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A previous example of inverting a gradient equation was the electrostatics equation. We can do the same for the magnetostatics equation, which has the following Geometric Algebra form in linear media

\begin{equation}\label{eqn:biotSavartGreens:20}

\spacegrad I \BB = – \mu \BJ.

\end{equation}

The Green’s inversion of this is

\begin{equation}\label{eqn:biotSavartGreens:40}

\begin{aligned}

I \BB(\Bx)

&= \int_V dV’ G(\Bx, \Bx’) \spacegrad’ I \BB(\Bx’) \\

&= \int_V dV’ G(\Bx, \Bx’) (-\mu \BJ(\Bx’)) \\

&= \inv{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } (-\mu \BJ(\Bx’)).

\end{aligned}

\end{equation}

We expect the LHS to be a bivector, so the scalar component of this should be zero. That can be demonstrated with some of the usual trickery

\begin{equation}\label{eqn:biotSavartGreens:60}

\begin{aligned}

-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)

&= \frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\

&= -\frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad’ \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\

&= -\frac{\mu}{4\pi} \int_V dV’ \lr{

\spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }

–

\frac{\spacegrad’ \cdot \BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }

}.

\end{aligned}

\end{equation}

The current \( \BJ \) is not unconstrained. This can be seen by premultiplying \ref{eqn:biotSavartGreens:20} by the gradient

\begin{equation}\label{eqn:biotSavartGreens:80}

\spacegrad^2 I \BB = -\mu \spacegrad \BJ.

\end{equation}

On the LHS we have a bivector so must have \( \spacegrad \BJ = \spacegrad \wedge \BJ \), or \( \spacegrad \cdot \BJ = 0 \). This kills the \( \spacegrad’ \cdot \BJ(\Bx’) \) integrand numerator in \ref{eqn:biotSavartGreens:60}, leaving

\begin{equation}\label{eqn:biotSavartGreens:100}

\begin{aligned}

-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)

&= -\frac{\mu}{4\pi} \int_V dV’ \spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} } \\

&= -\frac{\mu}{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }.

\end{aligned}

\end{equation}

This shows that the scalar part of the equation is zero, provided the normal component of \( \BJ/\Abs{\Bx – \Bx’} \) vanishes on the boundary of the infinite sphere. This leaves the Biot-Savart law as a bivector equation

\begin{equation}\label{eqn:biotSavartGreens:120}

I \BB(\Bx)

= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \wedge \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.

\end{equation}

Observe that the traditional vector form of the Biot-Savart law can be obtained by premultiplying both sides with \( -I \), leaving

\begin{equation}\label{eqn:biotSavartGreens:140}

\BB(\Bx)

= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \cross \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.

\end{equation}

This checks against a trusted source such as [1] (eq. 5.39).

# References

[1] David Jeffrey Griffiths and Reed College. *Introduction to electrodynamics*. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.