From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 14 Oct 2022 23:25:28 +0200
Subject: Initial commit after migration from Hugo

---
 source/know/concept/maxwells-equations/index.md | 259 ++++++++++++++++++++++++
 1 file changed, 259 insertions(+)
 create mode 100644 source/know/concept/maxwells-equations/index.md

(limited to 'source/know/concept/maxwells-equations')

diff --git a/source/know/concept/maxwells-equations/index.md b/source/know/concept/maxwells-equations/index.md
new file mode 100644
index 0000000..033ebf7
--- /dev/null
+++ b/source/know/concept/maxwells-equations/index.md
@@ -0,0 +1,259 @@
+---
+title: "Maxwell's equations"
+date: 2021-09-09
+categories:
+- Physics
+- Electromagnetism
+layout: "concept"
+---
+
+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** $\vb{J}_P = \ipdv{\vb{P}}{t}$:
+
+$$\begin{aligned}
+    \vb{J}_\mathrm{bound}
+    = \vb{J}_M + \vb{J}_P
+    = \nabla \cross \vb{M} + \pdv{\vb{P}}{t}
+\end{aligned}$$
+
+
+## Redundancy of Gauss' laws
+
+In fact, both of Gauss' laws are redundant,
+because they are already implied by Faraday's and Ampère's laws.
+Suppose we take the divergence of Faraday's law:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \nabla \cross \vb{E}
+    = - \nabla \cdot \pdv{\vb{B}}{t}
+    = - \pdv{}{t}(\nabla \cdot \vb{B})
+\end{aligned}$$
+
+Since the divergence of a curl is always zero,
+the right-hand side must vanish.
+We know that $\vb{B}$ can vary in time,
+so our only option to satisfy this is to demand that $\nabla \cdot \vb{B} = 0$.
+We thus arrive arrive at Gauss' law for magnetism from Faraday's law.
+
+The same technique works for Ampère's law.
+Taking its divergence gives us:
+
+$$\begin{aligned}
+    0
+    = \frac{1}{\mu_0} \nabla \cdot \nabla \cross \vb{B}
+    = \nabla \cdot \vb{J} + \varepsilon_0 \pdv{}{t}(\nabla \cdot \vb{E})
+\end{aligned}$$
+
+We integrate this over an arbitrary volume $V$,
+and apply the divergence theorem:
+
+$$\begin{aligned}
+    0
+    &= \int_V \nabla \cdot \vb{J} \dd{V} + \pdv{}{t}\int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V}
+    \\
+    &= \oint_S \vb{J} \cdot \dd{S} + \pdv{}{t}\int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V}
+\end{aligned}$$
+
+The first integral represents the current (charge flux)
+through the surface of $V$.
+Electric charge is not created or destroyed,
+so the second integral *must* be the total charge in $V$:
+
+$$\begin{aligned}
+    Q
+    = \int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V}
+    \quad \implies \quad
+    \nabla \cdot \vb{E}
+    = \frac{\rho}{\varepsilon_0}
+\end{aligned}$$
+
+And we thus arrive at Gauss' law from Ampère's law and charge conservation.
-- 
cgit v1.2.3