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

\[G(\mathbf{r} - \mathbf{r}', t - t') = -\frac{1}{4\pi |\mathbf{r} - \mathbf{r}'|} \delta(t - t' + \frac{1}{c}|\mathbf{r} - \mathbf{r}'|)\]

Section 11.2 - Potentials and Fields of Time-Dependent Electric Charge Distributions

Section 11.2.1 - Potentials of Continuous Charge and Current Distributions

From the properties of the Green function, we can calculate that

\[\begin{align} V_E(\mathbf{r}, t) &= \frac{1}{4\pi} \int \int \frac{\delta(t - t' - \frac{|\mathbf{r} - \mathbf{r}'|}{c})}{|\mathbf{r} - \mathbf{r}'|} \frac{\rho_e(\mathbf{r'}, t')}{\varepsilon_0} dV' dt' \\ \mathbf{A}_H(\mathbf{r}, t) &= \frac{1}{4\pi} \int \int \frac{\delta(t - t' - \frac{|\mathbf{r} - \mathbf{r}'|}{c})}{|\mathbf{r} - \mathbf{r}'|} \mathbf{J}_e(\mathbf{r'}, t') dV' dt \end{align}\]

Definition. Retarded time \(t_r\) is defined as \(t-c^{-1}|\mathbf{r} - \mathbf{r}'|\), which allows the integrals above to vanish for \(t' = t_r\).

\[\begin{align} V_E(\mathbf{r}, t) &= \frac{1}{4\pi\varepsilon_0} \int \frac{\rho_e(\mathbf{r'}, t_r)}{|\mathbf{r} - \mathbf{r}'|} dV' \\ \mathbf{A}_H(\mathbf{r}, t) &= \frac{1}{4\pi} \int \frac{\mathbf{J}_e(\mathbf{r'}, t_r)}{|\mathbf{r} - \mathbf{r}'|} dV' \end{align}\]

Section 11.2.2 - Fields of Continuous Charge and Current Distributions

We also assume no magnetic objects, so \(V_E = 0\), and no magnetic currents, so \(\mathbf{A}_E = 0\). Then, we know that

\[\mathbf{E} = -\nabla E - \mu_0 \frac{\partial}{\partial t} \mathbf{A}_H;\; \mathbf{H} = \nabla \times \mathbf{A}_H\]

This tells us that the magnetic field is

\[\begin{align} \mathbf{H} &= \nabla \times \mathbf{A}_H \\ &= \nabla \times \frac{1}{4\pi} \int \frac{\mathbf{J}_e(\mathbf{r'}, t_r)}{|\mathbf{r} - \mathbf{r}'|} dV' \\ &= \frac{1}{4\pi} \int \nabla \times (\frac{\mathbf{J}_e(\mathbf{r'}, t_r)}{|\mathbf{r} - \mathbf{r}'|}) dV' \end{align}\]

Now, let us apply \(\nabla \times (\mathbf{A}f) = -\mathbf{A} \times \nabla f + f(\nabla \times \mathbf{A})\), as well as \(\nabla \times \mathbf{A}(g(\mathbf{r})) = -\frac{\partial \mathbf{A}}{\partial g} \times \nabla g\)

\[\nabla \times (\frac{\mathbf{J}_e(\mathbf{r'}, t_r)}{|\mathbf{r} - \mathbf{r}'|}) = -\mathbf{J}(\mathbf{r}', t_r) \times \nabla(\frac{1}{|\mathbf{r} - \mathbf{r}'|}) - \frac{\dot{\mathbf{J}}(\mathbf{r}', t_r) \times \nabla t_r}{|\mathbf{r} - \mathbf{r}'|}\]

This lets us write \(\mathbf{H}\) as

\[\mathbf{H} = \frac{1}{4\pi} \int [( \frac{\mathbf{J}_(\mathbf{r}', t_r)}{|\mathbf{r} - \mathbf{r}'|^3} + \frac{\dot{\mathbf{J}}(\mathbf{r}', t_r)}{c |\mathbf{r} - \mathbf{r}'|^2}) \times (\mathbf{r} - \mathbf{r}')] dV'\]

Similarly, we can write \(-\nabla V_E\) as

\[-\nabla V_E = \frac{1}{4\pi \varepsilon_0} \int[\rho_e(\mathbf{r}', t_r) \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} + \frac{\dot{\rho}_e(\mathbf{r}', t_r)}{c |\mathbf{r} - \mathbf{r}'|} \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|}] dV'\]

We can then compute \(-\mu_0 \frac{\partial \mathbf{A}_H(\mathbf{r})}{\partial t}\) as

\[-\mu_0 \frac{\partial \mathbf{A}_H(\mathbf{r})}{\partial t} = -\frac{\mu_0}{4\pi} \int \frac{\dot{\mathbf{J}}_e(\mathbf{r}', t_r)}{|\mathbf{r} - \mathbf{r}'|} dV'\]

Summing these, we can write

\[\mathbf{E}(\mathbf{r}, t) = \frac{1}{4\pi \varepsilon_0} \int[\rho_e(\mathbf{r}', t_r) \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} + \frac{\dot{\rho}_e(\mathbf{r}', t_r)}{c} \frac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^2} - \frac{\dot{\mathbf{J}}_e(\mathbf{r}', t_r)}{c^2 |\mathbf{r} - \mathbf{r}'|}] dV'\]

The first term tells us that if current density is independent of time, then we recover Coulomb's law. The remaining terms are then painful to work through, and thus we will skip the calculation.

Now, for \(\mathbf{R} = \mathbf{r} - \mathbf{r}'\), and \(R = |\mathbf{R}|\), we see our final equation for the electric field becomes

\[\mathbf{E}(\mathbf{r}, t) = \frac{1}{4\pi\varepsilon_0} \int[\frac{\rho_e(\mathbf{r}', t_r)}{R^2} \hat{\mathbf{R}} + (\frac{2 \hat{\mathbf{R}}(\mathbf{J}(\mathbf{r}', t_r) \cdot \hat{\mathbf{R}}) - \mathbf{J}(\mathbf{r}', t_r)}{cR^2}) + \frac{\hat{\mathbf{R}} \times (\hat{\mathbf{R}} \times \dot{\mathbf{J}}(\mathbf{r}', t_r))}{c^2R}] dV\]

Section 11.2.3 - Fields of a Moving Electric Point Charge

There's a lot of math here. Just know that (directly from the textbook):

Summary of Properties of Fields of a Uniformly Moving Electric Point Charge 1. Electric Field reduced along direction of charge motion 2. Electric Field enhanced perpendicular to charge motion 3. Total electric flux conserved 4. Electric Field points from present position of charge 5. H field is perpendicular to direction of motion and E field. 6. E and H fields are related by \(\mathbf{H}_v(\mathbf{r}, t) = c \varepsilon_0 \mathbf{\beta} \times \mathbf{E}_v(\mathbf{r}, t)\) 7. E and H fields decrease as \(R_p^{-2}\) 8. There is no radiation. Energy does not propagate to infinity.

If the point charge is accelerating, simply cry.

Section 11.3 - Radiation

Section 11.3.1 - Radiation from an Oscillating Charge

Let a charge move such that \(z(t) = z_0 \cos \omega t\). Then, after a lot of math, \(\mathbf{E}\) is in the \(\hat{\mathbf{\theta}}\) direction and \(\mathbf{H}\) is in the \(\hat{\mathbf{\varphi}}\) direction. Then, \(\mathbf{S}\) is in the \(\hat{\mathbf{R}}\) direction, and falls off as \(R^{-2}\). Note that both the electric and magnetic fields depend on \(\sin\theta\), so the Poynting vector depends on \(\sin^2 \theta\). Average radiated power is proportional to \(\omega^4\) and \(p_0^2\), where \(p_0 = q_e z_0\).

The sky is blue as we see scattered light, which is dominated by blue frequencies due to the \(\omega^4\) term. During sunset, we see non-scattered (red) light. It is also polarized, as the electric field is perpendicular to the direction the wave travels.

Section 11.3.2 - Radiation from a Current Loop

Now, if the current loops in a circle with radius \(a\), the electric and magnetic fields are in the \(\hat{\mathbf{\phi}}\) and \(\hat{\mathbf{\theta}}\) directions respectively. Both depend on \(\sin\theta\) again, and fall off as \(r\), so the Poynting vector falls off as \(r^2\).

Section 11.3.3 - Radiation from a Linear Antenna

Not included in the textbook.