The stretched membrane
Now let us consider a completely different physical situation which, nevertheless, gives the same equations again. Consider a thin rubber sheet—a membrane—which has been stretched over a large horizontal frame (like a drumhead). Suppose now that the membrane is pushed up in one place and down in another; as shown in Fig. 12–3. Can we describe the shape of the surface? We will show how the problem can be solved when the deflections of the membrane are not too large. There are forces in the sheet because it is stretched. If we were to make a small cut anywhere, the two sides of the cut would pull apart (see Fig. 12–4). So there is a surface tension in the sheet, analogous to the one-dimensional tension in a stretched string. We define the magnitude of the surface tension $\tau$ as the force per unit length which will just hold together the two sides of a cut such as one of those shown in Fig. 12–4. Suppose now that we look at a vertical cross section of the membrane. It will appear as a curve, like the one in Fig. 12–5. Let $u$ be the vertical displacement of the membrane from its normal position, and $x$ and $y$ the coordinates in the horizontal plane. (The cross section shown is parallel to the $x$-axis.) Consider a little piece of the surface of length $\Delta x$ and width $\Delta y$. There will be forces on the piece from the surface tension along each edge. The force along edge $1$ of the figure will be $\tau_1\,\Delta y$, directed tangent to the surface—that is, at the angle $\theta_1$ from the horizontal. Along edge $2$, the force will be $\tau_2\,\Delta y$ at the angle $\theta_2$. (There will be similar forces on the other two edges of the piece, but we will forget them for the moment.) The net upward force on the piece from edges $1$ and $2$ is \begin{equation*} \Delta F=\tau_2\,\Delta y\sin\theta_2-\tau_1\,\Delta y\sin\theta_1. \end{equation*}  We will limit our considerations to small distortions of the membrane, i.e., to small slopes: we can then replace $\sin\theta$ by $\tan\theta$, which can be written as $\ddpl{u}{x}$. The force is then \begin{equation*} \Delta F= \biggl[\tau_2\,\biggl(\ddp{u}{x}\biggr)_2- \tau_1\,\biggl(\ddp{u}{x}\biggr)_1\biggr]\Delta y. \end{equation*}  The quantity in brackets can be equally well written (for small $\Delta x$) as \begin{equation*} \ddp{}{x}\biggl(\tau\,\ddp{u}{x}\biggr)\Delta x; \end{equation*}  then \begin{equation*} \Delta F= \ddp{}{x}\biggl(\tau\,\ddp{u}{x}\biggr)\Delta x\,\Delta y. \end{equation*}  There will be another contribution to $\Delta F$ from the forces on the other two edges; the total is evidently \begin{equation} \label{Eq:II:12:16} \Delta F=\biggl[ \ddp{}{x}\biggl(\tau\,\ddp{u}{x}\biggr)+ \ddp{}{y}\biggl(\tau\,\ddp{u}{y}\biggr) \biggr]\Delta x\,\Delta y. \end{equation}  The distortions of the diaphragm are caused by external forces. Let’s let $f$ represent the upward force per unit area on the sheet (a kind of “pressure”) from the external forces. When the membrane is in equilibrium (the static case), this force must be balanced by the internal force we have just computed, Eq. (12.16). That is \begin{equation*} f=-\frac{\Delta F}{\Delta x\,\Delta y}. \end{equation*}  Equation (12.16) can then be written \begin{equation} \label{Eq:II:12:17} f=-\FLPdiv{(\tau\,\FLPgrad{u})}, \end{equation}  where by $\FLPnabla$ we now mean, of course, the two-dimensional gradient operator $(\ddpl{}{x},\ddpl{}{y})$. We have the differential equation that relates $u(x,y)$ to the applied forces $f(x,y)$ and the surface tension $\tau(x,y)$, which may, in general, vary from place to place in the sheet. (The distortions of a three-dimensional elastic body are also governed by similar equations, but we will stick to two-dimensions.) We will worry only about the case in which the tension $\tau$ is constant throughout the sheet. We can then write for Eq. (12.17), \begin{equation} \label{Eq:II:12:18} \nabla^2u=-\frac{f}{\tau}. \end{equation}  We have another equation that is the same as for electrostatics!—only this time, limited to two-dimensions. The displacement $u$ corresponds to $\phi$, and $f/\tau$ corresponds to $\rho/\epsO$. So all the work we have done for infinite plane charged sheets, or long parallel wires, or charged cylinders is directly applicable to the stretched membrane. Suppose we push the membrane at some points up to a definite height—that is, we fix the value of $u$ at some places. That is the analog of having a definite potential at the corresponding places in an electrical situation. So, for instance, we may make a positive “potential” by pushing up on the membrane with an object having the cross-sectional shape of the corresponding cylindrical conductor. For example, if we push the sheet up with a round rod, the surface will take on the shape shown in Fig. 12–6. The height $u$ is the same as the electrostatic potential $\phi$ of a charged cylindrical rod. It falls off as $\ln(1/r)$. (The slope, which corresponds to the electric field $E$, drops off as $1/r$.) The stretched rubber sheet has often been used as a way of solving complicated electrical problems experimentally. The analogy is used backwards! Various rods and bars are pushed against the sheet to heights that correspond to the potentials of a set of electrodes. Measurements of the height then give the electrical potential for the electrical situation. The analogy has been carried even further. If little balls are placed on the membrane, their motion corresponds approximately to the motion of electrons in the corresponding electric field. One can actually watch the “electrons” move on their trajectories. This method was used to design the complicated geometry of many photomultiplier tubes (such as the ones used for scintillation counters, and the one used for controlling the headlight beams on Cadillacs). The method is still used, but the accuracy is limited. For the most accurate work, it is better to determine the fields by numerical methods, using the large electronic computing machines.