Pre-Quantum Electrodynamics

{\bf Example: separation of variables (Cartesian coordinates)}
Two infinite grounded metal plates parallel to the \(xz\) plane, one at \(y = 0\) and the other at \(y = a\). End at \(x = 0\) closed off with infinite insulated strip maintained at potential \(V_0 (y)\). Find potential inside the slot. \paragraph{Solution:} indep of \(z\), so 2d problem. Solve \[ \frac{\partial^2 V}{\partial^2 x} + \frac{\partial^2 V}{\partial^2 y} = 0 \label{Gr(3.20)} \] \[ (i) V(x, y = 0) = 0, \hspace{5mm} (ii)V(x, y = a) = 0, \hspace{5mm} (iii) V(0, y) = V_0 (y), \hspace{5mm} (iv) V (x \rightarrow \infty) \rightarrow 0. \label{Gr(3.21)} \] Look for solutions of form \[ V(x,y) = X(x) Y(y), \hspace{1cm} \frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} = 0 \label{Gr(3.23)} \] Choose \[ \frac{d^2 X_n}{dx^2} = k_n^2 X_n, \hspace{1cm} \frac{d^2Y_n}{dy^2} = -k_n^2 Y_n \label{Gr(3.26)} \] where \(k_n\) is some real number. We can linearly combine solutions of (\ref{Gr(3.26)}) for different \(k_n\) and still get a solution to (\ref{Gr(3.23)}). Let's look first of all at the solutions of (\ref{Gr(3.26)}) for a given \(k_n\). Since this is a second-order linear differential equation, there are two linearly independent solutions. Most general solution: \[ X_n(x) = Ae^{k_nx} + Be^{-k_nx}, \hspace{1cm} Y_n(y) = C \sin k_ny + D \cos k_ny \label{Gr(3.27)} \] Fix constants: from \((iv)\), \(A = 0\). From \((i)\), D = 0. Left with \[ V(x,y) = C_n e^{-k_nx} \sin k_n y \label{Gr(3.28)} \] Then, \((ii)\) requires \[ k_n = \frac{n\pi}{a}, \hspace{1cm} n = 1, 2, 3, ... \label{Gr(3.29)} \] But we can use as solution any linear combination of the functions defined by these momenta. Fix coefficients with Fourier series: \[ V(x,y) = \sum_{n=1}^{\infty} C_n e^{-n\pi x/a} \sin (n\pi x/a) \label{Gr(3.30)} \] Needed: \[ \int_0^a dy \sin(n\pi y/a) \sin (n' \pi y/a) = \delta_{n n'} \frac{a}{2} \label{Gr(3.33)} \] so \[ C_n = \frac{2}{a} \int_0^a dy V_0(y) \sin(n\pi y/a) \label{Gr(3.34)} \]

{\bf Example: rectangular pipe}
Two infinitely long grounded plates at \(y = 0,a\) are connected at \(x = \pm b\) to metal strips maintained at constant \(V = V_0\). Find the potential in the resulting rectangular pipe. \paragraph{Solution:} indep of \(z\). Laplace: \(\partial^2 V/\partial x^2 + \partial^2 V/\partial y^2 = 0\), boundary conditions \[ (i) V (y = 0) = 0, \hspace{5mm} (ii) V (y = a) = 0, \hspace{5mm} (iii) V(x = b) = 0, \hspace{5mm} (iv) V(x = -b) = 0. \label{Gr(3.40)} \] Generic solution: as (\ref{Gr(3.27)}), \[ V(x,y) = (Ae^{kx} + Be^{-kx}) (C\sin ky + D\cos ky). \] By symmetry, \(V(x,y) = V(-x,y)\) so \(A = B\). Now \(e^{kx} + e^{-kx} = 2\cosh kx\). Generic solution becomes (redefining \(C\) and \(D\)) \[ V(x,y) = \cosh kx (C\sin ky + D\cos ky). \] Boundary conditions \((i)\) and \((ii)\) require \(D = 0\), \(k = n\pi/a\) so \[ V(x,y) = C \cosh(n\pi x/a) \sin(n\pi y/a) \label{Gr(3.41)} \] with \((iv)\) already satisfied if \((iii)\) is. Full solution is linear combination of complete set of functions, \[ V(x,y) = \sum_{n=1}^{\infty} C_n \cosh(n\pi x/a) \sin(n\pi y/a). \] Coefficients: chosen such that \((iii)\) is fulfilled, \(V(b,y) = V_0\). From (\ref{Gr(3.35)}): \[ V(x,y) = \frac{4V_0}{\pi} \sum_{n = 1, 3, 5, ...} \frac{1}{n} \frac{\cosh (n\pi x/a)}{\cosh(n\pi b/a)} \sin(n\pi y/a). \label{Gr(3.42)} \]