Water wave analog of Casimir effect — Does it involve the zeta function? If not, why do QED calculations involve the zeta function? It is known that the Casimir effect has a water wave analog from classical wave theory. See also this video for a demonstration.
In what ways are the calculations for the effect in QED and the effect in classical wave mechanics different?
In particular, I am wondering whether the water wave calculations involve zeta-regularization. Do they at any point make use of the fact that $\zeta(-1) = -1/12$? If the answer is no, then why does QED require the use of zeta functions while water wave calculations don't?
 A: At some point in the calculation of the Casimir effect, you have to sum over the energy contained in each of mode of the cavity
\begin{equation}
E = \sum_n E_n
\end{equation}
where $n$ labels the mode number.
In electrodynamics, Maxwell's equations imply the dispersion relationship $E = p c$. Quantum mechanics implies $p = \hbar k$. For simplicity, let's just suppose we are in one spatial dimension. Then the boundary conditions imposed by the plates imply that $k = \pi n/L$ (where $L$ is the separation between the plates), and $n=1, 2, \cdots$. Putting this together, the Casimir energy will involve a sum like
\begin{equation}
E = \frac{\pi \hbar c}{L} \sum_{n=1}^\infty n
\end{equation}
where you see the famous sum over integers that can be evaluated using zeta function regularization (although, it can also be evaluated in other ways).
The calculation for water waves will be a different, since $E \sim \omega^2$ instead of $E\sim \omega$. Actually, (crucially) $E\sim A^2 \omega^2$, where $A$ is the amplitude of the water wave. Quantum mechanically, in the vacuum state, $A$ undergoes zero-point oscillations. Classically, $A$ is a free parameter of the system. To make a connection with the quantum calculation, I will assume that each mode of the water is excited with the same amplitude, but I suspect that this is a very hard condition to satisfy in practice, and I'll return to this important point below.
According to wikipedia, for small amplitude water waves (which we can treat as linear) that are shallow, the relationship $\omega = c_w k$ holds, where $c_w=\sqrt{g h}$ is the speed of sound in shallow water ($g$ is the acceleration due to gravity and $h$ is the depth of the water). In this limit (again focusing on the one-dimensional case for simplicity), you would therefore expect a sum like
\begin{equation}
E \sim \sum_n n^2
\end{equation}
Meanwhile, for deep water waves, $\omega \sim \sqrt{k}$, which would lead to a sum more like
\begin{equation}
E \sim \sum_n n
\end{equation}
Having said all of that, here are some further thoughts.

*

*In the QED case, there is no known cutoff for the sum over modes, which leads to an infinite result requiring regularization. For water waves, there is a known cutoff -- at some point, the approximation that water is a smooth fluid breaks down because water is made of atoms.


*There was that $A$ parameter we swept under the rug -- the amplitude of oscillations each mode. I suspect that when an experimenter excites a water tank, there is really a frequency dependence to $A$ such that low frequency modes have a higher amplitude than high frequency modes. This will introduce additional $n$ dependence in the sum that I didn't account for above, by replacing
\begin{equation}
\sum_n n \rightarrow \sum_n A_n^2 n
\end{equation}
where $A_n$ is some function that decreases with $n$ that you'd have to try to measure empirically, or with a simulation.  I suspect that $A_n$ actually plays the role of making the sum finite in the water case, and is the main difference between the water analog Casimir effect, and the actual QED Casimir effect.


*The paper you are citing is making an analogy between an experiment you can do with water, and the Casimir effect in QED. There are some limits (small amplitudes, deep water) where the math looks similar to QED. But, when these approximations break down, the water tank will simply have different behavior than QED. There is no analog of water turbulence in QED in the vacuum state, for example.


*There is a more general notion of zeta function than the Riemann zeta function, where one is summing over eigenvalues of an operator to some power, instead of specifically the integers $n$ to some power. You would expect this generalized form of the zeta function to appear in quantum mechanical Casimir effect calculations, since you are always summing the energy present in the different modes, and in quantum mechanics the energies of each mode are eigenvalues of a Hamiltonian operator.


*There's nothing particularly physical about the use of zeta function regularization. My friendly advice would not be to get too excited about the appearance of the zeta function, which is just a mathematical tool, and instead to focus on the physical similarities and differences between these cases that are invariant under choices of how to do the calculation.
