I'm a mathematics student trying to grasp some basics about wave propagation. A sentence I find very often in introductive physics textbooks is the following:
In a wave, energy is proportional to amplitude squared.
This is something I would like to understand better in the case of mechanical (linear) waves.
The simplest model is the vibrating string of mass density $\mu$ and tension $T$: here an element of string of rest length $dx$ and vertical displacement $y(x, t)$ possesses a kinetic energy $\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2$ and a potential elastic energy $\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2$. So we have the total energy
$$dE=\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2+\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2,$$
which completely explains the sentence. Can we obtain a similar formula for higher-dimensional waves? How should I modify the above formula for a (simplified model of an) elastic membrane, for example? I would expect something like
$$dE=\text{some constant}\left(\frac{\partial z}{\partial t}\right)^2+\text{some other constant}\left\lvert\nabla z\right\rvert^2,$$
where $z=z(x, y, t)$ is the vertical displacement. Am I right?
Best Answer
It's not true in general that the energy of a wave is always proportional to the square of its amplitude, but there are good reasons to expect this to be true in most cases, in the limit of small amplitudes. This follows simply from expanding the energy in a Taylor series, $E=a_0+a_1 A+a_2 A^2+\ldots$ We can take the $a_0$ term to be zero, since it would just represent some potential energy already present in the medium when there was no wave excitation. The $a_1$ term has to vanish, because otherwise it would dominate the sum for sufficiently small values of $A$, and you could then have waves with negative energy for an appropriately chosen sign of $A$. That means that the first nonvanishing term should be $A^2$. Since we don't expect the energy of the wave to depend on phase, we expect that only the even terms should occur, $E=a_2A^2+a_4A^4+\ldots$ So it's only in the limit of small amplitudes that we expect $E\propto A^2$.
The other issue to consider is that we had to assume that $E$ was a sufficiently smooth function of $A$ to allow it to be calculated using a Taylor series. This doesn't have to be true in general. As an easy example involving an oscillating particle, rather than a wave, consider a pointlike particle in a gravitational field, bouncing up and down elastically on an inflexible floor. If we define the amplitude as the height of the bounce, then we have $E \propto |A|$. But a realistic ball deforms, so the small-amplitude limit consists of the ball vibrating while remaining in contact with the floor, and we regain $E\propto A^2$.
You could also make up examples where $a_2$ vanishes and the first nonvanishing coefficient is $a_4$.