The Plane Electromagnetic Wave in Vacuum

[Anamitra]

A plane electromagnetic wave[traveling in the x-direction in an inertial frame]in vacuum is usually represented by an equation of the form:

$${E}{=}{E_{0}}{exp{[}{i}{(}{k}{x}{-}{\omega}{t}{)}{]}}$$

The wave velocity[phase velocity] is given by:

$${c}{=}{\frac{\omega}{k}}$$

We can perform Lorentz transformation on t and x[and y,z] to some other inertial frame moving in the x-x' direction.

The value $${\frac{\omega}{k}}$$ changes after transformation.But the value of c [wave velocity=phase velocity]should not change.
How does one explain this?
[If one considers a wave packet in vacuum each component should have the same speed=c]

 

[Dale]

The value $${\frac{\omega}{k}}$$ changes after transformation.

No it doesn't. You must have made a mistake in your math.

 

[Anamitra]

$${t}{=}{\gamma}{(}{t^{'}}{+}{\frac{vx^{'}}{c^{2}}{)}$$
$${x}{=}{\gamma}{(}{x^{'}}{+}{vt^{'}}{)}$$

$${{k}{x}{-}{\omega}{t}}{=}{x^{'}}{\gamma}{(}{k}{-}{v}{/}{c^{2}}{)}{-}{{t}^{'}}{\gamma}{(}{\omega}{-}{kv}{)}$$

$${\frac{{\omega}^{'}}{k^{'}}}{=}{\frac{{\omega}{-}{kv}}{{k}{-}{{v}{\omega}{/}{c^{2}}}$$

 

[Dale]

Yes, you almost have it. Now just make the substitution:
$$\omega=ck$$
And simplify.

 

[Anamitra]

Dalespam is of course correct. Thanks for that.

In fact we can get the formulas for relativistic Doppler Effect from the transformations of "omega" and "k" in the formulas of thread #3, without any reference to the position/location of the source[there is a small typing error in the second last formula of thread #3]

 

[Dale]

In fact we can get the formulas for relativistic Doppler Effect from the transformations of "omega" and "k" in the formulas of thread #3, without any reference to the position/location of the source

Yes, that is a very useful formula. I am glad you know how to derive it. I think that is a deeper level of understanding than just using the formula.