Expand knowledge base

6 files changed, 717 insertions, 6 deletions

diff --git a/content/know/concept/electric-field/index.pdc b/content/know/concept/electric-field/index.pdc
index ce2c4fc..6162e0b 100644
--- a/content/know/concept/electric-field/index.pdc
+++ b/content/know/concept/electric-field/index.pdc
@@ -15,8 +15,8 @@ markup: pandoc
 
 The **electric field** $\vb{E}$ is a vector field
 that describes electric effects,
-and is defined as the field that
-correctly predicts the Lorentz force
+and is defined as the field that correctly predicts
+the [Lorentz force](/know/concept/lorentz-force/)
 on a particle with electric charge $q$:
 
 $$\begin{aligned}
diff --git a/content/know/concept/electromagnetic-wave-equation/index.pdc b/content/know/concept/electromagnetic-wave-equation/index.pdc
new file mode 100644
index 0000000..68fe062
--- /dev/null
+++ b/content/know/concept/electromagnetic-wave-equation/index.pdc
@@ -0,0 +1,252 @@
+---
+title: "Electromagnetic wave equation"
+firstLetter: "E"
+publishDate: 2021-09-09
+categories:
+- Physics
+- Electromagnetism
+- Optics
+
+date: 2021-09-09T21:20:31+02:00
+draft: false
+markup: pandoc
+---
+
+# Electromagnetic wave equation
+
+The electromagnetic wave equation describes
+the propagation of light through various media.
+Since an electromagnetic (light) wave consists of
+an [electric field](/know/concept/electric-field/)
+and a [magnetic field](/know/concept/magnetic-field/),
+we need [Maxwell's equations](/know/concept/maxwells-equations/)
+in order to derive the wave equation.
+
+
+## Uniform medium
+
+We will use all of Maxwell's equations,
+but we start with Ampère's circuital law for the "free" fields $\vb{H}$ and $\vb{D}$,
+in the absence of a free current $\vb{J}_\mathrm{free} = 0$:
+
+$$\begin{aligned}
+    \nabla \cross \vb{H}
+    = \pdv{\vb{D}}{t}
+\end{aligned}$$
+
+We assume that the medium is isotropic, linear,
+and uniform in all of space, such that:
+
+$$\begin{aligned}
+    \vb{D} = \varepsilon_0 \varepsilon_r \vb{E}
+    \qquad \quad
+    \vb{H} = \frac{1}{\mu_0 \mu_r} \vb{B}
+\end{aligned}$$
+
+Which, upon insertion into Ampère's law,
+yields an equation relating $\vb{B}$ and $\vb{E}$.
+This may seem to contradict Ampère's "total" law,
+but keep in mind that $\vb{J}_\mathrm{bound} \neq 0$ here:
+
+$$\begin{aligned}
+    \nabla \cross \vb{B}
+    = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t}
+\end{aligned}$$
+
+Now we take the curl, rearrange,
+and substitute $\nabla \cross \vb{E}$ according to Faraday's law:
+
+$$\begin{aligned}
+    \nabla \cross (\nabla \cross \vb{B})
+    %= \nabla \cross \Big( \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \Big)
+    = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{t} (\nabla \cross \vb{E})
+    %= \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{t} \Big( \!-\! \pdv{\vb{B}}{t} \Big)
+    = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv[2]{\vb{B}}{t}
+\end{aligned}$$
+
+Using a vector identity, we rewrite the leftmost expression,
+which can then be reduced thanks to Gauss' law for magnetism $\nabla \cdot \vb{B} = 0$:
+
+$$\begin{aligned}
+    - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv[2]{\vb{B}}{t}
+    &= \nabla (\nabla \cdot \vb{B}) - \nabla^2 \vb{B}
+    = - \nabla^2 \vb{B}
+\end{aligned}$$
+
+This describes $\vb{B}$.
+Next, we repeat the process for $\vb{E}$:
+taking the curl of Faraday's law yields:
+
+$$\begin{aligned}
+    \nabla \cross (\nabla \cross \vb{E})
+    %= - \nabla \cross \pdv{\vb{B}}{t}
+    = - \pdv{t} (\nabla \cross \vb{B})
+    %= - \pdv{t} \Big( \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \Big)
+    = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv[2]{\vb{E}}{t}
+\end{aligned}$$
+
+Which can be rewritten using same vector identity as before,
+and then reduced by assuming that there is no net charge density $\rho = 0$
+in Gauss' law, such that $\nabla \cdot \vb{E} = 0$:
+
+$$\begin{aligned}
+    - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv[2]{\vb{E}}{t}
+    &= \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E}
+    = - \nabla^2 \vb{E}
+\end{aligned}$$
+
+We thus arrive at the following two (implicitly coupled)
+wave equations for $\vb{E}$ and $\vb{B}$,
+where we have defined the phase velocity $v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv[2]{\vb{E}}{t} - \frac{1}{v^2} \nabla^2 \vb{E}
+        = 0
+    }
+    \qquad \quad
+    \boxed{
+        \pdv[2]{\vb{B}}{t} - \frac{1}{v^2} \nabla^2 \vb{B}
+        = 0
+    }
+\end{aligned}$$
+
+Traditionally, it is said that the solutions are as follows,
+where the wavenumber $|\vb{k}| = \omega / v$:
+
+$$\begin{aligned}
+    \vb{E}(\vb{r}, t)
+    &= \vb{E}_0 \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+    \\
+    \vb{H}(\vb{r}, t)
+    &= \vb{H}_0 \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+\end{aligned}$$
+
+In fact, thanks to linearity, these solutions can be treated as
+terms in a Fourier series, meaning that virtually
+*any* function $f(\vb{k} \cdot \vb{r} - \omega t)$ is a valid solution.
+
+
+## Non-uniform medium
+
+A useful generalization is to allow spatial change
+in the relative permittivity $\varepsilon_r(\vb{r})$
+and the relative permeability $\mu_r(\vb{r})$.
+We still assume that the medium is linear and isotropic, so:
+
+$$\begin{aligned}
+    \vb{D}
+    = \varepsilon_0 \varepsilon_r(\vb{r}) \vb{E}
+    \qquad \quad
+    \vb{B}
+    = \mu_0 \mu_r(\vb{r}) \vb{H}
+\end{aligned}$$
+
+Inserting these expressions into Faraday's and Ampère's laws
+respectively yields:
+
+$$\begin{aligned}
+    \nabla \cross \vb{E}
+    = - \mu_0 \mu_r(\vb{r}) \pdv{\vb{H}}{t}
+    \qquad \quad
+    \nabla \cross \vb{H}
+    = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t}
+\end{aligned}$$
+
+We then divide Ampère's law by $\varepsilon_r(\vb{r})$,
+take the curl, and substitute Faraday's law, giving:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big)
+    = \varepsilon_0 \pdv{t} (\nabla \cross \vb{E})
+    = - \mu_0 \mu_r \varepsilon_0 \pdv[2]{\vb{H}}{t}
+\end{aligned}$$
+
+Next, we exploit linearity by decomposing $\vb{H}$ and $\vb{E}$
+into Fourier series, with terms given by:
+
+$$\begin{aligned}
+    \vb{H}(\vb{r}, t)
+    = \vb{H}(\vb{r}) \exp\!(- i \omega t)
+    \qquad \quad
+    \vb{E}(\vb{r}, t)
+    = \vb{E}(\vb{r}) \exp\!(- i \omega t)
+\end{aligned}$$
+
+By inserting this ansatz into the equation,
+we can remove the explicit time dependence:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) \exp\!(- i \omega t)
+    = \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp\!(- i \omega t)
+\end{aligned}$$
+
+Dividing out $\exp\!(- i \omega t)$,
+we arrive at an eigenvalue problem for $\omega^2$,
+with $c = 1 / \sqrt{\mu_0 \varepsilon_0}$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cross \Big( \frac{1}{\varepsilon_r(\vb{r})} \nabla \cross \vb{H}(\vb{r}) \Big)
+        = \Big( \frac{\omega}{c} \Big)^2 \mu_r(\vb{r}) \vb{H}(\vb{r})
+    }
+\end{aligned}$$
+
+Compared to a uniform medium, $\omega$ is often not arbitrary here:
+there are discrete eigenvalues $\omega$,
+corresponding to discrete **modes** $\vb{H}(\vb{r})$.
+
+Next, we go through the same process to find an equation for $\vb{E}$.
+Starting from Faraday's law, we divide by $\mu_r(\vb{r})$,
+take the curl, and insert Ampère's law:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big)
+    = - \mu_0 \pdv{t} (\nabla \cross \vb{H})
+    = - \mu_0 \varepsilon_0 \varepsilon_r \pdv[2]{\vb{E}}{t}
+\end{aligned}$$
+
+Then, by replacing $\vb{E}(\vb{r}, t)$ with our plane-wave ansatz,
+we remove the time dependence:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) \exp\!(- i \omega t)
+    = - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp\!(- i \omega t)
+\end{aligned}$$
+
+Which, after dividing out $\exp\!(- i \omega t)$,
+yields an analogous eigenvalue problem with $\vb{E}(r)$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cross \Big( \frac{1}{\mu_r(\vb{r})} \nabla \cross \vb{E}(\vb{r}) \Big)
+        = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r(\vb{r}) \vb{E}(\vb{r})
+    }
+\end{aligned}$$
+
+Usually, it is a reasonable approximation
+to say $\mu_r(\vb{r}) = 1$,
+in which case the equation for $\vb{H}(\vb{r})$
+becomes a Hermitian eigenvalue problem,
+and is thus easier to solve than for $\vb{E}(\vb{r})$.
+
+Keep in mind, however, that in any case,
+the solutions $\vb{H}(\vb{r})$ and/or $\vb{E}(\vb{r})$
+must satisfy the two Maxwell's equations that were not explicitly used:
+
+$$\begin{aligned}
+    \nabla \cdot (\varepsilon_r \vb{E}) = 0
+    \qquad \quad
+    \nabla \cdot (\mu_r \vb{H}) = 0
+\end{aligned}$$
+
+This is equivalent to demanding that the resulting waves are *transverse*,
+or in other words,
+the wavevector $\vb{k}$ must be perpendicular to
+the amplitudes $\vb{H}_0$ and $\vb{E}_0$.
+
+
+## References
+1.  J.D. Joannopoulos, S.G. Johnson, J.N. Winn, R.D. Meade,
+    *Photonic crystals: molding the flow of light*,
+    2nd edition, Princeton.
diff --git a/content/know/concept/heisenberg-picture/index.pdc b/content/know/concept/heisenberg-picture/index.pdc
index 2dd4118..9e4887d 100644
--- a/content/know/concept/heisenberg-picture/index.pdc
+++ b/content/know/concept/heisenberg-picture/index.pdc
@@ -37,9 +37,9 @@ converted to the Heisenberg picture by the following change of basis:
 
 $$\begin{aligned}
     \boxed{
-        \ket{\psi_H} = \ket{\psi_S(0)}
+        \ket{\psi_H} \equiv \ket{\psi_S(0)}
         \qquad
-        \hat{L}_H(t) = \hat{U}^\dagger(t) \hat{L}_S(t) \hat{U}(t)
+        \hat{L}_H(t) \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
     }
 \end{aligned}$$
 
diff --git a/content/know/concept/lorentz-force/index.pdc b/content/know/concept/lorentz-force/index.pdc
new file mode 100644
index 0000000..f0e9850
--- /dev/null
+++ b/content/know/concept/lorentz-force/index.pdc
@@ -0,0 +1,248 @@
+---
+title: "Lorentz force"
+firstLetter: "L"
+publishDate: 2021-09-08
+categories:
+- Physics
+- Electromagnetism
+
+date: 2021-09-08T17:00:32+02:00
+draft: false
+markup: pandoc
+---
+
+# Lorentz force
+
+The **Lorentz force** is an empirical force used to define
+the [electric field](/know/concept/electric-field/) $\vb{E}$
+and [magnetic field](/know/concept/magnetic-field/) $\vb{B}$.
+For a particle with charge $q$ moving with velocity $\vb{u}$,
+the Lorentz force $\vb{F}$ is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{F}
+        = q (\vb{E} + \vb{u} \cross \vb{B})
+    }
+\end{aligned}$$
+
+
+## Uniform magnetic field
+
+Consider the simple case of a uniform magnetic field
+$\vb{B} = (0, 0, B)$ in the $z$-direction,
+without an electric field $\vb{E} = 0$.
+If there are no other forces,
+Newton's second law states:
+
+$$\begin{aligned}
+    \vb{F}
+    = m \dv{\vb{u}}{t}
+    = q \vb{u} \cross \vb{B}
+\end{aligned}$$
+
+Evaluating the cross product yields
+three coupled equations for the components of $\vb{u}$:
+
+$$\begin{aligned}
+    \dv{u_x}{t}
+    = \frac{q B}{m} u_y
+    \qquad \quad
+    \dv{u_y}{t}
+    = - \frac{q B}{m} u_x
+    \qquad \quad
+    \dv{u_z}{t}
+    = 0
+\end{aligned}$$
+
+Differentiating the first equation with respect to $t$,
+and substituting $\dv*{u_y}{t}$ from the second,
+we arrive at the following harmonic oscillator:
+
+$$\begin{aligned}
+    \dv[2]{u_x}{t} = - \omega_c^2 u_x
+\end{aligned}$$
+
+Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
+which is always positive:
+
+$$\begin{aligned}
+    \boxed{
+        \omega_c
+        \equiv \frac{|q| B}{m}
+    }
+\end{aligned}$$
+
+Suppose we choose our initial conditions so that
+the solution for $u_x(t)$ is given by:
+
+$$\begin{aligned}
+    u_x(t)
+    = - u_\perp \sin\!(\omega_c t)
+\end{aligned}$$
+
+Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity.
+Then $u_y(t)$ is found to be:
+
+$$\begin{aligned}
+    u_y(t)
+    = \frac{m}{q B} \dv{u_x}{t}
+    = - \frac{m \omega_c}{q B} u_\perp \cos\!(\omega_c t)
+    = - \mathrm{sgn}(q) \: u_\perp \cos\!(\omega_c t)
+\end{aligned}$$
+
+Where $\mathrm{sgn}$ is the signum function.
+This tells us that the particle moves in a circular orbit,
+and that the direction of rotation is determined by $q$.
+
+Integrating the velocity yields the position,
+where we refer to the integration constants $x_{gc}$ and $y_{gc}$
+as the **guiding center**, around which the particle orbits or **gyrates**:
+
+$$\begin{aligned}
+    x(t)
+    = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + x_{gc}
+    \qquad \quad
+    y(t)
+    = - \mathrm{sgn}(q) \: \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + y_{gc}
+\end{aligned}$$
+
+The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by:
+
+$$\begin{aligned}
+    \boxed{
+        r_L
+        \equiv \frac{u_\perp}{\omega_c}
+        = \frac{m u_\perp}{|q| B}
+    }
+\end{aligned}$$
+
+Finally, it is trivial to integrate the equation for the $z$-direction velocity $u_z$:
+
+$$\begin{aligned}
+    z(t)
+    = u_z t + z_{gc}
+\end{aligned}$$
+
+In conclusion, the particle's motion parallel to $\vb{B}$
+is not affected by the magnetic field,
+while its motion perpendicular to $\vb{B}$
+is circular around an imaginary guiding center.
+The end result is that particles follow a helical path
+when moving through a uniform magnetic field.
+
+
+## Uniform electric and magnetic field
+
+Let us now consider a more general case,
+with constant uniform electric and magnetic fields $\vb{E}$ and $\vb{B}$,
+which may or may not be perpendicular.
+The equation of motion is then:
+
+$$\begin{aligned}
+    \vb{F}
+    = m \dv{\vb{u}}{t}
+    = q (\vb{E} + \vb{u} \cross \vb{B})
+\end{aligned}$$
+
+If we take the dot product with the unit vector $\vu{B}$,
+the cross product vanishes, leaving:
+
+$$\begin{aligned}
+    \dv{\vb{u}_\parallel}{t}
+    = \frac{q}{m} \vb{E}_\parallel
+\end{aligned}$$
+
+Where $\vb{u}_\parallel$ and $\vb{E}_\parallel$ are
+the components of $\vb{u}$ and $\vb{E}$
+that are parallel to $\vb{B}$.
+This equation is easy to integrate:
+the guiding center accelerates according to $(q/m) \vb{E}_\parallel$.
+
+Next, let us define the perpendicular component $\vb{u}_\perp$
+such that $\vb{u} = \vb{u}_\parallel \vu{B} + \vb{u}_\perp$.
+Its equation of motion is found by
+subtracting $\vb{u}_\parallel$'s equation from the original:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_\perp}{t}
+    = q (\vb{E} + \vb{u} \cross \vb{B}) - q \vb{E}_\parallel
+    = q (\vb{E}_\perp + \vb{u}_\perp \cross \vb{B})
+\end{aligned}$$
+
+To solve this, we go to a moving coordinate system
+by defining $\vb{u}_\perp = \vb{v}_\perp + \vb{w}_\perp$,
+where $\vb{v}_\perp$ is a constant of our choice.
+The equation is now as follows:
+
+$$\begin{aligned}
+    m \dv{t} (\vb{v}_\perp + \vb{w}_\perp)
+    = m \dv{\vb{w}_\perp}{t}
+    = q (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B} + \vb{w}_\perp \cross \vb{B})
+\end{aligned}$$
+
+We want to choose $\vb{v}_\perp$ such that the first two terms vanish,
+or in other words:
+
+$$\begin{aligned}
+    0
+    = \vb{E}_\perp + \vb{v}_\perp \cross \vb{B}
+\end{aligned}$$
+
+To find $\vb{v}_\perp$, we take the cross product with $\vb{B}$,
+and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$:
+
+$$\begin{aligned}
+    0
+    = \vb{B} \cross (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B})
+    = \vb{B} \cross \vb{E} + \vb{v}_\perp B^2
+    \quad \implies \quad
+    \boxed{
+        \vb{v}_\perp
+        = \frac{\vb{E} \cross \vb{B}}{B^2}
+    }
+\end{aligned}$$
+
+When $\vb{v}_\perp$ is chosen like this,
+the perpendicular equation of motion is reduced to:
+
+$$\begin{aligned}
+    m \dv{\vb{w}_\perp}{t}
+    = q \vb{w}_\perp \cross \vb{B}
+\end{aligned}$$
+
+Which is simply the case we treated previously with $\vb{E} = 0$,
+with a known solution
+(assuming $\vb{B}$ still points in the positive $z$-direction):
+
+$$\begin{aligned}
+    w_x(t)
+    = - w_\perp \sin\!(\omega_c t)
+    \qquad
+    w_y(t)
+    = - \mathrm{sgn}(q) \: w_\perp \cos\!(\omega_c t)
+\end{aligned}$$
+
+However, this result is shifted by a constant $\vb{v}_\perp$,
+often called the **drift velocity** $\vb{v}_d$,
+at which the guiding center moves transversely.
+Curiously, $\vb{v}_d$ is independent of $q$.
+
+Such a drift is not specific to an electric field.
+In the equations above, $\vb{E}$ can be replaced
+by a general force $\vb{F}/q$ (e.g. gravity) without issues.
+In that case, $\vb{v}_d$ does depend on $q$.
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_d
+        = \frac{\vb{F} \cross \vb{B}}{q B^2}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  F.F. Chen,
+    *Introduction to plasma physics and controlled fusion*,
+    3rd edition, Springer.
diff --git a/content/know/concept/magnetic-field/index.pdc b/content/know/concept/magnetic-field/index.pdc
index 2ad5fbf..3bfa90e 100644
--- a/content/know/concept/magnetic-field/index.pdc
+++ b/content/know/concept/magnetic-field/index.pdc
@@ -15,8 +15,8 @@ markup: pandoc
 
 The **magnetic field** $\vb{B}$ is a vector field
 that describes magnetic effects,
-and is defined as the field
-that correctly predicts the Lorentz force
+and is defined as the field that correctly predicts
+the [Lorentz force](/know/concept/lorentz-force/)
 on a particle with electric charge $q$:
 
 $$\begin{aligned}
diff --git a/content/know/concept/maxwells-equations/index.pdc b/content/know/concept/maxwells-equations/index.pdc
new file mode 100644
index 0000000..1551311
--- /dev/null
+++ b/content/know/concept/maxwells-equations/index.pdc
@@ -0,0 +1,211 @@
+---
+title: "Maxwell's equations"
+firstLetter: "M"
+publishDate: 2021-09-09
+categories:
+- Physics
+- Electromagnetism
+
+date: 2021-09-09T21:20:18+02:00
+draft: false
+markup: pandoc
+---
+
+# Maxwell's equations
+
+In physics, **Maxwell's equations** govern
+all macroscopic electromagnetism,
+and notably lead to the
+[electromagnetic wave equation](/know/concept/electromagnetic-wave-equation/),
+which describes the existence of light.
+
+
+## Gauss' law
+
+**Gauss' law** states that the electric flux $\Phi_E$ through
+a closed surface $S(V)$ is equal to the total charge $Q$
+contained in the enclosed volume $V$,
+divided by the vacuum permittivity $\varepsilon_0$:
+
+$$\begin{aligned}
+    \Phi_E
+    = \oint_{S(V)} \vb{E} \cdot \dd{\vb{A}}
+    = \frac{1}{\varepsilon_0} \int_{V} \rho \dd{V}
+    = \frac{Q}{\varepsilon_0}
+\end{aligned}$$
+
+Where $\vb{E}$ is the [electric field](/know/concept/electric-field/),
+and $\rho$ is the charge density in $V$.
+Gauss' law is usually more useful when written in its vector form,
+which can be found by applying the divergence theorem
+to the surface integral above.
+It states that the divergence of $\vb{E}$ is proportional to $\rho$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{E} = \frac{\rho}{\varepsilon_0}
+    }
+\end{aligned}$$
+
+This law can just as well be expressed for
+the displacement field $\vb{D}$
+and polarization density $\vb{P}$.
+We insert $\vb{E} = (\vb{D} - \vb{P}) / \varepsilon_0$
+into Gauss' law for $\vb{E}$, multiplied by $\varepsilon_0$:
+
+$$\begin{aligned}
+    \rho
+    = \nabla \cdot \big( \vb{D} - \vb{P} \big)
+    = \nabla \cdot \vb{D} - \nabla \cdot \vb{P}
+\end{aligned}$$
+
+To proceed, we split the net charge density $\rho$
+into a "free" part $\rho_\mathrm{free}$
+and a "bound" part $\rho_\mathrm{bound}$,
+respectively corresponding to $\vb{D}$ and $\vb{P}$,
+such that $\rho = \rho_\mathrm{free} + \rho_\mathrm{bound}$.
+This yields:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{D} = \rho_{\mathrm{free}}
+    }
+    \qquad \quad
+    \boxed{
+        \nabla \cdot \vb{P} = - \rho_{\mathrm{bound}}
+    }
+\end{aligned}$$
+
+By integrating over an arbitrary volume $V$
+we can get integral forms of these equations:
+
+$$\begin{aligned}
+    \Phi_D
+    &= \oint_{S(V)} \vb{D} \cdot \dd{\vb{A}}
+    = \int_{V} \rho_{\mathrm{free}} \dd{V}
+    = Q_{\mathrm{free}}
+    \\
+    \Phi_P
+    &= \oint_{S(V)} \vb{P} \cdot \dd{\vb{A}}
+    = - \int_{V} \rho_{\mathrm{bound}} \dd{V}
+    = - Q_{\mathrm{bound}}
+\end{aligned}$$
+
+
+## Gauss' law for magnetism
+
+**Gauss' law for magnetism** states that magnetic flux $\Phi_B$
+through a closed surface $S(V)$ is zero.
+In other words, all magnetic field lines entering
+the volume $V$ must leave it too:
+
+$$\begin{aligned}
+    \Phi_B
+    = \oint_{S(V)} \vb{B} \cdot \dd{\vb{A}}
+    = 0
+\end{aligned}$$
+
+Where $\vb{B}$ is the [magnetic field](/know/concept/magnetic-field/).
+Thanks to the divergence theorem,
+this can equivalently be stated in vector form as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{B} = 0
+    }
+\end{aligned}$$
+
+A consequence of this law is the fact that magnetic monopoles cannot exist,
+i.e. there is no such thing as "magnetic charge",
+in contrast to electric charge.
+
+
+## Faraday's law of induction
+
+**Faraday's law of induction** states that a magnetic field $\vb{B}$
+that changes with time will induce an electric field $E$.
+Specifically, the change in magnetic flux through a non-closed surface $S$
+creates an electromotive force around the contour $C(S)$.
+This is written as:
+
+$$\begin{aligned}
+    \oint_{C(S)} \vb{E} \cdot \dd{\vb{l}}
+    = - \dv{t} \int_{S} \vb{B} \cdot \dd{\vb{A}}
+\end{aligned}$$
+
+By using Stokes' theorem on the contour integral,
+the vector form of this law is found to be:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \times \vb{E} = - \pdv{\vb{B}}{t}
+    }
+\end{aligned}$$
+
+
+## Ampère's circuital law
+
+**Ampère's circuital law**, with Maxwell's correction,
+states that a magnetic field $\vb{B}$
+can be induced along a contour $C(S)$ by two things:
+a current density $\vb{J}$ through the enclosed surface $S$,
+and a change of the electric field flux $\Phi_E$ through $S$:
+
+$$\begin{aligned}
+    \oint_{C(S)} \vb{B} \cdot d\vb{l}
+    = \mu_0 \Big( \int_S \vb{J} \cdot d\vb{A} + \varepsilon_0 \dv{t} \int_S \vb{E} \cdot d\vb{A} \Big)
+\end{aligned}$$
+$$\begin{aligned}
+    \boxed{
+        \nabla \times \vb{B} = \mu_0 \Big( \vb{J} + \varepsilon_0 \pdv{\vb{E}}{t} \Big)
+    }
+\end{aligned}$$
+
+Where $\mu_0$ is the vacuum permeability.
+This relation also exists for the "bound" fields $\vb{H}$ and $\vb{D}$,
+and for $\vb{M}$ and $\vb{P}$.
+We insert $\vb{B} = \mu_0 (\vb{H} + \vb{M})$
+and $\vb{E} = (\vb{D} - \vb{P})/\varepsilon_0$
+into Ampère's law, after dividing it by $\mu_0$ for simplicity:
+
+$$\begin{aligned}
+    \nabla \cross \big( \vb{H} + \vb{M} \big)
+    &= \vb{J} + \pdv{t} \big( \vb{D} - \vb{P} \big)
+\end{aligned}$$
+
+To proceed, we split the net current density $\vb{J}$
+into a "free" part $\vb{J}_\mathrm{free}$
+and a "bound" part $\vb{J}_\mathrm{bound}$,
+such that $\vb{J} = \vb{J}_\mathrm{free} + \vb{J}_\mathrm{bound}$.
+This leads us to:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \times \vb{H} = \vb{J}_{\mathrm{free}} + \pdv{\vb{D}}{t}
+    }
+    \qquad \quad
+    \boxed{
+        \nabla \times \vb{M} = \vb{J}_{\mathrm{bound}} - \pdv{\vb{P}}{t}
+    }
+\end{aligned}$$
+
+By integrating over an arbitrary surface $S$
+we can get integral forms of these equations:
+
+$$\begin{aligned}
+    \oint_{C(S)} \vb{H} \cdot d\vb{l}
+    &= \int_S \vb{J}_{\mathrm{free}} \cdot \dd{\vb{A}} + \dv{t} \int_S \vb{D} \cdot \dd{\vb{A}}
+    \\
+    \oint_{C(S)} \vb{M} \cdot d\vb{l}
+    &= \int_S \vb{J}_{\mathrm{bound}} \cdot \dd{\vb{A}} - \dv{t} \int_S \vb{P} \cdot \dd{\vb{A}}
+\end{aligned}$$
+
+Note that $\vb{J}_\mathrm{bound}$ can be split into
+the **magnetization current density** $\vb{J}_M = \nabla \cross \vb{M}$
+and the **polarization current density** $\vb{J}_P = \pdv*{\vb{P}}{t}$:
+
+$$\begin{aligned}
+    \vb{J}_\mathrm{bound}
+    = \vb{J}_M + \vb{J}_P
+    = \nabla \cross \vb{M} + \pdv{\vb{P}}{t}
+\end{aligned}$$