Catchup 1

.gitignore

@@ -0,0 +1 @@
+.venv/

docs/index.md

@@ -0,0 +1,17 @@
+# Welcome to MkDocs
+
+For full documentation visit [mkdocs.org](https://www.mkdocs.org).
+
+## Commands
+
+* `mkdocs new [dir-name]` - Create a new project.
+* `mkdocs serve` - Start the live-reloading docs server.
+* `mkdocs build` - Build the documentation site.
+* `mkdocs -h` - Print help message and exit.
+
+## Project layout
+
+    mkdocs.yml    # The configuration file.
+    docs/
+        index.md  # The documentation homepage.
+        ...       # Other markdown pages, images and other files.

docs/javascripts/mathjax.js

@@ -0,0 +1,14 @@
+window.MathJax = {
+    loader: {load: ['[tex]/physics']},
+    tex: {
+        inlineMath: [["\\(", "\\)"], ["\\$", "\\$"]],
+        displayMath: [["\\[", "\\]"]],
+        processEscapes: true,
+        processEnvironments: true,
+        packages: {'[+]': ['physics']}
+    },
+    options: {
+        ignoreHtmlClass: ".*|",
+        processHtmlClass: "arithmatex"
+    }
+};

docs/math/diffeq/1-intro.md

@@ -0,0 +1,39 @@
+# Section 1 - Basic Concepts
+
+## Section 1.1 - Definitions
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/classes/de/Definitions.aspx).
+
+**Definition**. A *differential equation* is an equation that describes a function in terms of its derivatives. Examples of differential equations include Newton's Laws, among others.
+
+**Definition**. The *order* of a differential equation is the largest derivative present in the equation with a non-zero constant.
+
+**Definition**. A differential equation that only involves derivatives with respect to one variable is called an *ordinary* differential equation (ODE).
+
+**Definition**. A differential equation that describes a function in terms of derivatives with respect to more than one linearly-independent variable is called a *partial* equation.
+
+**Definition**. A *linear* differential equation is any differential equation that cn be written in the following form:
+
+\[ a_n(t)y^{(n)}(t) + a_{n-1}(t)+y^{n-1}(t) + . . . + a_1(t)y'(t) + a_0(t)y(t) = g(t) \]
+
+Note that $a_n(t)$ does not depeond on any derivative of $y$, so the presence of terms such as $e^y$ or $\sqrt{y'}$ signal that the equation is *nonlinear*.
+
+**Definition**. The *solution(s)* to a differential equation over an inverval $\alpha < t < \beta$ are any funcion(s) $y(t)$ that satisfy the differential equation.
+
+**Definition**. The *initial conditions* are a condition or set of conditions that constrain the possible solution sets.
+
+**Definition**. An *Initial Value Problem* is a differential equation along with the appropriate boundary or initial conditions.
+
+**Definition**. The *integral of validity* for a solution to a differential equation is the largest possible interval containing the initial coniditions for which the solution is valid.
+
+**Definition**. The *general solution* to a differential equation is the most general form a solution to a differential equation can take without requiring the initial conditions.
+
+**Definition**. The *actual solution* to a differential equation is the specific solution that satisfies the differential equation and the boundary conditions.
+
+**Definition**. A solution is said to be *explicit* if it can be written in the form $y = y(t)$. Otherwise, it is said to be *implicit*.
+
+## Section 1.2 - Directional Fields
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/DirectionFields.aspx).
+
+**Definition**. A directional field is the graph of a $t$ vs. $y(t)$, with vectors drawn at each point with a slope corresponding to $y'(t)$. Notably, each arrow will be pointed right (towards increasing $t$).

docs/math/diffeq/2-1st-order.md

@@ -0,0 +1,147 @@
+# Section 2 - First Order Differential Equations
+
+## Section 2.1 - Linear Differential Equations
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/Linear.aspx).
+
+Let the following first-order linear differential equation be given, with $p(t)$ and $g(t)$ continuos.
+
+$$
+\frac{dy}{dt} + p(t)y = g(t)
+$$
+
+### Deriving the Solution
+
+Next, we let $\mu(t)$ be our *integrating factor*. Multiply both sides of the equation through by $\mu(t)$.
+
+$$
+\mu(t)\frac{dy}{dt} + \mu(t)p(t)y = \mu(t)g(t)
+$$
+
+Now, define $\mu(t)$ so that $\mu(t)p(t) = \mu'(t)$. Thus, we can state the following:
+
+$$
+\mu(t)\frac{dy}{dt} + \mu'(t)y = \mu(t)g(t)
+$$
+
+The left of the preceeding equation is simply the product rule, so we can write $(\mu(t)y(t))' = \mu(t)g(t)$. Take the integral of both sides.
+
+\begin{align}
+    \int (\mu(t)y(t))' dt &= \int \mu(t)g(t) \\
+    \mu(t)y(t) + C &= \int \mu(t)g(t) dt \\
+    y(t) &= \frac{\int \mu(t)g(t) dt - C}{\mu(t)}
+\end{align}
+
+Let $C$ absorb the negative sig, and we see the following.
+
+$$
+y(t) = \frac{\int \mu(t)g(t) dt + C}{\mu(t)}
+$$
+
+This is the general solution to the differential equation. However, it is incomplete, as we do not know $\mu(t)$
+
+To derive the function, recall that we defined $\mu(t)p(t) = \mu'(t)$. Thus, we can rewrite this equation.
+
+\begin{align}
+    \frac{\mu'(t)}{\mu(t)} &= p(t) \\
+    (\ln \mu(t))' &= p(t) \\
+\end{align}
+
+Integrate both sides.
+
+\begin{align}
+    \ln \mu(t) + k = \int p(t) dt \\
+    \mu(t) &= e^{\int p(t) dt + k} \\
+    &= e^k e^{\int p(t) dt}
+\end{align}
+
+As $k$ is an unknown constant, rewrite this as $\mu(t) = k \exp(\int p(t) dt)$.
+
+### Summary
+
+The following differential equation is given.
+
+$$
+\frac{dy}{dt} + p(t)y = g(t)
+$$
+
+To find a solution to this differential equation, construct the **integrating factor** $\mu(t)$.
+
+$$\mu(t) = k \exp(\int p(t) dt)$$
+
+Thus, the solution to the differential equation can be written as the following.
+
+$$
+y(t) = \frac{\int \mu(t)g(t) dt + C}{\mu(t)}
+$$
+
+## Section 2.2 - Separable Differential Equations
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/Separable.aspx).
+
+Let the following differential equation of the following forms be given.
+
+\begin{align}
+    N(y) \frac{dy}{dx} &= M(x)
+    \frac{dy}{dx} &= \frac{M(x)}{N(y)} \\
+    \frac{dy}{dx} &= \frac{N(y)}{M(x)} \\
+    \frac{dy}{dx} &= N(y)M(x) \\
+\end{align}.
+
+For the sake of simplicty, select the following form:
+
+$$
+N(y) \frac{dy}{dx} = M(x)
+$$
+
+Thus, integrate both sides with respect to $x$.
+
+$$
+\int N(y) \frac{dy}{dx} dx = \int M(x) dx
+$$
+
+Since $y$ is really $y(x)$, we can make the following substitution:
+
+$$
+u = y(x) \text{ and } du = y'(x)dx = \frac{dy}{dx}{dx}
+$$
+
+This reduces the integral to the following:
+
+$$
+\int N(u) du = \int M(x) dx
+$$
+
+This is solvable from here.
+
+## Section 2.4 - Bernoulli Equations
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx).
+
+Let a differential equation of the following form be given, with $n \in \mathbb{N}; n \geq 2$
+
+$$
+y' + p(x)y = q(x)y^n
+$$
+
+This is a *Bernoulli equation*.
+
+Divide by $y^n$.
+
+$$
+y^{-n}y' + p(x)y^{1-n} = q(x)
+$$
+
+Now, make the substitution $v = y^{1-n}$. Thus, the derivative is as follows.
+
+$$
+v' = (1-n)y^{-n}y'
+$$
+
+Substituting into the first equation yields the following.
+
+$$
+\frac{1}{1-n}v' + p(x)v = q(x)
+$$
+
+After solving, be sure to rewrite in terms of $y$.

docs/math/diffeq/3-2nd-order.md

@@ -0,0 +1,167 @@
+# Section 3 - Second Order Differential Equations
+
+## Section 3.1 - Basic Concepts
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx).
+
+All second-order differential equations can be written in the following form:
+
+$$ p(t) y'' + q(t) y' + r(t) y = g(t) $$
+
+In the case where $p(t)$, $q(t)$, and $r(t)$ are constants, we write the equation as the following:
+
+$$ ay'' + by' + cy = g(t) $$
+
+This is a second-order differential equation with constant coefficients.
+
+**Definition**. In the event that $g(t) = 0$, we say the equation is *homogenous*. Otherwise, the equation is *nonhomogenous*.
+
+**Definition**. Principal of Superposition. Let $y_1(t)$ and $y_2(t)$ be solutions to a linear, homogenous differential equation. Then, any linear combination of said solutions is also a solution to the differential equation. In other words, with $c_1, c_2 \in \mathbb{R}$, the following is a solution to a differential equation.
+
+$$ y(t) = c_1 y_1(t) + c_2 y_2(t) $$
+
+Given a second-order homogenous differential equation with constant coeffictions, we assume solutions of the following form:
+
+$$ y(t) = e^{rt} $$
+
+Substituting this equation into the differential equationm, we see the following:
+
+$$ e^{rt}(ar^2 + br + c) = 0 $$
+
+Thus, we allow the *charactaristic equation* of the differential equation to be as follows:
+
+$$ ar^2 + br + c = 0 $$
+
+# Section 3.2 - Real & Distinct Roots
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx).
+
+When the two roots to the charactaristic equation are discrete roots in the real numbers, we see the following solutions.
+
+$$ y_1(t) = e^{r_1 t} $$
+
+$$ y_2(t) = e^{r_2 t} $$
+
+Thus, 
+
+$$ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} $$
+
+# Section 3.3 - Complex Roots
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx).
+
+Let the solutions to the charactaristic equation be of the following form:
+
+$$ r_{1,2} = \lambda \pm \mu i $$
+
+Thus, our two solutions are
+
+$$ y_1(t) = e^{(\lambda + \mu i)t} $$
+
+$$ y_2(t) = e^{(\lambda - \mu i)t} $$
+
+Recall Euler's Formula:
+
+$$ e^{i \theta} = \cos \theta + i \sin \theta $$
+
+A colliloquy of Euler's formula is the following:
+
+$$ e^{-i \theta} = \cos(-\theta) + i \sin(-\theta) = \cos \theta - i \sin \theta $$
+
+Thus, we can write our solutions as the following:
+
+\begin{align}
+    y_1(t) &= e^{(\lambda + \mu i)t} &= e^{\lambda t} e^{i \mu t} &= e^{\lambda t}(\cos(\mu t) + i \sin(\mu t)) \\
+    y_2(t) &= e^{(\lambda - \mu i)t} &= e^{\lambda t} e^{-i \mu t} &= e^{\lambda t}(\cos(\mu t) - i \sin(\mu t))
+\end{align}
+
+A linear combination of the two solutions can be written as the following:
+
+$$ y(t) = c_1 e^{\lambda t} \cos(\mu t) + c_2 e^{\lambda t} \sin(\mu t) $$
+
+## Section 3.4 - Repeated Roots
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx).
+
+Assume the solutions to the charactaristic equations are $r = r_1 = r_2$. Thus, the two equations $y_t(t)$ and $y_2(t)$ are not linearly independent.
+
+After a *lot* of algebra, we see that
+
+$$y_1(t) = e^{rt}$$
+
+$$y_2(t) = t e^{rt}$$
+
+## Section 3.5 - Reduction of Order
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx).
+
+Skipped.
+
+## Section 3.6 - Fundamental Set of Solutions, Wronskian
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/FundamentalSetsofSolutions.aspx).
+
+**Definition**. Given two functions $f(t)$, $g(t)$, the *Wronskian* is defined as
+
+$$
+W(f,g) = \det \begin{vmatrix}
+  f(t) & g(t) \\
+  f'(t) & g'(t)
+\end{vmatrix}
+$$
+
+**Definition**. If $W(f, g) \neq 0$, then $f(t)$ and $g(t)$ are said to form a *fundamental set of solutions*, and can be superimposed to form the general solution.
+
+## Section 3.8 - Nonhomogenous Differential Equations
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx).
+
+Assume we have the differential equation as follows:
+
+$$ y'' + p(t) y' + q(t) y = g(t) $$
+
+The equivilent homogenous differential equation is 
+
+$$ y'' + p(t) y' + q(t) y = 0 $$
+
+**Theorem**. Assume $Y_1(t)$, $Y_2(t)$ are solutions to the nonhomogenous differential equations. Then, $Y_1(t) - Y_2(t)$ is a solution to the homogenous differential equation. This can be proved by substitution.
+
+Thus, with $y_h(t)$ the solution to the homogenous problem, and $y_p(t)$ the solution to this particular problem, we can say that the general form of the solution to this differential equation is
+
+$$ y(t) = y_h(t) + y_p(t) $$
+
+## Section 3.9 - Undetermined Coefficients
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx).
+
+We know the following guesses for functions.
+
+| $g(t)$ | $y_p$ guess |
+| :- | -: |
+| $\alpha e^{\beta t}$ | $A e^{\beta t}$ |
+| $a \cos(\beta t)$ | $A \cos(\beta t) + B \sin(\beta t)$ |
+| $b \sin(\beta t)$ | $A \cos(\beta t) + B \sin(\beta t)$ |
+| $a \cos(\beta t) + \sin(\beta t)$ | $A \cos(\beta t) + B \sin(\beta t)$ |
+| n-th degree polynomial | $A_nt^n + A_{n-1}t^{n-1} + A_1 t + A_0$ |
+
+Combine this with the following:
+
+**Theorem**. Given $y_{p_1}(t)$ is a solution to $y'' + p(t)y' + q(t)y = g_1(t)$ and $y_{p_2}(t)$ is a solution to $y'' + p(t)y' + q(t)y = g_2(t)$, then the function $y_{p_1}(t) + y_{p_2}(t)$ is a solution to $y'' + p(t)y' + q(t)y = g_1(t) + g_2(t)$
+
+## Section 3.10 - Variation of Parameters
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx).
+
+Assume we have the differential equation as follows:
+
+$$ y'' + p(t) y' + q(t) y = g(t) $$
+
+The equivilent homogenous differential equation is 
+
+$$ y'' + p(t) y' + q(t) y = 0 $$
+
+For this method, we must have $y_1(t)$ and $y_2(t)$ known. Through a lot of math, we see that
+
+$$
+y_p = -y_1 \int \frac{y_2(t)g(t)}{W(y_1, y_2)} dt + y_2 \int \frac{y_1(t)g(t)}{W(y_1, y_2)} dt
+$$

docs/math/diffeq/4-laplace.md

@@ -0,0 +1,81 @@
+# Section 4 - Laplace Transformations
+## Section 4.1 - Definition
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx).
+
+**Definition**. The *Laplace transform* of a function is given by the following:
+
+$$
+\mathcal{L} \{f(t)\}(s) = F(s) = \int_0^{\infty} e^{-st}f(t) dt
+$$
+
+## Section 4.2 - Properties
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/LaplaceTransforms.aspx).
+
+The Laplace Transformation is a linear transformation over functions in $\mathbb{R}[t]$. That is, given $a, b \in \mathbb{R}, f(t), g(t) \in \mathbb{R}[t]$, we know that
+
+$$
+\mathcal{L} \{a f(t)\ + b g(t) \}(s) = a F(s) + b G(s)
+$$
+
+## Section 4.3 - Inverse Laplace Transformation
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/InverseTransforms.aspx).
+
+Given $F(s)$, we define the Inverse Laplace Transformation as the following;
+
+$$
+f(t) = \mathcal{L}^{-1} \{F(s)\}
+$$
+
+## Section 4.4 - Step Function
+
+The step/Heaviside function $u_c(t)$ is defined as 0 if $t < c$, and 1 if $t > c$.
+
+Alternatively, $u(t - c) = H(t - c)$ is 0 if $t < c$, and 1 if $t > c$.
+
+Applying this to the Laplace transform,
+
+$$
+\begin{align}
+    \mathcal{L} \{ u_c(t) f(t-c) \} &= \int_0^{\infty} e^{-st}u_c(t)f(t) dt \\
+    &= \int_c^{\infty} e^{-st}f(t) dt
+\end{align}
+$$
+
+If we let $u = t - c$,
+
+$$
+\begin{align}
+    \mathcal{L} \{ u_c(t) f(t-c) \} &= \int_0^{\infty} e^{-s(u+c)}f(u) du \\
+    &= \int_0^{\infty} e^{-su}e^{-cs}f(u) du \\
+    &= e^{-cs} \int_0^{\infty} e^{-su}f(u) du \\
+    &= e^{-cs} F(s)
+\end{align}
+$$
+
+## Section 4.5 - Laplace Transformation applied to IVPs
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx).
+
+**Theorum**. Given a function $f(t)$ with $C^n$ continuity, then
+
+$$
+\mathcal{L} \{ f^{(n)} (t) \} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - ... - s f^{(n-2)} (0) - f^{(n-1)} (0)
+$$
+
+For $n=1, 2$ we see that
+
+$$
+\begin{align}
+    \mathcal{L} \{ y' \} &= sY(s) - y(0) \\
+    \mathcal{L} \{ y'' \} &= s^2 Y(s) - s y(0) - y'(0)
+\end{align}
+$$
+
+We can take the Laplace transformation of an IVP, solve for $Y(s)$, then take the inverse to find the solution.
+
+## Section 4.6 - Nonconstant Coefficient IVPs
+
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx).

docs/physics/electrostatics/1-math.md

@@ -0,0 +1,72 @@
+# 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

@@ -0,0 +1,135 @@
+# 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

@@ -0,0 +1,68 @@
+# 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,
+
+$$
+V_e(\vb{r}) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
+$$
+
+
+$$
+V_m(\vb{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
+$$

mkdocs.yml

@@ -0,0 +1,15 @@
+site_name: Notes
+theme:
+  name: readthedocs
+
+
+markdown_extensions:
+  - pymdownx.arithmatex:
+      generic: true
+
+plugins:
+  - search
+
+extra_javascript:
+  - javascripts/mathjax.js
+  - https://unpkg.com/mathjax@3/es5/tex-mml-chtml.js