The simple wave equation is a really great equation, and it’s the starting point for lots of my fascinating stories in physics. Everyone studying physics and applied mathematics should be intimately familiar with it. It’s the quintessential equation used to describe waves, and in fact what you learn about it remains useful and nearly unchanged even if you’re describing sound waves, water waves, electromagnetic waves, or gravitational waves. But I recently found out there was something I didn’t know about this equation which I otherwise know like the back of my hand!

Let’s consider the simple wave equation for a long spring pulled tight. The height of the spring at a point $x$ is denoted $h(x,t)$, the spring has mass density $\mu$, and the spring is under tension $T$. Then the simple wave equation is a partial differential equation (PDE) which states the following:

A common project in studying this equation is to analyze the situation where we glue two different springs together with two different mass densities. Essentially, we let $\mu$ be a function of space, so our equation is now:

We can use this equation to describe the evolution of a wave pulse on two connected springs. In the following picture, the green spring is higher density than the blue spring. When the wave pulse goes from the higher density spring (which has a slower wave speed) to a lower density spring (higher wave speed), the wave pulse is partially transmitted and partially reflected. A series of still images looks as follows:

Great. This is something that every physics student learns and there are many textbooks out there with sections titled “transmission and reflection coefficients” where you work out this fact and find the magnitudes of the transmitted and reflected waves. I myself learned it in chapter 7 of A. P. French’s book “Vibrations and Waves”, and even though it seems very basic (“yes dear, I’m a serious physicist. Those slinkies are for a serious experiment!”) this same transmission-reflection calculation is widely applicable to acoustics, electromagnetic waves, seismic waves, and so on. For a better picture of why physicists care about slinkies I recommend this excellent 1959 presentation on simple waves by N. German.

But what happens if we allow the tension to vary over the length of the spring? Well, if a piece of the spring had a greater tension pulling it to the left than to the right, it would accelerate to the left or right instead of only in the vertical direction. Therefore we can’t just glue two different springs together “of a different tension” or anything like that. Instead, we have to have a physical board or pole which prevents any horizontal acceleration, but which is frictionless and so doesn’t impede vertical motion. Such a setup would look like the following diagram:

The grey pole in the middle makes sure that forces can still be transmitted through vertical displacement of the spring, but that horizontal forces can’t be. Can we just replace the constant $T$ with a function $T(x)$? No! The correct wave equation is…

This has a rather funny implication. In the first figure, when a wave pulse went from a medium with a slower wave speed to a medium with the faster wave speed, the reflected pulse had the same sign as the incident pulse. It started as an upwards pointing lump, and it ended as an upwards pointing lump. But this second equation predicts a different scenario: If the mass density stays the same but the tension changes then reflected pulse will have a negative sign! That is, the reflected pulse will point in the opposite direction that it started!

This was a bit of a shock to me, I thought that the phase of the reflected wave only depended on the change in wave speed, but in fact it can depend on whether that change was due to density or due to other factors!

(Note: I haven’t performed this experiment with real slinkies yet, but I did go rifling through various closets to try to find a slinky. I also ran a few simulations confirming things, but that’s not as much fun as having real pictures to show you!)

How the Wave Equation Comes About

Why is equation 1 the correct equation? Or, phrased differently, why do we have a partial derivative that hits the function $T(x)$ but not the function $\mu(x)$? Well, I have to think of it from an energy perspective. The energy of our system is as follows:

We can use this energy density to get equations of motions by considering Hamilton’s equations of motion for a field. If I weren’t so lazy, I could come up with some good exposition where I derived the equations of motion without ever using the word “Hamiltonian”, and indeed many good books on partial differential equations do this! Every PDE course I’ve seen has a section on deriving PDEs through different methods. Two good references here are Strauss’s book “Partial Differential Equations: An Introduction”, and “Nonlinear Partial Differential Equations for Scientists and Engineers” by Debnath. Chapter 2 of Debnath’s book discusses deriving nonlinear PDEs through the methods I describe here.

The basic idea is to take our energy density and write it as $\mathcal{H}(h_t,h_x,h,x)=\frac{1}{2}\mu(x) h_t^2+\frac{1}{2}T(x)h_x^2$, where I have written $h_t=\frac{\partial h}{\partial t}$ and $h_x=\frac{\partial h}{\partial x}$ as shorthand. The equation of motion is then:

The first term on the right-hand side is zero for our case, but I included it because it shows this equation’s relation to Newton’s $F=ma$. Newton’s physics predicts $m\ddot{h}=-V’(h)$, which is exactly our $-\partial\mathcal{H}/\partial h$ term. If Debnath’s book doesn’t cover this adequately enough for you, you can also refer to Goldstein’s book “Classical Mechanics” 3rd ed. where this is equation 13.62. This equation works for any $\mathcal{H}$ of the form $\mathcal{H}=\frac{1}{2}\mu(x) h_t^2+F(h_x,h,x)$ and gives us equations of motion that conserve total energy!

Granted, I pulled that equation out of a hat. But the point is that it gives us a framework where we can see that $\mu(x)$, being attached to the kinetic term, never gets differentiated with respect to $x$. The special thing about the tension $T(x)$ is that it’s part of the potential term, and this does get differentiated with respect to $x$.

Okay, so equation 1 is the right equation, let’s repeat the problem of transmission-reflection.

Transmission-Reflection for the wave equation

Transmission-reflection is a bit of a tedious calculation. We assume that we start with a wave moving to the right, which can be described by the wave which moves to the right with velocity $v_1$, $h_1=e^{i\omega(t-x/v_1)}$. This wave hits a change in a medium and it is partially reflected, giving a left-moving component of the wave $h_2=A e^{i\omega (t+x/v_1)}$. It’s also partially transmitted, giving a right-moving component of the wave in the new medium with velocity $v_2$, $h_3=B e^{i\omega (t-x/v_2)}$.

Then we plug this whole piecewise statement into the wave equation. My favorite way to solve this problem is actually to define the functions piecewise, and wherever there’s a discontinuity we get Dirac delta functions! I describe the code used to do this in Mathematica in the appendix. But from our PDE, we find that $B=1+A$ (which we could also have found by considering conservation of energy). If the tension were constant along the spring, we would impose the condition that the derivative of our height function must be constant, but that’s not the case any more! Solving $A$ for our new problem gives:

If the mass density is the same throughout, then this equation easily simplifies to:

If the tension is the same throughout, then with a little algebra this equation simplifies to:

How confusing is that?!!

Interestingly, it implies that if we choose $T_2=T_1 (v_2/v_1)$ and $\mu_2=\mu_1 (v_1/v_2)$, we get total transmission of the signal even though the wave speed in the medium changes! This is called impedance matching, and though I knew about it in circuits I didn’t know you could do it with springs by varying the tension (although like I said, “varying the tension” is quite a weird thing to do).

I believe that if we did this whole procedure for the case of electromagnetic waves, we would find that the electric permittivity $\varepsilon$ plays the role of the mass density, and the magnetic permeability $\mu$ plays the role of the tension. This makes me wonder if there are devices out there that make use of tuning $\mu$ and $\varepsilon$ together in order to get total transmission! But the permeability of most materials is very close to the permeability of the vacuum, so it might be impossible to tune, cf. section 9.3.2 of Griffiths Introduction to Electrodynamics, 3rd ed.

Appendix 1, Mathematica source code

I didn’t do the algebra by hand, I used Mathematica to solve it. I define mu and T in terms of the Heaviside theta function, e.g. $\mu(x)=\mu_1(1-\theta(x))+\mu_2\theta(x)$, and at the origin we have to substitute $\theta(0)=1/2$ (for us, this amounts to the assertion $\int (1/2-\theta(x))\delta(x)dx=0$).

Clear[w,v1,v2,mu,T,u1,u2,u3,u,a,b];
mu[x_]=mu1(1-HeavisideTheta[x])+mu2 HeavisideTheta[x];
T[x_]=T1(1-HeavisideTheta[x])+T2 HeavisideTheta[x];
b=1+a;
u1[x_,t_]=E^(I (w/v1 x-w t));
u2[x_,t_]=a E^(I (-w/v1 x-w t));
u3[x_,t_]=b E^(I (w/v2 x-w t));
u[x_,t_]=(1-HeavisideTheta[x])(u1[x,t]+u2[x,t])+HeavisideTheta[x]u3[x,t];
assumptions={v1^2 mu1==T1,v2^2 mu2==T2};
expression = Simplify[Table[Coefficient[FullSimplify[mu[x]Derivative[0,2][u][x,t]-D[T[x]D[u[x,t],x],x],Assumptions->assumptions],DiracDelta[x],i],{i,0,2}],Assumptions->assumptions]

The core PDE going into the problem is the formula mu[x]Derivative[0,2][u][x,t]-D[T[x]D[u[x,t],x],x], which should be zero. If we plug our functions defined in terms of products of HeavisideTheta into this equation, we get a bunch of terms which I organize in a power series in $\delta(x)$. The term proportional to $\delta(x)^0$ is zero almost everywhere because of the way we’ve constructed our functions. There is a nonzero term proportional to $\theta(x)(1-\theta(x))$, but this is equal to zero (weakly). Next, there’s a term $\delta(x)^2$ which is proportional to $1+a-b$, so we must choose b such that this vanishes. And finally there is the term proportional to $\delta(x)$, which we can evaluate by substituting expr2=expression[[2]] //. {x->0,HeavisideTheta[0]->1/2}, and finally solving for our a coefficient.

In[]:= Solve[expr2==0,a]
Out[]:= { { a->(-T2 v1+T1 v2)/(T2 v1+T1 v2) } }