Green's Theorem

Consider the vector identity

$$\nabla\cdot(\psi\,\nabla\phi) \equiv \psi\,\nabla^{\,2}\phi + \nabla\psi\cdot\nabla\phi,$$ (216)

where $\phi({\bf r})$ and $\psi({\bf r})$ are two arbitrary (but differentiable) vector fields. We can also write

$$\nabla\cdot(\phi\,\nabla\psi) \equiv \phi\,\nabla^{\,2}\psi + \nabla\phi\cdot\nabla\psi.$$ (217)

Forming the difference between the previous two equation, we get

$$\nabla\cdot(\psi\,\nabla\phi-\phi\,\nabla\psi) = \psi\,\nabla^{\,2}\phi - \phi\,\nabla^{\,2}\psi.$$ (218)

Finally, integrating this expression over some volume $V$ , bounded by the closed surface $S$ , and making use of the divergence theorem, we obtain

$$\int_V (\psi\,\nabla^{\,2}\phi-\phi\,\nabla^{\,2}\psi)\,dV = \int_S (\psi\,\nabla\phi-\phi\,\nabla\psi)\cdot d{\bf S}.$$ (219)

This result is known as Green's theorem.

Changing the variable of integration, the above result can be rewritten

$$\int_V \left[\psi({\bf r}')\,\nabla'^{\,2}\phi({\bf r}')-\phi({\b... ...tial n'}-\phi({\bf r}')\,\frac{\partial\psi({\bf r}')}{\partial n'}\right] dS',$$ (220)

where $\partial\phi({\bf r}')/\partial n'$ is shorthand for ${\bf n}'\cdot\nabla'\phi({\bf r}')$ , et cetera. Suppose that $\phi({\bf r})$ is a solution to Poisson's equation,

$$\nabla^{\,2}\phi = -\frac{\rho}{\epsilon_0},$$ (221)

associated with some charge distribution, $\rho({\bf r})$ , that extends over all space, subject to the boundary condition

$$\phi({\bf r}) \rightarrow 0 \vert{\bf r}\vert\rightarrow \infty.$$ (222)

Suppose, further, that

$$\psi({\bf r},{\bf r}') = \frac{1}{4\pi\,\vert{\bf r}-{\bf r}'\vert}.$$ (223)

It follows from Equation (25), and symmetry, that

$$\nabla^{\,2}\psi=\nabla'^{\,2}\psi = -\delta({\bf r}-{\bf r}').$$ (224)

The previous five equations can be combined to give

$$\int_V \phi({\bf r}')\,\delta({\bf r}-{\bf r}')\,dV' = \frac{1}{4\pi\,\epsilon_0}\int_V \frac{\rho({\bf r}')}{\vert{\b... ...\pi\,\epsilon_0}\int_S \frac{\sigma({\bf r}')}{\vert{\bf r}-{\bf r}'\vert}\,dS'$$

$$~~~~+\frac{1}{4\pi\,\epsilon_0}\int_S D({\bf r}')\,\frac{\partial}{\partial n'}\!\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dS',$$ (225)

where

$$\sigma = \epsilon_0\,{\bf n}\cdot\nabla\phi =-\epsilon_0\,{\bf n}\cdot{\bf E},$$ (226)

$$D = -\epsilon_0\,\phi.$$ (227)

It follows, by comparison with Equations (162), (202), and (205), that the three terms on the right-hand side of Equation (225) are the electric potential generated by the charges distributed within $S$ , the potential generated by the surface charge distribution, $\sigma({\bf r})$ , on $S$ , and the potential generated by the surface dipole distribution, $D({\bf r})$ , on $S$ , respectively.

Suppose that the point ${\bf r}$ lies within $S$ . In this case, Equation (225) yields

$$\phi({\bf r})= \frac{1}{4\pi\,\epsilon_0}\int_V \frac{\rho({\bf r... ...c{\partial}{\partial n'}\!\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dS'.$$ (228)

However, we know that the general solution to Equation (221), subject to the boundary condition (222), is

$$\phi({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\int_V \frac{\rho({\bf ... ...}\int_{\skew{3}\bar{V}} \frac{\rho({\bf r}')}{\vert{\bf r}-{\bf r}'\vert}\,dV'.$$ (229)

Here, $V$ denotes the region of space lying within the closed surface $S$ , whereas $\skew{3}\bar{V}$ denotes the region lying outside $S$ . A comparison of the previous two equations reveals that, for a point ${\bf r}$ lying within $S$ ,

$$\frac{1}{4\pi\,\epsilon_0}\int_{\skew{3}\bar{V}} \frac{\rho({\bf ... ...c{\partial}{\partial n'}\!\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dS'.$$ (230)

In other words, the electric potential (and, hence, the electric field) generated within $S$ by the charges external to $S$ is equivalent to that generated by the charge sheet $\sigma({\bf r})$ , and the dipole sheet $D({\bf r})$ , distributed on $S$ . Furthermore, because $\sigma$ depends on the normal derivative of the potential at $S$ , whereas $D$ depends on the potential at $S$ , it follows that we can completely determine the potential within $S$ once we known the distribution of charges within $S$ , and the values of the potential and its normal derivative on $S$ . In fact, this is an overstatement, because the potential and its normal derivative are not independent of one another, but are related via Poisson's equation. In other words, a knowledge of the potential on $S$ also implies a knowledge of its normal derivative, and vice versa. Hence, we can, actually, determine the potential within $S$ from a knowledge of the distribution of charges inside $S$ , and the distribution of either the potential, or its normal derivative, on $S$ . The specification of the potential on $S$ is known as a Dirichlet boundary condition. On the other hand, the specification of the normal derivative of the potential on $S$ is called a Neumann boundary condition.

Suppose that the point ${\bf r}$ lies outside $S$ . In this case, Equation (225) yields

$$0= \frac{1}{4\pi\,\epsilon_0}\int_V \frac{\rho({\bf r}')}{\vert{\... ...c{\partial}{\partial n'}\!\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dS'.$$ (231)

In other words, outside $S$ , the electric potential generated by the surface charge distribution $\sigma({\bf r})$ , combined with that generated by the surface dipole distribution $D({\bf r})$ , completely cancels out the electric potential (and, hence, the electric field) produced by the charges distributed within $S$ . As an example of this type of cancellation, suppose that $S$ is a spherical surface of radius $a$ , centered on the origin. Within $S$ , let there be a single charge $q$ , located at the origin. The electric field and potential generated by this charge are

$${\bf E} = \frac{q}{4\pi\,\epsilon_0\,r^{\,2}}\,{\bf e}_r,$$ (232)

and

$$\phi = \frac{q}{4\pi\,\epsilon_0\,r},$$ (233)

respectively. It follows from Equations (226) and (227) that the densities of the charge and dipole sheets at $r=a$ that are needed to cancel out the effect of the central charge in the region $r>a$ are

$$\sigma = - \frac{q}{4\pi\,a^{\,2}},$$ (234)

$$D = -\frac{q}{4\pi\,a},$$ (235)

respectively. Making use of Equations (212)-(215), and (232)-(235), the electric field and potential generated by the combination of the charge at the origin, the charge sheet at $r=a$ , and the dipole sheet at $r=a$ , are

$${\bf E} = \left\{ \begin{array}{lll} [q/(4\pi\,\epsilon_0\,r^{\,2... ...,{\bf e}_r&\mbox{\hspace{0.5cm}}& r<a\\ [0.5ex] {\bf0}&&r>a \end{array}\right.,$$ (236)

and

$$\phi = \left\{ \begin{array}{lll} q/(4\pi\,\epsilon_0\,r)&\mbox{\hspace{0.5cm}}& r<a\\ [0.5ex] 0&&r>a \end{array}\right.,$$ (237)

respectively. We can see that the charge and dipole sheet at $r=a$ do not affect the electric field, or the potential, due to the central charge in the region $r<a$ , but completely cancel out this charge's field and potential in the region $r>a$ .