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Travelling waves are of the form $f(x-ct)$, where $c$ is the speed.

Now if we have something like $$ u(x,t)=e^{i(kx-\omega t)} \tag{$*$} $$ when I see it right, we can write this as $$ u(x,t)=e^{\frac{i}{k}(x-ct)},\quad c:=\omega/k. $$

Hence, am I right to say that $(*)$ is a travelling wave, winding around the $x-$axis to the right with speed $\omega/k$?

What I am a bit confused about is that we now have the factor $1/k$. Moreover, when considering $f(x-ct)$ we have one speed $c$ which is multiplicatd with $t$; here, we seem to have two kind of speeds (some spatial, namely $k$, which is multiplied with space $x$ and some temporal, namely $\omega$, which is multiplied with time $t$. In other words, I am not sure what $kx-\omega t$ actually means.

mathfemi
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    Yes that's the phase speed – aw80 Aug 16 '16 at 12:56
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    The peaks occur at $kx - wt = 2 \pi j$ where $j \in \mathbb{Z}$ , integer. If you consider this peak to be moving in time then $\displaystyle k \frac{dx}{dt} - w = 0$ which gives $\displaystyle \frac{dx}{dt} = \frac{w}{k}$ –  Aug 16 '16 at 13:00
  • You're on the right track, but: (a) What do you mean by "winding around the $x$-axis"? (b) Your second form for $u(x,t)$ is inconsistent with your first: $e^{i(kx - \omega t)} = e^{i k(x - ct)} \neq e^{k} e^{i(x - ct)}.$ – Michael Seifert Aug 16 '16 at 13:15
  • @MichaelSeifert (a) When plotting $$e^{i/k(x-ct)},~~c:=\omega/k$$ this is a "helix" winding around the x-axis, isn't it? (b) you are right. (c) I have some problems to understand what $kx-\omega t$ "means". – mathfemi Aug 16 '16 at 13:28

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Let $\phi=kx-\omega t$ be the phase of the wave. Then, note that when the phase $\phi$ is constant, then $kx-\omega t$ is constant also.

Suppose now that we move a small distance $\Delta x$. What is the commensurate change in time $\Delta t$ such that $\phi=kx-\omega t$ remains constant? That is, what is $\Delta t$ such that

$$kx-\omega t=k(x+\Delta x)-\omega (t+\Delta t) \tag 1$$

Solving $(1)$, we find that $\Delta t=\frac{k}{\omega }\Delta x$ or

$$\frac{\Delta x}{\Delta t}=\frac{\omega}{k } \tag 2$$

Hence, we need to move with a speed equal to $\frac{\omega }{k}$ in order to move along with constant phase $\phi$.

Mark Viola
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