- #36
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It's quite easy to explain. Take for simplicity a plane wave (any em. wave can be built by plane waves in the sense of a Fourier integral anyway). Then the electromagnetic field is of the form
$$(\vec{E},\vec{B})(t,\vec{x})=\text{Re}[(\vec{E}_0,\vec{B}_0) \exp(-\mathrm{i} x \cdot k)].$$
This means the propagation direction is ##\vec{k}##, and you have ##k^0=|\vec{k}|## from Maxwell's equations, i.e., the phase velocity is ##c=1## (natural units).
So to know the direction of propagation in another frame you just have to use the Lorentz-transformation properties. The phase under the exponential is manfestly covariant and thus it's very easy to determine ##k'##; it's simply
$$k'=\Lambda k.$$
Suppose you have a boost in ##x## direction. Then, with ##\gamma=(1-v^2)^{-1/2}##,
$$\Lambda =\begin{pmatrix} \gamma & -v \gamma & 0 & 0\\ -v \gamma & \gamma & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0&0 &0 &1 \end{pmatrix}.$$
So you get
$$k'=\Lambda \begin{pmatrix} |\vec{k}| \\ \vec{k} \end{pmatrix}=\begin{pmatrix} \gamma (|\vec{k}|-v k^1) \\ \gamma (k^1-v |\vec{k}|) \\ k^2 \\ k^3 \end{pmatrix}.$$
As you see, the frequency (0-component) is not the same, which is the relativistic Doppler shift, and also the direction is not the same in the new frame (aberration effect).
Of course, all the effects together conspire such that the physics doesn't change at all due to Lorentz boosts, as it must be due to the principle of special relativity (no change of physics due to change of the inertial frame).
$$(\vec{E},\vec{B})(t,\vec{x})=\text{Re}[(\vec{E}_0,\vec{B}_0) \exp(-\mathrm{i} x \cdot k)].$$
This means the propagation direction is ##\vec{k}##, and you have ##k^0=|\vec{k}|## from Maxwell's equations, i.e., the phase velocity is ##c=1## (natural units).
So to know the direction of propagation in another frame you just have to use the Lorentz-transformation properties. The phase under the exponential is manfestly covariant and thus it's very easy to determine ##k'##; it's simply
$$k'=\Lambda k.$$
Suppose you have a boost in ##x## direction. Then, with ##\gamma=(1-v^2)^{-1/2}##,
$$\Lambda =\begin{pmatrix} \gamma & -v \gamma & 0 & 0\\ -v \gamma & \gamma & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0&0 &0 &1 \end{pmatrix}.$$
So you get
$$k'=\Lambda \begin{pmatrix} |\vec{k}| \\ \vec{k} \end{pmatrix}=\begin{pmatrix} \gamma (|\vec{k}|-v k^1) \\ \gamma (k^1-v |\vec{k}|) \\ k^2 \\ k^3 \end{pmatrix}.$$
As you see, the frequency (0-component) is not the same, which is the relativistic Doppler shift, and also the direction is not the same in the new frame (aberration effect).
Of course, all the effects together conspire such that the physics doesn't change at all due to Lorentz boosts, as it must be due to the principle of special relativity (no change of physics due to change of the inertial frame).
Agree the rest, though.