From ad62f9ec9524ad47c2330fc74824c25f2805d1ef Mon Sep 17 00:00:00 2001 From: Indigo5684 <159226326+Indigo5684@users.noreply.github.com> Date: Tue, 30 Sep 2025 13:19:42 -0500 Subject: [PATCH] Electrodynamics Section 11.1 --- .../11-potential-formulation.md | 180 +++++++++++++++++- 1 file changed, 179 insertions(+), 1 deletion(-) diff --git a/docs/physics/electrodynamics/11-potential-formulation.md b/docs/physics/electrodynamics/11-potential-formulation.md index 9f75559..fdbf983 100644 --- a/docs/physics/electrodynamics/11-potential-formulation.md +++ b/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