Four-wavevector $k$

Question

In special relativity the four-vector $k$ is defined: $$ k = (\mathbf{k},\frac{\omega}{c})$$ We can then write: $$\mathbf{k}\cdot{x} - \omega t = k\cdot x$$ Where of course: $$ x = (\mathbf{x},ct)$$

My question is how do we know $k$ is indeed a four vector? I'm asking because before using its property of transforming by Lorentz Transformation from one reference frame to another, we need to know it is indeed a four vector. In the textbook I read it is said that $k\cdot x$ is a phase of a wave that determines the position on the wave relative to the crests of the wave, and that has to be the same in any frame (i.e it is a four scalar). This explanation is not clear to me, I will be glad for clarification or another way of seeing this.

Answer 1

How do we define a four-vector?

Given two frames of reference, a four-vector $V$ is defined as a quantity which transforms according to the Lorentz transformation matrix $\Lambda$:

\begin{align} \mathbf{V'} &= \Lambda \mathbf{V} \tag{1}\\ V^{\mu} &= \Lambda^{\mu}_{\nu}V^{\nu} \\ V_{\mu} &= \Lambda^{\nu}_{\mu}V_{\nu} \end{align}

How do we know $k_{\mu}$ is a four-vector?

Because the phase is Lorentz invariant. It just means that $$\phi \equiv \mathbf{k}\cdot\mathbf{r} - \omega t = \mathbf{k'}\cdot\mathbf{r'} - \omega' t'. \tag{2}$$

Why is $\phi$ invariant?

Jackson gives the standard argument that the elapsed phase of the wave is proportional to the number of wave-crests that have passed the observer, and thus it must be frame-independent.

Also, interference effects are due to differences in the phase of waves. Notice that at a given space-time point, whether there is light or not cannot depend on the velocity of the observer looking there.

So we conclude that the phase $\phi = \mathbf{k}\cdot\mathbf{r} - \omega t$ is a Lorentz invariant.

What are the components of $k_{\mu}$?

You know the Lorentz transformations for spacetime vector $r = (\mathbf{r}, ct)$. Apply them in eqn. $(2)$ to recover the forms for $k_{\mu}$ which you have already mentioned in the question.

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Also, if you know that $p=(\mathbf{p},E/c)$ forms a four-vector, and since $E =\hbar \omega$, $\mathbf{p}=\hbar \mathbf{k}$, then $(\mathbf{k}, \omega/c)$ must be a four-vector too.

Answer 2

Is it enough to say that the phase:

$$ \phi(x^{\mu}) $$

is a scalar field, so that:

$$ k_{\mu} \equiv \partial_{\mu}\phi(x^{\mu})$$

is a 4-vector by manifest covariance?

Then, in any reference frame:

$$ \partial_{\mu}\phi(x^{\mu}) = (\frac 1 c \frac{\partial \phi}{dt}, \vec \nabla \phi)=(\frac{\omega} c, \vec k)$$