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How to prove that $$k=(\frac{\omega}{c} , \vec{k})$$ is a four-vector?

Where $\omega$ is the frequency and $\vec{k}$ is a wave vector.

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    $\begingroup$ Your question is not very clear. Also: What do you know? Multilinear algebra? tensorcalculus? Every 4 component vector can be a four-vector... $\endgroup$ Commented Jun 23, 2020 at 18:21
  • $\begingroup$ @NathanaelNoir is that true? A four vector is something whose components transform in a particular way $\endgroup$
    – jim
    Commented Feb 20 at 20:43

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We know, the four-position $$x^\mu=(ct,\vec{r})$$ is a four-vector. We also know, for a plane wave its phase at any certain point in spacetime $$\phi=-\omega t+\vec{k}\cdot\vec{r},$$ must be a four-scalar (i.e. the same value in every reference frame). To convince yourself about this, consider a scalar plane wave like $$\psi(t,\vec{r})=A\cos(-\omega t+\vec{k}\cdot\vec{r})$$ and a particular point-like event (like an atom emitting an $\alpha$ particle at a certain position and time). Then all observers (regardless which reference frame they use) can agree on whether this event happened at a $\psi$-wave crest (i.e. $\phi=0$), at a $\psi$-wave trough (i.e. $\phi=180°$), or at a certain phase $\phi$ in between.

Let us rewrite this phase: $$\begin{align} \phi&=-\omega t+\vec{k}\cdot\vec{r} \\ &=-\left(\frac{\omega}{c},-\vec{k}\right)\cdot(ct,\vec{r}) && \text{use }x^\mu=(ct,\vec{r}) \text{ and define } k_\mu=\left(\frac{\omega}{c},-\vec{k}\right) \\ &= -k_\mu x^\mu \end{align}$$

The left side ($\phi$) of this equation is a four-scalar, so the expression on the right side ($-k_\mu x^\mu$) must be a four-scalar too. This means, since $x^\mu$ is a four-vector, $k_\mu$ must also be a four-vector to make their product a four-scalar. So we found $$k_\mu=\left(\frac{\omega}{c},-\vec{k}\right)$$ is a covariant (i.e. lower index) four-vector. And the corresponding contravariant (i.e. upper index) four-vector is $$k^\mu=\left(\frac{\omega}{c},\vec{k}\right).$$

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  • $\begingroup$ Correct, but to complete the proof, I suggest, you would need to give an argument that the phase $\phi$ is Lorentz-invariant (without assuming that $k^\mu$ is a 4-vector, so as to avoid circularity). $\endgroup$ Commented Feb 20 at 14:39
  • $\begingroup$ @AndrewSteane A valid point. I have tried to add a convincing argument for this. $\endgroup$ Commented Feb 20 at 15:08
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A four-vector is an element of a four-dimensional vector space considered as a representation space of the standard representation of the Lorentz group, the $(\frac{1}{2},\frac{1}{2})$ representation. Therefore a four-vector transform in a specific way under Lorentz transformation.

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