# 9. Green’s Reciprocation Theorem

*Michael Fowler,
University of Virginia*

## What It Is

One simple theorem George Green published in his 1828 paper
is his *Reciprocation Theorem*. (This is Jackson's term, Wikipedia calls it reciprocity.)
It seems almost trivial, but often leads to surprising
results with very little effort.
Here it is:

Consider two different volume and surface charge distributions, in the same identical geometry.

Charge distribution* A*
has volume charge density ${\rho}_{A}\left(\overrightarrow{r}\right)$ and boundary surface charge densities ${\sigma}_{A}\left(\overrightarrow{r}\right),$ generating electrostatic potential ${\phi}_{A}\left(\overrightarrow{r}\right).$

In that *same
space* (meaning with the same surfaces) charge distribution B, with densities ${\rho}_{B}$,
${\sigma}_{B}$ would give potential ${\phi}_{B}\left(\overrightarrow{r}\right).$

Then the Reciprocation Theorem is:

$$\underset{V}{\int}{\rho}_{A}{\phi}_{B}dV+{\displaystyle \underset{S}{\int}{\sigma}_{A}{\phi}_{B}da}}={\displaystyle \underset{V}{\int}{\rho}_{B}{\phi}_{A}dV+{\displaystyle \underset{S}{\int}{\sigma}_{B}{\phi}_{A}da}}.$$

*In words*: If the *A*
charge densities were at rest in the *B*
potential, the total potential energy would be the same as that of the *B *charge densities held at rest in the *A* potential.

To prove it, we’ll just consider volume charges (the surface charges could be taken as volume charges in a thin layer limit anyway).

Then

$${\phi}_{B}\left(\overrightarrow{r}\right)=\frac{1}{4\pi {\epsilon}_{0}}{\displaystyle \underset{V}{\int}\frac{{\rho}_{B}\left({\overrightarrow{r}}^{\prime}\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime}\right|}}$$

and

$$\underset{V}{\int}{\rho}_{A}\left(\overrightarrow{r}\right){\phi}_{B}\left(\overrightarrow{r}\right){d}^{3}r}=\frac{1}{4\pi {\epsilon}_{0}}{\displaystyle \underset{V}{\int}\frac{{\rho}_{A}\left(\overrightarrow{r}\right){\rho}_{B}\left({\overrightarrow{r}}^{\prime}\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime}\right|}}{d}^{3}r{d}^{3}{r}^{\prime}.$$

But this is symmetric in *A*,
*B*, proving the theorem: it’s that
simple!

## Earnshaw’s Theorem from the Reciprocation Theorem

Earnshaw’s theorem is often stated simply as: *In a charge-free region, the potential
cannot have a maximum or minimum.* (Because locally the field in all
directions would have to point all inwards or all outwards, so nonzero
divergence, meaning there must be charge present.)

A more informative version is: *in a charge-free region, the potential at a point is the same as the
average potential on any spherical surface centered at that point*, provided
only
that the spherical surface is itself within the charge-free region.

To prove this using the Reciprocation Theorem, take:

** System A**: the existing
setup, with charge distribution ${\rho}_{A}\left(\overrightarrow{r}\right)$ all outside the region of interest, generating
potential ${\phi}_{A}\left(\overrightarrow{r}\right).$

** System B**: a spherical
surface, radius $R,$ with

*uniform*surface charge density $\sigma $, the total surface charge being $4\pi {R}^{2}\sigma =-Q,$ and a point charge $+Q$ at the center of the sphere.

The spherical surface is taken to be in the charge-free
region of system *A*.

The *B* system, the
charged sphere plus the equal but negative central point charge, gives *zero* potential outside the sphere (which
is where all the *A *system charge is),
so

$\int {\rho}_{A}\left(\overrightarrow{r}\right)}\text{\hspace{0.05em}}{\phi}_{B}\left(\overrightarrow{r}\right)=0.$

Therefore from the theorem $\int {\phi}_{A}\left(\overrightarrow{r}\right){\rho}_{B}\left(\overrightarrow{r}\right)}\text{\hspace{0.05em}}\text{\hspace{0.17em}}=0,$ and taking the center of the sphere as the origin for convenience, this becomes, integrating over the area of the spherical surface (and cancelling out $Q$ from both sides):

$$\frac{1}{4\pi {R}^{2}}{\displaystyle \int {\phi}_{A}\left(\overrightarrow{r}\right){R}^{2}d\Omega}\text{\hspace{0.17em}}\text{\hspace{0.05em}}={\phi}_{A}\left(0\right).$$

*Note*: if this looks too easy, there are many more
difficult proofs on the web.

*Exercise*:
Suppose we take a new system *B*:
a sphere centered at the origin with surface charge density proportional to the
coordinate $z.$ We replace the center point charge with a
point dipole, such that the potential from this system is zero outside the
sphere. System *A* is as before: what does
the Reciprocation Theorem tell you this time? *Hint*: you can think of that surface charge as two equally-sized solid spheres of charge of opposite sign, their centers a very small distance apart.

## Revisiting an Earlier Result Using the Reciprocation Theorem

We established in the Electrostatics II lecture that if different parts of a spherical surface are held at different potentials, the consequent potential at any point P in space outside the sphere can be found by integrating over the spherical surface with a weighting factor equal to the density of induced charge on a perfectly conducting grounded sphere with a unit charge at the point P.

In fact, this result follows immediately from the reciprocation theorem!

For

$\underset{V}{\int}{\rho}_{A}{\phi}_{B}dV+{\displaystyle \underset{S}{\int}{\sigma}_{A}{\phi}_{B}da}}={\displaystyle \underset{V}{\int}{\rho}_{B}{\phi}_{A}dV+{\displaystyle \underset{S}{\int}{\sigma}_{B}{\phi}_{A}da}},$

consider two systems *A*,
*B* with identical geometry: just a
sphere plus one point outside it.

System *A* is our
spherical surface divided into parts held at different potentials, labeled by
the variable potential ${\phi}_{A}\left({\overrightarrow{r}}^{\prime}\right),$ ${\overrightarrow{r}}^{\prime}$ being any point on the surface of the sphere, and
system *A* also includes the point P,
position $\overrightarrow{r}$ say, at which we want to find the potential from the charge distribution on the spherical surface (but there is no volume charge in *A*).

System *B* is the
geometrically identical sphere, but now a fully connected conducting surface,
grounded and so at zero potential, and now there is a unit charge at the point
P, that is, ${\rho}_{B}\left(\overrightarrow{r}\right)=\delta \left(\overrightarrow{r}-{\overrightarrow{r}}_{\text{P}}\right).$

The left-hand side of the equation above is identically zero: ${\rho}_{A}=0$ in the volume, ${\phi}_{B}=0$ on the surface.

The right-hand side, using ${\rho}_{B}\left(\overrightarrow{r}\right)=\delta \left(\overrightarrow{r}-{\overrightarrow{r}}_{\text{P}}\right),$ gives ${\phi}_{A}\left(\overrightarrow{r}\right)=-{\displaystyle \underset{S}{\int}{\sigma}_{B}\left({\overrightarrow{r}}^{\prime}\right){\phi}_{A}\left({\overrightarrow{r}}^{\prime}\right)da},$ where ${\sigma}_{B}\left({\overrightarrow{r}}^{\prime}\right)$ would be the surface charge density induced at ${\overrightarrow{r}}^{\prime}$ on a grounded conducting sphere by unit charge at $\overrightarrow{r}.$

Note also that this proof generalizes from a conducting sphere to any closed conducting surface.

## Symmetry of the Dirichlet Green's Function

We’ve shown the Dirichlet Green’s function is *symmetric*,

${G}_{D}\left(\overrightarrow{r},{\overrightarrow{r}}^{\prime}\right)={G}_{D}\left({\overrightarrow{r}}^{\prime},\overrightarrow{r}\right).$

This also follows easily from the Reciprocation Theorem:
take two systems *A*, *B* having the same set of *grounded* conducting surfaces, one with a
single unit charge at $\overrightarrow{r},$ the other with a single unit charge at ${\overrightarrow{r}}^{\prime}$.
Now, by definition, ${G}_{D}\left(\overrightarrow{r},{\overrightarrow{r}}^{\prime}\right)$ is the potential at $\overrightarrow{r}$ from the single unit charge at ${\overrightarrow{r}}^{\prime}$ plus the charges induced on the grounded
surfaces, and vice versa. Symmetry
follows from The Theorem. (The result is certainly not intuitively obvious:
think of an odd shaped conductor, say a sphere but with a tall thin conical
"mountain" somewhere. Now put $\overrightarrow{r}$ just above the mountain peak, ${\overrightarrow{r}}^{\prime}$ above the plane on the other side.)

*Exercise*:
Write this out explicitly, in terms of ${\rho}_{A},{\phi}_{A},{\rho}_{B},{\phi}_{B}.$