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

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docs/physics/electrostatics/2-coulomb.md
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## 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.
Consider two point charges, $q$ and $Q$, with the latter being at the origin of the coordinate system. Let $q$ be located at point $\mathbf{r}$ relative to the origin.
Thus, according to Coulomb's Law,
$$
\begin{align}
F^e_{qQ}(\vb{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
F^m_{qQ}(\vb{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
F^e_{qQ}(\mathbf{r}) &= \frac{q_e Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
F^m_{qQ}(\mathbf{r}) &= \frac{q_m Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
\end{align}
$$
Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\vb{r}$.
Divide by the charge $q$ to obtain the *electric or magnetic field* at point $\mathbf{r}$.
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vu{r}}{\abs{\vb{r}}^2} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vu{r}}{\abs{\vb{r}}^2}
E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2} \\
H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}
\end{align}
$$
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.
Now, let $Q$ be at point $\mathbf{r'}$. Then, the unit vector becomes $\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}}|$, and we see the following.
$$
\begin{align}
E(\vb{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
E(\mathbf{r}) &= \frac{Q_e}{4 \pi \varepsilon_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
H(\mathbf{r}) &= \frac{Q_m}{4 \pi \mu_0} \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
\end{align}
$$
@@ -35,97 +35,89 @@ With multiple charges, we can apply the *superposition principal* to see the fol
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}
E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^N Q_e \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} \\
H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \sum_{i=1}^N Q_m \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
\end{align}
$$
We can convert this to an integral as $N$ goes to infinity.
$$
\begin{align}
E(\vb{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V' \\
H(\vb{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V'
\end{align}
$$
$$\begin{align}
E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V' \\
H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V'
\end{align}$$
## Section 2.3 - Divergence and Curl of the Electrostatic or Magnetostatic Field
From a lot of advanced math, we know that
$$
\div{\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3}} = 4 \pi \delta(\vb{r}-\vb{r'})
$$
$$\nabla \cdot \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = 4 \pi \delta(\mathbf{r}-\mathbf{r'})$$
Now, apply the divergence operator over $\vb{r}$ to the electrostatic and magnetostatic fields.
Now, apply the divergence operator over $\mathbf{r}$ to the electrostatic and magnetostatic fields.
$$
\begin{align}
\div{E(\vb{r})} &= \div{(\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')} \\
\div{H(\vb{r})} &= \div{(\frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3} \dd V')}
\end{align}
$$
$$\begin{align}
\nabla \cdot E(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V') \\
\nabla \cdot H(\mathbf{r}) &= \nabla \cdot (\frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} d V')
\end{align}$$
As the divergence operator does not operate on $\vb{r'}$, we see that
As the divergence operator does not operate on $\mathbf{r'}$, we see that
$$
\begin{align}
\div{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_e(\vb{r})}{\varepsilon_0} \\
\div{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \div{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
&= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\vb{r'}) \delta(\vb{r}-\vb{r'}) \dd V' \\
&= \frac{\rho_m(\vb{r})}{\mu_0}
\end{align}
$$
$$\begin{align}
\nabla \cdot E(\mathbf{r}) &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
&= \frac{1}{4 \pi \varepsilon_0} 4 \pi \int_V \rho_e(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
&= \frac{\rho_e(\mathbf{r})}{\varepsilon_0} \\
\nabla \cdot H(\mathbf{r}) &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \cdot (\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}) d V' \\
&= \frac{1}{4 \pi \mu_0} 4 \pi \int_V \rho_m(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'}) d V' \\
&= \frac{\rho_m(\mathbf{r})}{\mu_0}
\end{align}$$
The curl of an electrostatic or magnetostatic is relatively simple.
$$
\begin{align}
\curl{E(\vb{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
\curl{H(\vb{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
\nabla \times{E(\mathbf{r})} &= \frac{1}{4 \pi \varepsilon_0} \int_V \rho_e(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
\nabla \times{H(\mathbf{r})} &= \frac{1}{4 \pi \mu_0} \int_V \rho_m(\mathbf{r'}) \nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} d V' \\
\end{align}
$$
Additionally, we know $\curl{f\vb{A}} = f \curl{\vb{A}} + \grad{f}\cross\vb{A}$. Thus,
Additionally, we know $\nabla \times{f\mathbf{A}} = f \nabla \times{\mathbf{A}} + \nabla{f}\times\mathbf{A}$. Thus,
$$
\begin{align}
\curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} &= \frac{1}{\abs{\vb{r}-\vb{r'}}^3} \curl{(\vb{r}-\vb{r'})} + (\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}}) \cross (\vb{r}-\vb{r'}) \\
\nabla \times{(\frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3})} &= \frac{1}{|\mathbf{r}-\mathbf{r'}|^3} \nabla \times{(\mathbf{r}-\mathbf{r'})} + (\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}}) \times (\mathbf{r}-\mathbf{r'}) \\
\end{align}
$$
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.
We can verify that $\nabla \times{(\mathbf{r}-\mathbf{r'})} = 0$, cancelling the first term. Additionally, $\nabla \times{\frac{1}{|\mathbf{r}-\mathbf{r'}|^3}} = -3 \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^5}$, which when crossed with $\mathbf{r}-\mathbf{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
## Section 2.4 - Electric and Magnetic Flux Densities
The electric and magnetic flux density vectors are given by $\varepsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
The electric and magnetic flux density vectors are given by $\varepsilon_0 \mathbf{E}$ and $\mu_0 \mathbf{H}$.
Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
$$
\Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
\Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d S = Q_m
$$
Thus, applying divergence theorem,
$$
Q_e = \Phi_e = \varepsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{E}} \dd V
Q_e = \Phi_e = \varepsilon_0 \int_S \mathbf{E} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V
$$
$$
Q_m = \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd = \varepsilon_0 \int_V \div{\vb{H}} \dd V
Q_m = \Phi_m = \mu_0 \int_S \mathbf{H} \cdot \hat{\mathbf{n}} d = \varepsilon_0 \int_V \nabla \cdot \mathbf{H} d V
$$
Since $Q_e = \int_V \rho_e \dd V$ and $Q_m = \int_V \rho_m \dd V$, we see that
Since $Q_e = \int_V \rho_e d V$ and $Q_m = \int_V \rho_m d V$, we see that
$$
\begin{align}
\int_V \rho_e \dd V &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
\rho_e &= \varepsilon_0 \int_V \div{\vb{E}} \dd V \\
\int_V \rho_m \dd V &= \mu_0 \int_V \div{\vb{H}} \dd V \\
\rho_m &= \mu_0 \int_V \div{\vb{H}} \dd V \\
\int_V \rho_e d V &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
\rho_e &= \varepsilon_0 \int_V \nabla \cdot \mathbf{E} d V \\
\int_V \rho_m d V &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
\rho_m &= \mu_0 \int_V \nabla \cdot \mathbf{H} d V \\
\end{align}
$$

114
docs/physics/electrostatics/3-electro-magnetic-potentials.md
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@@ -2,22 +2,22 @@
## 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
The force on charge $q$ is given by $\mathbf{F}(\mathbf{r}) = q_e \mathbf{E}(\mathbf{r})$ or $\mathbf{F}(\mathbf{r}) = q_m \mathbf{H}(\mathbf{r})$. If this charge is moved $d{\mathbf{l}} = d x \hat{\mathbf{x}} + d y \hat{\mathbf{y}} + d z \hat{\mathbf{z}}$, the change in internal energy (work) this produces can be written as
$$
\dd{U}= - \vb{F} \vdot \dd{\vb{l}}
d{U}= - \mathbf{F} \cdot d{\mathbf{l}}
$$
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:
Rewriting this, $\mathbf{F} = -\nabla{U}$, with $U$ as potential energy. Now, we can denote this change in internal energy in terms of $q$ as follows:
$$
\vb{E}(\vb{r}) = \frac{1}{q_e} \vb{F_e}(\vb{r}) = - \frac{1}{q_e} \grad{U_e(\vb{r})} = -\grad{V_e(\vb{r})}
\mathbf{E}(\mathbf{r}) = \frac{1}{q_e} \mathbf{F_e}(\mathbf{r}) = - \frac{1}{q_e} \nabla{U_e(\mathbf{r})} = -\nabla{V_e(\mathbf{r})}
$$
The units of electrostatic potential is Joule/Coulomb, also known as a Volt. Thus, the units of the electric field should be expressed in Volts/meter. Similarly,
$$
\vb{H}(\vb{r}) = \frac{1}{q_m} \vb{F_m}(\vb{r}) = - \frac{1}{q_m} \grad{U_m(\vb{r})} = -\grad{V_m(\vb{r})}
\mathbf{H}(\mathbf{r}) = \frac{1}{q_m} \mathbf{F_m}(\mathbf{r}) = - \frac{1}{q_m} \nabla{U_m(\mathbf{r})} = -\nabla{V_m(\mathbf{r})}
$$
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.
@@ -25,45 +25,45 @@ The units of magnetostatic potential is Joule/Weber, also known as an Ampere. Th
With this, we can calculate work. Moving a charge $q$ from $A$ to $B$, we see that
$$
\delta W = \int_A^B \vb{F} \vdot \dd{\vb{l}} = q_e \int_A^B \vb{E} \vdot \dd{\vb{l}} = -q_e \int_A^B \grad{\vb{V}} \vdot \dd{\vb{l}} = -q_e \delta V_e
\delta W = \int_A^B \mathbf{F} \cdot d{\mathbf{l}} = q_e \int_A^B \mathbf{E} \cdot d{\mathbf{l}} = -q_e \int_A^B \nabla{\mathbf{V}} \cdot d{\mathbf{l}} = -q_e \delta V_e
$$
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,
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 $\mathbf{r'} = \mathbf{\infty}$ to $\mathbf{r'} = \mathbf{r}$. Thus,
$$
W = -q_e \frac{Q_e}{4 \pi \varepsilon_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_e \frac{Q_e}{4 \pi \varepsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \varepsilon_0} \frac{1}{r}
W = -q_e \frac{Q_e}{4 \pi \varepsilon_0} \int_{\infty}^0 \frac{\hat{\mathbf{r'}}}{r'^2} \cdot (\hat{\mathbf{r'}}) d{r'} = -q_e \frac{Q_e}{4 \pi \varepsilon_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_e \frac{Q_e}{4 \pi \varepsilon_0} \frac{1}{r}
$$
$$
W = -q_m \frac{Q_m}{4 \pi \mu_0} \int_{\infty}^0 \frac{\vu{r'}}{r'^2} \vdot (\vu{r'}) \dd{r'} = -q_m \frac{Q_m}{4 \pi \mu_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_m \frac{Q_m}{4 \pi \mu_0} \frac{1}{r}
W = -q_m \frac{Q_m}{4 \pi \mu_0} \int_{\infty}^0 \frac{\hat{\mathbf{r'}}}{r'^2} \cdot (\hat{\mathbf{r'}}) d{r'} = -q_m \frac{Q_m}{4 \pi \mu_0} [\frac{-1}{r'}]_{\infty}^{r'} = q_m \frac{Q_m}{4 \pi \mu_0} \frac{1}{r}
$$
Letting the potential as $\vb{r} \rightarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following:
Letting the potential as $\mathbf{r} \rightarrow \infty$ equal $0$ be our reference and dividing out $q$, we find that the voltage for arrangement is the following:
$$
V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 r} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 r}
V_e(\mathbf{r}) = \frac{Q_e}{4 \pi \varepsilon_0 r} \text{ and } V_m(\mathbf{r}) = \frac{Q_m}{4 \pi \mu_0 r}
$$
Now, if we let the stationary charge $Q$ be located at $\vb{r'}$, we see that
Now, if we let the stationary charge $Q$ be located at $\mathbf{r'}$, we see that
$$
V_e(\vb{r}) = \frac{Q_e}{4 \pi \varepsilon_0 \abs{\vb{r} - \vb{r'}}} \text{ and } V_m(\vb{r}) = \frac{Q_m}{4 \pi \mu_0 \abs{\vb{r} - \vb{r'}}}
V_e(\mathbf{r}) = \frac{Q_e}{4 \pi \varepsilon_0}{|\mathbf{r}-\mathbf{r'}|} \text{ and } V_m(\mathbf{r}) = \frac{Q_m}{4 \pi \mu_0}{|\mathbf{r}-\mathbf{r'}|}
$$
If we allow multiple charges, this becomes
$$
V_e(\vb{r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^N \frac{Q_ei}{\abs{\vb{r}-\vb{r_i}}}
V_e(\mathbf{r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^N \frac{Q_{ei}}{|\mathbf{r}-\mathbf{r_i}|}
$$
Taking this to its natural limit,
$$
V_e(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
V_e(\mathbf{r}) = \frac{1}{4 \pi \varepsilon_0} \int_{V'} \frac{\rho_e(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}}| d{V'}
$$
$$
V_m(\vb{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
V_m(\mathbf{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}}| d{V'}
$$
## Section 3.2 - Energy of a Charge Distribution
@@ -71,43 +71,43 @@ $$
Given two point charges $Q_{e1}, Q_{e2}$ we know the work to bring them together is
$$
W_2 = W_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}}
W_2 = W_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_{e1} Q_{e2}}{|\mathbf{r_2} - \mathbf{r_1}|}
$$
Superposition applies here. The energy to create $N$ charges is
$$
W_n = \frac{1}{2} \frac{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
W_n = \frac{1}{2} \frac{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{|\mathbf{r_i}-\mathbf{r_j}|}
$$
For the sake of symmetry, sum overall charges and divide by 2.
$$
W_n = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
W_n = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{|\mathbf{r_i}-\mathbf{r_j}|}
$$
Rearranging, we see the following:
$$
W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \varepsilon_0} \frac{Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
= \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\vb{r_i})
W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \varepsilon_0} \frac{Q_{ej}}{|\mathbf{r_i}-\mathbf{r_j}|}
= \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\mathbf{r_i})
$$
We can rewrite this as a Riemann sum and convert to an integral.
$$
W_e = \frac{1}{2} \int_V p_e(\vb{r}) V_e(\vb{r}) \dd V ; \quad
W_m = \frac{1}{2} \int_V p_m(\vb{r}) V_m(\vb{r}) \dd V
W_e = \frac{1}{2} \int_V p_e(\mathbf{r}) V_e(\mathbf{r}) d V ; \quad
W_m = \frac{1}{2} \int_V p_m(\mathbf{r}) V_m(\mathbf{r}) d V
$$
We can also express this as
$$
W_e = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \int_V \int_{V'} \frac{\rho_e(\vb{r})\rho_e(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V}
W_e = \frac{1}{2} \frac{1}{4 \pi \varepsilon_0} \int_V \int_{V'} \frac{\rho_e(\mathbf{r})\rho_e(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} d{V'} d{V}
$$
$$
W_m = \frac{1}{2} \frac{1}{4 \pi \mu_0} \int_V \int_{V'} \frac{\rho_m(\vb{r})\rho_m(\vb{r'})}{\abs{\vb{r} - \vb{r'}}} \dd{V'} \dd{V}
W_m = \frac{1}{2} \frac{1}{4 \pi \mu_0} \int_V \int_{V'} \frac{\rho_m(\mathbf{r})\rho_m(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} d{V'} d{V}
$$
Note the $\frac{1}{2}$ is the same anti-double-counting factor introduced previously. If we were to determine the potential based on a different set of charges, the factor would be absent.
@@ -115,28 +115,28 @@ Note the $\frac{1}{2}$ is the same anti-double-counting factor introduced previo
We can now write an expression for energy of a charge density in terms of the field that it produces.
$$
W = \frac{\varepsilon_0}{2} \int_V (\div{\vb{E}(\vb{r})}) V(\vb{r}) \dd V
W = \frac{\varepsilon_0}{2} \int_V (\nabla \cdot \mathbf{E}(\mathbf{r})) V(\mathbf{r}) d V
$$
Simplifying, we see that
$$
W_e = \frac{\varepsilon_0}{2} \int_{V} E^2(\vb{r}) \dd V ; \quad
W_m = \frac{\mu_0}{2} \int_{V} H^2(\vb{r}) \dd V
W_e = \frac{\varepsilon_0}{2} \int_{V} E^2(\mathbf{r}) d V ; \quad
W_m = \frac{\mu_0}{2} \int_{V} H^2(\mathbf{r}) d V
$$
## Section 3.3 - The Poisson and Laplace Equations
We know that $\vb{E}(\vb{r}) = -\div{V_e(\vb{r})}$ and $\vb{H}(\vb{r}) = -\div{V_m(\vb{r})}$
We know that $\mathbf{E}(\mathbf{r}) = -\nabla \cdot V_e(\mathbf{r})$ and $\mathbf{H}(\mathbf{r}) = -\nabla \cdot V_m(\mathbf{r})$
Combined this, as well as the first of the Maxwell equations, we see that
$$
\div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\varepsilon_0}
\nabla \cdot \mathbf{E} = -\nabla \cdot \nabla V_e = - \nabla^2{V_e} = \frac{\rho_e}{\varepsilon_0}
$$
$$
\div{\vb{H}} = -\div{\grad{V_m}} = - \laplacian{V_m} = \frac{\rho_m}{\mu_0}
\nabla \cdot \mathbf{H} = -\nabla \cdot \nabla V_m = - \nabla^2{V_m} = \frac{\rho_m}{\mu_0}
$$
The last inequality is called the Poisson Equation, or the inhomogeneous Laplace equation.
@@ -144,19 +144,19 @@ The last inequality is called the Poisson Equation, or the inhomogeneous Laplace
To solve this equation, we define a Green function as follows:
$$
\laplacian G(\vb{r}, \vb{r'}) = \delta(\vb{r} - \vb{r'})
\nabla^2 G(\mathbf{r}, \mathbf{r'}) = \delta(\mathbf{r} - \mathbf{r'})
$$
Now, we can construct a potential function in terms of said green function that satisfies the Laplace equation.
$$
V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'}
V_e(\mathbf{r}) = - \int_V G(\mathbf{r}, \mathbf{r'}) \frac{\rho_e(\mathbf{r'})}{\varepsilon_0} d{V'}
$$
This is the specific solution. Let $\psi(\vb{r})$ be a solution to the homogenous equation. We can state the following:
This is the specific solution. Let $\psi(\mathbf{r})$ be a solution to the homogenous equation. We can state the following:
$$
V_e(\vb{r}) = \psi(\vb{r}) - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\varepsilon_0} \dd{V'}
V_e(\mathbf{r}) = \psi(\mathbf{r}) - \int_V G(\mathbf{r}, \mathbf{r'}) \frac{\rho_e(\mathbf{r'})}{\varepsilon_0} d{V'}
$$
We will consider the potential of a point charge. THat is, the limit of potential is zero as distance approaches infinity.
@@ -164,36 +164,36 @@ We will consider the potential of a point charge. THat is, the limit of potentia
Recall the potential of a point charge:
$$
V_e(\vb{r}) = \frac{Q_e}{\varepsilon_0} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}}
V_e(\mathbf{r}) = \frac{Q_e}{\varepsilon_0} \frac{1}{4 \pi |\mathbf{r} - \mathbf{r}|}
$$
We know that $- \laplacian{V(\vb{r})} = \div{\vb{E}(\vb{r})}$. Thus, recall the electric field of a point charge.
We know that $- \nabla^2{V(\mathbf{r})} = \nabla \cdot \mathbf{E}(\mathbf{r})$. Thus, recall the electric field of a point charge.
$$
\vb{E}(\vb{r}) = -\grad{V(\vb{r})} = \frac{Q_e}{\varepsilon_0} \grad({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) = \frac{Q_e}{\varepsilon_0} \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3}
\mathbf{E}(\mathbf{r}) = -\nabla{V(\mathbf{r})} = \frac{Q_e}{\varepsilon_0} \nabla({\frac{-1}{4\pi|\mathbf{r}-\mathbf{r'}|}}) = \frac{Q_e}{\varepsilon_0} \frac{\mathbf{r} - \mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}
$$
Taking the divergence, we find that
$$
- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\varepsilon_0} = \frac{Q_e}{\varepsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}})
= \frac{Q_e}{\varepsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\varepsilon_0} \delta(\vb{r} - \vb{r'})
- \nabla^2{V(\mathbf{r})} = G(\mathbf{r}, \mathbf{r'}) \frac{Q_e}{\varepsilon_0} = \frac{Q_e}{\varepsilon_0} \nabla^2({\frac{-1}{4\pi|\mathbf{r}-\mathbf{r'}|}})
= \frac{Q_e}{\varepsilon_0} \nabla \cdot \frac{\mathbf{r} - \mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} = \frac{Q_e}{\varepsilon_0} \delta(\mathbf{r} - \mathbf{r'})
$$
Thus, we see that
$$
\laplacian {\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}} = \delta(\vb{r} - \vb{r'}) \quad \Rightarrow \quad G(\vb{r}, \vb{r'}) = {\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}
\nabla^2 {\frac{-1}{4\pi}}{|\mathbf{r}-\mathbf{r'}|} = \delta(\mathbf{r} - \mathbf{r'}) \quad \Rightarrow \quad G(\mathbf{r}, \mathbf{r'}) = {\frac{-1}{4\pi}}{|\mathbf{r}-\mathbf{r'}|}
$$
Finally,
$$
V_e(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_e}{\varepsilon_0} \dd{V'}
V_e(\mathbf{r}) = \int_{V'} \frac{1}{4 \pi |\mathbf{r} - \mathbf{r}|} \frac{\rho_e}{\varepsilon_0} d{V'}
$$
$$
V_m(\vb{r}) = \int_{V'} \frac{1}{4 \pi \abs{\vb{r} - \vb{r}}} \frac{\rho_m}{\mu_0} \dd{V'}
V_m(\mathbf{r}) = \int_{V'} \frac{1}{4 \pi |\mathbf{r} - \mathbf{r}|} \frac{\rho_m}{\mu_0} d{V'}
$$
## Section 3.4 - The Laplace and Poisson Equations with Boundary Conditions
@@ -215,52 +215,52 @@ We want a simple way to write the Green function.
Let us assume all charge is contained in a sphere with radius $R$ centered at the origin. Then, for points $r$ far from the origin, the Green function can be written as
$$
\frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} = \frac{1}{4\pi \sqrt{r^2 - 2\vb{r} \vdot \vb{r'} + r'^2}}
= \frac{1}{4\pi r}(1 - 2 \vu{r} \vdot \vu{r'} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}
\frac{1}{4 \pi|\mathbf{r}-\mathbf{r'}|} = \frac{1}{4\pi \sqrt{r^2 - 2\mathbf{r} \cdot \mathbf{r'} + r'^2}}
= \frac{1}{4\pi r}(1 - 2 \hat{\mathbf{r}} \cdot \hat{\mathbf{r'}} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}
$$
This inverse square root term $(1 - 2 \vu{r} \vdot \vu{r'} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}$ can be expanded as a power series in $\frac{r'}{r}$.
This inverse square root term $(1 - 2 \hat{\mathbf{r}} \cdot \hat{\mathbf{r'}} + \frac{r'}{r} + \frac{r'^2}{r^2})^{-\frac{1}{2}}$ can be expanded as a power series in $\frac{r'}{r}$.
The first two terms of this power series are simple enough.
$$
G(\vb{r}, \vb{r'}) = \frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} \approx \frac{1}{4 \pi r} ( 1 + \frac{\vu{r} \vdot \vb{r'}}{r}); \quad \text{ for} r > r'
G(\mathbf{r}, \mathbf{r'}) = \frac{1}{4 \pi |\mathbf{r}-\mathbf{r'}|} \approx \frac{1}{4 \pi r} ( 1 + \frac{\hat{\mathbf{r}} \cdot \mathbf{r'}}{r}); \quad \text{ for} r > r'
$$
Applying this to the equation for voltage, we see that
$$
V_e(r) = \frac{1}{\varepsilon_0} \int_{V'} G(\vb{r}, \vb{r'}) p_e(\vb{r'}) \dd{V'} \approx
\frac{1}{4 \pi \varepsilon_0 r} \int_{V'} (1 + \frac{\vu{r} \vdot \vb{r'}}{r}) p_e(\vb{r'}) \dd{V'}
= \frac{Q_e}{4 \pi \varepsilon_0 r} + \frac{\vu{r} \vdot \vb{p}}{4 \pi \varepsilon_0 r^2}
V_e(r) = \frac{1}{\varepsilon_0} \int_{V'} G(\mathbf{r}, \mathbf{r'}) p_e(\mathbf{r'}) d{V'} \approx
\frac{1}{4 \pi \varepsilon_0 r} \int_{V'} (1 + \frac{\hat{\mathbf{r}} \cdot \mathbf{r'}}{r}) p_e(\mathbf{r'}) d{V'}
= \frac{Q_e}{4 \pi \varepsilon_0 r} + \frac{\hat{\mathbf{r}} \cdot \mathbf{p}}{4 \pi \varepsilon_0 r^2}
$$
**Definition**. The first and second terms of this equation are the *monopole* and *dipole* terms respectively.
**Definition**. We define $\vb{p}$ as the *electric dipole moment*, and in the magnetic version, $\vb{m}$ as the *magnetic dipole moment* as follows:
**Definition**. We define $\mathbf{p}$ as the *electric dipole moment*, and in the magnetic version, $\mathbf{m}$ as the *magnetic dipole moment* as follows:
$$
\vb{p} = \int_{V'} \vb{r'} \rho_e(\vb{r'}) \dd{V'}
\mathbf{p} = \int_{V'} \mathbf{r'} \rho_e(\mathbf{r'}) d{V'}
$$
Notably, the moments only depend on the charge density, not the point at which the field is being examined. That is, this integral only needs to be computed once.
To compute higher-order terms, let $\epsilon = 2\frac{r'}{r}\vu{r}\vdot\vu{r'}-(\frac{r'}{r})^2$. Now we can expand $(1-\epsilon)^{-\frac{1}{2}}$.
To compute higher-order terms, let $\varepsilon = 2\frac{r'}{r}\hat{\mathbf{r}}\cdot\hat{\mathbf{r'}}-(\frac{r'}{r})^2$. Now we can expand $(1-\varepsilon)^{-\frac{1}{2}}$.
$$
(1-\epsilon)^{-\frac{1}{2}} = 1 + \frac{1}{2}\epsilon + \frac{3}{8}\epsilon^2 + \frac{5}{16}\epsilon^3 + \ldots
(1-\varepsilon)^{-\frac{1}{2}} = 1 + \frac{1}{2}\varepsilon + \frac{3}{8}\varepsilon^2 + \frac{5}{16}\varepsilon^3 + \ldots
$$
However, we want an expansion in terms of $t = \frac{r'}{r}$. To do this, we write the expansion as
$$
\frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} = \frac{1}{4 \pi r} \sum_{n=0}^{\infty} (\frac{r'}{r}) P_n(\vu{r} \vdot \vu{r'})
\frac{1}{4 \pi |\mathbf{r}-\mathbf{r'}|} = \frac{1}{4 \pi r} \sum_{n=0}^{\infty} (\frac{r'}{r}) P_n(\hat{\mathbf{r}} \cdot \hat{\mathbf{r'}})
$$
Here, $P_n(\vu{r} \vdot \vu{r'})$ is a polynomial. Because $\abs{\vb{r} - \vb{r'}}$ is symmetric, we can say that if $r' > r$ instead, simply switch the two. Thus the equation becomes
Here, $P_n(\hat{\mathbf{r}} \cdot \hat{\mathbf{r'}})$ is a polynomial. Because $|\mathbf{r}-\mathbf{r'}|$ is symmetric, we can say that if $r' > r$ instead, simply switch the two. Thus the equation becomes
$$
\frac{1}{4 \pi \abs{\vb{r} - \vb{r'}}} = \frac{1}{4 \pi} \sum_{n=0}^{\infty} (\frac{r'^n_{<}}{r^{n+1}_{>}}) P_n(\vu{r} \vdot \vu{r'})
\frac{1}{4 \pi |\mathbf{r}-\mathbf{r'}|} = \frac{1}{4 \pi} \sum_{n=0}^{\infty} (\frac{r'^n_{<}}{r^{n+1}_{>}}) P_n(\hat{\mathbf{r}} \cdot \hat{\mathbf{r'}})
$$
Where $r_>$ is the greater of $r, r'$, and $r_<$ the lesser.

30
docs/physics/electrostatics/4-conductors.md
View File

@@ -12,11 +12,11 @@ We note that there is no electric fiend inside a conductor, as charges internal
Additionally, as the field is zero, it follows from Maxwell's equations that there is no charge inside a conductor. However, charge may be present at the surface. For sufficiently symmetric charges, this charge may be calculated.
Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \vb{E} \vdot \dd{\vb{l}}$. Since $\vb{E} = 0$ inside the conductor, the voltage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential.
Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \mathbf{E} \cdot d{\mathbf{l}}$. Since $\mathbf{E} = 0$ inside the conductor, the voltage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential.
The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The parallel voltage difference is zero, so the electric field must be zero.
The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $d{\mathbf{l}}$. Now, $\mathbf{E} \cdot d{\mathbf{l}} = \mathbf{E}_s \cdot d{\mathbf{l}}_s + \mathbf{E}_p \cdot d{\mathbf{l}}_p = d{V_s} + d{V_p}$, in terms of parallel and perpendicular components. The parallel voltage difference is zero, so the electric field must be zero.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \nabla \cdot \mathbf{E} d{V} = \frac{Q_e}{\varepsilon_0} = \frac{A\sigma_e}{\varepsilon_0}$. Thus, $\sigma_e = \varepsilon_0 E$.
## Section 4.3 - Exercises involving conductors at fixed potentials
@@ -24,23 +24,23 @@ Consider a square with left and right potentials $V(0, y) = V(l, y) = V_1$ and $
In spherical polar coordinates, we see that with azimuthal symmetry, $V(r, \theta) = \sum_{l=0}^\infty a_l r^l P_l(cos\theta)$ where $P_l(x)$ are Legendre polynomials.
**Theorem**. 4.3.3: A Laplace equation's solution must be unique inside a volume $\Omega$ if $\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S} = 0]$. With this, consider a surface $\dd{\Omega}$ that surrounds conductors. The integral vanishes if a) the potential is specified on each conductor or b) the total charge on each conductor is specified.
**Theorem**. 4.3.3: A Laplace equation's solution must be unique inside a volume $\Omega$ if $\int_{d{\Omega}}[\Phi(\mathbf{r})\nabla{\Phi{\mathbf{r}}} \cdot \hat{\mathbf{n}} d{S} = 0]$. With this, consider a surface $d{\Omega}$ that surrounds conductors. The integral vanishes if a) the potential is specified on each conductor or b) the total charge on each conductor is specified.
Now, define $\Phi(\vb{r})$ as the difference between any two potential solutions to the Laplace equation at point $\vb{r}$. Since potential must be a constant,
Now, define $\Phi(\mathbf{r})$ as the difference between any two potential solutions to the Laplace equation at point $\mathbf{r}$. Since potential must be a constant,
$$
\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}}] \vdot \vu{n} \dd{S}
= \sum_{i=1}^N \Phi_i \int_{\dd{\Omega_i}} \grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S}
\int_{d{\Omega}}[\Phi(\mathbf{r})\nabla{\Phi{\mathbf{r}}}] \cdot \hat{\mathbf{n}} d{S}
= \sum_{i=1}^N \Phi_i \int_{d{\Omega_i}} \nabla{\Phi{\mathbf{r}}} \cdot \hat{\mathbf{n}} d{S}
$$
Thus, if potential is specified, $\Phi_i$ vanishes for that conductor. If the total charge is instead specified, the gradient vanishes because there is no difference in charge between any two points.
## Section 4.4 - Electric Field, Polarization Field, and Flux Density in the Presence of Conductors
**Definition**. A *bound charge* is any charge in a conductor that is bound to an atom and not free to be redistributed at the surface. We say that bound charges are the source of the polarization field $\vb{P}$. Additionally, we note the charge density of bound charges is $\rho_{eb}$. Thus,
**Definition**. A *bound charge* is any charge in a conductor that is bound to an atom and not free to be redistributed at the surface. We say that bound charges are the source of the polarization field $\mathbf{P}$. Additionally, we note the charge density of bound charges is $\rho_{eb}$. Thus,
$$
\div \vb{P} = - \rho_{eb}
\nabla \cdot \mathbf{P} = - \rho_{eb}
$$
This field is zero outside of a material, and if non-zero inside a material, will drop to zero at the surface discontinuously. If there is a component perpendicular to the surface, the discontinuity will generate curl. If there is a component parallel to the surface, it will generate divergence.
@@ -48,19 +48,19 @@ This field is zero outside of a material, and if non-zero inside a material, wil
**Definition**. Charges not bound are called *free*, with density denoted as $\rho_{ef}$. Combined with $\rho_{eb}$, they form the basis of the electric field. THat is,
$$
\varepsilon_0 \div \vb{E} = \rho_{ef} + \rho_{eb}
\varepsilon_0 \nabla \cdot \mathbf{E} = \rho_{ef} + \rho_{eb}
$$
**Definition**. The electric flux density field $\vb{D}$ is defined as
**Definition**. The electric flux density field $\mathbf{D}$ is defined as
$$
\vb{D} = \varepsilon_0 \vb{E} + \vb{P}
\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}
$$
Both $\vb{D}$ and $\vb{P}$ have units of Coulombs/m^2. Additionally, we see that
Both $\mathbf{D}$ and $\mathbf{P}$ have units of Coulombs/m^2. Additionally, we see that
$$
\div \vb{D} = \div (\varepsilon_0 \vb{E} + \vb{P}) = \div \varepsilon_0 \vb{E} + \div \vb{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
\nabla \cdot \mathbf{D} = \nabla \cdot (\varepsilon_0 \mathbf{E} + \mathbf{P}) = \nabla \cdot \varepsilon_0 \mathbf{E} + \nabla \cdot \mathbf{P} = (\rho_{ef} + \rho_{eb}) - \rho_{eb} = \rho_{ef}
$$
## Section 4.5 - Induced Electric Charges, their Potentials and Fields
@@ -79,4 +79,4 @@ This unit, $\frac{C}{V}$, is known as a Farad. For a sphere, $C = 4 \pi \varepsi
## Section 4.7 - Forces on Charged Conductors in Electric Fields
We know that $\vb{F} = \int \vb{E}_{ext}(\vb{r}) \rho_e(\vb{r}) dV$, where $\vb{E}_{ext}(\vb{r})$ is the external electric field and $\rho_e(\vb{r})$ is the charge density of the object.
We know that $\mathbf{F} = \int \mathbf{E}_{ext}(\mathbf{r}) \rho_e(\mathbf{r}) dV$, where $\mathbf{E}_{ext}(\mathbf{r})$ is the external electric field and $\rho_e(\mathbf{r})$ is the charge density of the object.

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docs/physics/electrostatics/5-moving-charges.md
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@@ -4,69 +4,67 @@
We want to work in a steady-state system. Thus, we restrict ourselves to currents that do not change in time.
With math, we see that $\div \vb{J}(\vb{r}) = -\frac{\partial \rho(\vb{r})}{\partial t}$. Since we are only considering a steady-state system, $\div \vb{J}_e = \div \vb{J}_m = 0$.
With math, we see that $\nabla \cdot \mathbf{J}(\mathbf{r}) = -\frac{\partial \rho(\mathbf{r})}{\partial t}$. Since we are only considering a steady-state system, $\nabla \cdot \mathbf{J}_e = \nabla \cdot \mathbf{J}_m = 0$.
**Definition**. The *conductance* of a material is $G = \frac{1}{R}$, where $R$ is the resistance of a material.
For a wire of uniform cross-sectional area, we see that $G = \sigma \frac{A}{L}$, where $A$ is the cross-sectional area, $L$ is the length of the wire, and $\sigma$ is the conductivity of a wire. Inverted, we see that $R$ = $\rho \frac{L}{A}$, where $\rho = \frac{1}{\sigma}$ is the resistivity of the wire.
**Definition**. *Ohm's Law* can be written as $I = G V$, or inverted, $V = IR$. In a wire, we see that current density $\vb{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \vb{E}$
**Definition**. *Ohm's Law* can be written as $I = G V$, or inverted, $V = IR$. In a wire, we see that current density $\mathbf{} = \frac{I}{A} = \sigma \frac{V}{L} = \sigma \mathbf{E}$
## Section 5.2 - Currents and Curling Fields
We know that $\vb{J}_e = \curl{\vb{H}}$ and $\vb{J}_m = -\curl{\vb{E}}$. That is, current densities cause the opposing field to curl.
We know that $\mathbf{J}_e = \nabla \times{\mathbf{H}}$ and $\mathbf{J}_m = -\nabla \times{\mathbf{E}}$. That is, current densities cause the opposing field to curl.
For a wire with current $I_e$, we see that applying Stoke's theorem to the first equation,
$ \int_S \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_{\partial S} = \vb{H} \vdot \dd{\vb{l}}$. Apply the identity $\curl{\vb{H}} = \vb{J}_e$ to the left side to see that $\int_S \curl{\vb{H}} \vdot \vu{n} \dd{S} = \int_S \vb{J}_e \vdot \vu{n} \dd{S} = (I_e)_S$, or the current passing through the cross-sectional area. By the original equation, we see that $(I_e)_S = \vb{H} \vdot \dd{\vb{l}}$.
$ \int_S \nabla \times{\mathbf{H}} \cdot \hat{\mathbf{n}} d{S} = \int_{\partial S} = \mathbf{H} \cdot d{\mathbf{l}}$. Apply the identity $\nabla \times{\mathbf{H}} = \mathbf{J}_e$ to the left side to see that $\int_S \nabla \times{\mathbf{H}} \cdot \hat{\mathbf{n}} d{S} = \int_S \mathbf{J}_e \cdot \hat{\mathbf{n}} d{S} = (I_e)_S$, or the current passing through the cross-sectional area. By the original equation, we see that $(I_e)_S = \mathbf{H} \cdot d{\mathbf{l}}$.
If we assume cylindrical coordinates and that $\vb{H}(vb{r}) = H_\varphi(s) \vu{\varphi}$, then $\vb{H} \vdot \dd{\vb{l}} = \int_0^{2\pi} H_\varphi(S) s \dd{\varphi}$, so then $(I_e)_S = \int_0^{2\pi} H_\varphi(S) s \dd{\varphi}$. Thus, for $s > a$ (where $a$ is the radius of the wire), $2\pi s H_\varphi = I_e$, and for $s < a$, $2\pi s H_\varphi = I_e \frac{s^2}{a^2}$.
If we assume cylindrical coordinates and that $\mathbf{H}(vb{r}) = H_\varphi(s) \hat{\mathbf{\varphi}}$, then $\mathbf{H} \cdot d{\mathbf{l}} = \int_0^{2\pi} H_\varphi(S) s d{\varphi}$, so then $(I_e)_S = \int_0^{2\pi} H_\varphi(S) s d{\varphi}$. Thus, for $s > a$ (where $a$ is the radius of the wire), $2\pi s H_\varphi = I_e$, and for $s < a$, $2\pi s H_\varphi = I_e \frac{s^2}{a^2}$.
---
By Helmholtz Theorem, we know that $\vb{H}(\vb{r}) = \curl{\vb{A}(\vb{r})}$. For a current-carrying wire, $\vb{A}(\vb{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{\dd{\vb{l'}}}{|\vb{r}-\vb{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
By Helmholtz Theorem, we know that $\mathbf{H}(\mathbf{r}) = \nabla \times{\mathbf{A}(\mathbf{r})}$. For a current-carrying wire, $\mathbf{A}(\mathbf{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{d{\mathbf{l'}}}{|\mathbf{r}-\mathbf{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
$$
\vb{H}(\vb{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\vb{r}-\vb{r'}) \cross \dd{\vb{l'}}}{|\vb{r}-\vb{r'}|^3}
\mathbf{H}(\mathbf{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\mathbf{r}-\mathbf{r'}) \times d{\mathbf{l'}}}{|\mathbf{r}-\mathbf{r'}|^3}
$$
---
Consider a current loop instead, on the $x-y$ plane and current $I$. Then, $r = z \vu{z}$ and $\dd{\vb{l'}} = R \vu{\varphi'} \dd{\phi'}$, and the magnetic field collapses to $\vb{H}(s = 0, z) = \frac{I_e R^2}{2(R^2 + z^2)^{\frac{3}{2}}} \vu{z}$
Consider a current loop instead, on the $x-y$ plane and current $I$. Then, $r = z \hat{\mathbf{z}}$ and $d{\mathbf{l'}} = R \hat{\mathbf{\varphi'}} d\phi'$, and the magnetic field collapses to $\mathbf{H}(s = 0, z) = \frac{I_e R^2}{2(R^2 + z^2)^{\frac{3}{2}}} \hat{\mathbf{z}}$
---
Consider some infinite bar magnet with height $h$ and width $w$. Then, the top and bottom surfaces will have a magnetic charge with density $\vb{J}_m^+ = M_0 \vb{b} \delta(z - h)$ and $\vb{J}_m^- = -M_0 \vb{v} \delta(z)$ respectively. By definition, $I_m = M_0 w v$.
Consider some infinite bar magnet with height $h$ and width $w$. Then, the top and bottom surfaces will have a magnetic charge with density $\mathbf{J}_m^+ = M_0 \mathbf{b} \delta(z - h)$ and $\mathbf{J}_m^- = -M_0 \mathbf{v} \delta(z)$ respectively. By definition, $I_m = M_0 w v$.
Now, consider a loop around only the top of the conductor. Then,
$$
\int_S \vb{J}_m \vdot \vu{n} \dd{S} = I_m = M_o w v
\int_S \mathbf{J}_m \cdot \hat{\mathbf{n}} d{S} = I_m = M_o w v
$$
By definition,
$$
\int_S \vb{J}_m \vdot \vu{n} \dd{S} = -\int_S (\curl{\vb{E}}) \vdot \vu{n} \dd{S}
\int_S \mathbf{J}_m \cdot \hat{\mathbf{n}} d{S} = -\int_S (\nabla \times{\mathbf{E}}) \cdot \hat{\mathbf{n}} d{S}
$$
Applying Stokes theorem,
$$
\int_S (\curl{\vb{E}}) \vdot \vu{n} \dd{S} = M_0 w v
\int_S (\nabla \times{\mathbf{E}}) \cdot \hat{\mathbf{n}} d{S} = M_0 w v
$$
## Section 5.3 - Forces on Moving Charges and Current
Consider an electric charge moving with velocity $\vb{v}$ in a magnetic parallel plate capacitor with charge densities $\plusminus \sigma_m$. That is, $\mu_0 \vb{H} = \sigma_m \vu{z}$. Then, we can apply theorems to see the resulting force.
Consider an electric charge moving with velocity $\mathbf{v}$ in a magnetic parallel plate capacitor with charge densities $\pm \sigma_m$. That is, $\mu_0 \mathbf{H} = \sigma_m \hat{\mathbf{z}}$. Then, we can apply theorems to see the resulting force.
**Theorem**. *Lorentz Force Law* states that $\vb{F} = q_e \vb{v} \cross \u_0 \vb{H}$ in the presence of a magnetic field. In the presence of both an electric and magnetic field, $\vb{F} = q_e (\vb{E} + \vb{v} \cross \u_0 \vb{H})$.
**Theorem**. *Lorentz Force Law* states that $\mathbf{F} = q_e \mathbf{v} \times \mu_0 \mathbf{H}$ in the presence of a magnetic field. In the presence of both an electric and magnetic field, $\mathbf{F} = q_e (\mathbf{E} + \mathbf{v} \times \mu_0 \mathbf{H})$.
**Theorem**. *Ampere's Force Law* states that generalizing the previous theorem, we can see that
$$
\dd{\vb{F}} = I_e \dd{\vb{L}} \cross \u_0 \vb{H}(\vb{r})
$$
$$d\mathbf{F} = I_e d{\mathbf{L}} \times \mu_0 \mathbf{H}(\mathbf{r})$$
## Section 5.4 - Multipole Expansion of a Vector Potential