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Two inertial frames $\mathrm{K}$ and $\mathrm{k’}$ are considered. They are in relative uniform motion along the $x-x’$ direction with relative speed equals to $v$. In the frame $\mathrm{K’}$ we have a cuboidal piece of dielectric [at rest wrt $\mathrm{K’}$]with a flat face perpendicular to the $x-x’$ direction, that is, this particular face is parallel to the $y-z$ direction.The dielectric is homogeneous and isotropic within itself.

We now consider Maxwell’s equations [in a medium] wrt the dielectric in the rest frame of the dielectric,ie, $\mathrm{K’}$.

If these equations are transformed, they should retain their form in $K$ [according to the first postulate of SR]. But the individual values of the variables may change

With this information we proceed into the paradox.

Speed of light in the dielectric as observed from $k’=nc$ [$n$ is a positive fraction less than $1$]

Relative speed between the frames, $v=cn’$[$n’$ is also a positive fraction less than $1$]

For normal incidence:

Speed of light in the dielectric as observed from $K$ [From Velocity-Addition Rule of SR]: $$v=\frac{nc+n’c}{1+nn’} \tag1$$

For oblique ray inside the medium at $\theta$ degrees degrees with respect to the $x’-$axis in the $K’$ frame:

$$v’_x=nc\ \cos (\theta)$$

$$v’_y=nc \ \sin(\theta)$$

$$v’_z=0 \ \ \ \ \ \ \ \ \ \ \ \ \ $$

[$v'_z$ has been taken to be zero for the convenience of calculations]

Observations from $K$ :

$$v_x=\frac{nc \ \cos(\theta)+n’c}{1+nn’\ \cos (\theta)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$

$$v_y=\frac{nc \ \sin(\theta)}{1+nn’\ \cos(\theta)}\sqrt{1-n’^2}$$

$v_z=0$ , therefore :

$$v=\sqrt{ \left[\frac{nc \ \cos(\theta)+n’c}{1+nn’\ \cos (\theta)} \right]^2+ \left[\frac{nc \ \sin(\theta)} {1+nn’\ \cos(\theta)}\sqrt{1-n’^2}\right]^2} \tag2$$

The results from $(1)$ and $(2)$ are not identical , though from the invariant Maxwell’s equations [in a medium] we understand that the speed of light should be the same in all directions inside the dielectric as observed from $K$. What would be your answer to this paradox.

[My assessment:This paradoxical situation arises from the fact that we have applied SR in an incorrect context.It has been applied in an anisotropic and inhomogeneous configuration. You could of course have a different assessment]

[The dielectric within itself is homogeneous and isotropic. But the overall space being considered is not homogeneous and isotropic]

If Maxwell's equations change their form wrt the Lorentz transformations,Gauss Law,Div $B=0$ etc may change if a piece of dielectric is loaded into a moving train!

Qmechanic
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Anamitra Palit
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    This is so complicated that the number of loopholes that need to be discussed seems way too huge and it is probably uneconomic to try to answer this question. First, Maxwell's equations in a moving medium are complicated and get extra terms so that Lorentz invariance - which is spontaneously broken by the medium and picks a preferred rest frame - is pretty much inconsequential. Maxwell's equations for moving medium were discussed in various SE articles linked here http://physics.stackexchange.com/questions/9847/moving-conductors-in-magnetic-fields-is-there-electric-field-or-not – Luboš Motl Nov 15 '11 at 13:21
  • (1) has the ray of light normal to the face in k' where it'll be normal in k, so are you setting theta to 0? – John McAndrew Nov 15 '11 at 19:23
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    @Anamitra: This question is absurd--- you are assuming that Maxwell's equations inside a dielectric are invariant to Lorentz transformations--- they are not. The moving dielectric is not even described by a dielectric constant anymore, because a static electric field produces magnetic fields in the interior of a moving dielectric. The whole question is doing too much formal mathematics considering that the physical picture is completely off, -1, and I vote to close. – Ron Maimon Nov 15 '11 at 20:36
  • If Maxwell's equations change their form wrt Lorentz transformations,them Gauss Law ,DivB=0,Ampere's circuital law etc will change if a dielectric is loaded into a moving train!That is even more absurd---- as absurd as your negative vote. – Anamitra Palit Nov 15 '11 at 23:15
  • Ron Miamon has suggested two conflicting ideas in the same breath:(1)The first postulate of relativity should not hold for Maxwell's equations in a medium(2)He has used field transformations given by relativity----a static electric field produces magnetic fields in the interior of a moving dielectric.Incidentally these field transformations follow from SR where the First Postulate is given due regard. – Anamitra Palit Nov 15 '11 at 23:39
  • @ron do you mean covariant or invariant? – John McAndrew Nov 16 '11 at 01:25
  • The question as originally written was a mess, but there's a good physical concept to get at under there, so I edited it to improve it. FYI I may be deleting these comments later. – David Z Nov 16 '11 at 02:52
  • I'm missing something. Why should we expect it to be invariant in the first place? Regardless of the correctness of your equations, the medium is a reference frame, and as such I don't see why I should be bothered by the fact that the speed of light isn't isotropic. Could someone clearly answer what the case is to begin with? – Alan Rominger Nov 16 '11 at 03:25
  • @John: I mean "covariant". The description using a dielectric constant is not covariant, because the dielectric constant is part of a general linear 4-index constitutive tensor which relates F in the material to F outside. There have been a host of questions here about this. – Ron Maimon Nov 16 '11 at 20:18
  • @David it has the potential to be a great question since it hints at relativistically-induced optical anisotropy. – John McAndrew Nov 18 '11 at 02:46

3 Answers3

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There's nothing wrong with the application of SR in this context. SR is where reference frames move at constant velocities, regardless to anisotropy or whatever.

I believe there's nothing wrong with the fact that the speed of light in a dielectric in motion is not the same in all the directions. In reference frame K the "cuboidal piece of dielectric" is shortened in X direction. Hence - it's not actually isotropic. One can say that electric susceptibility of such a moving dielectric is a tensor.

valdo
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  • If Maxwell's equations in a medium are considered to be invariant wrt transformation between inertial frames then the speed of light in the medium should be constant in all directions when viewed from any inertial frame:

    Speed of light in the medium=1/sqrt[epsilon*mu}

     – Anamitra Palit Nov 15 '11 at 14:19
  • The angle theta has been used wrt ot K' where the dielectric is at rest."cnCos(theta)" is the x-component of the velocity of light – Anamitra Palit Nov 15 '11 at 15:35
  • The angle theta has been used wrt ot K' where the dielectric is at rest."cnCos(theta)" is the x-component of the velocity of light in the dielectric[wrt K'] while cnSin(theta) is the corresponding y-component in K'.All these quantities appear on the RHS of the transformations.Incidentally my basic aim is to evaluate the magnitude of the speed of light as observed from K in two ways--(1)By applying SR(2) from Maxwell's equations in a medium assuming them to retain their form on transformation between inertial frames.I am getting unequal values – Anamitra Palit Nov 15 '11 at 15:41
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The angle $\theta$ is used in both reference frames. As I recall from discussions of such SR paradoxi, the angle can change when you switch reference frames. Try to derive the values without explicit use of $\theta$.

Although, looking at the calculation, everything seems correct, with the angle used only in K'.

David Gonzales
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  • The angle theta has been used wrt ot K' where the dielectric is at rest."cnCos(theta)" is the x-component of the velocity of light in the dielectric[wrt K'] while cnSin(theta) is the corresponding y-component in K'.All these quantities appear on the RHS of the transformations.Incidentally my basic aim is to evaluate the magnitude of the speed of light[in the dielectric] as observed from K in two ways--(1)By applying SR(2) from Maxwell's equations in a medium assuming them to retain their form on transformation between inertial frames.I am getting unequal values – Anamitra Palit Nov 15 '11 at 15:51
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in K' we have $\;\; v'_{x}=v'\cos(\theta')=cn'\cos(\theta'){n'}\;\;,\;v'_{y}=v'\sin(\theta')=cn'\sin(\theta')$

we replace in the velocity composition equation in K :

$$v_{x}=v\cos(\theta)=c\,\frac{n\cos(\theta')+n'}{1+nn'\cos(\theta')}$$

$$v_{y}=v\sin(\theta)=\,\frac{cn\sin(\theta')\sqrt{1-n'^{2}}}{1+nn'\cos(\theta')}$$

a simple calculation gives: $$\;\;v=c\;\left[\frac{(\vec{u}+\vec{u'})^{2}-(\vec{u'}\times\vec{u})^{2}}{(1+\vec{u'}.{\vec{u}})^{2}}\right]^{\frac{1}{2}}$$

with : $\;n\,\vec{e}=\vec{u}\;\;,n'\,\vec{e'}=\vec{u'}\;\;,||\vec{e}||=||\vec{e'}||=1 $

if $\;\;\vec{e}\parallel \vec{e'} \;\;\Rightarrow \;\;\vec{u'}\times\vec{u}=\vec{0}\;\;$ i.e $\;\;\theta'=0$

we find equation (1)

There is a similarity with the fizeau experiment: https://en.wikipedia.org/wiki/Fizeau_experiment ps: My first answer was wrong because I used the refractive index.

The Tiler
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