# Proving a given vector to be a four-vector

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.

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

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 $$k^\mu=\left(\frac{\omega}{c},\vec{k}\right).$$
• 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). Commented Feb 20 at 14:39
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.