There is a very old equation known as the "wave equation". It's an ordinary classical non-relativistic differential equation which applies to just about every kind of ordinary wave you can imagine (fluid dynamics, acoustics, mechanical waves, etc). It can be derived from Hooke's law simply by imagining a sequence of beads on a string, where each bead has the same mass. In the continuum limit of a string with constant tension and mass per unit length, such as that on a violin, you get this:
$$
\displaystyle {\frac {\partial ^{2}u}{\partial t^{2}}}=c^{2}{\frac {\partial ^{2}u}{\partial x^{2}}}
$$
where $c = \sqrt{T/\rho}$ is the velocity of traveling waves on the string, T is the tension of the string, and $\rho$ is the mass per unit length of the string.
This equation was discovered by d’Alembert in 1746 (see, eg. https://en.wikipedia.org/wiki/Wave_equation)
The 3D version of this, for ordinary fluids, mechanical waves, etc. is:
$$
\displaystyle {\frac {\partial ^{2}}{\partial t^{2}}}u -c^2\mathbf {\nabla } ^{2}u =0.
$$
Fast forward 150+ years, and we have Einstein's theory of special relativity, which says that the energy of any body in motion is: $$\displaystyle E = \sqrt{(pc)^2 + (mc^2)^2}$$
Moving forward a few more decades, we can combine this with the assumption from quantum mechanics that momentum should not be viewed as an ordinary real number, but instead a non-commuting functional operator in a complex Hilbert space $ \displaystyle p = -i\hbar \frac{d}{dx} $. This gives us the Klein-Gordon equation, the relativistic version of the Schrodinger Equation:
$$
\displaystyle {\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\psi -\mathbf {\nabla } ^{2}\psi +{\frac {m^{2}c^{2}}{\hbar ^{2}}}\psi =0
$$
For the ultra-relativistic case when $E >> m$, this reduces to exactly the classic wave equation we derived from Hooke's law:
$$
\displaystyle {\frac {\partial ^{2}\psi}{\partial t^{2}}}=c^{2}{\frac {\nabla ^{2}\psi}{\partial x^{2}}}
$$
My question is: why does the non-relativistic non-quantum wave equation for a string on a violin look more like the wave equation for a relativistic quantum particle than a non-relativistic quantum particle? Is that purely coincidence, or could it be seen as an early hint of the equations of relativity showing up in the 1700's? Without working out the math, a priori I would have guessed that the wave equation for a non-relativistic string would look more like the Schrodinger Equation than Klein-Gordon, since relativity is not involved, ie there is no reason to expect it to have a Lorentz invariant form:
$$
\displaystyle i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi.
$$
Also, it seems interesting that this equation was first derived when thinking about the motion of a 1-dimensional string. This is a long shot, but… does this by any chance relate to the fact that quantum gravity turned out to have a consistent embedding in a quantum mechanics of strings, but not in a quantum mechanics of particles?
Best Answer
With one time dimension and one space dimension, the wave equation's most general solution is $f(t-x/c)+g(t+x/c)$ with $f,\,g$ twice-differentiable, i.e. each solution is a sum of two speed-$c$ waves, one moving right, the other left. Equivalently, the former's value at a time $t$ depends on the behaviour of the wave's source a time $x/c$ earlier, so is a "retarded" solution, while the latter is an "advanced" solution.
I've moved from the maths to the physics, but we can just as easily go the other way. If we only want the retarded them, a solution is of the form $f(t-x/c)$, and the PDE this fits is $u_t=-u_x/c$. But as soon as we work in multiple space dimensions, e.g. because we're thumping a drum skin, the requirement that both sides are scalars effects a generalization such as $u_t=-\hat{e}\cdot\nabla u/c$, with $\hat{e}$ an arbitrary unit vector. Too arbitrary. It picks out a privileged direction in space. (We can fix that with a spacetime-unifying idea like $\gamma^\mu u_\mu=0$, but let's put that aside for a moment.) If we use second-order derivatives too we avoid this problem, viz. $u_{tt}=u_{ii}$. The only downside is there are now advanced solutions too. (The general selection in multiple space dimensions looks a fair bit more complicated, but the principle is similar; we just need a $k$-space integral instead of a $2$-term sum.)
But one of the most important ideas in science is there's more than one place you can start your theory. We're not choosing axioms, then seeing which theorems they give us, like a mathematician with a preferred set theory. The universe just "is", all at once. So let's do something very ahistorical: suppose you tried to write down a relativistic quantum field theory without first knowing what classical relativity looks like. I know that sounds crazy, but bear with me.
Uniting time and space means actions become spacetime integrals of Lagrangian densities expressible in terms of fields and spacetime derivatives, rather than just time integrals of Lagrangians expressible in terms of coordinates and their time derivatives. Even before you work out ideas like Lorentz-invariance, you already realize a field ought to vibrate in a similar retarded-plus-maybe-advanced way. This quickly gives you the Klein-Gordon equation before you know Einstein's energy-momentum relation! Or if you work with a spinor with enough components, the privileged-direction idea comes off as more workable, because of a $\gamma^\mu\partial_\mu$ operator. (Hilariously, though, it actually doesn't get rid of the advanced solutions; or on a related point, it doesn't get rid of $E<0$ solutions to $E=m^2c^4+p^2c^2$.) But Lagrangians tend to give us even-order time derivatives in our EOMs, although there's a way round that.
So the real question is why does plucking a string look like Klein-Gordon rather than Dirac? (Schrödinger is out of the running because of the $f(t-x/c)$ Ansatz; we need $\nabla$ to have the same degree as $\partial_t$.) Well, the unhelpful answer is "because this derivation says so". The intuitive summary of that unhelpful answer is the string's behaviour is built on top of Newton's second law, so you need a second-order time derivative. If you think we should make our inference in the opposite direction, that won't satisfy you, in which case I'll need to motivate why the string amplitude shouldn't be a spinor with at least $4$ components. However, I suspect you don't find that negative fact surprising enough to warrant an "explanation".