I read in a textbook any kind of wave of frequency $\nu$ and wave numer $\vec{k}$ has a spacetime dependence $\exp(i\vec{k}\cdot\vec{x}-iwt)$ where $w=2\pi\nu$. It then said that Lorentz invariance requires ($\vec{k}$, $w$) to transform like a four vector.

I know that the Lorentz invariance of the four vector ($\vec{x}$, ct) is due to speed of light being same in all inertial reference frames. Similarly, Lorentz invariance of the four vector ($\vec{p}$, E) is due to the invariance of rest mass in all frames.

When that ($\vec{k}$, $w$) is Lorentz invariant, exactly what is he referring to being invariant? What is the physics that is invariant?


Based on your description, the textbook doesn't say that the four-vector $(\vec{k},\omega)$ is Lorentz-invariant. It says that $(\vec{k},\omega)$ transforms as a Lorentz vector.

When you apply a Lorentz transformation, the vectors $(\vec{x},ct)$ and $(\vec{p},E/c)$ change. The position of an object in spacetime is not the same in every frame, and its momentum and energy are not the same either. Instead, the way that those vectors change is well-defined; in particular, for a Lorentz transformation with boost velocity $v$ and Lorentz factor $\gamma$ in the $x$ direction, we have that the new four-vectors $(\vec{x}',ct')$ and $(\vec{p}',E'/c)$ are given by:

$$\vec{x}'=\gamma(x-vt)\hat{x}+y\hat{y}+z\hat{z}$$ $$ct'=\gamma\left(ct-\frac{vx}{c}\right)$$ $$\vec{p}'=\gamma\left(p_x-\frac{vE}{c^2}\right)\hat{x}+p_y\hat{y}+p_z\hat{z}$$ $$\frac{E'}{c}=\gamma\left(\frac{E}{c}-\frac{vp_x}{c}\right)$$

The Lorentz-invariant quantities that are derived from these vectors are Lorentz scalars. In particular, the dot product of two Lorentz vectors is a Lorentz scalar. So, from these two Lorentz vectors, we get the following three Lorentz-invariant quantities:

  • $(\vec{x},ct)\cdot(\vec{x},ct)=c^2t^2-\vec{x}^2$, which is the squared spacetime interval between the origin and the point described by the four-vector;
  • $\left(\vec{p},\frac{E}{c}\right)\cdot\left(\vec{p},\frac{E}{c}\right)=\frac{E^2}{c^2}-\vec{p}^2$, which is the squared rest mass; and
  • $\left(\vec{p},\frac{E}{c}\right)\cdot(\vec{x},ct)=Et-\vec{p}\cdot\vec{x}$, which as far as I know does not have a common name.

The exact same thing applies to the $(\vec{k},\omega/c)$ four-vector. If you apply that same Lorentz transform, this new four-vector $(\vec{k}',\omega'/c)$ is given by:

$$\vec{k}'=\gamma\left(k_x-\frac{v\omega}{c^2}\right)\hat{x}+k_y\hat{y}+k_z\hat{z}$$ $$\frac{\omega'}{c}=\gamma\left(\frac{\omega}{c}-\frac{vk_x}{c}\right)$$

Adding this to the other four-vectors, we can have a total of six Lorentz scalars. In addition to the three described above, we also have:

  • $\left(\vec{k},\frac{\omega}{c}\right)\cdot\left(\vec{k},\frac{\omega}{c}\right)=\frac{\omega^2}{c^2}-\vec{k}^2$, which describes the dispersion relation of the wave;
  • $\left(\vec{k},\frac{\omega}{c}\right)\cdot(\vec{x},ct)=\omega t-\vec{k}\cdot\vec{x}$, which is the phase of the wave; and
  • $\left(\vec{k},\frac{\omega}{c}\right)\cdot\left(\vec{p},\frac{E}{c}\right)=\frac{\omega E}{c^2}-\vec{k}\cdot\vec{p}$, which as far as I know doesn't have a common name.
  • $\begingroup$ Thank you for this answer. I learned a lot! $\endgroup$
    – TaeNyFan
    May 4 '20 at 15:45

The phase $\theta=\vec k \cdot \vec x -\omega t$ is invariant,
where $(\vec k, \omega)$ is a 4-vector describing the propagation of the wave.
(In a spacetime diagram, it is 4-vector that is orthogonal to the wavefronts.)

[updated inspired by @Dvij D.C. 's comment]

  • $\begingroup$ I think the OP is asking for the physical interpretation of the invariant object constructed out of $(\vec{k},\omega)$ itself, i.e., that of $|\vec{k}|^2-\omega^2$. Not sure. $\endgroup$
    – Dvij D.C.
    May 4 '20 at 14:52
  • $\begingroup$ I see your point. That may be. $\endgroup$
    – robphy
    May 4 '20 at 15:01
  • $\begingroup$ Thank you for the answer. The first line is exactly what I was looking for. $\endgroup$
    – TaeNyFan
    May 4 '20 at 15:44

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