# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### The Laplace Equation
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Of course, the simplest situation is to start by looking at the region of space where there is no charge density. The potential then solves Laplace's equation. How can it possibly look ?

###### The Laplace Equation in One Dimension

\[ \frac{d^2 V(x)}{dx^2} = 0 \Longrightarrow V(x) = mx + b \label{Gr(3.6)} \] Properties: \paragraph{1.} \(V(x)\) is the average of \(V(x + a)\) and \(V(x - a)\) for any \(a\). \paragraph{2.} Solutions to Laplace's equation have no local maxima or minima.

Boundary conditions: always work: two end values, one end value + same end derivative value. Not always: one end value + derivative value at other end, two end derivative values.

###### The Laplace Equation in Two Dimensions

\[ \frac{d^2 V}{dx^2} + \frac{d^2 V}{dy^2} = 0. \] Properties: \paragraph{1.} The value of \(V(x,y)\) equals the average value around the point: \[ V(x,y) = \frac{1}{2\pi R} \oint V dl \] \paragraph{2.} \(V\) has no local maxima or minima. All extrema occur at the boundaries.

###### The Laplace Equation in Three Dimensions

\[ {\boldsymbol \nabla}^2 V = 0 \] Properties: \paragraph{1.} \(V({\bf r})\) is the average value of \(V\) over any spherical surface centered at \({\bf r}\): \[ V({\bf r}) = \frac{1}{4\pi R^2} \oint V da \] \paragraph{2.} \(V\) can have no local maxima or minima. All extrema occur at the boundaries.

\paragraph{Another way of seeing this} is to write the second derivatives as \[ \frac{\partial^2 V({\bf r})}{\partial x^2} = f_x ({\bf r}), \hspace{5mm} \frac{\partial^2 V({\bf r})}{\partial y^2} = f_y ({\bf r}), \hspace{5mm} \frac{\partial^2 V({\bf r})}{\partial z^2} = f_z ({\bf r}), \hspace{5mm} f_x + f_y + f_z = 0. \] The \(f_a ({\bf r})\) represent the three components of the curvature of \(V({\bf r})\). An extremum of \(V\) at \({\bf r}_e\) would be characterized by \({\boldsymbol \nabla} V |_{{\bf r}_e} \cdot \delta{\bf r} = 0\) for any infinitesimal displacement \(\delta{\bf r}\) around the extremum point. For a local minimum, the second derivative form should be greater than zero, \(\sum_{i,j} \frac{\partial^2 V}{\partial r_i \partial r_j} \delta r_i \delta r_j > 0\) for any displacement vector. Choosing alternately displacements along the three axes, the form becomes \(f_x (\delta x)^2\), \(f_y (\delta y)^2\) or \(f_z (\delta z)^2\). Since the squared displacements are necessarily positive, we thus require \(f_x > 0\), \(f_y > 0\) and \(f_z > 0\). This is impossible in view of the \(f_x + f_y + f_z = 0\) condition above.

Earnshaw's theorem
Since solutions to Laplace's equation have no local minimum, it is impossible to find a static distribution of charges which generates an electrostatic field with a stable equilibrium position for a test charge.

Going back to Poisson's equation, we can make a few comments:

- representation (\ref{eq:Poisson}) highlights the 'local' nature of the coupling between electrostatic fields and charges: fields are 'created' where the charges 'sit'. This is also seen by looking at the integrand of (\ref{eq:V_from_rho}). If electrostatics was nonlocal, a modified representation like (\ref{eq:V_from_rho}) would still exist, but not a local differential one like (\ref{eq:Poisson}).
- as written, representations (\ref{eq:E_from_rho}) and (\ref{eq:V_from_rho}) require the knowledge of the charge density distribution \(\rho({\bf r})\) throughout space to determine the potential at any given point.
- (\ref{eq:Poisson}), being purely local, might allow to determine the potential at a specified point, provided we know the charge density distribution around this specified point, and at some set of other reference points (to make the solution unique).

We therefore want to ask the question: under what conditions can an electrostatic problem be fully defined by solving Poisson's equation ? We start by mentioning some cases, and interpreting them thereafter.

First case: the electrostatic potential is uniquely determined if \(\rho({\bf r})\) is given throughout all space. In this case, Poisson's equation is explicitly solved by (\ref{eq:V_from_rho}). Explicit check: \[ {\boldsymbol \nabla}^2 V ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' \rho({\bf r}') {\boldsymbol \nabla}^2 \frac{1}{|{\bf r} - {\bf r}'|} = \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' (-4\pi) \delta ({\bf r} - {\bf r}') = -\frac{\rho ({\bf r})}{\varepsilon_0}. \] where we have used (\ref{Gr(1.102)}), and the fact that the delta function is always resolved since we integrate over all space. Note: it is implicitly assumed that the integral in (\ref{eq:V_from_rho}) converges, i.e. that the charge density \(\rho({\bf r})\) is sufficiently well-behaved (does not become singular).

First case (corollary): the electrostatic potential is uniquely determined in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density \(\rho ({\bf x})\) is given everywhere within \({\cal V}\), vanishes outside of \({\cal V}\), and the value of the surface charge density \(\sigma\) is given everywhere on the boundary \({\cal S}\). Of course, \({\cal S}\) need not be a connected surface.

This is obvious: we know where all the charges are, so this is really the same as the first case.

Second case: the electrostatic potential is uniquely determined in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density \(\rho ({\bf x})\) is given everywhere within \({\cal V}\), and the value of \(V\) is given everywhere on the boundary \({\cal S}\). Of course, \({\cal S}\) need not be a connected surface.

Here, the logic is quite simple: since the electrostatic potential is known on all the surface enclosing the space \({\cal V}\), and since Poisson's equation is local, we need not consider anything outside of \({\cal V}\) to obtain \(V\) within \({\cal V}\).

Given a solution \(V_1 ({\bf r})\), we can easily show that it is unique. Suppose there was another solution \(V_2 ({\bf r})\). Look at the difference, \(U \equiv V_1 - V_2\). In the bulk, \(U\) obeys the Laplace equation \[ {\boldsymbol \nabla}^2 U = {\boldsymbol \nabla}^2 V_1 - {\boldsymbol \nabla}^2 V_2 = -\frac{\rho}{\varepsilon_0} + \frac{\rho}{\varepsilon_0} = 0. \] Moreover, \(U ({\bf r}) = 0\) for \({\bf r} \in {\cal S}\). Since solutions to the Laplace equation take their maximal and minimal value on the boundary, we must have \(U = 0\) \(\forall {\bf r} \in {\cal V}\) (Griffiths' proof).

This all feels a bit amateurish and not very systematic. Can we be more precise and general? What kinds of boundary information do we really need to specify the solution uniquely ?

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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