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docs/physics/electrostatics/1-math.md

@@ -5,69 +5,69 @@
 **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.
 
 $$
-\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}
+\mathbf{\Phi}(\mathbf{r}) = \frac{1}{4 \pi} \int_V \frac{D(\mathbf{r}')}{|{\mathbf{r}-\mathbf{r}'}|} d{V'}
 $$
 
 $$
-\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}
+\mathbf{A}(\mathbf{r}) = \frac{1}{4 \pi} \int_V \frac{C(\mathbf{r}')}{|{\mathbf{r}-\mathbf{r}'}|} d{V'}
 $$