From 3170fc4b5c915669cf209a521e551115a9bd0809 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 20 Oct 2021 11:50:20 +0200
Subject: Expand knowledge base

---
 content/know/concept/maxwells-equations/index.pdc | 54 +++++++++++++++++++++++
 1 file changed, 54 insertions(+)

(limited to 'content/know/concept/maxwells-equations/index.pdc')

diff --git a/content/know/concept/maxwells-equations/index.pdc b/content/know/concept/maxwells-equations/index.pdc
index 1551311..967372d 100644
--- a/content/know/concept/maxwells-equations/index.pdc
+++ b/content/know/concept/maxwells-equations/index.pdc
@@ -209,3 +209,57 @@ $$\begin{aligned}
     = \vb{J}_M + \vb{J}_P
     = \nabla \cross \vb{M} + \pdv{\vb{P}}{t}
 \end{aligned}$$
+
+
+## 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.
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