Motion satisfies the wave equation if the following holds true $$\nu^2 * \frac {\partial ^2 \psi} {\partial x^2} = \frac {\partial ^2\psi} {\partial t^2}$$ where $\psi$ is the vertical displacement of the medium (if a transverse wave), $x$ is the position of each particle whose displacement is being considered / the position at which the displacement is considered, $t$ is the time elapsed, and $\nu$ is the velocity of the wave in the transverse direction.

Also, $$\nu^2 =\left(\frac {\partial x}{\partial t}\right)^2$$

from most general considerations of velocity.

My question is this:

I think that the velocity of the wave can be interpreted as 'the speed at which the $x$-value at which a certain displacement can be found changes'. That would make sense, given that the velocity is also $\frac {\partial x}{\partial t}$, and this is, literally, the rate of change of a position on a string / in a medium, wrt time. Why is it that this, the velocity, is commensurate with $\frac {\partial x}{\partial t}$, when $\frac {\partial x}{\partial t}$ implies, I think, that position of a particle / part of the medium is changing with respect to time, i.e that $x$ and $t$ are not independent? They seem from prior consideration to be independent.

In the wave equation, $\psi$ is varying with respect to time and position, which is why the derivatives make sense when taken wrt those values. The positions $x$, are positions in the medium (say, along a string), and they are fixed points, which are used as reference points for calculating amplitude ($\psi$).

So how does it come that there is a formula, and we work with it, where $x$ and $t$ don't seem to be independent variables, because we take the derivative of one with respect to the other, and it has meaning?

Is it just a mathematical formality, because we have something that can be written in terms of $x$ and in terms of $t$, and so they can be written in terms of each other (and the function is differentiable), so a derivative exists?

How comes it (if it does) to have that meaning I ascribed to it earlier, of the change in position, $x$ with respect to $t$, where a particular amplitude of the wave can be found (which is what the velocity is)?

  • 2
    $\begingroup$ $x$ and $t$ are both independent variables; taking $\frac{\partial x}{\partial t}$ doesn't make much sense in the first place. $\endgroup$ Dec 9, 2018 at 11:02
  • $\begingroup$ I think so... But I don't understand what v can be if not that. They look very independent. But, how do I understand the horizontal velocity of the wave - it seems as if the varying thing, for the horizontal velocity, is the place a displacement $\psi$ can be found. How do I relate this to the original wave equation and the quantities there? $\endgroup$
    – Basican
    Dec 9, 2018 at 11:03
  • $\begingroup$ re "and ν is the velocity of the wave in the transverse direction" --should that not be longitudinal? v is in the direction of wave propagation. $\endgroup$
    – user45664
    Dec 9, 2018 at 18:01

1 Answer 1


Your description of the horizontal velocity as "the speed at which the $x$-value at which a certain displacement can be found changes" is perfectly fine, and is usually termed the "phase velocity" when discussing solutions of wave equations.

It comes directly from the fact that the solutions may be written as functions of a combined variable like this: $\psi(x,t)=\psi(\phi)$ where $\phi=x-vt$, or $\phi=x+vt$. More generally, any linear combination of $\psi(x-vt)$ and $\psi(x+vt)$ will be a solution.

If we fix $\phi$, the value of $\psi$ remains constant. The positions $x$ at which $\phi$ (the phase) remains constant satisfy $x=vt$, or $x=-vt$, if we have one of these elementary solutions.

Although $x$ and $t$ are indeed independent variables, if one focuses on a point of constant $\phi$, such as the crest of a wave, as a function of time, a velocity $v=\pm x/t$ emerges from the equation $\phi=$ constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.