next up previous
Next: One-dimensional solution of Poisson's Up: Electrostatics Previous: Poisson's equation

The uniqueness theorem

We have already seen the great value of the uniqueness theorem for Poisson's equation (or Laplace's equation) in our discussion of Helmholtz's theorem (see Sect. 3.11). Let us now examine this theorem in detail.

Consider a volume $V$ bounded by some surface $S$. Suppose that we are given the charge density $\rho$ throughout $V$, and the value of the scalar potential $\phi_S$ on $S$. Is this sufficient information to uniquely specify the scalar potential throughout $V$? Suppose, for the sake of argument, that the solution is not unique. Let there be two potentials $\phi_1$ and $\phi_2$ which satisfy

$\displaystyle \nabla^2 \phi_1$ $\textstyle =$ $\displaystyle - \frac{\rho}{\epsilon_0},$ (666)
$\displaystyle \nabla^2 \phi_2$ $\textstyle =$ $\displaystyle - \frac{\rho}{\epsilon_0}$ (667)

throughout $V$, and
$\displaystyle \phi_1$ $\textstyle =$ $\displaystyle \phi_S,$ (668)
$\displaystyle \phi_2$ $\textstyle =$ $\displaystyle \phi_S$ (669)

on $S$. We can form the difference between these two potentials:
\phi_3 = \phi_1 - \phi_2.
\end{displaymath} (670)

The potential $\phi_3$ clearly satisfies
\nabla^2\phi_3 = 0
\end{displaymath} (671)

throughout $V$, and
\phi_3 =0
\end{displaymath} (672)

on $S$.

According to vector field theory,

\nabla\cdot(\phi_3 \nabla\phi_3) \equiv (\nabla \phi_3)^2 +\phi_3\nabla^2\phi_3.
\end{displaymath} (673)

Thus, using Gauss' theorem
\int_V \left\{ (\nabla\phi_3)^2 +\phi_3 \nabla^2\phi_3\right\} dV =
\oint_S \phi_3 \nabla \phi_3\cdot d{\bf S}.
\end{displaymath} (674)

But, $\nabla^2\phi_3 = 0$ throughout $V$, and $\phi_3=0$ on $S$, so the above equation reduces to
\int_V (\nabla \phi_3)^2 dV = 0.
\end{displaymath} (675)

Note that $(\nabla \phi_3)^2$ is a positive definite quantity. The only way in which the volume integral of a positive definite quantity can be zero is if that quantity itself is zero throughout the volume. This is not necessarily the case for a non-positive definite quantity: we could have positive and negative contributions from various regions inside the volume which cancel one another out. Thus, since $(\nabla \phi_3)^2$ is positive definite, it follows that
\phi_3 = {\rm constant}
\end{displaymath} (676)

throughout $V$. However, we know that $\phi_3=0$ on $S$, so we get
\phi_3 =0
\end{displaymath} (677)

throughout $V$. In other words,
\phi_1 = \phi_2
\end{displaymath} (678)

throughout $V$ and on $S$. Our initial assumption that $\phi_1$ and $\phi_2$ are two different solutions of Poisson's equation, satisfying the same boundary conditions, turns out to be incorrect.

The fact that the solutions to Poisson's equation are unique is very useful. It means that if we find a solution to this equation--no matter how contrived the derivation--then this is the only possible solution. One immediate use of the uniqueness theorem is to prove that the electric field inside an empty cavity in a conductor is zero. Recall that our previous proof of this was rather involved, and was also not particularly rigorous (see Sect. 5.4). We know that the interior surface of the conductor is at some constant potential $V$, say. So, we have $\phi=V$ on the boundary of the cavity, and $\nabla^2 \phi
=0 $ inside the cavity (since it contains no charges). One rather obvious solution to these equations is $\phi=V$ throughout the cavity. Since the solutions to Poisson's equation are unique, this is the only solution. Thus,

{\bf E} = - \nabla\phi = -\nabla V = {\bf0}
\end{displaymath} (679)

inside the cavity.

Suppose that some volume $V$ contains a number of conductors. We know that the surface of each conductor is an equipotential surface, but, in general, we do not know what potential each surface is at (unless we are specifically told that it is earthed, etc.). However, if the conductors are insulated it is plausible that we might know the charge on each conductor. Suppose that there are $N$ conductors, each carrying a charge $Q_i$ ($i=1$ to $N$), and suppose that the region $V$ containing these conductors is filled by a known charge density $\rho$, and bounded by some surface $S$ which is either infinity or an enclosing conductor. Is this enough information to uniquely specify the electric field throughout $V$?

Well, suppose that it is not enough information, so that there are two fields ${\bf E}_1$ and ${\bf E}_2$ which satisfy

$\displaystyle \nabla\cdot{\bf E}_1$ $\textstyle =$ $\displaystyle \frac{\rho}{\epsilon_0},$ (680)
$\displaystyle \nabla\cdot{\bf E}_2$ $\textstyle =$ $\displaystyle \frac{\rho}{\epsilon_0}$ (681)

throughout $V$, with
$\displaystyle \oint_{S_i} {\bf E}_1\cdot d{\bf S}_i$ $\textstyle =$ $\displaystyle \frac{Q_i}{\epsilon_0},$ (682)
$\displaystyle \oint_{S_i} {\bf E}_2\cdot d{\bf S}_i$ $\textstyle =$ $\displaystyle \frac{Q_i} {\epsilon_0}$ (683)

on the surface of the $i$th conductor, and, finally,
$\displaystyle \oint_{S} {\bf E}_1\cdot d{\bf S}_i$ $\textstyle =$ $\displaystyle \frac{Q_{\rm total}}{\epsilon_0},$ (684)
$\displaystyle \oint_{S} {\bf E}_2\cdot d{\bf S}_i$ $\textstyle =$ $\displaystyle \frac{Q_{\rm total}} {\epsilon_0}$ (685)

over the bounding surface, where
Q_{\rm total} = \sum_{i=1}^N Q_i + \int_V \rho dV
\end{displaymath} (686)

is the total charge contained in volume $V$.

Let us form the difference field

{\bf E}_3 = {\bf E}_1 - {\bf E}_2.
\end{displaymath} (687)

It is clear that
\nabla\cdot {\bf E}_3 = 0
\end{displaymath} (688)

throughout $V$, and
\oint_{S_i} {\bf E}_3 \cdot d{\bf S}_i = 0
\end{displaymath} (689)

for all $i$, with
\oint_S {\bf E}_3 \cdot d{\bf S} = 0.
\end{displaymath} (690)

Now, we know that each conductor is at a constant potential, so if

{\bf E}_3 = -\nabla\phi_3,
\end{displaymath} (691)

then $\phi_3$ is a constant on the surface of each conductor. Furthermore, if the outer surface $S$ is infinity then $\phi_1=\phi_2 = \phi_3 =0$ on this surface. If the outer surface is an enclosing conductor then $\phi_3$ is a constant on this surface. Either way, $\phi_3$ is constant on $S$.

Consider the vector identity

\nabla\cdot(\phi_3 {\bf E}_3) \equiv \phi_3 \nabla \cdot{\bf E}_3 + {\bf E}_3\cdot
\end{displaymath} (692)

We have $\nabla\cdot {\bf E}_3 = 0$ throughout $V$, and $\nabla\phi_3 = -{\bf E}_3$, so the above identity reduces to
\nabla\cdot (\phi_3  {\bf E}_3) = - E_3^{ 2}
\end{displaymath} (693)

throughout $V$. Integrating over $V$, and making use of Gauss' theorem, yields
\int_V E_3^{ 2} dV = - \sum_{i=1}^N \oint_{S_i} \phi_3 {\bf E}_3\cdot d{\bf S}_i
-\oint_S \phi_3 {\bf E}_3\cdot d{\bf S}.
\end{displaymath} (694)

However, $\phi_3$ is a constant on the surfaces $S_i$ and $S$. So, making use of Eqs. (689) and (690), we obtain
\int_V E_3^{ 2}  dV =0.
\end{displaymath} (695)

Of course, $E_3^{ 2}$ is a positive definite quantity, so the above relation implies that
{\bf E}_3 = {\bf0}
\end{displaymath} (696)

throughout $V$: i.e., the fields ${\bf E}_1$ and ${\bf E}_2$ are identical throughout $V$.

For a general electrostatic problem involving charges and conductors, it is clear that if we are given either the potential at the surface of each conductor or the charge carried by each conductor (plus the charge density throughout the volume, etc.) then we can uniquely determine the electric field. There are many other uniqueness theorems which generalize this result still further: i.e., we could be given the potential of some of the conductors and the charge carried by the others, and the solution would still be unique.

At this point, it is worth noting that there are also uniqueness theorems associated with magnetostatics. For instance, if the current density, ${\bf j}$, is specified throughout some volume $V$, and either the magnetic field, ${\bf B}$, or the vector potential, ${\bf A}$, is specified on the bounding surface $S$, then the magnetic field is uniquely determined throughout $V$ and on $S$. The proof of this proposition proceeds along the usual lines. Suppose that the magnetic field is not uniquely determined. In other words, suppose there are two magnetic fields, ${\bf B}_1$ and ${\bf B}_2$, satisfying

$\displaystyle \nabla\times{\bf B}_1$ $\textstyle =$ $\displaystyle \mu_0 {\bf j},$ (697)
$\displaystyle \nabla\times{\bf B}_2$ $\textstyle =$ $\displaystyle \mu_0 {\bf j},$ (698)

throughout $V$. Suppose, further, that either ${\bf B}_1={\bf B}_2={\bf B}_S$ or ${\bf A}_1={\bf A}_2={\bf A}_S$ on $S$. Forming the difference field, ${\bf B}_3={\bf B}_1-{\bf B}_2$, we have
\nabla\times{\bf B}_3={\bf0}
\end{displaymath} (699)

throughout $V$, and either ${\bf B}_3={\bf0}$ or ${\bf A}_3={\bf0}$ on $S$. Now, according to vector field theory,
\int\left[(\nabla\times{\bf U})^2 - {\bf U}\cdot\nabla\times...
... \equiv \int {\bf U}\times(\nabla\times{\bf U})\cdot {\bf dS}.
\end{displaymath} (700)

Setting ${\bf U}={\bf A}_3$, and using ${\bf B}_3=\nabla\times{\bf A}_3$ and Eq. (699), we obtain
\int B_3^{ 2} dV = \int {\bf A}_3\times{\bf B}_3\cdot{\bf dS}.
\end{displaymath} (701)

However, we know that either ${\bf B}_3$ or ${\bf A}_3$ is zero on $S$. Hence, we obtain
\int B_3^{ 2} dV = 0.
\end{displaymath} (702)

Since, $B_3^{ 2}$ is positive definite, the only way in which the above equation can be satisfied is if $B_3$ is zero throughout $V$. Hence, ${\bf B}_1={\bf B}_2$ throughout $V$, and the solution is therefore unique.

next up previous
Next: One-dimensional solution of Poisson's Up: Electrostatics Previous: Poisson's equation
Richard Fitzpatrick 2006-02-02