14 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.
Now, we can see that \mathbf{H} = \int_V (-\frac{1}{4|\mathbf{r} - \mathbf{r'}|}) \mathbf{J}_e(\mathbf{r'}) dV' = \int_V G(\mathbf{r} - \mathbf{r'}) \mathbf{J}_e(\mathbf{r'}) dV'.
From this, by knowing \nabla G(\mathbf{r} - \mathbf{r'}) = -\nabla' G(\mathbf{r} - \mathbf{r'}) we can find the divergence is
$$\begin{align} \nabla \cdot \mathbf{A}_h &= \int_V \nabla \cdot(G(\mathbf{r} - \mathbf{r'}) \mathbf{J}_e(\mathbf{r'})) dV' \ &= \int_V \mathbf{J}_e(\mathbf{r}') (-\nabla G(\mathbf{r} - \mathbf{r'})) dV \ &= \int_V G(\mathbf{r} - \mathbf{r'}) \nabla' \cdot \mathbf{J}_e(\mathbf{r}') dV \ \end{align}$$
We can then apply the continuity equation to see that
\nabla \cdot \mathbf{A}_H = \int_V G(\mathbf{r} - \mathbf{r'}) (-\frac{\partial \rho_e(\mathbf{r}')}{\partial t})
However, as we are working with time-independent fields, we set \frac{\partial rho}{\partial t} = 0, which forces the divergence to zero.
Conventional Approach
Using the constructive equations, and in the absence of magnetic current density, we see that \nabla \times \mathbf{E} = 0 in the time-independent case. Additionally, we can see that \nabla \cdot \mathbf{B} = 0 and \nabla \times \mathbf{B} = \mu_0 \mathbf{J}_e + \nabla \times \mathbf{M}.
This lets us write that \mathbf{E} = -\nabla V_E and \mathbf{B} = \nabla \times \mathbf{A}_B by the Helmholtz theorem.
We can now see that in the electric case, the Maxwell equations tell us that
\nabla \cdot \mathbf{E} = \frac{\rho_e}{\varepsilon_0} \rightarrow V_E(\mathbf{r}) = \int_{V'} \frac{1}{4\pi |\mathbf{r} - \mathbf{r'}|} \frac{\rho_e(\mathbf{r})}{\varepsilon_0} dV'
In the magnetic case, we again see that as \nabla \times \mathbf{A_B} = 0 under gauge transformation, we can write
\mathbf{A}_B(\mathbf{r}) = \int_{V'} \frac{1}{4\pi |\mathbf{r} - \mathbf{r'}|}(\mu_0 \mathbf{J}_e(\mathbf{r}) + \nabla' \times \mathbf{M}(\mathbf{r}')) dV'
Section 11.1.2 - Potentials for Time-Dependent Fields
Let us now consider the time-dependent cases, in which Maxwell's equations can be written as
$$\begin{align} \nabla \times \mathbf{E} &= -\mathbf{J}_m - \frac{\partial}{\partial t}(\mu_0 \mathbf{H} + \mathbf{M}) \ \nabla \times \mathbf{H} &= \mathbf{J}_e - \frac{\partial}{\partial t}(\varepsilon_0 \mathbf{E} + \mathbf{P}) \end{align}$$
We can now write
$$\begin{align} \mathbf{E} &= -\nabla V_E + \nabla \times \mathbf{A}_E - \mu_0 \frac{\partial}{\partial t} \mathbf{A}_H \ \mathbf{H} &= -\nabla V_H + \nabla \times \mathbf{A}_H + \varepsilon_0 \frac{\partial}{\partial t} \mathbf{A}_E \end{align}$$
We can then take the divergence of both sides to see that
$$\begin{align} \nabla \cdot \mathbf{E} &= -\nabla^2 V_E - \mu_0 \frac{\partial}{\partial t} \nabla \cdot \mathbf{A}_H = \frac{\rho_e}{\varepsilon_0} \ \nabla \cdot \mathbf{H} &= -\nabla^2 V_H + \varepsilon_0 \frac{\partial}{\partial t} \nabla \cdot \mathbf{A}_E = \frac{\rho_m}{\mu_0} \end{align}$$
Meanwhile, the curl equations tell us that
$$\begin{align} \nabla \times \mathbf{E} &= \nabla \times (\nabla \times \mathbf{A}_E) - \mu_0 \frac{\partial}{\partial t} (\nabla \times \mathbf{A}_H) + \mu_0 \frac{\partial}{\partial t} \mathbf{H} &= -\mathbf{J}_m - \frac{\partial}{\partial t} \mathbf{M} \ \nabla \times \mathbf{H} &= \nabla \times (\nabla \times \mathbf{A}_H) + \varepsilon_0 \frac{\partial}{\partial t}(\nabla \times \mathbf{A}_E) - \varepsilon_0 \frac{\partial}{\partial t} \mathbf{E} &= \mathbf{J}_e + \frac{\partial}{\partial t} \mathbf{P} \end{align}$$
We can expand the double curl and substitute the fields to see that
$$\begin{align} \nabla (\nabla \cdot \mathbf{A}_E) - \nabla^2 \mathbf{A}_E - \mu_0 \frac{\partial}{\partial t} (\nabla \times \mathbf{A}_H) + \mu_0 \frac{\partial}{\partial t} (-\nabla V_H + \nabla \times \mathbf{A}_H + \varepsilon_0 \frac{\partial}{\partial t} \mathbf{A}_E) &= -\mathbf{J}_m - \frac{\partial}{\partial t} \mathbf{M} \ \nabla(\nabla \cdot \mathbf{A}_H) - \nabla^2 \mathbf{A}_H + \varepsilon_0 \frac{\partial}{\partial t}(\nabla \times \mathbf{A}_E) - \varepsilon_0 \frac{\partial}{\partial t} (-\nabla V_E + \nabla \times \mathbf{A}_E - \mu_0 \frac{\partial}{\partial t} \mathbf{A}_H) &= \mathbf{J}_e + \frac{\partial}{\partial t} \mathbf{P} \end{align}$$
We can simplify this - the remaining curl terms cancel out.
$$\begin{align} \nabla (\nabla \cdot \mathbf{A}_E) - \nabla^2 \mathbf{A}_E + \mu_0 \frac{\partial}{\partial t} (-\nabla V_H + \varepsilon_0 \frac{\partial}{\partial t} \mathbf{A}_E) &= -\mathbf{J}_m - \frac{\partial}{\partial t} \mathbf{M} \ \nabla(\nabla \cdot \mathbf{A}_H) - \nabla^2 \mathbf{A}_H - \varepsilon_0 \frac{\partial}{\partial t} (-\nabla V_E - \mu_0 \frac{\partial}{\partial t} \mathbf{A}_H) &= \mathbf{J}_e + \frac{\partial}{\partial t} \mathbf{P} \end{align}$$
Now, rearrange terms.
$$\begin{align} -\nabla^2 \mathbf{A}_E + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{A}_E}{\partial t^2} + \nabla (\nabla \cdot \mathbf{A}_E - \mu_0 \frac{\partial V_H}{\partial t}) &= -\mathbf{J}_m - \frac{\partial}{\partial t} \mathbf{M} \
- \nabla^2 \mathbf{A}_H + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{A}_H}{\partial t^2} + \nabla(\nabla \cdot \mathbf{A}_H + \varepsilon_0 \frac{\partial V_E}{\partial t}) &= \mathbf{J}_e + \frac{\partial}{\partial t} \mathbf{P} \end{align}$$
Now, we can choose a gauge such that \nabla \cdot \mathbf{A_E} = \mu_0 \frac{\partial V_H}{\partial t} and \nabla \cdot \mathbf{A_H} = - \varepsilon_0 \frac{\partial V_E}{\partial t}. This allows us to write V_E \mapsto V_E - \mu_0 \frac{\partial \lambda}{\partial t}, V_H \mapsto V_H + \varepsilon_0 \frac{\partial \psi}{\partial t}. If we then say that \mathbf{A}_E \mapsto \mathbf{A}_E + \nabla \psi and \mathbf{A}_H \mapsto \mathbf{A}+H + \nabla \lambda, we can then solve \nabla \cdot \mathbf{A}_E - \mu_0 \frac{\partial V_H}{\partial t} = f(\mathbf{r}, t) and \nabla \cdot \mathbf{A}_H - \varepsilon_0 \frac{\partial V_E}{\partial t} = g(\mathbf{r}, t) to find wave equations, which have a Green function and thus have guaranteed solutions.
Substituting this gauge into the previous equations, we see that
$$\begin{align} (-\nabla^2 + \mu_0 \varepsilon_0 \frac{\partial^2}{\partial t^2}) \mathbf{A}_E &= -\mathbf{J}_m - \frac{\partial}{\partial t} \mathbf{M} \ (-\nabla^2 + \mu_0 \varepsilon_0 \frac{\partial^2}{\partial t^2}) \mathbf{A}_H &= \mathbf{J}_e + \frac{\partial}{\partial t} \mathbf{P} \end{align}$$
We also know previously that
$$\begin{align} -\nabla^2 V_E - \mu_0 \frac{\partial}{\partial t} \nabla \cdot \mathbf{A}_H = \frac{\rho_e}{\varepsilon_0} \ -\nabla^2 V_H + \varepsilon_0 \frac{\partial}{\partial t} \nabla \cdot \mathbf{A}_E = \frac{\rho_m}{\mu_0} \end{align}$$
We can substitute our gauge to see that
$$\begin{align} -\nabla^2 V_E + \mu_0 \frac{\partial}{\partial t} \varepsilon_0 \frac{\partial V_E}{\partial t} = \frac{\rho_e}{\varepsilon_0} \ -\nabla^2 V_H + \varepsilon_0 \frac{\partial}{\partial t} \mu_0 \frac{\partial V_H}{\partial t} = \frac{\rho_m}{\mu_0} \end{align}$$
Simplifying, we see that
$$\begin{align} -(\nabla^2 + \mu_0 \varepsilon_0 \frac{\partial^2}{\partial t^2}) V_E = \frac{\rho_e}{\varepsilon_0} \ -(\nabla^2 + \mu_0 \varepsilon_0 \frac{\partial^2}{\partial t^2}) V_H = \frac{\rho_m}{\mu_0} \end{align}$$
Conventional Approach to Potentials for Time-Dependent Fields
In the conventional approach, we use the $\mathbf{E}$-\mathbf{B} form of our Maxwell's equations, and presume no magnetic current nor charge to see that
$$\begin{align} \nabla \cdot \mathbf{E} &= \frac{\rho_e}{\varepsilon_0} \ \nabla \cdot \mathbf{B} &= 0 \ \nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} &= 0 \ \nabla \times \mathbf{B} - \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} &= \mu_0 \mathbf{J}_e + \mu_0 \frac{\partial \mathbf{P}}{\partial t} + \nabla \times \mathbf{M} \end{align}$$
This allows us to cancel terms in our Helmholtz construction, leading to
$$\begin{align} \mathbf{B} &= \nabla \times \mathbf{A}_B \ \mathbf{E} &= -\nabla V_E - \frac{\partial \mathbf{A}_B}{\partial t} \end{align}$$
Taking the divergence of \mathbf{E}, we see that
\nabla \cdot(-\nabla V_E - \frac{\partial}{\partial t} \mathbf{A_B}) = \frac{\rho_e}{\varepsilon_0}
Taking the curl of \mathbf{B} and rearranging, we see that
$$\begin{align} \nabla \times (\nabla \times \mathbf{A}_B) - \mu_0 \varepsilon_0 \frac{\partial}{\partial t}[-\nabla V_E - \frac{\partial \mathbf{A}_B}{\partial t}] &= \mu_0 \mathbf{J}_e + \mu_0 \frac{\partial \mathbf{P}}{\partial t} + \nabla \times \mathbf{M} \ -\nabla^2 \mathbf{A}_B + \nabla(\nabla \cdot \mathbf{A}_B) + \mu_0 \varepsilon_0 \frac{\partial}{\partial t} \nabla V_E + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{A}_B}{\partial t^2} &= \mu_0 \mathbf{J}_e + \mu_0 \frac{\partial \mathbf{P}}{\partial t} + \nabla \times \mathbf{M} \ -\nabla^2 \mathbf{A}_B + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{A}_B}{\partial t^2} + \nabla(\nabla \cdot \mathbf{A}_B + \mu_0 \varepsilon_0 \frac{\partial V_E}{\partial t}) &= \mu_0 \mathbf{J}_e + \mu_0 \frac{\partial \mathbf{P}}{\partial t} + \nabla \times \mathbf{M} \ \end{align}$$
Now, choose a gauge that sets \nabla \cdot \mathbf{A}_B = -\mu_0 \varepsilon_0 \frac{\partial V_E}{\partial t}.
This will give us the wave equation
(-\nabla^2 + \mu_0 \varepsilon_0 \frac{\partial^2}{\partial t^2}) \mathbf{A}_B = \mu_0 \mathbf{J}_e + \mu_0 \frac{\partial \mathbf{P}}{\partial t} + \nabla \times \mathbf{M}
Applying the gauge to the divergence equation shows us that
(-\nabla^2 + \mu_0 \varepsilon_0 \frac{\partial^2}{\partial t^2}) V_E = \frac{\rho_e}{\varepsilon_0}
Section 11.1.3 - Green Function for the Wave Equation
We know the Green function for the Laplacian is
G(\mathbf{r} - \mathbf{r}') = -\frac{1}{4 \pi |\mathbf{r} - \mathbf{r}'}
That is, \nabla^2 G(\mathbf{r} - \mathbf{r}') = \delta(\mathbf{r} - \mathbf{r}').
How, we want to find the Green function for the wave equation. That is, we want to find the function that satisfies
(\nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}) G(\mathbf{r} - \mathbf{r}', t - t') = \delta(\mathbf{r} - \mathbf{r}', t - t')
We know that the wave equation shares values upon characteristic lines through $\mathbf{r}$-t space. That is, if t - \frac{1}{c} |\mathbf{r} - \mathbf{r}'| is a constant, we are on a characteristic line and the function shares a value. We can call this value t_r and define it as
t_r = t - \frac{1}{c} |\mathbf{r} - \mathbf{r}'|
Now, we will make a guess for the Green function as
G(\mathbf{r} - \mathbf{r}', t - t') = -\frac{1}{4\pi |\mathbf{r} - \mathbf{r}'|} \delta(t - t' + \frac{1}{c}|\mathbf{r} - \mathbf{r}'|)