Question: how is a (subluminal) wave's wave-vector related to the 4-velocity (phase velocity)?
Wikipedia says that $$k^\mu \propto v^\mu \quad\quad\quad (1)$$ as stated in https://en.wikipedia.org/wiki/Wave_vector. I'm pretty sure this is wrong.
Here $k^\mu = (k_0, {\bf{k}})$ is the wave vector, $v^\mu$ is the four-velocity (phase velocity) of the wave. (Here the speed of light $c=1$.)
Assume, of course, that $v<1$, i.e., a subluminal wave. If $v=1$ then the wave vector is null, and there is no 4-velocity.
The wave equation is $$\frac{1}{v^2}\frac{\partial f}{\partial (x^0)^2} - ∇^2f=0.\quad\quad\quad(2)$$
A general solution of $(2)$ is $f(x)=f(k_\sigma x^\sigma) = f(k_0 x^0 - \bf{k \cdot x})$. Plugging this into (2) we see that the wave vector must satisfy $$k_0^2-v^2 (k_1^2+k_2^2+k_3^2 )=0,\quad\quad v=k_0/|\bf{k}|.$$
I believe $(1)$ is false except when $v\approx 1$. I think the following analysis is correct (writing $k = |\bf{k}|$):
$$k^σ = (k_0, {\bf{k}}) = k_0 \left( 1,\frac{ k\bf{v} }{ k_0 v}\right) = k_0 \left( 1,\frac{\bf{v}}{v^2} \right) = k_0 \gamma^{-1} \left(γ,\frac{γ \bf{v}}{v^2} \right) = v(-k_σ k^σ)^{1/2} \left(γ, \frac{γ\bf{v}}{v^2} \right).$$
When $v≈1$, we have $$k^σ ≈ (-k_\rho k^\rho )^{1/2} v^σ. $$ If $v=1$, however, then $k_σ k^σ=0$ and $v^σ$ is undefined.
Agreed?
