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.


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, $$-2\,\pi\,N = \Delta\mathbf{R} \cdot \mathbf{K}^\mu .$$


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.


[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.


1 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 $\vec k$.

  • $\begingroup$ Waht do you mean by, "can be contracted with a four vector..." Do you just mean to indicate that it can be multiplied by another four-vector using a dot product? $\endgroup$ Commented Aug 1, 2022 at 17:24
  • $\begingroup$ @MichaelLevy yes, although I personally would only call it a “dot product” in three spatial dimensions. In any case, it is the linear operation that takes two vectors and produces a scalar. $\endgroup$
    – Dale
    Commented Aug 1, 2022 at 18:42

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