Refractive index of dielectric in different frames of reference 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.

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!
 A: 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.
A: 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 $n'=1$.
