Catchup 1

docs/physics/electrostatics/1-math.md

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+# Chapter 1 - Mathematics
+## 1.5 - Dyads and Tensors
+
+**Definition**. A *dyadic* is a representation of two-ish vectors.
+
+$$
+\stackrel{\leftrightarrow}{\vb{D}} = \begin{matrix}
+    D_{xx} \vu{x}\vu{x} &+ D_{xy} \vu{x}\vu{y} &+ D{xz} \vu{x}\vu{z} \\
+    + D_{yx} \vu{y}\vu{x} &+ D_{yy} \vu{y}\vu{y} &+ D{yz} \vu{y}\vu{z} \\
+    + D_{zx} \vu{z}\vu{x} &+ D_{zy} \vu{z}\vu{y} &+ D{zz} \vu{z}\vu{z}
+\end{matrix}
+$$
+
+**Definition**. If a dyadic can be written as a composition of two vectors $\vb{A}$ and $\vb{B}$, it is called a *dyad*.
+
+$$
+\vb{AB} = \begin{matrix}
+    A_x B_x \vu{x}\vu{x} &+ A_x B_y \vu{x}\vu{y} &+ A_x B_z \vu{x}\vu{z} \\
+    + A_y B_x \vu{y}\vu{x} &+ A_y B_y \vu{y}\vu{y} &+ A_y B_z \vu{y}\vu{z} \\
+    + A_z B_x \vu{z}\vu{x} &+ A_z B_y \vu{z}\vu{y} &+ A_z B_z \vu{z}\vu{z}
+\end{matrix}
+$$
+
+The dot product of a dyad $\stackrel{\leftrightarrow}{\vb{D}} = \vb{AB}$ and vector $\va{v}$ can be written as follows:
+
+$$
+(\vb{AB}) \vdot \va{v} = \vb{A} (\vb{B} \vdot \va{v})
+$$
+
+**Definition**. A *symmetric/antisymmetric* dyadic is defined the same way that a matrix is.
+
+**Definition**. The *identity dyadic* is $\stackrel{\leftrightarrow}{\vb{I}} = \vu{x}\vu{x} + \vu{y}\vu{y} + \vu{z}\vu{z}$.
+
+**Definition**. FOr a *tensor*, with coordinates $u^i$, we have two sets of basis vectors:
+
+$$
+\vb{e}_i = \pdv{\vb{r}}{u^i}
+$$
+
+$$
+\vb{e}^i = \grad{u^i}
+$$
+
+## 1.9 - Helmholtz Theorem
+
+Given an arbitrary vector field $\vb{F}(\vb(r))$, we can write said field as a composition of a curl-free component $\vb{\Phi}(\vb{r})$ and a divergence-free component $\vb{A}(\vb{r})$ as follows:
+
+$$
+\vb{F}(\vb{r}) = - \grad{\vb{\Phi}(\vb{r})} + \curl{\vb{A}(\vb{r})}
+$$
+
+**Definition**. Here, the gradient of the scalar potential is $\grad{\vb{\Phi}(\vb{r})}$ and the curl of the vector potential is $\curl{\vb{A}(\vb{r})}$. Thus, the scalar potential is $\vb{\Phi}(\vb{r})$ and the vector potential is $\vb{A}(\vb{r})$.
+
+Letting said field be over bounded volume $V$ with closed surface $\partial V$, and the functions $\vb{C}(\vb{r}) = \curl{\vb{F}(\vb{r})}$ and $\vb{D}(\vb{r}) = \div{\vb{F}(\vb{r})}$ are known, we can say that
+
+$$
+\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} - \frac{1}{4 \pi} \int_{\partial V} \frac{\vb{F}(\vb{r}') \vdot \va{n}'}{\abs{\vb{r}-\vb{r}'}} \dd{S'}
+$$
+
+$$
+\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} - \frac{1}{4 \pi} \int_{\partial V} \va{n}' \cross \frac{\vb{F}(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{S'}
+$$
+
+Now, assume that $\lim(\frac{\vb{F}(\vb{r})}{\vb{r}}) = 0$ as $\vb{r} \rightarrow \infty$, with a large enough volume, we see that the second terms vanish.
+
+$$
+\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}
+$$
+
+$$
+\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}
+$$

docs/physics/electrostatics/2-coulomb.md

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+# Chapter 2 - Coulomb's Laws, Electric and Magnetic Fields
+
+## Section 2.2 - Parallel Treatment of Electric and Magnetic Fields
+
+Consider two point charges, $q$ and $Q$, with the latter being at the origin of the coordinate system. Let $q$ be located at point $\vb{r}$ relative to the origin.
+
+Thus, according to Coulomb's Law,
+
+$$
+\begin{align}
+    F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
+    F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
+\end{align}
+$$
+
+Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\vb{r}$.
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
+    H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
+\end{align}
+$$
+
+Now, let $Q$ be at point $\vb{r'}$. Then, the unit vector becomes $\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}}$, and we see the following.
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{Q_e}{4 \pi \epsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
+    H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
+\end{align}
+$$
+
+With multiple charges, we can apply the *superposition principal* to see the following:
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
+    H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
+\end{align}
+$$
+
+We can convert this to an integral as $N$ goes to infinity.
+
+$$
+\begin{align}
+    E(\vb{r}) &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
+    H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V'
+\end{align}
+$$
+
+## Section 2.3 - Divergence and Curl of the Electrostatic or Magnetostatic Field
+
+From a lot of advanced math, we know that
+
+$$
+\div{\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}} = 4 \pi \delta(\vb{r}-\vb{r'})
+$$
+
+Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnetostatic fields.
+
+$$
+\begin{align}
+    \div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
+    \div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')}
+\end{align}
+$$
+
+As the divergence operator does not operate on $\vb{r'}$, we see that
+
+$$
+\begin{align}
+    \div{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+        &= \frac{1}{4 \pi \epsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
+        &= \frac{\rho_e(\vb{r})}{\epsilon_0} \\
+    \div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+        &= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
+        &= \frac{\rho_m(\vb{r})}{\mu_0}
+\end{align}
+$$
+
+The curl of an electrostatic or magnetostatic is relatively simple.
+
+
+$$
+\begin{align}
+    \curl{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+    \curl{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
+\end{align}
+$$
+
+Additionally, we know $\curl{f\vb{A}} = f \curl{\vb{A}} + \grad{f}\cross\vb{A}$. Thus,
+
+$$
+\begin{align}
+    \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} &= \frac{1}{\abs{\vb{r}-\vb{r'}}^3} \curl{(\vb{r}-\vb{r'})} + (\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}}) \cross (\vb{r}-\vb{r'}) \\
+\end{align}
+$$
+
+We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Additionally, $\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}} = -3 \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^5}$, which when crossed with $\vb{r}-\vb{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
+
+## Section 2.4 - Eletric and Magnetic Flux Densities
+
+The electric and magnetic flux density vectors are given by $\epsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
+
+Now, given $S$ is a surfance enclosing $Q_e$ or $Q_m$ total charge, we denotate flux as following:
+
+$$
+\Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
+$$
+
+Thus, applying divergence theorem,
+
+$$
+Q_e = \Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{E}} \dd V
+$$
+
+$$
+Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \epsilon_0 \int_V \div{\vb{H}} \dd V
+$$
+
+Since $Q_e = \int_V \rho_e \dd V$ and $Q_m = \int_V \rho_m \dd V$, we see that
+
+$$
+\begin{align}
+    \int_V \rho_e \dd V &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\
+    \rho_e &= \epsilon_0 \int_V \div{\vb{E}} \dd V \\
+    \int_V \rho_m \dd V &= \mu_0 \int_V \div{\vb{H}} \dd V \\
+    \rho_m &= \mu_0 \int_V \div{\vb{H}} \dd V \\
+\end{align}
+$$
+
+**Definition**. This is known as *Gauss' Law*.
+
+With applicable symnetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.

docs/physics/electrostatics/3-electro-magnetic-potentials.md

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+# Chapter 3 - Eletric and Magnetic Scalar Potentials
+
+# Section 3.1 - Work and Energy in Electrostatics and Magnetostatics
+
+The force on charge $q$ is given by $\vb{F}(\vb{r}) = q_e \vb{E}(\vb{r})$ or $\vb{F}(\vb{r}) = q_m \vb{H}(\vb{r})$. If this charge is moved $\dd{\vb{l}} = \dd x \vu{x} + \dd y \vu{y} + \dd z \vu{z}$, the change in internal energy (work) this produces can be written as
+
+$$
+\dd{U}= - \vb{F} \vdot \dd{\vb{l}}
+$$
+
+Rewriting this, $\vb{F} = -\grad{U}$, with $U$ as potential energy. Now, we can denote this change in internal energy in terms of $q$ as follows:
+
+$$
+\vb{E}(\vb{r}) = \frac{1}{q_e} \vb{F_e}(\vb{r}) = - \frac{1}{q_e} \grad{U_e(\vb{r})} = -\grad{V_e(\vb{r})}
+$$
+
+The units of electrostatic potential is Joule/Coublomb, also known as a Volt. Thus, the units of the electric field should be expressed in Volts/meter. Similarly,
+
+$$
+\vb{H}(\vb{r}) = \frac{1}{q_m} \vb{F_m}(\vb{r}) = - \frac{1}{q_m} \grad{U_m(\vb{r})} = -\grad{V_m(\vb{r})}
+$$
+
+The units of magnetostatic potential is Joule/Weber, also known as an Ampere. Thus, the units of the magnetic field can be written as Amperes/meter.
+
+With this, we can calculate work. Moving a charge $q$ from $A$ to $B$, we see that
+
+$$
+\delta W = \int_A^B \vb{F} \vdot \dd{\vb{l}} = q_e \int_A^B \vb{E} \vdot \dd{\vb{l}} = -q_e \int_A^B \grad{\vb{V}} \vdot \dd{\vb{l}} = -q_e \delta V_e 
+$$
+
+Strictly speaking, this is a potential difference. To find the absolute potential, assume a point charge $Q$ at the origin, and a charge $q$. We take the work as $q$ moves from $\vb{r'} = \vb{\infty}$ to $\vb{r'} = \vb{r}$. Thus,
+
+$$
+W = -q_e \frac{Q_e}{4 \pi \epsilon_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_e \frac{Q_e}{4 \pi \epsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \epsilon_0} \frac{1}{r}
+$$
+
+$$
+W = -q_m \frac{Q_m}{4 \pi \mu_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_m \frac{Q_m}{4 \pi \mu_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_m \frac{Q_m}{4 \pi \mu_0} \frac{1}{r}
+$$
+
+Letting the potential as $\vb{r} \leftarrow \infty$ equal $0$ be our reference and dividing out `q`, we find that the voltage for arrangement is the following:
+
+$$
+V_e(\vb{r}) = \frac{Q_e}{4 \pi \epsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r}
+$$
+
+Now, if we let the stationary charge $Q$ be located at $\vb{r'}$, we see that
+
+$$
+V_e(\vb{r}) = \frac{Q_e}{4 \pi \epsilon_0 \abs{\vb{r} - \vb{r'}}} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 \abs{\vb{r} - \vb{r'}}}
+$$
+
+If we allow multiple charges, this becomes
+
+$$
+V_e(\vb{r}) = \frac{1}{4\pi \epsilon_0} \sum_{i=1}^N \frac{Q_ei}{\abs{\vb{r}-\vb{r_i}}}
+$$
+
+Taking this to its natural limit,
+
+$$
+V_e(\vb{r}) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
+$$
+
+
+$$
+V_m(\vb{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
+$$