The setup

A transparent isotropic dielectric medium moving in the negative $x'$ direction at speed $v$ in frame $S'$ is stationary in frame $S$, where it has refractive index $n$. In other words, frame $S'$ is moving at speed $v$ in the positive $x$ direction relative to frame $S$. We take the two frames' origins to coincide at $t = 0$.

Calculate the refractive index $n'$ in frame $S'$ experienced by light travelling inside the dielectric along the $y$-axis in frame $S$.

I have drawn a schematic of the setup below.

enter image description here

I have found two approaches to this question, both of which give me different answers. I was hoping somebody would be able to point out the flaw in one of these methods.

Method 1

The 4-wavevector of a photon moving at an angle $\theta$ to the $x$-axis in the $x$-$y$ plane is

$$ K = (\omega/c, k \cos \theta, k \sin \theta, 0 )$$

By applying a Lorentz transformation to this vector we arrive at the following relations between frequency, wavenumber and angle in $S'$ and those quantities in $S$:

$$ \omega'/c = \gamma\omega/c - \gamma\beta k \cos \theta $$ $$ k' \cos \theta' = \gamma k \cos \theta - \gamma \beta \omega /c $$ $$ k' \sin \theta' = k \sin \theta$$

The refractive index in $S'$ is defined by

$$ \omega'/k' = c/n' \qquad \implies \qquad n' = ck'/\omega'$$

We set $\theta = \pi/2$, corresponding to motion in the $y$-direction. Then summing the squares of the bottom two equations to eliminate $\theta'$, square-rooting, and then dividing the result by the first equation, we find:

$$ ck'/\omega' = \frac{c}{\omega\gamma} \sqrt{k^2 + \frac{\gamma^2\beta^2 \omega^2}{c^2}}$$

If we pull a factor of $k^2$ outside we can write this thus:

$$ ck'/\omega' = n' = \frac{n}{\gamma} \sqrt{1 + \frac{\gamma^2\beta^2 }{n^2}}$$

Method 2

Here I take a more 'first-principles' approach. We can identify two events in spacetime --- the point at which a particular photon enters the dielectric medium, and the point at which it leaves. Let us define the co-ordinates in frame $S$ of the first of these events to be $(0,0,0,0)$. Then in $S$ we know that the co-ordinates of the point at which the photon leaves the medium are $(ct, 0, y, 0)$, where $t$ and $y$ are related by $c/n = y/t$ --- this is just speed = distance / time. Applying a Lorentz transformation to these two points to find the co-ordinates in $S'$:

$$ \mathrm{entrance}_{S'} = (0,0,0,0) \qquad \mathrm{exit}_{S'} = (\gamma ct, -\gamma \beta ct, y, 0)$$

The total time for the photon to move through the block is hence $\gamma t$, whilst the total distance traveled is, by Pythagoras',

$$ \sqrt{\gamma^2 \beta^2 c^2 t^2 + y^2} $$

Dividing these quantities should give us $c/n'$, the speed of the photon. Hence we can write

$$n' = \frac{c \gamma t }{ \sqrt{\gamma^2 \beta^2 c^2 t^2 + y^2}} = \frac{c \gamma t }{ \sqrt{\gamma^2 \beta^2 c^2 t^2 + c^2 t^2 / n^2}}$$

So we have

$$ n' = \frac{\gamma}{\sqrt{\gamma^2 \beta^2 + 1/n^2}}$$

which is not the same as the first expression. I anticipate I've made a very silly mistake somewhere along the line here, but I can't for the life of me spot it!

  • $\begingroup$ Great question, a great deal of correct work showing the "error" - fantastic! Red Act's answer is a correct one. $\endgroup$ Commented Sep 2, 2014 at 23:55

2 Answers 2


The problem here isn't a simple algebra error, but rather an issue with the physics. A medium which at rest is isotropic no longer behaves as an isotropic medium when it is moving relativistically. Instead, it behaves as a nonreciprocal bianisotropic material.

In particular, the phase velocity of light at a particular frequency in a medium is no longer the same in all directions if the medium is moving. This effect can be detected even if the medium is moving at a non-relativistic speed, as in the Fizeau experiment.

Because of this, any attempt to determine one consistent number for what a refractive index turns into under a Lorentz transformation is doomed to failure, because the equivalent to the refractive index is no longer a single number in a frame in which the medium is moving. Instead, it's necessary to treat the refractive index (or really the relative permittivity, which is closely related) as a tensor.

Unfortunately, I'm unable to find a full tensor treatment of refraction in moving media online. I did find this paper on relativistic optics in moving media, but it uses Clifford algebra as an alternative to a tensor treatment, and unfortunately the paper is behind a paywall. However, the non-paywalled abstract for that paper is my source for the above claim that a medium that's isotropic at rest behaves as a nonreciprocal bianisotropic medium when moving.

  • $\begingroup$ Surprisingly, there aren't many good treatments on this topic around, and indeed I have seen incorrect treatments make it into peer reviewed journals like JOSA even. It would be good matter for a review paper in a journal like that. $\endgroup$ Commented Sep 2, 2014 at 23:59
  • $\begingroup$ Thanks for the reply. An earlier part of the problem was to calculate the refractive index experienced by a photon moving in the x-direction, and indeed the expression is different. So I'm familiar with this idea of anistropy. But I still don't quite see why 'Method 2' fails --- I suppose the question is how we define the refractive index in a given direction/how we define the permittivity tensor. I would have thought that my definition of 'refractive index in the y-direction' in the second method is legitimate, but evidently not. $\endgroup$
    – gj255
    Commented Sep 3, 2014 at 12:54
  • $\begingroup$ I just saw this answer and I agree with this answer. Method 2 failed because the dielectric was moving (as well as the electric dipoles that reradiate the transmitted field). $\endgroup$
    – Taiben
    Commented Sep 3, 2014 at 17:44

Method 1 is correct.

Method 2 makes the mistake that you use the distance travelled in the reference frame, but the correct distance must take into account that the dielectric also moves.

You can check the limiting case of $n=1$ which should give $n'=1$.

  • 1
    $\begingroup$ I gave this a down vote because I don't think the statement about method 2 making a mistake makes sense. The Lorentz transformation captures everything there is to say about the distance between the two events as measured in $S'$. If after a careful re-examination your answer still makes sense to you, please edit this answer to point out in clearer detail the way that you think that method 2 is making a mistake. $\endgroup$
    – Red Act
    Commented Sep 2, 2014 at 4:26
  • $\begingroup$ Yes, I'm with Red Act here, but thank you for the answer Taiben. I sort of see where you're coming from --- is it the case that the relevant distance is not that travelled by the photon in S', but that travelled relative to the block in S'? Or something like that? But in the x' direction the photon hasn't travelled relative to the block at all (in S'), and if we write the distance travelled as $y$, we certainly don't get the right expression... $\endgroup$
    – gj255
    Commented Sep 2, 2014 at 10:58
  • $\begingroup$ My answer was actually correct, please think twice and I apologize if I should have been more specific. Basically, the following statement in your method 2 was incorrect: Then in S we know that the co-ordinates of the point at which the photon leaves the medium are (ct,0,y,0), where t and y are related by c/n=y/t --- this is just speed = distance / time. $\endgroup$
    – Taiben
    Commented Sep 3, 2014 at 17:40
  • $\begingroup$ @Taiben What about that quoted statement looks like a mistake to you? Keep in mind that the medium is stationary in S, so motion of the medium isn't an issue in S, and the question specifies that the light travels along the y-axis in frame S, so the angle of the beam isn't an issue in S. Or are you perhaps being thrown by the choice of conventions being used in expressing the four-vector? $\endgroup$
    – Red Act
    Commented Sep 3, 2014 at 23:17

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