2.8 KiB
Chapter 11 - Potential Formulation of Electrodynamics
Section 11.1 - Forces, Fields, Potentials, and Greens Functions
Section 11.1.1 - Potentials for Time-independent Fields
We know that when the electric and magnetic fields are time-independent, \nabla \times \mathbf{E} = -\mathbf{J}_m and \nabla \times \mathbf{H} = \mathbf{J}_e.
We also know that Helmholtz theorem tells us that we can write the fields such that
$$\begin{align} \mathbf{E} &= -\nabla V_E + \nabla \times \mathbf{A}_E \ \mathbf{H} &= -\nabla V_H + \nabla \times \mathbf{A}_H \end{align}$$
We can then recall that
$$\begin{align} V_E(\mathbf{r}) &= \frac{1}{4\pi} \int_{V'} \frac{1}{|\mathbf{r} - \mathbf{r}'|} \frac{\rho_e(\mathbf{r}')}{\varepsilon_0} dV' \ V_H(\mathbf{r}) &= \frac{1}{4\pi} \int_{V'} \frac{1}{|\mathbf{r} - \mathbf{r}'|} \frac{\rho_e(\mathbf{r}')}{\mu_0} dV' \ A_E(\mathbf{r}) &= \frac{1}{4\pi} \int_{V'} \frac{1}{|\mathbf{r} - \mathbf{r}'|} (-s\mathbf{J}m(\mathbf{r})) dV' \ A_H(\mathbf{r}) &= \frac{1}{4\pi} \int{V'} \frac{1}{|\mathbf{r} - \mathbf{r}'|} (\mathbf{J}_e(\mathbf{r})) dV' \end{align}$$
We can verify this by taking the curl or divergence of the fields and seeing that Maxwell's equations hold true.
A notable result is that for the Laplace operator \nabla^2, the function -\frac{1}{4\pi |\mathbf{r} - \mathbf{r}'|} is the Green function for the operator. That is, \nabla^2 G = \delta(\mathbf{r} - \mathbf{r}').
We can show \nabla \times (\nabla \times \mathbf{A}_E) = -\mathbf{J}_m and likewise as \nabla \times (\nabla \times \mathbf{A}) = -\nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A}. We now claim that \nabla \cdot \mathbf{A} = 0
This would them imply that \nabla \times (\nabla \times \mathbf{A}_E) = -\nabla^2 \mathbf{A}_E = -\mathbf{J}_m.
To justify setting \nabla \cdot \mathbf{A} = 0, let us assume otherwise. Assume that \nabla \cdot \mathbf{A} = \psi(\mathbf{r}). Then, there exists some \mathbf{A}'(\mathbf{r}) = \mathbf{A}(\mathbf{r}) + \nabla f(\mathbf{r}). It is clear that \nabla \times \mathbf{A}'(\mathbf{r}) = \nabla \times \mathbf{A}(\mathbf{r}).
We can see that \nabla \cdot \mathbf{A}(\mathbf{r}) = \nabla \cdot \mathbf{A}(\mathbf{r}) + \nabla^2 f(\mathbf{r}) = \psi(\mathbf{r}) + \nabla^2 f(\mathbf{r}). Now, if we assert -\nabla^2 f(\mathbf{r}) = \psi(\mathbf{r}), that is,
f(\mathbf{r}) = \frac{1}{4\pi} \int_V \frac{\psi(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|} dV'
Under this assumption, we see that \nabla \cdot \mathbf{A}' = 0. Notably, we do not need to calculate f(\mathbf{r}). it is sufficient to show that it exists. By the Helmholtz theorem, it does,a s \nabla f is a vector field with divergence -\phi and no curl, so \nabla f is the gradient of a scalar potential given by the above integral.