Special Relativity – How to Derive the Four-Wavevector from First Principles

Question

Context

From [1], I understand that, "The four-wavevector is a wave four-vector that is defined, in Minkowski coordinates, as
$$\mathbf{K}^\mu = \left(\frac{\omega}{c}, \mathbf{k}\right).\tag{1}"$$

I am studying special relativity using [2]. In [2] Gray, there is derivation of the four-wavevector. The expression in [2] is given as
$$\mathbf{L} = \left(f, \frac{l}{\lambda}, \frac{m}{\lambda}, \frac{n}{\lambda}\right),$$
where the wave moving in direction $\mathbf{n} = (l,m,n)$ subject to $\left\|\mathbf{n}\right\| =1$.
The derivation does not work for me for two reasons. Firstly, I can not ascertain the constant of proportionality from $\mathbf{n}$ and $\mathbf{k}$. Secondly, there is a factor $2\,\pi$ difference between $\mathbf{L}$ and $\mathbf{K}^\mu$. Thirdly, there is detail in Gray's derivation that I can't wrap my head around.

Based on this background, I have asked myself how to derive the four-wavevector in (1). Yet, I do not succeed. So, I pose the questions to the community.

Question

From first principles, how can one derive the four-wavevector in (1)?

My Attempt

Imagine a series of waves viewed from reference frame $S$. The waves move in a direction given by the unit wave vector, $\mathbf{k}$, and with speed $ \left\| \mathbf{u}\right\|$. Imagine following a point on the crest of one of these waves. The point on the wave has a displacement $$\left(\frac{ k^{(1)}}{\left\|\mathbf{k}\right\|} \Delta{x}
+
\frac{ k^{(2)}}{\left\|\mathbf{k}\right\|} \Delta{y} +
\frac{ k^{(3)}}{\left\|\mathbf{k}\right\|} \Delta{z}
\right)\tag{2}$$ in time $\Delta t$. Thus, the motion of this point on the wave is subject to
$$\frac{ k^{(1)}}{\left\|\mathbf{k}\right\|} \Delta{x}
+
\frac{ k^{(2)}}{\left\|\mathbf{k}\right\|} \Delta{y} +
\frac{ k^{(3)}}{\left\|\mathbf{k}\right\|} \Delta{z}
= \left\| \mathbf{u}\right\| \,\Delta t . $$

Now, imagine a train of such waves, each separated by wavelength $\lambda$. Consider two events. The first event occurs at position $\mathbf{r}_1$, which is on the crest of a wave, at time $t_1$; while the second event occurs at time a later time, $t_2$, and at position $\mathbf{r}_1 + \left[N\,\lambda+ \left\|\mathbf{u}\right\|\,\left(t_2 – t_1 \right)\right]\,\frac{\mathbf{k}_1}{\left\|\mathbf{k}_1\right\| } $. The displacment 4-vector, $\Delta{\mathbf{R}}$, is $\Delta{\mathbf{R}} = \left(\Delta{x^{(0)}},\Delta{x^{(1)}},\Delta{x^{(2)}},\Delta{x^{(3)}}\right)$.

To be clear, since the two events happen at different times, and since the wave train is in motion, the spatial distance between the two events is $\left\| \mathbf{u}\right\| \left(t_2-t_1\right) + N\,\lambda$. In my reading of [2], I understand that Gray also gives this distance as equal to
$$\frac{ k^{(1)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(1)}
+
\frac{ k^{(2)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(2)} +
\frac{ k^{(3)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(3)}\tag{3}.$$
Gray than uses the transitive law and sets the two distances equal to each other. [The crux of the matter is that I simply do not see how these two distances are equal]. Ultimately, if these distances are equal we have that
$$\frac{ k^{(1)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(1)}
+
\frac{ k^{(2)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(2)} +
\frac{ k^{(3)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(3)} = \left\| \mathbf{u}\right\| \left(t_2-t_1\right) + N\,\lambda.\tag{4}$$ After adjusting (4), we have
$$-N\,\lambda= \left\| \mathbf{u}\right\| \left(t_2-t_1\right)
–
\frac{ k^{(1)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(1)}
–
\frac{ k^{(2)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(2)}
–
\frac{ k^{(3)}}{\left\|\mathbf{k}\right\|} \Delta{x}^{(3)}
\tag{5}.$$
Next, baring in mind Minkowski metric, we can rewrite (5) as
$$-N = \begin{bmatrix}\Delta{x}^{(0)}\\\Delta{x}^{(1)}\\\Delta{x}^{(2)}\\\Delta{x}^{(3)}\end{bmatrix}\cdot \begin{bmatrix}\frac{\left\| \mathbf{u}\right\|}{\lambda\,c}
\\
\frac{ k^{(1)}}{\lambda\,\left\|\mathbf{k}\right\|} \Delta{x}^{(1)}
\\
\frac{ k^{(2)}}{\lambda\,\left\|\mathbf{k}\right\|} \Delta{x}^{(2)}
\\
\frac{ k^{(3)}}{\lambda\,\left\|\mathbf{k}\right\|} \Delta{x}^{(3)}
\end{bmatrix}
\tag{6}.$$
Using the relation $\left\| \mathbf{u}\right\| = \frac{\omega\,\lambda}{2\,\pi }$ and the relation $\lambda \left\|\mathbf{k}\right\| = 2\,\pi$, I rewrite (6) as
$$-2\,\pi\,N =
\begin{bmatrix}
\Delta{x}^{(0)}\\\Delta{x}^{(1)}\\\Delta{x}^{(2)}\\\Delta{x}^{(3)}\end{bmatrix}
\cdot
\begin{bmatrix}\frac{\omega}{ c}
\\
k^{(1)}
\\
k^{(2)}
\\
k^{(3)}
\end{bmatrix}
\tag{7}.$$
I now see that the rightmost vector on the right side of (7) is the four-wavevector, while the the leftmost vector on the right side of (7) is the displacement four vector. Thus,
$$-2\,\pi\,N =
\Delta\mathbf{R}
\cdot
\mathbf{K}^\mu
.$$

Discussion

Even though I do not understand one crucial part of the derivation, I am able to derive the correct form of the four-wavevector. This is troubling to me. I am hoping that either someone one will explain to me how come the equality given in (4) holds true; or, that someone will present a better derivation of the four-wavevector from first principles.

Bibliography

[1] https://en.wikipedia.org/wiki/Wave_vector

[2] N. Gray, A Student's Guide to Special Relativity, Cambridge University Press, 2022, p. 115-6.

Answer 1

That derivation is very much the long way around and misses the point of the wave four-vector.

We start with the fact that the phase $\phi$ is a scalar. Then in a local inertial frame we have $$\phi= \omega t - \vec k \cdot \vec x = \eta_{\mu\nu} K^\mu X^\nu$$ where in the local inertial coordinates $X^\nu=(ct,\vec x)$ and $K^\mu=(\omega/c,\vec k)$.

Then, since $X^\nu$ is a known four-vector then $K^\mu$ is necessarily also a four-vector because it can be contracted with a four-vector to produce a scalar.

Since $K^\mu$ is an important quantity we give it a name, the wave four-vector, highlighting its relationship with the wave-vector $\vec k$.