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author | Prefetch | 2021-10-20 11:50:20 +0200 |
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committer | Prefetch | 2021-10-20 11:50:20 +0200 |
commit | 3170fc4b5c915669cf209a521e551115a9bd0809 (patch) | |
tree | 4cb70f6715be60b383fd2a23662d9795f56fef62 /content/know/concept/maxwells-equations | |
parent | 2a5024543ed99f569fbade8744e6c8001f2edb02 (diff) |
Expand knowledge base
Diffstat (limited to 'content/know/concept/maxwells-equations')
-rw-r--r-- | content/know/concept/maxwells-equations/index.pdc | 54 |
1 files changed, 54 insertions, 0 deletions
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. |