Electrodynamics Section 11.1

docs/physics/electrodynamics/11-potential-formulation.md

@@ -2,7 +2,7 @@
 
 ## Section 11.1 - Forces, Fields, Potentials, and Greens Functions
 
-## Section 11.1.1 - Potentials for Time-independent Fields
+### 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 $\nabla \times \mathbf{H} = \mathbf{J}_e$.
 
@@ -38,3 +38,181 @@ $$f(\mathbf{r}) = \frac{1}{4\pi} \int_V \frac{\psi(\mathbf{r}')}{|\mathbf{r} - \
 
 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
+
+### Section 11.2.2 - Fields of Continuous Charge and Current Distributions
+
+### Section 11.2.3 - Fields of a Moving Electric Point Charge