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From The Feynman Lectures:

"The next subject we shall discuss is the interference of waves in both space and time. Suppose that we have two waves travelling in space. We know, of course, that we can represent a wave travelling in space by $e^{i(ωt−kx)}$. This might be, for example, the displacement in a sound wave. This is a solution of the wave equation provided that $ω^2=k^2c^2$."

https://www.feynmanlectures.caltech.edu/I_48.html

Is that simply because we want to use positive numbers here?

He goes on to say:

"In this case we can write it as $e^{−ik(x−ct)}$, which is of the general form $f(x−ct)$. Therefore this must be a wave which is travelling at this velocity, $ω/k$, and that is $c$ and everything is all right."

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  • $\begingroup$ What do you mean by "want to use posivite numbers"? $\endgroup$
    – noah
    Feb 9, 2021 at 7:12
  • $\begingroup$ i didn't really know tbh. i should've said +/-. $\endgroup$
    – user288251
    Feb 10, 2021 at 3:41

2 Answers 2

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Well, the wave equation is $$ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} $$ where $c$ is just a constant, and has the interpretation of wave velocity (by simple dimensional analysis). Putting in $u(t, x) = e^{i(\omega t−kx)}$: $$ \frac{\partial^2}{\partial t^2} e^{i(\omega t−kx)} = c^2 \frac{\partial^2}{\partial x^2}e^{i(\omega t−kx)}\\ -\omega^2 e^{i(\omega t−kx)}=c^2(-k^2)e^{i(\omega t−kx)} $$ Since this should hold for arbitrary $t$ and $x$, $$ -\omega^2=-c^2k^2\implies\omega^2=c^2k^2 $$

The square is important here. This allows $c=\pm \frac\omega k$, which tells us that the solutions to the linear wave equation are a combination ("superposition") of left-moving waves (with velocity $-\frac\omega k$) and right-moving waves (with velocity $+\frac\omega k$).

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    $\begingroup$ Just a small comment: From the wave equation you can indeed deduce that $c$ must be a velocity, because its dimension must be $LT^{-1}$. $\endgroup$ Feb 9, 2021 at 10:09
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It helps to also recall the four-momentum invariant relation $p_\mu p^\mu = m^2$ which for the above case, m=0 gives, $(p^0)^2 - \textbf{p}^2 = 0$ where $\textbf{p}$ is the spatial momentum. The zeroth component is energy/c, $\hbar \omega/c$ and the momentum is given by $]textbf{p} = \hbar k$. Plug in, to get the above result.

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