Electrodynamics Chapter 11

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

@@ -213,6 +213,88 @@ $$G(\mathbf{r} - \mathbf{r}', t - t') = -\frac{1}{4\pi |\mathbf{r} - \mathbf{r}'
 
 ### 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.