The wave equation is given by


I am not interested in the derivation ,I am more interested in the physical meaning behind this equation ,

Why should the second derivative of the displacement of the particle with respect to time be equal to some constant times the second derivative of the displacement of the particle with respect to position.

  • $\begingroup$ From $x = vt$ we have $\frac{\partial}{\partial t} = \frac{\partial}{\partial (\frac{x}{v})} = v \frac{\partial}{\partial x}$. So solutions of $\frac{\partial}{\partial t} f(t,x)= v \frac{\partial}{\partial x}f(t,x)$ will have $x$ and $t$ related by $x = vt$, i.e. be functions $f(x - vt)$, positively traveling waves. But what about $x = - vt$? We now have $\frac{\partial}{\partial t} = - v \frac{\partial}{\partial x}$, i.e. negatively travelling waves $f(x + vt)$. What equation would naturally descibe situations with positively and negatively traveling waves simultaneously? You guessed it. $\endgroup$
    – bolbteppa
    Dec 7, 2021 at 18:42
  • $\begingroup$ See this answer. $\endgroup$
    – march
    Dec 7, 2021 at 19:55
  • $\begingroup$ @march ,your answer was good ; but I have a question the curvature is given by ,$\dfrac{dT}{ds}$ ,,where $T$ is the unit $Tangent$ vector ,but the curvature according to you is $\dfrac{d²u}{dx²}$,I don't get it . $\endgroup$
    – Harry Case
    Dec 8, 2021 at 4:08
  • $\begingroup$ I was using curvature in the colloquial sense, not in the "differential geometry of curves" sense: I just meant that how the function is curved determines how much it pulls back. The correct term to use there is the "concavity", which is exactly what I use later on. $\endgroup$
    – march
    Dec 8, 2021 at 4:11
  • 1
    $\begingroup$ Well, it needs to be there for the units to work out, and then you solve the equation and see that it happens to be the speed of the waves. Frankly, I think the best intuition for the wave equation come from the derivation for it. My explanation of how you take the difference between the derivatives on either side to get the net force is part of the derivation. I've just extracted that bit to explain the spatial derivatives. A heuristic explanation will never explain all the constants in an equation; that requires the derivation. $\endgroup$
    – march
    Dec 8, 2021 at 16:18

1 Answer 1


The physical meaning of different terms in wave equations depends on the context, we are studying. Supposing this to mechanics. It's easy to see that the term $$\partial^2_t\psi(x,t)$$ is the acceleration of the particle. Here $\psi$ is the displacement of the particle. Furthermore, consider the system to be the chain of particles connected with spring. The acceleration now comes from spring force which in term depends on relative displacement from the mean.

Since a particle is connected to two-particle, one on the left and the other on the right, we have two differences (single derivative) from the mean which in turn take the form of a double derivative. $$(\Delta \psi)_\text{left}-(\Delta \psi)_\text{right}=\Delta^2\psi$$

That's the reason why there's a double derivative with respect to space coordinate.

  • $\begingroup$ Shouldn't that be $\nabla^2 \psi$? "which in turn take the form of a double derivative" $\endgroup$
    – Gert
    Dec 7, 2021 at 21:09
  • $\begingroup$ I am referring to a single dimension $\endgroup$
    – Harry Case
    Dec 8, 2021 at 2:38

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.