Laplaces equation is:

\begin{displaymath}
\nabla^{2}\phi=\frac{\partial^{2}\phi}{\partial x^{2}}+\frac...
...^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}=0\cdots(Eqn. 1)
\end{displaymath}

The statement of the theorem goes: A scalar field $\phi(\vec{r})$ obeying the Laplace equation does not have any local maxima or minima; all its stationary points are saddle points. So, it is impossible to keep a charged particle or body in stable static equilibrium by means of electrostatic forces alone. How to distinguish between minimum, maximum or saddle point? Use the matrix of second derivatives:


\begin{displaymath}
M=\left(\begin{array}{ccc}
\frac{\partial^{2}\phi}{\partial ...
...ial^{2}\phi}{\partial z^{2}}
\end{array}\right)\cdots(Eqn. 2)
\end{displaymath}

This has to be evaluated at the stationary point (first derivative zero). Being a symmetric matrix we have three real eigenvalues $M_{1}$, $M_{2}$ and $M_{3}$. If all of them are positive, we have a minimum. If all are negative, we have a maximum. If there is a mixture of signs, we have a saddle point. Now the laplacian is just the sum of the diagonal elements of the matrix:


\begin{displaymath}
\nabla^{2}\phi=tr(M)\cdots(Eqn. 3)
\end{displaymath}

Since trace does not change due to similarity transformations, after diagonalisation the sum of the eigenvalues will be equal to the trace.


\begin{displaymath}
tr(M)=M_{1}+M_{2}+M_{3}=\nabla^{2}\phi=0\cdots(Eqn. 4)
\end{displaymath}

Now this zero sum rules out all three eigenvalues being positive or negative at the same time. So there cannot be maximum or minimum anywhere. In the special case where the matrix is zero, all three eigenvalues vanish. In that case the distinction between maximum, minimum and saddle point follows from higher derivatives. They form a tensor which has highly restricted form due to the requirement of Laplace's equation being satisfied. There also, the same result holds.

Stirling Engine 2020-01-02