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I was re-reading Eugene Hecht's textbook on Optics, and he derived the one-dimensional differential wave equation:

$$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2} \ ,$$

using the "foreknowledge", what I want to know is from where does this equality find its basis, from physical interpretation alone.

I understand the mathematics behind it, but I am unable to imagine how this equality would work, when looking at a specific point at a certain time instant on a wave.

Also, what's the role of $\frac{1}{v^2}$ in that physical interpretation?

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Aug 4 at 10:28
  • $\begingroup$ As you said in the comment bellow, in mechanical waves this equation arises from newton's second law, a=F/m, where the temporal derivative is the acceleration and the spatial derivative comes from calculation the forces acting on each point of the field. It makes sense that as psi is most of time defined as the deviation from the steady state, in most cases you will see that the force is related to how much the field in the particular point deviates from yhe average of its surroundings, meaning d2/dx2 $\endgroup$
    – Ofek Gillon
    Commented Aug 4 at 15:16

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Have you ever flicked the end of a string or cord up and down to produce a "bump" that travels along the cord? (If you haven't, see here for something along these lines.) The travel of that bump is what the wave equation describes. More precisely, the wave equation forces its solutions to take either the form \begin{align} \psi(x,t) = f(x - vt) \end{align} or \begin{align} \psi(x,t) = g(x + vt) \end{align} or a sum of the two forms, $\psi(x,t) = f(x-vt) + g(x+vt)$. $f$ and $g$ have, respectively, the interpretation of a waveform (which can take any shape) traveling either in the positive or negative $x$ direction at speed $v$. (If you're having trouble seeing this, recall that adding to or subtracting from the argument of a function shifts the graph of the function. In this case, the waveform is being shifted to the right or left by a distance $vt$ that increases with time at rate $v$.)

Since the wave equation is all about enforcing the dynamical travel of its solutions, it's not really possible to interpret it at a given point and at a given time. Instead, one should think about a "video" (like the YouTube video I linked above) showing how the wave pulse moves in time.

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  • $\begingroup$ I think I should have been more specific in my question about what I am not able to understand, I understand the wave equation you provided above and how it represents a wave on a string like you said, what I don't understand is how to interpret one-dimensional differential equation where there's equality between second derivative with respect to time (which might indicate acceleration) and second derivation with respect to position (which suggests the curvature at that point). Kindly let me know if I need to be more specific or if you need anymore information. $\endgroup$
    – Harsh
    Commented Aug 4 at 10:55
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    $\begingroup$ @Harsh I see. Thanks for clarifying. Indeed, the derivatives have their usual meaning here: the second time derivative is the acceleration, and the second position derivative is the curvature of the waveform. I will try to elaborate on why this makes sense when I get some time (and if someone else doesn't beat me to it). $\endgroup$
    – d_b
    Commented Aug 4 at 19:38

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