2025-04-16 Notes

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.

docs/physics/mechanics/14-collision-theory.md

@@ -0,0 +1,87 @@
+# Chapter 14 - Collision Theory
+
+## Section 14.1  - The Scattering Angle and Impact Parameter
+
+**Definition**. The angle between incoming and outgoing velocities is the *scattering angle* $\theta$. Note that $\theta = 0$ corresponds to no scattering and $\theta = \pi$ is a maximal value for $\theta$.
+
+**Definition**. The impact parameter $b$ is the perpendicular distance from the projectile's incoming path to a parallel axis through the center of the target. $b = 0$ implies a head-on collision. Note that $\theta = \theta(b)$.
+
+## Section 14.2 - The Collision Cross Section
+
+**Definition**. Consider multiple targets. Then, the *target density* $n_{tar}$ is the number of targets per unit area as viewed from the incident direction.
+
+If $A$ is the total area of the target assembly, the total number of targets becomes $n_{tar}A$.
+
+**Definition**. The *cross sectional area* or *cross-section* is defined as $\sigma = \pi R^2$, where $R$ is the radius of one target as seen from the front.
+
+Now, the total area of all targets is $n_{tar}A\sigma$. Then, we can see the probability of a hit is simply the area of all targets divided by area, or  $n_{tar} \sigma$. Naturally, if we send a test beam with $N_{inc}$ particles, we expect some fraction $N_{sc}$ to be scattered. Then,
+
+$$N_{sc} = N_{inc} n_{tar} \sigma$$
+
+If we then let $R_{inc} = N_{inc} / \Delta t$ for some time $\Delta t$, we find the rate of incoming particles per unit time. We can do the same to $N_{sc}$ to see that
+
+$$R_{sc} = R_{inc} n_{tar} \sigma$$
+
+**Definition**. Typical nuclear dimensions are about $10^{-14} \text{m}$, so the cross-sections are measured in units of $10^{-28} \text{m}^2$. This unit is known as $1 \text{barn}$.
+
+## Section 14.3 - Generalizations of the Cross Section
+
+Consider an incident sphere with radius $R_1$ and a target sphere of radius $R_2$. Then, we only care for $b \leq R_1 + R_2$. We know that $A = \pi(R_1 + R_2)^2$. Then, $\sigma = A = \pi(R_1 + R_2)^2$ (as any interaction in the area results in a collision). So, $N_{sc} = N_{inc} n_{tar} \sigma$
+
+Now, consider an example in which the particle may be captured or absorbed as well. Then, we can repeat the previous logic to see that $N_{cap} = N_{inc} n_{tar} \sigma$. If a target can both deflect and capture particles, we see both $N_{cap}$ and $N_{sc}$, where $\sigma_{cap} + \sigma_{sc} = \sigma_{tot}$.
+
+**Definition**. $\sigma_{cap}$ and $\sigma_{sc}$ are the *capture cross section* and *scattering cross section* respectively. Additionally, we can define the *ionization cross section* $\sigma_{ion}$ as the effective area of the target atom for an ionizing electron, and the *fission cross section* $\sigma_{ris}$ as the effective area of a $U^235$ nucleus for fission by neutron bombardment.
+
+**Definition**. A collision is said to be *elastic* if the internal motion of the target is left unchanged. Otherwise, the collision is *elastic*.
+
+**Definition**. The *ground state* of an atom is its lowest possible energy level. If an incident electron scatters elastically, it will leave the target in its ground state. Otherwise, *atomic excitation* will be seen.
+
+Note that we can differentiate the types of collisions as $\sigma_{sc} = \sigma_{el} + \sigma_{inel}$. Then, the total cross section $\sigma_{tot} = \sigma_{sc} + \sigma_{cap} + \sigma_{ion}$, which is the total cross section for any interaction with the target particle.
+
+## Section 14.4 - The Differential Scattering Cross Section
+
+**Definition**. For a cylinder on a circle with radius $r$, with an arc length of $s$, we define the angle $\delta \theta = s/r$, which comes from the definition of radians.
+
+**Definition**. For a cone on a sphere with radius $r$ and area $A$, we define the solid angle $\delta \Omega = A / r^2$, with units called *steradians* (abbreviated as sr), and ranges from $0$ to $4\pi$ (due to the maximum surface area of a sphere). Note this works for any shape of cone (eg. cones with non-rectangular bases).
+
+We will work in modified spherical polar coordinates, with the target on the origin and $z = \rho$. For a cone in the range $\theta$ to $\theta + d\theta$ and $\phi$ to $\phi + d\phi$, that is, cones with a rectangular base, we see that based on $A = r^2 \sin \theta d\theta + d\phi$,
+
+$$d\Omega = \sin \theta d\theta d\phi$$
+
+We now define
+
+$$N_{sc} (\text{into } d\Omega) = N_{inc} n_{tar} d\sigma (\text{into } d\Omega)$$
+
+**Definition**. Here, $d\simga = \frac{d\sigma}{d\Omega} d\Omega$, where we define the *differential scattering cross section* as $d\sigma / d\Omega$. This lets us say
+
+$$N_{sc} (\text{into } d\Omega) = N_{inc} n_{tar} \frac{d\sigma}{d\Omega} d\Omega$$
+
+We then see that
+
+$$\omega = \int \frac{d\sigma}{d\Omega}(\theta, \phi) d\Omega = \int_0^\pi \sin \theta \int_0^{2\pi} \frac{d\sigma}{d\Omega}(\theta, \phi) d\phi d\theta$$
+
+## Section 14.5 - Calculating the Differential Cross Section
+
+Consider the case of axial symmetry, that is, the differential cross section is independent of $\phi$. Then, we  can see that for $\theta = \theta(b)$, we can consider particles approaching in the range $b$ to $b + db$. The annulus created by  this has a cross-sectional area of $d\sigma = 2\pi b db$.
+
+We then see that the particles emerge between angles $\theta$ and $\theta + d\theta$ with solid angle $d\Omega = 2\pi \sin \theta d\theta$
+
+We can thus compute the differential cross section as
+
+$$\frac{d\sigma}{d\Omega} = \frac{b}{\sin \theta} |\frac{db}{d\theta}|$$
+
+## Section 14.6 - Rutherford Scattering
+
+Consider scattering electrons of off nuclei. Then, we know that
+
+$$F = \frac{kqQ}{r^2} = \frac{\gamma}{r^2}$$
+
+The rest of this section is complicated.
+
+## Section 14.7 - Cross Sections in Various Frames
+
+Skipped
+
+## Section 14.8 - Relation of the CM and Lab Scattering Angles
+
+Skipped