# 22.  Constructing Cylindrical Green’s Functions

Note: the point of this shortish lecture is to try to make clear how the different components of the Green’s function, corresponding to different variables, work together to generate the overall delta function when $-{\nabla }^{2}$ is applied. Some of Jackson’s Chapter 3 problems are used as examples.

## The Two Representations

The Green’s function is the solution of ${\nabla }^{2}G\left(\stackrel{\to }{r},{\stackrel{\to }{r}}^{\prime }\right)=-\delta \left(\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }\right),$ in electrostatics it’s the potential at $\stackrel{\to }{r}$ generated by a single unit charge at ${\stackrel{\to }{r}}^{\prime }$ in the given space with the given boundary conditions.

For the three-dimensional systems we consider, the eigenstates of the Laplacian can be written as products of three functions, each depending on one of the coordinate variables, and the delta function can also be written as a product of three delta functions, one for each variable.  The Green’s function can be written as a sum of terms, each of which is a product of functions of the separate variables, but these factors are not independent of each other: we saw in the spherical case that the radial solutions went as powers of the radius, but the powers were dependent on the angular solutions. (In quantum mechanics, this corresponds to the probability of being close to the origin depending on the angular momentum.)

The simplest one-dimensional system (discussed in lecture 11) illustrates a possible choice in representing the Green’s function.

We constructed Green’s functions ${G}_{D}\left(x,{x}^{\prime }\right)$ satisfying $-{\partial }^{2}{G}_{D}\left(x,{x}^{\prime }\right)/\partial {x}^{2}=\delta \left(x-{x}^{\prime }\right)$ in the one-dimensional space $0\le x\le 1$ having boundary conditions ${G}_{D}\left(0,{x}^{\prime }\right)={G}_{D}\left(1,{x}^{\prime }\right)=0.$

Choice A:  The complete set of eigenfunctions of the Laplacian in this space is:   $\sqrt{2}\mathrm{sin}n\pi x,\text{ }n=±1,±2,±3,\dots$ and the delta function is $2\sum _{n}\mathrm{sin}n\pi x\mathrm{sin}n\pi {x}^{\prime }.$

It follows that the (Dirichlet) Green’s function can be written

${G}_{D}\left(x,{x}^{\prime }\right)=2\sum _{n}\frac{\mathrm{sin}n\pi x\mathrm{sin}n\pi {x}^{\prime }}{{\pi }^{2}{n}^{2}},$

a sum over all eigenstates of the differential operator $-{\nabla }^{2}$ , which acting here, $-{\nabla }^{2}{G}_{D}\left(x,{x}^{\prime }\right),$ removes the denominators, leaving the delta function.

Choice B:  The other approach is to take two zero energy eigenstates of the differential operator (here just straight lines) which satisfy the boundary condition at opposite ends, and multiply them together:

${G}_{D}\left(x,{x}^{\prime }\right)={x}_{<}\left(1-{x}_{>}\right).$

As a function of $x,\text{\hspace{0.17em}}{G}_{D}$ has a slope discontinuity at $x={x}^{\prime }$ and this generates the delta function $\delta \left(x-{x}^{\prime }\right)$ when the differential operator is applied.  (The slope discontinuity is independent of ${x}^{\prime }$ because it’s the Wronskian of the two solutions).

## Which to Use?

### Cartesian

For the Cartesian problem of a charge inside a grounded conducting cube, option A can be used for all three directions, although as Jackson points out (p 129), you could, for example, use A for two and use B for the third direction.

### Spherical

For the angular variables in spherical polar coordinates, there are no endpoints, so option A is necessary.

The radial function has eigenstates labeled by a nonnegative integer $l,$ they are ${r}^{l}$ (well-behaved at the origin endpoint) and ${r}^{-\left(l+1\right)}$ (well-behaved at infinity), so option B naturally gives ${r}_{<}^{l}/{r}_{>}^{l+1}$ and this is to be multiplied by the complete set of angular states corresponding to this $l,$ that is, $\sum _{m=-l}^{l}{Y}_{l}^{m}*\left({\theta }^{\prime },{\varphi }^{\prime }\right){Y}_{l}^{m}\left(\theta ,\varphi \right),$ with a normalizing factor  $\frac{4\pi }{2l+1}.$  For details, see lecture 20.

The full spherical Green’s function is then given by summing over all $l$ these products of radial and angular functions.

### Cylindrical

There are several ways to construct the Green’s function in cylindrical coordinates.

${\nabla }^{2}G\left(\stackrel{\to }{r},{\stackrel{\to }{r}}^{\prime }\right)=-\frac{4\pi }{\rho }\delta \left(\rho -{\rho }^{\prime }\right)\delta \left(\varphi -{\varphi }^{\prime }\right)\delta \left(z-{z}^{\prime }\right).$

Here

$\delta \left(\varphi -{\varphi }^{\prime }\right)=2\pi \sum _{m=-\infty }^{\infty }{e}^{im\left(\varphi -{\varphi }^{\prime }\right)}$

and

$\delta \left(z-{z}^{\prime }\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}dk{e}^{ik\left(z-{z}^{\prime }\right)}=\frac{1}{\pi }\underset{0}{\overset{\infty }{\int }}dk\mathrm{cos}k\left(z-{z}^{\prime }\right).$

First, the azimuthal angle $\varphi$ has no endpoints, of course, so it must appear in the Green’s function as option A. That is,

$G\left(\stackrel{\to }{r},{\stackrel{\to }{r}}^{\prime }\right)=\sum _{m=-\infty }^{\infty }{e}^{im\left(\varphi -{\varphi }^{\prime }\right)}f\left(m,z,{z}^{\prime },\rho ,{\rho }^{\prime }\right).$

(Note for the observant: I’m just following Jackson here, going from $\nu$ to $m,$ presumably because in the problems we’ll consider it’s usually an integer.)

On applying $-{\nabla }^{2},$ the function $f$ must generate delta functions in $z,\rho ,$ then summing over all terms

$\sum _{m=-\infty }^{\infty }{e}^{im\left(\varphi -{\varphi }^{\prime }\right)}=2\pi \delta \left(\varphi -{\varphi }^{\prime }\right).$

Second, the linear along-the-axis variable $z.$ First, the range could be either infinite or finite: for example, the field inside a cylinder with closed ends. Furthermore, the variable ${k}^{2}$ can be positive or negative.  For positive ${k}^{2},$ the solutions are exponentials, so for a cylinder of length $L,$ the appropriate factor in the Green’s function (vanishing at both ends) will be $\frac{\mathrm{sinh}k{z}_{<}\mathrm{sinh}k\left(L-{z}_{>}\right)}{\mathrm{sinh}kL}.$ (Jackson problem 3.17.)

If the range is infinite, positive ${k}^{2}$ gives a factor ${e}^{-k\left({z}_{>}-{z}_{<}\right)}.$ For negative ${k}^{2},$ the solutions are sines and cosines, and for length $L$ in the $z$-direction, we recover the factor previously discussed in the one-dimensional example, $\frac{4}{L}\sum _{n=1}^{\infty }\mathrm{sin}\left(\frac{n\pi z}{L}\right)\mathrm{sin}\left(\frac{n\pi {z}^{\prime }}{L}\right).$

Third, the radial equation, just as in the spherical case, depends on the parameters from the other equations, in this case ${k}^{2}$ and $m$ (or $\nu$: Jackson moves around on this.)

As discussed in the previous lecture, for positive ${k}^{2}$ the variable change to $x=k\rho$ gives the Bessel equation in standard form, $\frac{{d}^{2}R}{d{x}^{2}}+\frac{1}{x}\frac{dR}{dx}+\left(1-\frac{{\nu }^{2}}{{x}^{2}}\right)R=0.$

In building the Green’s function, the radial delta function can be represented as

$\frac{1}{\rho }\delta \left(\rho -{\rho }^{\prime }\right)=\underset{0}{\overset{\infty }{\int }}k{J}_{\nu }\left(k\rho \right){J}_{\nu }\left(k{\rho }^{\prime }\right)dk$

for (quoting Jackson) any $\mathrm{Re}\left(\nu \right)>-1.$ The first equation in problem 3.23 uses this, the second uses the exponential Bessel radial functions arranged to give zero at the cylindrical wall, and to have the slope discontinuity at $\rho ={\rho }^{\prime }.$ The third equation has no slope discontinuity, instead it sums over all states to create the delta function.