As described by Jackson's Classical Eletrodynamics [1], the phase of a plane wave is an invariant quantity for all coordinate frames. Jackson justifies this statement using the fact that the phase of a plane wave is proportional to the number of wave crests that have passed by the observer.

I have several problems to find a link between these two statements. How is the invariance of the phase linked with the fact that it's a wave-crest counter?

Using the Lorentz transformation laws for the 4-wavevector $(\omega/c, \vec{k})$ I can reach a simple expression for the frequency seen by a rest frame,

$$\begin{eqnarray} \omega'=\omega \end{eqnarray} \gamma\left(1-\beta\cos\theta\right)$$

where $\theta$ is the angle between wave propagation's vector $\vec{k}$ and the velocity vector of the motion system $\vec{\beta}$. This equation states that the frequency varies between different reference systems, so it changes the number of wave crests that pass by the observer in function of whether he is placed in the rest reference system or in one that is in motion.

So, I don't understand the Jackson assertion.

  1. Jackson, John David. Classical Electrodynamics. John Wiley & Sons, 2007. APA p. 519.
  • $\begingroup$ The way Jackson's text sees things, "the number of wave crests that have passed by the observer" is what the phase is, at its core, so in some ways it's a tautology. That's not to say that it's trivial, but to be able to show the relevant connections, we need to know where you are starting from, so ─ how are you defining the phase of a plane wave, to begin with? $\endgroup$ – Emilio Pisanty Jul 6 '17 at 14:53
  • $\begingroup$ @Luc is right, i considered $\omega$ as counter of wave crests , but it's wrong because it's a number in time unity. $\endgroup$ – Algebrato Jul 6 '17 at 15:25
  • $\begingroup$ Ah, I see. No, the problem is that you're confusing the phase ($\phi=\omega t-\vec k\cdot\vec x$, dimensionless) with the angular frequency ($\omega$, with dimensions of inverse time). The counter of wave crests is $\phi$, not $\omega$. $\endgroup$ – Emilio Pisanty Jul 6 '17 at 15:53
  • $\begingroup$ Another simple way to see this, in the simple case of electromagnetic waves, is that there is a transformation law for the fields $(E,B)\rightarrow(E',B')$, and this transformation law is such that $(0,0)\rightarrow(0,0)$. Therefore all observers will agree on the events in spacetime at which the fields both vanish. $\endgroup$ – user4552 Jul 6 '17 at 17:08
  • $\begingroup$ Sorry @BenCrowell, your example is not clear. The most easy situation is for $E$, $B\perp \beta$. For this fields the trasformation law become: $E'_\parallel = E_\parallel$, $E'_\perp =\gamma(E_\perp+\beta \times B )$ and $B'_\parallel = B_\parallel$, $B'_\perp =\gamma(B_\perp-\beta \times E )$, so you can't direct conclude that there is a transformation law such that all field vanished. No? $\endgroup$ – Algebrato Jul 6 '17 at 21:49

Think of the unit of $\omega$. It's in $s^{-1}$, clearly showing that $\omega$ is not a count but a count per unit of time, if you want to interpret it as such. Now think of a short signal for which the wave has only a finite number of periods, let's say it propagates along the x-axis for simplicity, and that we consider observers moving along only along the x-axis, slower than the celerity of the wave. Any such observer will see the following. For a while, the medium is undisturbed. Then the medium will "rise" and reach a maximum, then "fall" and reach a minimum, and carry on do that for a while, after which point, the medium will be undisturbed again. Every single observer will agree about the number of maxima, and minima he has seen.


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