Phase invariance of an EM wave in special relativity

[quasar_4]

Homework Statement

So I'm trying to show for a specific, given EM plane wave in vacuum that

$$kx - \omega t = k' x' - \omega' t'$$

but I'm running into some difficulties. I'm hoping someone can show me where I'm going wrong. Here's the setup:

In the lab frame K, a plane EM wave traveling in vacuum has an electric field given by

$$\vec{E}(\vec{x},t) = \hat{z} E0 \cos{(\frac{k}{\sqrt{2}} (x+y) - \omega t)}$$

where E0, k are positive real constants, omega = ck

Homework Equations

Maxwell's equations
Lorentz transformation

The Attempt at a Solution

First I found the B field in K using Faraday's law. Then I used the standard transformations in Jackson to go from E, B to E', B' in the K' frame.

My electric field in the K' frame is

$$\vec{E'}(\vec{x},t) =\hat{z} \gamma (1-\beta) \cos{(\frac{k}{\sqrt{2}} (x+y)-\omega t)}$$

Now I want to use the Lorentz transformation,

$$x = \gamma (x' + \beta c t') \text{ , } ct = \gamma (c t' + \beta x')$$

to put E' in terms of the primed coordinates. If I understand Jackson correctly (he says phase of a plane wave is an invariant), then shouldn't we be able to demand that

$$\frac{k}{\sqrt{2}} (x+y) -\omega t = \frac{k'}{\sqrt{2}} (x'+y') -\omega' t'$$

?

I'm trying to prove this, doing simple substitution with x, y, t -> x', y', t', but I can't get the factors of k' and w' to come out correctly because of the square root of 2. I'm stuck at:
$$\frac{k}{\sqrt{2}} (x+y) -\omega t = \gamma \left[k x' (\frac{1}{\sqrt{2}}-\beta) + \omega t' (\frac{\beta}{\sqrt{2}}-1) \right] + \frac{y' k}{\sqrt{2}}$$

I don't know my initial assumption is wrong, or the algebra is just tricky. Any help appreciated!

 

[vela]

If I understand Jackson correctly (he says phase of a plane wave is an invariant), then shouldn't we be able to demand that

$$\frac{k}{\sqrt{2}} (x+y) -\omega t = \frac{k'}{\sqrt{2}} (x'+y') -\omega' t'$$

?

No, when he says the phase is invariant, he means

$$k_\mu x^\mu = k'_\mu x'^\mu$$

where $k'^\mu = \Lambda^\mu_\nu k^\nu$. Though the x and y components of k are equal in frame K, they generally won't be in K'.

 

[quasar_4]

Ok, thanks.