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