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:ID:       5c36d0f1-06ad-436a-a56f-5ecc198b9b3e
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#+title: magnetostatics
#+author: Preston Pan
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* Introduction
Magnetostatics is a little bit of an oxymoron; the magnetic field is created by a moving current of charges and a magnetic
field for a point charge is therefore hard to model or often wrong; because magnetostatics assumes a steady current, a point
charge moving cannot be replaced with another charge. However, for some continuous current distribution, the magnetic field is:
\begin{align*}
\vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J} \times \hat{r}}{r^{2}}dV
\end{align*}
Which is the Bio-Savart law. Later, we will derive this from the axioms of [[id:32f0b8b1-17bc-4c91-a824-2f2a3bbbdbd1][electrostatics]] and [[id:e38d94f2-8332-4811-b7bd-060f80fcfa9b][special relativity]].

* Divergence of B
The divergence of B is given by:
\begin{align*}
\vec{\nabla} \cdot \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{\nabla} \cdot (\vec{J} \times \hat{r})}{r^{2}}dV
= \frac{\mu_{0}}{4\pi}\int_{V}\vec{\nabla} \cdot (\vec{J} \times \frac{\hat{r}}{r^{2}})dV \\
\end{align*}
Now we want to evaluate $\vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}}$ using the [[id:4bfd6585-1305-4cf2-afc0-c0ba7de71896][del operator]] rules:
\begin{align*}
\vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}} = (\vec{\nabla} \times \vec{J}) \cdot \frac{\hat{r}}{r^{2}} - \vec{J} \cdot (\vec{\nabla} \times \frac{\hat{r}}{r^{2}})
\end{align*}
And while $B$, and by extension $\vec{\nabla}$ is dependent on $\vec{r}$, the radial distance between two charges,
$\vec{J}$ is a function of $r'$, the position vector to a given charge. This, of course, means that $J$ does not
depend on any of the variables that we are taking the derivative over. Thus:
\begin{align*}
\vec{\nabla} \times \vec{J} = 0
\end{align*}
Also, due to the [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law, the curl of $\frac{\hat{r}}{r^{2}}$ is zero, and therefore:
\begin{align*}
\vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}} = 0
\end{align*}
And therefore:
\begin{align*}
\vec{\nabla} \cdot \vec{B} = 0
\end{align*}
This is one of [[id:fde2f257-fa2e-469a-bc20-4d11714a515e][Maxwell's Equations]].
* Curl of B
The curl of $\vec{B}$ is given by:
\begin{align*}
\vec{\nabla} \times \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{\nabla} \times (\vec{J} \times \hat{r})}{r^{2}}dV \\
= \frac{\mu_{0}}{4\pi}\int_{V}\vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}})dV
\end{align*}
where $\vec{\nabla} \times \vec{J} \times \hat{r}$ is given by the [[id:4bfd6585-1305-4cf2-afc0-c0ba7de71896][del operator]] identity:
\begin{align*}
\vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = \vec{J}(\vec{\nabla} \cdot \frac{\hat{r}}{r^{2}}) - \frac{\hat{r}}{r^{2}}(\vec{\nabla} \cdot \vec{J}) + (\frac{\hat{r}}{r^{2}} \cdot \vec{\nabla})\vec{J} - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}}
\end{align*}
Due to the [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law, we know that $\vec{\nabla} \cdot \frac{\hat{r}}{r^{2}} = 4\pi\delta(\vec{r})$; From the section above we know that $\vec{\nabla} \cdot \vec{J} = 0$; hence:
\begin{align*}
\vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) + (\frac{\hat{r}}{r^{2}} \cdot \vec{\nabla})\vec{J} - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}}
\end{align*}
The first directional derivative is zero because $\vec{J}$ does not depend on the same coordinates as $\vec{\nabla}}$
with the same reasoning as for the divergence, so we have:
\begin{align*}
\vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}}
\end{align*}
The second term reduces to zero for some reason (reason coming later; I currently do not know why). Now plugging this back into the original equation:
\begin{align*}
\vec{\nabla} \times \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}4\pi\vec{J}(r')\delta(\vec{r})dV \\
= \mu_{0}\int_{V}\vec{J}(r')\delta(\vec{r'} - \vec{r''})dV
\end{align*}
Where $r''$ is the location of the test particle. Now the [[id:90574fea-88f4-4b80-9cda-32cff0bcb76d][dirac delta]] distribution propreties under an integral:
\begin{align*}
\vec{\nabla} \times \vec{B} = \mu_{0}\vec{J}(\vec{r''})
\end{align*}
or simply:
\begin{align*}
\vec{\nabla} \times \vec{B} = \mu_{0}\vec{J}
\end{align*}
This is for magnetostatics only; [[id:fde2f257-fa2e-469a-bc20-4d11714a515e][Maxwell's Equations]] offer a correction to this equation.
* The Vector Potential
We can define a /vector potential/ $\vec{A}(\vec{r})$ such that:
\begin{align*}
\vec{B} = \vec{\nabla} \times \vec{A}
\end{align*}
which is consistent with the fact that:
\begin{align*}
\vec{\nabla} \cdot \vec{B} = 0
\end{align*}
By taking the divergence of both sides in the first equation in this section. When $\vec{J}$ is zero at infinity:
\begin{align*}
\vec{A} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J}}{r}d\tau
\end{align*}
The reasoning for this is not obvious, even to me. One could analogize this to the scalar potential for [[id:32f0b8b1-17bc-4c91-a824-2f2a3bbbdbd1][electrostatics]].