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