A Paradox in Special Relativity

Question

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!

Answer 1

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.

Answer 2

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'.

Answer 3

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.