I'm reading the book "A Student's Guide to Waves" by Daniel A. Fleisch and Laura Kinnaman in an attempt to get quickly up to speed in some of the terminology and maths of waves.

At some point it mentions the following equation $$\psi = y = f(x,t)$$ And that the easiest way to think about the shape of the wave is by considering a snapsnot, i.e. $$y=f(x,0)$$ Then it goes on saying "Many waves maintain the same shape over time - the wave moves in the direction of propagation, but all peaks and troughs move in unison, so the shape does not change as the wave moves. For such "non-dispersive" waves, $f(x,0)$ can be written as $f(x)$, since the shape of the wave doesn't depend on when you take the snapshot."

I don't understand why this would be true, because the phase would still be dependent on when the snapshot was taken? And if it were to be considered true, couldn't you in the same way consider a dispersive wave as a function of only location if you are talking about a snapshot only?


1 Answer 1


A one-dimensional wave that keeps its form over time has the general form $y(x, t)= f(x\pm ct)$ where $c$ is the speed of the wave. At $t=0$ one has $y(x, 0)=f(x)$. For more complicated waves this is not true: the form of the wave will be some function of two variables $g(x, t)$. While it is true that you could write $g(x, 0) = g(x)$ for some function $g(x)$, at later times the function $g(x, t)$ does not need to resemble $g(x)$. You might like to play around with this graph to get a better idea of what this means.

  • $\begingroup$ Thanks for your answer and sorry for the late response, somehow it really didn't get through to me we were only talking about the shape of the wave. But I feel like I correctly understand in that case ;) $\endgroup$
    – reveance
    May 19, 2021 at 16:03

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