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

@@ -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.