From 72af3910f5e70622cd6f9c889130d17f31401872 Mon Sep 17 00:00:00 2001 From: Indigo5684 <159226326+Indigo5684@users.noreply.github.com> Date: Tue, 30 Sep 2025 13:19:25 -0500 Subject: [PATCH] Catchup 1 --- .gitignore | 1 + docs/index.md | 17 ++ docs/javascripts/mathjax.js | 14 ++ docs/math/diffeq/1-intro.md | 39 ++++ docs/math/diffeq/2-1st-order.md | 147 +++++++++++++++ docs/math/diffeq/3-2nd-order.md | 167 ++++++++++++++++++ docs/math/diffeq/4-laplace.md | 81 +++++++++ docs/physics/electrostatics/1-math.md | 72 ++++++++ docs/physics/electrostatics/2-coulomb.md | 135 ++++++++++++++ .../3-electro-magnetic-potentials.md | 68 +++++++ mkdocs.yml | 15 ++ 11 files changed, 756 insertions(+) create mode 100644 .gitignore create mode 100644 docs/index.md create mode 100644 docs/javascripts/mathjax.js create mode 100644 docs/math/diffeq/1-intro.md create mode 100644 docs/math/diffeq/2-1st-order.md create mode 100644 docs/math/diffeq/3-2nd-order.md create mode 100644 docs/math/diffeq/4-laplace.md create mode 100644 docs/physics/electrostatics/1-math.md create mode 100644 docs/physics/electrostatics/2-coulomb.md create mode 100644 docs/physics/electrostatics/3-electro-magnetic-potentials.md create mode 100644 mkdocs.yml diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..21d0b89 --- /dev/null +++ b/.gitignore @@ -0,0 +1 @@ +.venv/ diff --git a/docs/index.md b/docs/index.md new file mode 100644 index 0000000..000ea34 --- /dev/null +++ b/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. diff --git a/docs/javascripts/mathjax.js b/docs/javascripts/mathjax.js new file mode 100644 index 0000000..05843aa --- /dev/null +++ b/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" + } +}; \ No newline at end of file diff --git a/docs/math/diffeq/1-intro.md b/docs/math/diffeq/1-intro.md new file mode 100644 index 0000000..756a848 --- /dev/null +++ b/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 $t$). \ No newline at end of file diff --git a/docs/math/diffeq/2-1st-order.md b/docs/math/diffeq/2-1st-order.md new file mode 100644 index 0000000..c64971b --- /dev/null +++ b/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 $y$. \ No newline at end of file diff --git a/docs/math/diffeq/3-2nd-order.md b/docs/math/diffeq/3-2nd-order.md new file mode 100644 index 0000000..4d895dd --- /dev/null +++ b/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 + +$$ +y_p = -y_1 \int \frac{y_2(t)g(t)}{W(y_1, y_2)} dt + y_2 \int \frac{y_1(t)g(t)}{W(y_1, y_2)} dt +$$ \ No newline at end of file diff --git a/docs/math/diffeq/4-laplace.md b/docs/math/diffeq/4-laplace.md new file mode 100644 index 0000000..575679d --- /dev/null +++ b/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). \ No newline at end of file diff --git a/docs/physics/electrostatics/1-math.md b/docs/physics/electrostatics/1-math.md new file mode 100644 index 0000000..48047f6 --- /dev/null +++ b/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. + +$$ +\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} +$$ + +$$ +\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} +$$ diff --git a/docs/physics/electrostatics/2-coulomb.md b/docs/physics/electrostatics/2-coulomb.md new file mode 100644 index 0000000..dee2dd4 --- /dev/null +++ b/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 $r$. \ No newline at end of file diff --git a/docs/physics/electrostatics/3-electro-magnetic-potentials.md b/docs/physics/electrostatics/3-electro-magnetic-potentials.md new file mode 100644 index 0000000..915358a --- /dev/null +++ b/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, + +$$ +V_e(\vb{r}) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'} +$$ + + +$$ +V_m(\vb{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'} +$$ \ No newline at end of file diff --git a/mkdocs.yml b/mkdocs.yml new file mode 100644 index 0000000..330e78d --- /dev/null +++ b/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 \ No newline at end of file