BELIEVE   ME   NOT!    - -     A   SKEPTIC's   GUIDE  

Speed of Propagation

Neither of the images in Fig. 14.1 captures the most important qualitative feature of the wave: namely, that it propagates -- i.e. moves steadily along in the direction of $\Vec{k}$. If we were to let the snapshot in Fig. 14.1b become a movie, so that the time dependence could be seen vividly, what we would see would be the same wave pattern sliding along the graph to the right at a steady rate. What rate? Well, the answer is most easily given in simple qualitative terms:

The wave has a distance $\lambda$ (one wavelength) between "crests." Every period $T$, one full wavelength passes a fixed position. Therefore a given crest travels a distance $\lambda$ in a time $T$ so the velocity of propagation of the wave is just

$$c \; = \; {\lambda \over T} \qquad \hbox{\rm or} \qquad c \; = \; {\omega \over k}$$ (14.6)

where I have used $c$ as the symbol for the propagation velocity even though this is a completely general relationship between the frequency $\omega$, the wave vector magnitude $k$ and the propagation velocity $c$ of any sort of wave, not just electromagnetic waves (for which $c$ has its most familiar meaning, namely the speed of light).

This result can be obtained more easily by noting that $A$ is a function only of the phase $\theta$ of the oscillation,

$$\theta \; \equiv \; k x - \omega t$$ (14.7)

and that the criterion for "seeing the same waveform" is  $\theta =$ constant  or  $d\theta = 0$. If we take the differential of Eq. (7) and set it equal to zero, we get

$$d\theta \; = \; k \, dx \, - \, \omega \, dt \; = \; 0 \quad \mbox{\rm or} \quad k \, dx \; = \; \omega \, dt$$

$$\mbox{\rm or} \quad {dx \over dt} = {\omega \over k} .$$

But  $dx/dt = c$,  the propagation velocity of the waveform. Thus we reproduce Eq. (6). This treatment also shows why we chose   ${\displaystyle e^{-i \omega t}}$  for the time dependence so that Eq. (7) would describe the phase: if we used   ${\displaystyle e^{+i \omega t}}$  then the phase would be   $\theta \; \equiv \; k x + \omega t$  which gives  $dx/dt = -c$,  - i.e. a waveform propagating in the negative $x$ direction (to the left as drawn).

If we use the relationship (6) to write   $(kx - \omega t) = k(x - ct)$,  so that Eq. (4) becomes

$$A(x,t) \; = \; A_{_0} \; e^{ik(x - ct)} ,$$

we can extend the above argument to waveforms that are not of the ideal sinusoidal shape shown in Fig. 14.1; in fact it is more vivid if one imagines some special shape like (for instance) a pulse propagating down a string at velocity $c$. As long as $A(x,t)$ is a function only of  $x' = x - ct$, no matter what its shape, it will be static in time when viewed by an observer traveling along with the wave^14.5at velocity $c$. This doesn't require any elaborate derivation;  $x'$  is just the position measured in such an observer's reference frame!

Footnotes

. . . wave^14.5

Don't try this with an electromagnetic wave! The argument shown here is explicitly nonrelativistic, although a more mathematical proof reaches the same conclusion without such restrictions.