docs/physics/electromagnetism/electromagnetic-dynamics.md: add

docs/physics/electromagnetism/electromagnetic-dynamics.md

@@ -0,0 +1,274 @@
+# Electromagnetic dynamics
+
+We let loose the electrostatic and magnetostatic regime and consider the interplay of electric and magnetic fields as electromagnetic dynamics.
+
+## Electromagnetic dynamics in vacuum
+
+> *Axiom 1*: The electric $\mathbf{E}: (\mathbf{r},t) \mapsto \mathbf{E}(\mathbf{r},t)$ and magnetic field $\mathbf{B}: (\mathbf{r},t) \mapsto \mathbf{B}(\mathbf{r},t)$ are described by
+>
+> $$
+>   \begin{align*}
+>       \nabla \cdot \mathbf{E} &= \frac{\rho}{\epsilon_0},\\
+>       \nabla \cdot \mathbf{B} &= 0,\\
+>       \nabla \times \mathbf{E} + \partial_t \mathbf{B} &= \mathbf{0},\\
+>       \nabla \times \mathbf{B} - \mu_0 \epsilon_0 \partial_t \mathbf{E} &= \mu_0 \mathbf{J},   
+>   \end{align*}
+> $$
+>
+> with $\rho$ the volume charge density, $\mathbf{J}$ the volume current density and $\varepsilon_0$ and $\mu_0$ the permittivity and permeability of vacuum.
+
+The equations in *Axiom 1* are called the Maxwell equations and their integral form can be obtained from the curl and divergence theorem, respectively.
+
+It follows that from the definition of the electric force and magnetic force on a point charge $q$ at a field point $\mathbf{r}$ moving with a velocity $\mathbf{v}$ that the electromagnetic force $\mathbf{F}$ is given by
+
+$$
+    \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}),
+$$
+
+for all $\mathbf{E}: (\mathbf{r},t) \mapsto \mathbf{E}(\mathbf{r},t)$ and $\mathbf{B}: (\mathbf{r},t) \mapsto \mathbf{B}(\mathbf{r},t)$ the electric and magnetic field (Lorentz law).
+
+In the linear assumption we may express the volume current density $\mathbf{J}$ in terms of the electromagnetic force per unit charge
+
+$$
+    \mathbf{J} = \sigma (\mathbf{E} + \mathbf{v} \times \mathbf{B}),
+$$
+
+with $\sigma$ the conductivity of the medium. In the particular case that $\mathbf{v} = \mathbf{0}$ we obtain Ohm's law.
+
+> *Definition 1*: Let the **emf** $\epsilon$ be defined as
+>
+> $$
+>   \varepsilon = \oint_L \mathbf{E} \cdot d\mathbf{l},
+> $$
+>
+> for all $\mathbf{E}: (\mathbf{r},t) \mapsto \mathbf{E}(\mathbf{r},t)$.
+
+*Definition 1* imposes that
+
+$$
+    \varepsilon = - d_t \int_S \mathbf{B} \cdot d\mathbf{a},
+$$
+
+which is called the Flux rule. According to Lenz and *Axiom 1* nature abhors a change in flux.
+
+### Inductance
+
+Consider two loops of wire, by proportionality we may define $\Phi_2 = M_{21} I_1$ as the flux of the magnetic field created by loop 1 through loop 2 and $\Phi_1 = M_{12} I_2$ the opposite with $M_{ij}$ the mutual inductance.
+
+It then follows (without stating the proof) that 
+
+$$
+    M_{21} = M_{12} = \frac{\mu_0}{4\pi} \oiint \frac{1}{\mathfrak{r}} d\mathbf{l}_1 \cdot d\mathbf{l}_2,
+$$
+
+is the Neumann formula.
+
+By proportionality we may as well define $\Phi = L I$ with $L$ the self-inductance.
+
+### Energy in magnetic fields
+
+Consider 
+
+$$
+    d_t W = - \varepsilon I = L I d_t I,
+$$
+
+then $W = \frac{1}{2} L I^2$ is the work required to build up the line current density from zero to $I$.
+
+It follows then that for a volume current density $\mathbf{J}$ the work $W$ needed to construct the system is given by
+
+$$
+    W = \frac{1}{2} \int_\mathscr{V} \mathbf{A} \cdot \mathbf{J} d\tau.
+$$
+
+From this we may state:
+
+> *Theorem 1*: The work $W$ required to construct the system can be expressed in terms of the magnetic field $\mathbf{B}$ as
+>
+> $$
+>   W = \frac{1}{2\mu_0} \int_{\mathbb{R}^3} \|\mathbf{B}\|^2 d\tau.
+> $$
+
+??? note "Proof:"
+    
+    Rewrite the work in terms of the magnetic field $\mathbf{B}$:
+
+    $$
+        \begin{align*}
+            W &= \frac{1}{2\mu_0} \int_\mathscr{V} \mathbf{A} \cdot (\nabla \times \mathbf{B})d\tau,\\
+              &= \frac{1}{2\mu_0} \Bigg(\int_\mathscr{V} \|\mathbf{B}\|^2 d\tau - \int_\mathscr{V} \nabla \cdot (\mathbf{A} \times \mathbf{B}) d\tau \Bigg),\\
+              &= \frac{1}{2\mu_)} \Bigg(\int_\mathscr{V} \|\mathbf{B}\|^2 d\tau - \int_{\partial \mathscr{V}} (\mathbf{A} \times \mathbf{B}) \cdot d\mathbf{a} \Bigg).
+        \end{align*}
+    $$
+
+    If we now set $\mathscr{V} = \mathbb{R}^3$ then the integral over $\partial \mathscr{V}$ goes to zero and we are left with:
+
+    $$
+        W = \frac{1}{2\mu_0} \int_{\mathbb{R}^3} \|\mathbf{B}\|^2 d\tau.
+    $$
+
+That is performing the integral over all space.
+
+### Conservation of charge
+
+Formally, the charge in a domain $\mathscr{V}$ is
+
+$$
+    Q(t) = \int_\mathscr{V} \rho(\mathbf{r},t) d\tau,
+$$
+
+and the current out of the boundary of the domain $\partial \mathscr{V}$ is
+
+$$
+    Q'(t) = - \oint_{\partial \mathscr{V}} \mathbf{J}(\mathbf{r},t) \cdot d\mathbf{a},
+$$
+
+such that
+
+$$
+    \int_\mathscr{V} \partial_t \rho(\mathbf{r},t) d\tau = - \int_\mathscr{V} \nabla \cdot \mathbf{J}(\mathbf{r},t) d\tau,
+$$
+
+and since this is true for any domain $\mathscr{V}$, it follows that
+
+$$
+    \partial_t \rho + \nabla \cdot \mathbf{J} = 0,
+$$
+
+conservation of charge.
+
+> *Theorem 2*: *Axiom 1* imposes that the volume charge density $\rho: (\mathbf{r},t) \mapsto \rho(\mathbf{r},t)$ and volume current density $\mathbf{J}: (\mathbf{r},t) \mapsto \mathbf{J}(\mathbf{r},t)$ adhere to
+>
+> $$
+>   \partial_t \rho + \nabla \cdot \mathbf{J} = 0.
+> $$
+
+??? note "Proof:"
+
+    $$
+        \begin{align*}
+            \nabla \cdot \mathbf{J} &= \frac{1}{\mu_0} \nabla \cdot \Big(\nabla \times \mathbf{B} - \mu_0 \epsilon_0 \partial_t \mathbf{E}\Big),\\ 
+            &= - \epsilon_0 \partial_t (\nabla \cdot \mathbf{E}),\\
+            &= - \partial_t \rho.
+        \end{align*}
+    $$
+
+### Conservation of energy
+
+The work $W$ done by the electromagnetic force may be expressed as
+
+$$
+    \begin{align*}
+        W &= \mathbf{F} \cdot d\mathbf{l},\\
+          &= q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} dt,\\
+          &= q \mathbf{E} \cdot \mathbf{v} dt,
+    \end{align*}
+$$
+
+such that the rate at which work is done on a domain $\mathscr{V}$ is
+
+$$
+    d_t W = \int_\mathscr{V} \mathbf{E} \cdot \mathbf{J} d\tau.
+$$
+
+> *Theorem 3*: *Axiom 1* imposes that the electric field $\mathbf{E}: (\mathbf{r},t) \mapsto \mathbf{E}(\mathbf{r},t)$, magnetic field $\mathbf{B}: (\mathbf{r},t) \mapsto \mathbf{B}(\mathbf{r},t)$ and volume current density $\mathbf{J}: (\mathbf{r},t) \mapsto \mathbf{J}(\mathbf{r},t)$ adhere to
+>
+> $$
+>   \mathbf{E} \cdot \mathbf{J} + \frac{1}{2} \partial_t \bigg(\epsilon_0 \|\mathbf{E}\|^2 + \frac{1}{\mu_0} \|\mathbf{B}\|^2\bigg) + \frac{1}{\mu_0} \nabla \cdot \Big(\mathbf{E} \times \mathbf{B}\Big) = 0.
+> $$
+
+??? note "Proof:"
+    
+    $$
+        \begin{align*}
+            \mathbf{E} \cdot \mathbf{J} &= \frac{1}{\mu_0} \mathbf{E} \cdot (\nabla \times \mathbf{B}) - \epsilon_0 \mathbf{E} \cdot \partial_t \mathbf{E},\\
+            &= \frac{1}{\mu_0} \Big(-\mathbf{B} \cdot \partial_t \mathbf{B} - \nabla \cdot (\mathbf{E} \times \mathbf{B})\Big) -\epsilon_0 \mathbf{E} \cdot \partial_t \mathbf{E},\\
+            &= -\frac{1}{2} \partial_t \bigg(\epsilon_0 \|\mathbf{E}\|^2 + \frac{1}{\mu_0} \|\mathbf{B}\|^2\bigg) - \frac{1}{\mu_0} \nabla \cdot (\mathbf{E} \times \mathbf{B}).
+        \end{align*}
+    $$
+
+We may thus write this result as
+
+$$
+    d_t W + d_t \int_\mathscr{V} u d\tau + \oint_{\partial \mathscr{V}} \mathbf{S} \cdot d\mathbf{a} = 0,
+$$
+
+with $u = \frac{1}{2}\big(\epsilon_0 \|\mathbf{E}\|^2 + \frac{1}{\mu_0} \|\mathbf{B}\|^2\big)$ the **electromagnetic energy density** and $\mathbf{S} = \frac{1}{\mu_0} \big(\mathbf{E} \times \mathbf{B}\big)$ the **electromagnetic flux density**, called Poynting's theorem.
+
+### Conservation of momentum
+
+We may write the divergence of the **energy-momentum tensor** of the electromagnetic field $\nabla \cdot \mathbf{T}$ in terms of the electric and magnetic field:
+
+$$
+    \nabla \cdot \mathbf{T} = \epsilon_0 \bigg(\big(\nabla \cdot \mathbf{E}\big) \mathbf{E} + \big(\mathbf{E} \cdot \nabla\big) \mathbf{E} \bigg) + \frac{1}{\mu_0} \bigg(\big(\nabla \cdot \mathbf{B}\big) + \big(\mathbf{B} \cdot \nabla\big) \mathbf{B}\bigg) - \frac{1}{2} \nabla \bigg(\epsilon_0 \|\mathbf{E}\|^2 + \frac{1}{\mu_0}\|\mathbf{B}\|^2\bigg),
+$$
+
+and define a **momentum density** $\mathbf{g}$ in terms of $\mathbf{S}$:
+
+$$
+    \mathbf{g} = \mu_0 \epsilon_0 \mathbf{S},
+$$
+
+then it follows from Newton's second law that:
+
+> *Theorem 4*: *Axiom 1* imposes that the divergence of the energy-momentum tensor of the electromagnetic field $\nabla \cdot \mathbf{T}$ and the momentum density $\mathbf{g}$ adhere to
+>
+> $$
+>   \nabla \cdot \mathbf{T} - \partial_t \mathbf{g} = \mathbf{0}.
+> $$
+
+??? note "Proof:"
+
+    Should be rewritten/reconsidered in Lagrangian/Hamiltonian formalism.
+
+## Electromagnetic dynamics in matter
+
+Let $\mathbf{J}_p = \partial_t \mathbf{P}$ denote the polarisation volume current density such that
+
+$$
+    \begin{align*}
+        \nabla \cdot \mathbf{J} &= \nabla \cdot (\mathbf{J}_f + \mathbf{J}_b + \mathbf{J}_p),\\
+        &= \nabla \cdot (\mathbf{J}_f + \nabla \times \mathbf{M} + \partial_t \mathbf{P}),\\
+        &= -(\partial_t \rho_f + \partial_t \rho_p),\\
+        &= - \partial_t \rho,
+    \end{align*}
+$$
+
+complying to the conservation of charge.
+
+> *Theorem 5*: The electric field $\mathbf{E}: (\mathbf{r},t) \mapsto \mathbf{E}(\mathbf{r},t)$, electric displacement $\mathbf{D}: (\mathbf{r},t) \mapsto \mathbf{D}(\mathbf{r},t)$, magnetic field $\mathbf{B}: (\mathbf{r},t) \mapsto \mathbf{B}(\mathbf{r},t)$ and auxiliary field $\mathbf{H}: (\mathbf{r},t) \mapsto \mathbf{H}(\mathbf{r},t)$ in a medium are described by
+>
+> $$
+>   \begin{align*}
+>       \nabla \cdot \mathbf{D} &= \rho_f,\\
+>       \nabla \cdot \mathbf{B} &= 0,\\
+>       \nabla \times \mathbf{E} + \partial_t \mathbf{B} &= \mathbf{0},\\
+>       \nabla \times \mathbf{H} - \partial_t \mathbf{D} &= \mathbf{J}_f,
+>   \end{align*}
+> $$
+>
+> with $\rho_f$ the free volume charge density and $\mathbf{J}_f$ the free volume current density.
+
+??? note "Proof:"
+
+    In a medium we have
+
+    $$
+        \begin{align*}
+            \nabla \cdot \mathbf{E} &= \frac{1}{\epsilon_0} (\rho_f + \rho_b),\\
+            &= \frac{1}{\epsilon_0}(\rho_f - \nabla \cdot \mathbf{P}),
+        \end{align*}
+    $$
+
+    or $\nabla \cdot \mathbf{D} = \rho_f$, and
+
+    $$
+        \begin{align*}
+            \nabla \times - \mu_0 \epsilon_0 \partial_t \mathbf{E} &= \mu_0(\mathbf{J}_f + \mathbf{J}_b + \mathbf{J}_p),\\
+            &= \mu_0(\mathbf{J}_f + \nabla \times \mathbf{M} + \partial_t \mathbf{P}),
+        \end{align*}
+    $$
+
+    or $\nabla \times \mathbf{H} = \mathbf{J}_f + \partial_t \mathbf{D}$.
+
+Which require constitutive relations for closure.

docs/physics/electromagnetism/electrostatics.md

@@ -2,7 +2,7 @@
 
 A Notational remark, let the vector from origin to source point be given by $\mathbf{r}'$ and the vector from origin to field point by $\mathbf{r}$. The vector from source to field point is given by $\bm{\mathfrak{r}} = \mathbf{r} - \mathbf{r}'$.
 
-Let us start with electrostatics in vacuum, where no current, no magnetic field and no time dependence are taken into account. We call this the **electrostatic regime**.
+In the **electrostatic regime** there is no current, no magnetic fields and time dependence is not taken into account
 
 ## Electrostatics in vacuum
 
@@ -12,16 +12,16 @@ Electrostatics builds entirely on the following axiom:
 > 
 > $$
 >   \begin{align*}
->       \nabla \cdot \mathbf{E} &= \frac{\rho}{\varepsilon_0},\\
+>       \nabla \cdot \mathbf{E} &= \frac{\rho}{\epsilon_0},\\
 >       \nabla \times \mathbf{E} &= \mathbf{0},
 >   \end{align*}
 > $$
 >
-> with $\rho: \mathbf{r} \mapsto \rho(\mathbf{r})$ the **space charge density** and $\varepsilon_0$ the **permittivity of space**. 
+> with $\rho: \mathbf{r} \mapsto \rho(\mathbf{r})$ the **volume charge density** and $\epsilon_0$ the **vacuum permittivity**. 
 
 The following definition connects the electric field $\mathbf{E}: \mathbf{r} \mapsto \mathbf{E}(\mathbf{r})$ to the Newtonian formalism of mechanics:
 
-> *Definition 1*: The force field $\mathbf{F}$ on a point charge $q$ at a field point $\mathbf{r}$ is defined as
+> *Definition 1*: The electric force $\mathbf{F}$ on a point charge $q$ at a field point $\mathbf{r}$ is defined as
 >
 > $$
 >   \mathbf{F}(\mathbf{r}) = q \mathbf{E}(\mathbf{r}),
@@ -34,7 +34,7 @@ From *Axiom 1* we may proof that the following theorem holds (Coulomb law):
 > *Theorem 1*: The electric field $\mathbf{E}: \mathbf{r} \mapsto \mathbf{E}(\mathbf{r})$ may be described as
 >
 > $$
->   \mathbf{E}(\mathbf{r}) = \begin{cases} \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} \qquad &\mathrm{0D},\\ \frac{1}{4\pi\varepsilon_0} \int \frac{\lambda}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} dl \qquad &\mathrm{1D},\\ \frac{1}{4\pi\varepsilon_0} \iint \frac{\sigma}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} da \qquad &\mathrm{2D},\\ \frac{1}{4\pi\varepsilon_0} \iiint \frac{\rho}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} d\tau \qquad &\mathrm{3D}, \end{cases}
+>   \mathbf{E}(\mathbf{r}) = \begin{cases} \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} \qquad &\mathrm{0D},\\ \frac{1}{4\pi\epsilon_0} \int \frac{\lambda}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} dl' \qquad &\mathrm{1D},\\ \frac{1}{4\pi\epsilon_0} \iint \frac{\sigma}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} da' \qquad &\mathrm{2D},\\ \frac{1}{4\pi\epsilon_0} \iiint \frac{\rho}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} d\tau' \qquad &\mathrm{3D}, \end{cases}
 > $$
 >
 > with $\lambda$ and $\sigma$ the line and surface charge density.
@@ -46,7 +46,7 @@ From *Axiom 1* we may proof that the following theorem holds (Coulomb law):
 From the divergence theorem it follows as well that 
 
 $$
-    \oint_S \mathbf{E} \cdot d\mathbf{a} = \int_V \nabla \cdot \mathbf{E} d\tau = \int_V \frac{\rho}{\varepsilon_0} d\tau = \frac{Q}{\varepsilon_0},
+    \oint_S \mathbf{E} \cdot d\mathbf{a} = \int_V \nabla \cdot \mathbf{E} d\tau = \int_V \frac{\rho}{\epsilon_0} d\tau = \frac{Q}{\epsilon_0},
 $$
 
 with $Q$ the enclosed charge. This result is called Gauß' law. 
@@ -75,14 +75,14 @@ One may simply proof that from *Definition 2* it follows that:
 
     Follows from gradient theorem.
 
-Now from (*Axiom 1*) $\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$ it follows that $\nabla^2 V = - \frac{\rho}{\varepsilon_0}$.
+Now from (*Axiom 1*) $\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$ it follows that $\nabla^2 V = - \frac{\rho}{\epsilon_0}$.
 
 Similarly to *Theorem 1* we may state the following about the electric potential $V$:
 
 > *Theorem 3*: The electric potential $V: \mathbf{r} \mapsto V(\mathbf{r})$ may be described as
 >
 > $$
->   V(\mathbf{r}) = \begin{cases} \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{\mathfrak{r}} \qquad &\mathrm{0D},\\ \frac{1}{4\pi\varepsilon_0} \int \frac{\lambda}{\mathfrak{r}} dl \qquad &\mathrm{1D},\\ \frac{1}{4\pi\varepsilon_0} \iint \frac{\sigma}{\mathfrak{r}} da \qquad &\mathrm{2D},\\ \frac{1}{4\pi\varepsilon_0} \iiint \frac{\rho}{\mathfrak{r}} d\tau \qquad &\mathrm{3D}.\end{cases}
+>   V(\mathbf{r}) = \begin{cases} \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{\mathfrak{r}} \qquad &\mathrm{0D},\\ \frac{1}{4\pi\epsilon_0} \int \frac{\lambda}{\mathfrak{r}} dl' \qquad &\mathrm{1D},\\ \frac{1}{4\pi\epsilon_0} \iint \frac{\sigma}{\mathfrak{r}} da' \qquad &\mathrm{2D},\\ \frac{1}{4\pi\epsilon_0} \iiint \frac{\rho}{\mathfrak{r}} d\tau' \qquad &\mathrm{3D}.\end{cases}
 > $$
 
 ??? note "Proof:"
@@ -120,7 +120,7 @@ As we set the reference point at infinity, then $qV$ could be interpreted as the
     $$
         \begin{align*}
             W_i &= q_i \sum_j V_j(\mathbf{r}_i),\\ 
-                &= q_i \sum_j \frac{1}{4\pi\varepsilon_0} \frac{q_j}{\mathfrak{r}_{ij}},
+                &= q_i \sum_j \frac{1}{4\pi\epsilon_0} \frac{q_j}{\mathfrak{r}_{ij}},
         \end{align*}
     $$
 
@@ -129,14 +129,14 @@ As we set the reference point at infinity, then $qV$ could be interpreted as the
     $$
         \begin{align*}
             W &= \sum_{i=1}^N W_i,\\
-              &= \frac{1}{2} \sum_{i=1}^N \sum_{j \neq i} \frac{1}{4\pi\varepsilon_0} \frac{q_j}{\mathfrak{r}_{ij}},\\
+              &= \frac{1}{2} \sum_{i=1}^N \sum_{j \neq i} \frac{1}{4\pi\epsilon_0} \frac{q_j}{\mathfrak{r}_{ij}},\\
               &=\frac{1}{2} \sum_{i=1}^N q_i V(\mathbf{r}_i).
         \end{align*}
     $$
 
 Is the energy stored in the system.
 
-It follows now that for a space charge density $\rho$ the work $W$ needed to construct the system is given by
+It follows now that for a volume charge density $\rho$ the work $W$ needed to construct the system is given by
 
 $$
     W = \frac{1}{2} \int_\mathscr{V} \rho V d\tau,
@@ -147,7 +147,7 @@ and the integrals for line and surface charge densities ($\lambda,\sigma$) are o
 > *Theorem 5*: The work $W$ required to construct the system can be expressed in terms of the electric field $\mathbf{E}$ as
 >
 > $$
->   W = \frac{\varepsilon_0}{2} \int_{\mathbb{R}^3} \|\mathbf{E}\|^2 d\tau.
+>   W = \frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \|\mathbf{E}\|^2 d\tau.
 > $$
 
 ??? note "Proof:"
@@ -156,16 +156,16 @@ and the integrals for line and surface charge densities ($\lambda,\sigma$) are o
 
     $$
         \begin{align*}
-            W &= \frac{\varepsilon_0}{2} \int_\mathscr{V} (\nabla \cdot \mathbf{E}) V d\tau,\\
+            W &= \frac{\epsilon_0}{2} \int_\mathscr{V} (\nabla \cdot \mathbf{E}) V d\tau,\\
-              &= \frac{\varepsilon_0}{2} \Big(-\int_\mathscr{V} \mathbf{E} \cdot (\nabla V) d\tau + \oint_{\partial \mathscr{V}} V \mathbf{E} \cdot d\mathbf{a}\Big),\\
+              &= \frac{\epsilon_0}{2} \Big(-\int_\mathscr{V} \mathbf{E} \cdot (\nabla V) d\tau + \oint_{\partial \mathscr{V}} V \mathbf{E} \cdot d\mathbf{a}\Big),\\
-              &= \frac{\varepsilon_0}{2} \Big(\int_\mathscr{V} \|\mathbf{E}\|^2 d\tau + \oint_\mathscr{\partial \mathscr{V}} V \mathbf{E} \cdot d\mathbf{a}\Big).
+              &= \frac{\epsilon_0}{2} \Big(\int_\mathscr{V} \|\mathbf{E}\|^2 d\tau + \oint_\mathscr{\partial \mathscr{V}} V \mathbf{E} \cdot d\mathbf{a}\Big).
         \end{align*}
     $$
 
     If we now set $\mathscr{V} = \mathbb{R}^3$ then the integral over $\partial \mathscr{V}$ goes to zero and we are left with:
 
     $$
-        W = \frac{\varepsilon_0}{2} \int_{\mathbb{R}^3} \|\mathbf{E}\|^2 d\tau.
+        W = \frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \|\mathbf{E}\|^2 d\tau.
     $$
 
 That is performing the integral over all of space. Which is mathematically rather nice, but imposes that the construction of point charges requires an infinite amount of energy. A result of the introduction of infinities. 
@@ -177,11 +177,11 @@ As may be observed, this result does not obey the superposition principle.
 In the electrostatic regime the following properties of conductors are valid:
 
 1. $\mathbf{0} = \mathbf{J} = \sigma \mathbf{E} \implies \mathbf{E} = \mathbf{0}$ inside a conductor.
-2. $\mathbf{0} = \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} \implies \rho = 0$ inside a conductor.
+2. $\mathbf{0} = \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \implies \rho = 0$ inside a conductor.
 
 Which implies that any net charge resides on the surface, that a conductor is an equipotential and that the electric field is normal to the surface of the conductor.
 
-Since the electric field $\mathbf{E}$ is proportional to the space charge density $\rho$ or charge $Q$, so also is $V$. Which enables us to define a constant of proportionality $C = \frac{Q}{V}$, the capacitance. 
+Since the electric field $\mathbf{E}$ is proportional to the volume charge density $\rho$ or charge $Q$, so also is $V$. Which enables us to define a constant of proportionality $C = \frac{Q}{V}$, the capacitance. 
 
 The work required to charge a capacitor can then be expressed as
 
@@ -194,7 +194,7 @@ $$
 The primary task of electrostatics is to find the electric field $\mathbf{E}$ of a given stationary charge distribution $\rho$. In principle, this purpose is accomplished by Coulombs's law as by the result of *Theorem 1*. But these integrals are generally difficult to solve for any but the simplest charge distributions. In some case symmetries in the charge distribution allow Gauß' law to be used. But more often than not we require to determine the potential $V$ with the result in *Theorem 3* or with Poisson's equation:
 
 $$
-    \nabla^2 V = -\frac{\rho}{\varepsilon_0},
+    \nabla^2 V = -\frac{\rho}{\epsilon_0},
 $$
 
 where in the region where $\rho = 0$ this reduces to Laplace's equation:
@@ -223,13 +223,13 @@ This result brings us to the consideration of the boundary condition for which t
 
 and consequentially:
 
-> *Corollary 2*: An electric potential $V: \mathbf{r} \mapsto V(\mathbf{r})$ that satisfies $\nabla^2 V=-\frac{\rho}{\varepsilon_0}$ in some volume $\mathscr{V}$ is unique if $\rho: \mathbf{r} \mapsto \rho(\mathbf{r})$ is known $\forall \mathbf{r} \in \mathscr{V}$ and $V$ is specified on the boundary surface $\partial \mathscr{V}$.
+> *Corollary 2*: An electric potential $V: \mathbf{r} \mapsto V(\mathbf{r})$ that satisfies $\nabla^2 V=-\frac{\rho}{\epsilon_0}$ in some volume $\mathscr{V}$ is unique if $\rho: \mathbf{r} \mapsto \rho(\mathbf{r})$ is known $\forall \mathbf{r} \in \mathscr{V}$ and $V$ is specified on the boundary surface $\partial \mathscr{V}$.
 
 The combination of *Theorem 7* and *Corollary 2* is known as the **first uniqueness theorem**.
 
 The **second uniqueness theorem** is relevant when the boundary is defined by conductors:
 
-> *Theorem 8*: An electric potential $V: \mathbf{r} \mapsto V(\mathbf{r})$ that satisfies $\nabla^2 V = -\frac{\rho}{\varepsilon_0}$ in some volume $\mathscr{V}$ that is surrounded by conductors is unique if $\rho: \mathbf{r} \mapsto \rho(\mathbf{r})$ is known $\forall \mathbf{r} \in \mathscr{V}$ and the total charge on the conductors is known.
+> *Theorem 8*: An electric potential $V: \mathbf{r} \mapsto V(\mathbf{r})$ that satisfies $\nabla^2 V = -\frac{\rho}{\epsilon_0}$ in some volume $\mathscr{V}$ that is surrounded by conductors is unique if $\rho: \mathbf{r} \mapsto \rho(\mathbf{r})$ is known $\forall \mathbf{r} \in \mathscr{V}$ and the total charge on the conductors is known.
 
 The method of images to solve for the electric potential $V$ makes use of these uniqueness theorems. In essence this method uses a solveable image system that obeys the same boundary conditions, then by unicity the same solution should be obtained. Though, the electric work required to construct the system is not always the same.
 
@@ -242,7 +242,7 @@ Approximation is the de facto option if no exact solution of the electric potent
 > *Theorem 9*: The electric potential $V: \mathbf{r} \mapsto V(\mathbf{r})$ of a charge distribution $\rho: \mathbf{r} \mapsto \rho(\mathbf{r})$ may be described as
 >
 > $$
->   V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{n=0}^\infty \frac{1}{r^{n+1}} \iiint (r')^n P_n(\cos \alpha) \rho(\mathbf{r}') d\tau',
+>   V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \sum_{n=0}^\infty \frac{1}{r^{n+1}} \iiint (r')^n P_n(\cos \alpha) \rho(\mathbf{r}') d\tau',
 > $$
 >
 > with $\cos \alpha = \frac{\langle \mathbf{r}, \mathbf{r}'\rangle}{r r'}$.
@@ -252,7 +252,7 @@ Approximation is the de facto option if no exact solution of the electric potent
     We start with
 
     $$
-        V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \iiint \frac{\rho(\mathbf{r}')}{\mathfrak{r}} d\tau'.
+        V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \iiint \frac{\rho(\mathbf{r}')}{\mathfrak{r}} d\tau'.
     $$
 
     We may write $\mathfrak{r}$ as
@@ -282,13 +282,13 @@ Approximation is the de facto option if no exact solution of the electric potent
     which obtains
 
     $$
-        V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{n=0}^\infty \frac{1}{r^{n+1}} \iiint (r')^n P_n(\cos \alpha) \rho(\mathbf{r}') d\tau'.
+        V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \sum_{n=0}^\infty \frac{1}{r^{n+1}} \iiint (r')^n P_n(\cos \alpha) \rho(\mathbf{r}') d\tau'.
     $$
 
 The monopole $(n=0)$ term of $V$ is then given by
 
 $$
-    V_0(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{1}{r} \int \rho(\mathbf{r}') d\tau',
+    V_0(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{1}{r} \iiint \rho(\mathbf{r}') d\tau',
 $$
 
 is zero if the net charge is zero.
@@ -297,24 +297,24 @@ The dipole $(n=1)$ term of $V$ is given by
 
 $$
     \begin{align*}
-        V_1(\mathbf{r}) &= \frac{1}{4\pi\varepsilon_0} \cdot \frac{1}{r^2} \langle \frac{1}{r} \mathbf{r}, \int \mathbf{r}' \rho(\mathbf{r}') d\tau' \rangle,\\
+        V_1(\mathbf{r}) &= \frac{1}{4\pi\epsilon_0} \cdot \frac{1}{r^2} \Bigg\langle \frac{1}{r} \mathbf{r}, \iiint \mathbf{r}' \rho(\mathbf{r}') d\tau' \Bigg\rangle,\\
-        &= \frac{1}{4\pi\varepsilon_0} \cdot \frac{1}{r^2} \langle \mathbf{e}_r, \mathbf{p} \rangle,
+        &= \frac{1}{4\pi\epsilon_0} \cdot \frac{1}{r^2} \langle \mathbf{e}_r, \mathbf{p} \rangle,
     \end{align*}
 $$
 
-with $\mathbf{p} = \int \mathbf{r}' \rho(\mathbf{r}') d\tau'$ the **dipole moment**.
+with $\mathbf{p} = \iiint \mathbf{r}' \rho(\mathbf{r}') d\tau'$ the **electric dipole moment**.
 
-For a collection of $N \in \mathbb{N}$ point charges we have that the dipole moment $\mathbf{p}$ is given by
+For a collection of $N \in \mathbb{N}$ point charges we have that the electric dipole moment $\mathbf{p}$ is given by
 
 $$
     \mathbf{p} = \sum_{i=1}^N q_i \mathbf{r}_i'.
 $$
 
-Note that if the net charge of the charge distribution is zero then the dipole moment is independent of the choice of origin. 
+Note that if the net charge of the charge distribution is zero then the electric dipole moment is independent of the choice of origin. 
 
 ## Electrostatics in matter
 
-The dipole moment $\mathbf{p}$ of an induced dipole due to an external field $\mathbf{E}$ is given by
+The electric dipole moment $\mathbf{p}$ of an induced dipole due to an external field $\mathbf{E}$ is given by
 
 $$
     \mathbf{p} = \bm{\alpha} \lrcorner \mathbf{E},
@@ -322,7 +322,7 @@ $$
 
 with $\bm{\alpha} \in \mathscr{T}^2_0$ the polarisability.
 
-> *Definition 3*: The induced dipole moment $\mathbf{p}$ in a medium may be expressed by the **polarisation** $\mathbf{P}$ of the medium defined in terms of
+> *Definition 3*: The induced electric dipole moment $\mathbf{p}$ in a medium may be expressed by the **polarisation** $\mathbf{P}$ of the medium defined in terms of
 >
 > $$
 >   \mathbf{p} = \int_\mathscr{V} \mathbf{P} d\tau,
@@ -330,7 +330,7 @@ with $\bm{\alpha} \in \mathscr{T}^2_0$ the polarisability.
 >
 > with $\mathscr{V}$ the volume of the medium.
 
-The polarisation $\mathbf{P}$ is in essence a sort of dipole moment per unit volume of the medium.
+The polarisation $\mathbf{P}$ is in essence a sort of electric dipole moment per unit volume of the medium.
 
 > *Theorem 10*: The polarisation of the medium $\mathbf{P}$ adheres to
 >
@@ -341,42 +341,48 @@ The polarisation $\mathbf{P}$ is in essence a sort of dipole moment per unit vol
 >   \end{align*}
 > $$
 >
-> with $\rho_b: \mathbf{r} \mapsto \rho_b(\mathbf{r})$ the **bound space charge density** and $\sigma_b: \mathbf{r} \mapsto \sigma_b(\mathbf{r})$ the **bound surface charge density**.
+> with $\rho_b: \mathbf{r} \mapsto \rho_b(\mathbf{r})$ the **bound volume charge density** and $\sigma_b: \mathbf{r} \mapsto \sigma_b(\mathbf{r})$ the **bound surface charge density**.
 
 ??? note "Proof:"
 
     The dipole term of the potential in terms of the polarisation $\mathbf{P}$ is given by
 
     $$
-        V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int_\mathscr{V} \frac{\mathbf{P}(\mathbf{r}') \cdot \mathbf{e}_\mathfrak{r}}{\mathfrak{r}^2} d\tau'.
+        V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int_\mathscr{V} \frac{\mathbf{P}(\mathbf{r}') \cdot \mathbf{e}_\mathfrak{r}}{\mathfrak{r}^2} d\tau'.
     $$
     
-    Now observe that $\nabla' \frac{1}{\mathfrak{r}} = \frac{1}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r}, obtains
+    Now observe that $\nabla' \frac{1}{\mathfrak{r}} = \frac{1}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r}$, obtains
 
     $$
-        V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int_\mathscr{V} \mathbf{P}(\mathbf{r}') \cdot \nabla' \frac{1}{\mathfrak{r}} d\tau',
+        V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int_\mathscr{V} \mathbf{P}(\mathbf{r}') \cdot \nabla' \frac{1}{\mathfrak{r}} d\tau',
+    $$
+
+    and by integration by parts
+
+    $$
+        V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \Bigg(\int_\mathscr{V} \nabla' \cdot \frac{1}{\mathfrak{r}} \mathbf{P}(\mathbf{r}') d\tau' - \int_\mathscr{V} \frac{1}{\mathfrak{r}} \nabla' \cdot \mathbf{P}(\mathbf{r}') d\tau'\Bigg),
     $$
 
     such that by the divergence theorem we obtain
 
     $$
-        V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \Bigg(\oint_{\partial \mathscr{V}} \frac{1}{\mathfrak{r}} \mathbf{P}(\mathbf{r}') \cdot d\mathbf{a}' - \int_\mathscr{V} \frac{1}{\mathfrak{r}} \nabla' \cdot \mathbf{P}(\mathbf{r}') d\tau' \Bigg).
+        V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \Bigg(\oint_{\partial \mathscr{V}} \frac{1}{\mathfrak{r}} \mathbf{P}(\mathbf{r}') \cdot d\mathbf{a}' - \int_\mathscr{V} \frac{1}{\mathfrak{r}} \nabla' \cdot \mathbf{P}(\mathbf{r}') d\tau' \Bigg).
     $$
 
     Setting $\rho_b = - \nabla \cdot \mathbf{P}$ and $\sigma_b = \mathbf{P} \cdot \mathbf{e}_n$ we obtain
 
     $$
-        V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \Bigg(\oint_{\partial \mathscr{V}} \frac{\sigma_b(\mathbf{r}')}{\mathfrak{r}} da' + \int_\mathscr{V} \frac{\rho_b(\mathbf{r}')}{\mathfrak{r}} d\tau'\Bigg),
+        V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \Bigg(\oint_{\partial \mathscr{V}} \frac{\sigma_b(\mathbf{r}')}{\mathfrak{r}} da' + \int_\mathscr{V} \frac{\rho_b(\mathbf{r}')}{\mathfrak{r}} d\tau'\Bigg),
     $$
 
-    implying that the potential of a polarised object is the same as that produced by a space charge density $\rho_b$ plus a a surface charge density $\sigma_b$. 
+    implying that the potential of a polarised object is the same as that produced by a volume charge density $\rho_b$ plus a a surface charge density $\sigma_b$. 
 
-We may as well define the **free space charge density** $\rho_f$ in terms of the space charge density $\rho$ and the bound space charge density $\rho_b$ by notion of $\rho_f = \rho - \rho_b$. 
+We may as well define the **free volume charge density** $\rho_f$ in terms of the volume charge density $\rho$ and the bound volume charge density $\rho_b$ by notion of $\rho_f = \rho - \rho_b$. 
 
 > *Definition 4*: Let the electric displacement $\mathbf{D}$ of the medium be defined as
 >
 > $$
->   \mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P},
+>   \mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P},
 > $$
 >
 > with $\mathbf{E}$ the electric field and $\mathbf{P}$ the polarisation of the medium.
@@ -392,30 +398,38 @@ The usefullness of *Definition 4* may become apparent with the following result:
 >   \end{align*}
 > $$
 >
-> with $\rho_f$ the free space charge density of the medium and $\mathbf{P}$ the polarisation of the medium.
+> with $\rho_f$ the free volume charge density of the medium and $\mathbf{P}$ the polarisation of the medium.
 
 ??? note "Proof:"
     
     From *Definition 4* we may write the electric field in terms of the electric displacement $\mathbf{D}$ and the polarisation $\mathbf{P}$ of the medium
 
     $$
-        \mathbf{E} = \frac{1}{\varepsilon_0} \bigg(\mathbf{D} - \mathbf{P}\bigg),
+        \mathbf{E} = \frac{1}{\epsilon_0} \bigg(\mathbf{D} - \mathbf{P}\bigg),
     $$
 
     such that with *Axiom 1* we obtain
 
     $$
-        \varepsilon_0 \nabla \cdot \mathbf{E} = \nabla \cdot \mathbf{D} - \rho_b \implies \nabla \cdot \mathbf{D} = \rho_f,
+        \epsilon_0 \nabla \cdot \mathbf{E} = \nabla \cdot \mathbf{D} + \rho_b \implies \nabla \cdot \mathbf{D} = \rho_f,
     $$
 
     and
 
     $$
-        \varepsilon_0 \nabla \times \mathbf{E} = \nabla \times \mathbf{D} - \nabla \times \mathbf{P} \implies \nabla \times \mathbf{D} = \nabla \times \mathbf{P}.
+        \epsilon_0 \nabla \times \mathbf{E} = \nabla \times \mathbf{D} - \nabla \times \mathbf{P} \implies \nabla \times \mathbf{D} = \nabla \times \mathbf{P}.
     $$
 
+Recall that from the divergence theorem we have
+
+$$
+    \oint_{\partial \mathscr{V}} \mathbf{D} \cdot d\mathbf{a} = \int_\mathscr{V} \rho_f d\tau = Q_f,
+$$
+
+with $Q_f$ the enclosed free charge, if $\nabla \times \mathbf{D} = \mathbf{0}$.
+
 ### Linear media
 
-In linear media we have $\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E}$ with $\chi_e$ the electric susceptibility of the medium. 
+In linear media we have $\mathbf{P} = \epsilon_0 \chi_e \mathbf{E}$ with $\chi_e$ the electric susceptibility of the medium. 
 
-Furthermore $\mathbf{D} = \varepsilon_0 (1 + \chi_e) \mathbf{E} = \varepsilon \mathbf{E}$ with $\varepsilon$ the permittivity and $\varepsilon_r = 1 + \chi_e = \frac{\varepsilon}{\varepsilon_0}$ the dielectric constant.
+Furthermore $\mathbf{D} = \epsilon_0 (1 + \chi_e) \mathbf{E} = \epsilon \mathbf{E}$ with $\epsilon$ the permittivity and $\epsilon_r = 1 + \chi_e = \frac{\epsilon}{\epsilon_0}$ the dielectric constant.

docs/physics/electromagnetism/magnetostatics.md

@@ -0,0 +1,245 @@
+# Magnetostatics
+
+In the **magnetostatic regime** there are steady currents, no moving point charges and time dependence is not taken into account.
+
+You may notice that the magnetostatics section is almost like a mirror of the electrostatics section.
+
+## Magnetostatics in vacuum
+
+Magnetostatics builds entirely on the following axiom:
+
+> *Axiom 1*: In the magnetostatic regime the magnetic field $\mathbf{B}: \mathbf{r} \mapsto \mathbf{B}(\mathbf{r})$ is described by
+>
+> $$
+>   \begin{align*}
+>       \nabla \cdot \mathbf{B} &= \mathbf{0},\\
+>       \nabla \times \mathbf{B} &= \mu_0 \mathbf{J},
+>   \end{align*}
+> $$
+>
+> with $\mathbf{J}: \mathbf{r} \mapsto \mathbf{J}(\mathbf{r})$ the **volume current density** and $\mu_0$ the **vacuum permeability**. 
+
+The following definition connects the magnetic field $\mathbf{B}: \mathbf{r} \mapsto \mathbf{B}(\mathbf{r})$ to the Newtonian formalism of mechanics:
+
+> *Definition 1*: The magnetic force $\mathbf{F}$ on a moving point charge $q$ with velocity $\mathbf{v}$ at a field point $\mathbf{r}$ is defined as
+>
+> $$
+>   \mathbf{F}(\mathbf{r}) = q \mathbf{v} \times \mathbf{B}(\mathbf{r}),
+> $$
+>
+> for all $\mathbf{B}: \mathbf{r} \mapsto \mathbf{B}(\mathbf{r})$ the magnetic field.
+
+Note that magnetic forces do no work.
+
+From *Axiom 1* we may proof that the following theorem holds (Biot-Savart law):
+
+> *Theorem 1*: The magnetic field $\mathbf{B}: \mathbf{r} \mapsto \mathbf{B}(\mathbf{r})$ may be described as
+>
+> $$
+>   \begin{align*}
+>       \mathbf{B}(\mathbf{r}) = \begin{cases} \frac{\mu_0}{4\pi} \int \mathbf{I}(\mathbf{r}') \times \frac{1}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} dl' \qquad &\mathrm{1D},\\ \frac{\mu_0}{4\pi} \iint \mathbf{K}(\mathbf{r}') \times \frac{1}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} da' \qquad &\mathrm{2D},\\ \frac{\mu_0}{4\pi} \iiint \mathbf{J}(\mathbf{r}') \times \frac{1}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r} d\tau' \qquad &\mathrm{3D}, \end{cases}
+>   \end{align*}
+> $$
+>
+> with $\mathbf{I}$ and $\mathbf{K}$ the line and surface current density.
+
+??? note "Proof:"
+
+    Follows from curl theorem.
+
+From the curl theorem it follows as well that
+
+$$
+    \oint_L \mathbf{B} \cdot d\mathbf{l} = \int_S (\nabla \times \mathbf{B}) \cdot d\mathbf{a} = \mu_0 \int_S \mathbf{J} \cdot d\mathbf{a} = \mu I,
+$$
+
+with $I$ the enclosed line current. This result is called Ampères' law.
+
+Since the magnetic field $\mathbf{B}$ is solenoidal in the magnetostatic regime we may define the following:
+
+> *Definition 2*: The **magnetic potential** $\mathbf{A}: \mathbf{r} \mapsto \mathbf{A}(\mathbf{r})$ of the magnetic field $\mathbf{B}$ is defined as
+>
+> $$
+>   \begin{align*}
+>       \nabla \times \mathbf{A} &= \mathbf{B},\\
+>       \nabla \cdot \mathbf{A} &= \mathbf{0},
+>   \end{align*}
+> $$
+>
+> for all $\mathbf{r}$.
+
+From (*Axiom 1*) $\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$ it follows that $\nabla^2 \mathbf{A} = -\mu_0 \mathbf{J}$. 
+
+Similarly to *Theorem 1* we may state the following about the magnetic potential $\mathbf{A}$:
+
+> *Theorem 2*: The magnetic potential $\mathbf{A}: \mathbf{r} \mapsto \mathbf{A}(\mathbf{r})$ may be described as
+>
+> $$
+>   \begin{align*}
+>       \mathbf{A}(\mathbf{r}) = \begin{cases} \frac{\mu_0}{4\pi} \int \mathbf{I} \frac{1}{\mathfrak{r}} dl' \qquad &\mathrm{1D},\\ \frac{\mu_0}{4\pi} \iint \mathbf{K} \frac{1}{\mathfrak{r}} da' \qquad &\mathrm{2D},\\ \frac{\mu_0}{4\pi} \int \mathbf{J} \frac{1}{\mathfrak{r}} d\tau' \qquad &\mathrm{3D}. \end{cases}
+>   \end{align*}
+> $$
+
+??? note "Proof:"
+
+    Follows from Poisson equation.
+
+### Multipole expansion
+
+Approximation is the de facto option if no exact solution of the magnetic potential exists. The multipole expansion enables the approximation of the magnetic potential at large distances. Its fundament is the following result:
+
+> *Theorem 3*: The magnetic potential $\mathbf{A}: \mathbf{r} \mapsto \mathbf{A}(\mathbf{r})$ of a current distribution $\mathbf{J}: \mathbf{r} \mapsto \mathbf{J}(\mathbf{r})$ may be described as
+>
+> $$
+>   \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \sum_{n=0}^\infty \frac{1}{r^{n+1}} \iiint (r')^n P_n(\cos \alpha) \mathbf{J}(\mathbf{r}') d\tau',
+> $$
+>
+> with $\cos \alpha = \frac{\langle \mathbf{r}, \mathbf{r}'\rangle}{r r'}$. 
+
+??? note "Proof:"
+
+    Follows from electric case.
+
+The monopole $(n=0)$ term of $\mathbf{A}$ is then given by
+
+$$
+    \mathbf{A}_0(\mathbf{r}) = \frac{\mu_0}{4\pi} \iiint \mathbf{J}(\mathbf{r}') d\tau',
+$$
+
+is always $\mathbf{0}$. 
+
+The dipole $(n=1)$ term of $\mathbf{A}$ is given by
+
+$$
+    \begin{align*}
+        \mathbf{A}_1(\mathbf{r}) &= \frac{\mu_0}{4\pi} \cdot \frac{1}{r^2} \iiint r' \cos \alpha \mathbf{J}(\mathbf{r}') d\tau',\\
+        &= \frac{\mu_0}{4\pi} \cdot \frac{1}{r^2} \Bigg(\frac{1}{2} \iiint \mathbf{r}' \times \mathbf{J}(\mathbf{r}') d\tau' \Bigg) \times \mathbf{e}_r,\\
+        &= \frac{\mu_0}{4\pi} \cdot \frac{1}{r^2} \mathbf{m} \times \mathbf{e}_r,
+    \end{align*}
+$$
+
+with $\mathbf{m} = \frac{1}{2} \iiint \mathbf{r}' \times \mathbf{J}(\mathbf{r}') d\tau'$ the **magnetic dipole moment**.
+
+It is a clear fact that the magnetic dipole moment is independent of the choice of origin.
+
+## Magnetostatics in matter
+
+The magnetic dipole moment $\mathbf{m}$ of an induced dipole due to an external field $\mathbf{B}$ is given by
+
+$$
+    \mathbf{m} = \bm{\beta} \lrcorner \mathbf{B},
+$$
+
+with $\bm{\beta} \in \mathscr{T}_0^2$ the magnetisability.
+
+> *Definition 3*: The induced magnetic dipole moment $\mathbf{m}$ in a medium may be expressed by the **magnetisation** $\mathbf{M}$ of the medium defined in terms of
+>
+> $$
+>   \mathbf{m} = \int_\mathscr{V} \mathbf{M} d\tau,
+> $$
+>
+> with $\mathscr{V}$ the volume of the medium.
+
+The magnitisation $\mathbf{M}$ is in essence a sort of magnetic dipole moment per unit volume of the medium.
+
+> *Theorem 4*: The magnetisation of the medium $\mathbf{M}$ adheres to
+>
+> $$
+>   \begin{align*}
+>       \nabla \times \mathbf{M} &= \mathbf{J}_b,\\
+>       \mathbf{M} \times \mathbf{e}_r &= \mathbf{K}_b,
+>   \end{align*}
+> $$
+> 
+> with $\mathbf{J}_b: \mathbf{r} \mapsto \mathbf{J}_b(\mathbf{r})$ the **bound volume current density** and $\mathbf{K}_b: \mathbf{r} \mapsto \mathbf{K}_b(\mathbf{r})$ the **bound surface current density**.
+
+??? note "Proof:"
+
+    The dipole term of the potential in terms of the magnetisation $\mathbf{M}$ is given by
+
+    $$
+        \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int_\mathscr{V} \frac{\mathbf{M}(\mathbf{r}') \times \mathbf{e}_\mathfrak{r}}{\mathfrak{r}^2} d\tau'.
+    $$
+
+    Now observe that $\nabla \frac{1}{\mathfrak{r}} = \frac{1}{\mathfrak{r}^2} \mathbf{e}_\mathfrak{r}$, obtains
+
+    $$
+        \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int_\mathscr{V} \mathbf{M}(\mathbf{r}') \times \nabla' \frac{1}{\mathfrak{r}} d\tau',
+    $$
+
+    and by integration by parts
+
+    $$
+        \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \Bigg(\int_\mathscr{V} \frac{1}{\mathfrak{r}} \nabla' \times \mathbf{M}(\mathbf{r}') d\tau' - \int_\mathscr{V} \nabla' \times \frac{\mathbf{M}(\mathbf{r}')}{\mathfrak{r}} d\tau' \Bigg),
+    $$
+
+    such that by the (adapted) divergence theorem we obtain
+
+    $$
+        \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \Bigg(\oint_{\partial \mathscr{V}} \frac{1}{\mathfrak{r}} \mathbf{M}(\mathbf{r}') \times d\mathbf{a}' + \int_\mathscr{V} \frac{1}{\mathfrak{r}} \big(\nabla' \times \mathbf{M}(\mathbf{r}')\big) d\tau'\Bigg).
+    $$
+
+    Setting $\mathbf{J}_b = \nabla \times \mathbf{M}$ and $\mathbf{K}_b = \mathbf{M} \times \mathbf{e}_n$, we obtain
+
+    $$
+        \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \Bigg(\oint_{\partial \mathscr{V}} \frac{1}{\mathfrak{r}} \mathbf{K}_b(\mathbf{r}') da' + \int_\mathscr{V} \frac{1}{\mathfrak{r}} \mathbf{J}_b(\mathbf{r}') d\tau'\Bigg),
+    $$
+    
+    implying that the potential of a magnetised object is the same as that produced by a volume current density $\mathbf{J}_b$ plus a surface current density $\mathbf{K}_b$. 
+
+We may as well define the **free volume current density** $\mathbf{J}_f$ in terms of the volume current density $\mathbf{J}$ and the bound volume current density $\mathbf{J}_b$ by notion of $\mathbf{J}_f = \mathbf{J} - \mathbf{J}_b$.
+
+> *Definition 4*: Let the auxiliary field $\mathbf{H}$ of the medium be defined as
+>
+> $$
+>   \mathbf{H} = \frac{1}{\mu_0} \mathbf{B} - \mathbf{M},
+> $$
+>
+> with $\mathbf{B}$ the magnetic field and $\mathbf{M}$ the magnetisation of the medium.
+
+The usefullness of *Definition 4* may become apparent with the following result:
+
+> *Theorem 5*: The auxiliary field of the medium $\mathbf{H}$ adheres to
+>
+> $$
+>   \begin{align*}
+>       \nabla \times \mathbf{H} &= \mathbf{J}_f,\\
+>       \nabla \cdot \mathbf{H} &= - \nabla \cdot \mathbf{M},
+>   \end{align*}
+> $$
+>
+> with $\mathbf{J}_f$ the free volume current density of the medium and $\mathbf{M}$ the magnetisation of the medium. 
+
+??? note "Proof:"
+
+    From *Definition 4* we may write the magnetic field in terms of the auxiliary field $\mathbf{H}$ and the magnetisation $\mathbf{M}$ of the medium
+
+    $$
+        \mathbf{B} = \mu_0 \Big(\mathbf{H} + \mathbf{M}\Big)
+    $$
+
+    such that with *Axiom 1* we obtain
+    
+    $$
+        \frac{1}{\mu_0} \nabla \times \mathbf{B} = \nabla \times \mathbf{H} + \mathbf{J}_b \implies \mathbf{H} = \mathbf{J}_f,
+    $$
+
+    and
+
+    $$
+        \frac{1}{\mu_0} \nabla \cdot \mathbf{B} = \nabla \cdot \mathbf{H} + \nabla \cdot \mathbf{M} \implies \nabla \cdot \mathbf{H} = - \nabla \cdot \mathbf{M}.
+    $$
+
+Recall that from the curl theorem we have
+
+$$
+    \oint_L \mathbf{H} \cdot d\mathbf{l} = \int_S \mathbf{J}_f \cdot d\mathbf{a} = I_f,
+$$
+
+with $I_f$ the enclosed free line current, if $\nabla \cdot \mathbf{H} = \mathbf{0}$.
+
+### Linear media
+
+In linear media we have $\mathbf{M} = \chi_m \mathbf{H}$ with $\chi_m$ the magnetic susceptibility of the medium. 
+
+Furthermore $\mathbf{B} = \mu_0 (\mu + \mathbf{M}) = \mu_0 (1 + \chi_m) \mathbf{H} = \mu \mathbf{H}$ with $\mu$ the permeability and $\mu_r = 1 + \chi_m = \frac{\mu}{\mu_0}$ the relative permeability. 

mkdocs.yml

@@ -196,8 +196,8 @@ nav:
 #    - 'Statistical mechanics':
     - 'Electromagnetism': 
       - 'Electrostatics': physics/electromagnetism/electrostatics.md
-#      - 'Magnetostatics': 
+      - 'Magnetostatics': physics/electromagnetism/magnetostatics.md
-#      - 'Electrodynamics': 
+      - 'Electromagnetic dynamics': physics/electromagnetism/electromagnetic-dynamics.md
       - 'Optics':
         - 'Waves': physics/electromagnetism/optics/waves.md
         - 'Electromagnetic waves': physics/electromagnetism/optics/electromagnetic-waves.md