed in tensors (or tensor densities, depending upon where the squareroot of -g is attached) need only be composed of connection free antisymmetric tensors of lower indices.

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- #65

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In any case, your maxwell's equations used in post #14 are not the right ones to use. A dielectric medium may have no free charge, howwever it is not in general free of charge nor current under changing fields. Therefore the divergence of E is nonzero in Gauss law should be nonzero and you must include the term for current in Ampere's law.

The quantity

[tex]\int \rho {dV}[/tex]----- (1)

is not supposed to change on transformation. Vacuum equtions are being considered here.rho is the

Regarding the Maxwell's equations in a medium:

If the

If you apply a changing external field to the dielictric, it should be considered in all the frames .

Now in the equation:

[tex]\nabla E=\frac{\rho}{\epsilon}[/tex] ------------ (2)

rho represents FREE charge density[in the matter equations]

You simply cannot create free charges, even if you consider a varying external field. It is a sheer impossiblity.

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- #66

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Of course special relativity is an approximation, and so is all the rest. But as you should expect, people are suspicious of any ideas that say that maxwell's equations, insufficiently flat spacetimeseem to be incompatable with special relativity. This seems to be your thesis, correct?

It is important to understand/investigate the exact reasons why a law should become inaccurate.

"Sufficiently flat spacetime" is indicative of a weak field.

Violation of Special relativity can occur due to:

1.Effects of gravity[the weak field]

2.Inhomogenity and anisotropy of space[garvity can[rather it does] contribute to anisotropy and inhomogeneity]

The above two points are linked but they are not identical.

It is important to know the manner in which inhomogeneity and anisotropy of space can cause the violations[apart from the effects of gravity]

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A Thought Experiment to Perform:

There are two inertial frames K and K' in uniform relative motion wrt each other along the x-x' direction.

There is a transparent dielectric--a large cubical one--in the frame k' ,with a flat face perpendicular to the x axes. The observer is in the unprimed frame,k [and he is stationary wrt K]. He flashes light on the moving dielectric from his source[light source]

Now Maxwell's Equations[in matter] should retain their form in the frames K and K'.But the values of the individual variables may change.

There are two inertial frames K and K' in uniform relative motion wrt each other along the x-x' direction.Speed of K' wrt k =v

There is a transparent dielectric--a large cubical one--in the frame k' ,with a flat face perpendicular to the x axes. The observer is in the unprimed frame,k [and he is stationary wrt K]. He flashes light on the moving dielectric from his source[light source]

The speed of light in the dielectric as observed from K = C1

The speed of light in the dielectric as observed from K' = C2

[Asuming that the value 1/Sqrt[epsilon*mu] has changed: the form of the equations remaining preserved]

The speed of light in different obliqie directions inside the dielectric] as observed from K will be different in different directions according to the velocity addition rule of Special Relativity--it should not be a constant value C1

[One should note that time dialation affects all the directions though length contraction operates only the x-x' direction in this problem]

*This paradoxical result stems from the fact that we have used SR in an inhomoheneous , anisotropic situation----SR has been applied in an incorrect context.*

Now as I have said previously that violation of SR may be due to

1. Inhomogeneity and anisotropy of space

2. Gravity

The above points need to be investigated.The Gravity part we know--in the form of Genaral Relativity.But there in a serious incompleteness--effects other than gravity contributing to point (1) are totally ingnored.

There are two inertial frames K and K' in uniform relative motion wrt each other along the x-x' direction.

There is a transparent dielectric--a large cubical one--in the frame k' ,with a flat face perpendicular to the x axes. The observer is in the unprimed frame,k [and he is stationary wrt K]. He flashes light on the moving dielectric from his source[light source]

Now Maxwell's Equations[in matter] should retain their form in the frames K and K'.But the values of the individual variables may change.

There are two inertial frames K and K' in uniform relative motion wrt each other along the x-x' direction.Speed of K' wrt k =v

There is a transparent dielectric--a large cubical one--in the frame k' ,with a flat face perpendicular to the x axes. The observer is in the unprimed frame,k [and he is stationary wrt K]. He flashes light on the moving dielectric from his source[light source]

The speed of light in the dielectric as observed from K = C1

The speed of light in the dielectric as observed from K' = C2

[Asuming that the value 1/Sqrt[epsilon*mu] has changed: the form of the equations remaining preserved]

The speed of light in different obliqie directions inside the dielectric] as observed from K will be different in different directions according to the velocity addition rule of Special Relativity--it should not be a constant value C1

[One should note that time dialation affects all the directions though length contraction operates only the x-x' direction in this problem]

Now as I have said previously that violation of SR may be due to

1. Inhomogeneity and anisotropy of space

2. Gravity

The above points need to be investigated.The Gravity part we know--in the form of Genaral Relativity.But there in a serious incompleteness--effects other than gravity contributing to point (1) are totally ingnored.

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- #68

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What I mean to say is that this conversation is useless. You are incorrect, and I have explained why to the best of my ability.I simply don't understand what you mean to say.

If that is insufficient for you then you will have to look elsewhere, such as a good textbook, to fix your understanding. I don't have a deep enough practical understanding to explain it in multiple different ways.

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What I mean to say is that this conversation is useless. You are incorrect, and I have explained why to the best of my ability.

If that is insufficient for you then you will have to look elsewhere, such as a good textbook, to fix your understanding. I don't have a deep enough practical understanding to explain it in multiple different ways.

Too irrelevant to be understood.

You may concentrate on the following:

This paradoxical result stems from the fact that we have used SR in an inhomoheneous , anisotropic situation----SR has been applied in an incorrect context.

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Try this: go get a textbook and study it.Too irrelevant to be understood.

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Thats not going to help anybody in the context of the given problem.Try this: go get a textbook and study it.

Thank you very much!

- #72

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