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-rw-r--r--source/know/concept/maxwells-equations/index.md97
1 files changed, 49 insertions, 48 deletions
diff --git a/source/know/concept/maxwells-equations/index.md b/source/know/concept/maxwells-equations/index.md
index ea052ce..36eb7b2 100644
--- a/source/know/concept/maxwells-equations/index.md
+++ b/source/know/concept/maxwells-equations/index.md
@@ -17,10 +17,10 @@ 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 $\varepsilon_0$:
+**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 $$\varepsilon_0$$:
$$\begin{aligned}
\Phi_E
@@ -29,12 +29,12 @@ $$\begin{aligned}
= \frac{Q}{\varepsilon_0}
\end{aligned}$$
-Where $\vb{E}$ is the [electric field](/know/concept/electric-field/),
-and $\rho$ is the charge density in $V$.
+Where $$\vb{E}$$ is the [electric field](/know/concept/electric-field/),
+and $$\rho$$ is the charge density in $$V$$.
Gauss' law is usually more useful when written in its vector form,
which can be found by applying the divergence theorem
to the surface integral above.
-It states that the divergence of $\vb{E}$ is proportional to $\rho$:
+It states that the divergence of $$\vb{E}$$ is proportional to $$\rho$$:
$$\begin{aligned}
\boxed{
@@ -43,10 +43,10 @@ $$\begin{aligned}
\end{aligned}$$
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 $\varepsilon_0$:
+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 $$\varepsilon_0$$:
$$\begin{aligned}
\rho
@@ -54,11 +54,11 @@ $$\begin{aligned}
= \nabla \cdot \vb{D} - \nabla \cdot \vb{P}
\end{aligned}$$
-To proceed, we split the net charge density $\rho$
-into a "free" part $\rho_\mathrm{free}$
-and a "bound" part $\rho_\mathrm{bound}$,
-respectively corresponding to $\vb{D}$ and $\vb{P}$,
-such that $\rho = \rho_\mathrm{free} + \rho_\mathrm{bound}$.
+To proceed, we split the net charge density $$\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}
@@ -71,7 +71,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-By integrating over an arbitrary volume $V$
+By integrating over an arbitrary volume $$V$$
we can get integral forms of these equations:
$$\begin{aligned}
@@ -89,10 +89,10 @@ $$\begin{aligned}
## Gauss' law for magnetism
-**Gauss' law for magnetism** states that magnetic flux $\Phi_B$
-through a closed surface $S(V)$ is zero.
+**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:
+the volume $$V$$ must leave it too:
$$\begin{aligned}
\Phi_B
@@ -100,7 +100,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Where $\vb{B}$ is the [magnetic field](/know/concept/magnetic-field/).
+Where $$\vb{B}$$ is the [magnetic field](/know/concept/magnetic-field/).
Thanks to the divergence theorem,
this can equivalently be stated in vector form as follows:
@@ -117,10 +117,10 @@ in contrast to electric charge.
## Faraday's law of induction
-**Faraday's law of induction** states that a magnetic field $\vb{B}$
-that changes with time will induce an electric field $E$.
-Specifically, the change in magnetic flux through a non-closed surface $S$
-creates an electromotive force around the contour $C(S)$.
+**Faraday's law of induction** states that a magnetic field $$\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}
@@ -141,37 +141,38 @@ $$\begin{aligned}
## Ampère's circuital law
**Ampère's circuital law**, with Maxwell's correction,
-states that a magnetic field $\vb{B}$
-can be induced along a contour $C(S)$ by two things:
-a current density $\vb{J}$ through the enclosed surface $S$,
-and a change of the electric field flux $\Phi_E$ through $S$:
+states that a magnetic field $$\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}$$
-Where $\mu_0$ is the vacuum permeability.
-This relation also exists for the "bound" fields $\vb{H}$ and $\vb{D}$,
-and for $\vb{M}$ and $\vb{P}$.
-We insert $\vb{B} = \mu_0 (\vb{H} + \vb{M})$
-and $\vb{E} = (\vb{D} - \vb{P})/\varepsilon_0$
-into Ampère's law, after dividing it by $\mu_0$ for simplicity:
+Where $$\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}$$
-To proceed, we split the net current density $\vb{J}$
-into a "free" part $\vb{J}_\mathrm{free}$
-and a "bound" part $\vb{J}_\mathrm{bound}$,
-such that $\vb{J} = \vb{J}_\mathrm{free} + \vb{J}_\mathrm{bound}$.
+To proceed, we split the net current density $$\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}
@@ -184,7 +185,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-By integrating over an arbitrary surface $S$
+By integrating over an arbitrary surface $$S$$
we can get integral forms of these equations:
$$\begin{aligned}
@@ -195,9 +196,9 @@ $$\begin{aligned}
&= \int_S \vb{J}_{\mathrm{bound}} \cdot \dd{\vb{A}} - \dv{}{t}\int_S \vb{P} \cdot \dd{\vb{A}}
\end{aligned}$$
-Note that $\vb{J}_\mathrm{bound}$ can be split into
-the **magnetization current density** $\vb{J}_M = \nabla \cross \vb{M}$
-and the **polarization current density** $\vb{J}_P = \ipdv{\vb{P}}{t}$:
+Note that $$\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}
@@ -221,8 +222,8 @@ $$\begin{aligned}
Since the divergence of a curl is always zero,
the right-hand side must vanish.
-We know that $\vb{B}$ can vary in time,
-so our only option to satisfy this is to demand that $\nabla \cdot \vb{B} = 0$.
+We know that $$\vb{B}$$ can vary in time,
+so our only option to satisfy this is to demand that $$\nabla \cdot \vb{B} = 0$$.
We thus arrive arrive at Gauss' law for magnetism from Faraday's law.
The same technique works for Ampère's law.
@@ -234,7 +235,7 @@ $$\begin{aligned}
= \nabla \cdot \vb{J} + \varepsilon_0 \pdv{}{t}(\nabla \cdot \vb{E})
\end{aligned}$$
-We integrate this over an arbitrary volume $V$,
+We integrate this over an arbitrary volume $$V$$,
and apply the divergence theorem:
$$\begin{aligned}
@@ -245,9 +246,9 @@ $$\begin{aligned}
\end{aligned}$$
The first integral represents the current (charge flux)
-through the surface of $V$.
+through the surface of $$V$$.
Electric charge is not created or destroyed,
-so the second integral *must* be the total charge in $V$:
+so the second integral *must* be the total charge in $$V$$:
$$\begin{aligned}
Q