magnetostatics

Calculating the magnetostatic field from the moment

November 14, 2016 math and physics play No comments , , , , , ,

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The vector potential, to first order, for a magnetostatic localized current distribution was found to be

\begin{equation}\label{eqn:magneticFieldFromMoment:20}
\BA(\Bx) = \frac{\mu_0}{4 \pi} \frac{\Bm \cross \Bx}{\Abs{\Bx}^3}.
\end{equation}

Initially, I tried to calculate the magnetic field from this, but ran into trouble. Here’s a new try.

\begin{equation}\label{eqn:magneticFieldFromMoment:40}
\begin{aligned}
\BB
&=
\frac{\mu_0}{4 \pi}
\spacegrad \cross \lr{ \Bm \cross \frac{\Bx}{r^3} } \\
&=
-\frac{\mu_0}{4 \pi}
\spacegrad \cdot \lr{ \Bm \wedge \frac{\Bx}{r^3} } \\
&=
-\frac{\mu_0}{4 \pi}
\lr{
(\Bm \cdot \spacegrad) \frac{\Bx}{r^3}
-\Bm \spacegrad \cdot \frac{\Bx}{r^3}
} \\
&=
\frac{\mu_0}{4 \pi}
\lr{
-\frac{(\Bm \cdot \spacegrad) \Bx}{r^3}
– \lr{ \Bm \cdot \lr{\spacegrad \inv{r^3} }} \Bx
+\Bm (\spacegrad \cdot \Bx) \inv{r^3}
+\Bm \lr{\spacegrad \inv{r^3} } \cdot \Bx
}.
\end{aligned}
\end{equation}

Here I’ve used \( \Ba \cross \lr{ \Bb \cross \Bc } = -\Ba \cdot \lr{ \Bb \wedge \Bc } \), and then expanded that with \( \Ba \cdot \lr{ \Bb \wedge \Bc } = (\Ba \cdot \Bb) \Bc – (\Ba \cdot \Bc) \Bb \). Since one of these vectors is the gradient, care must be taken to have it operate on the appropriate terms in such an expansion.

Since we have \( \spacegrad \cdot \Bx = 3 \), \( (\Bm \cdot \spacegrad) \Bx = \Bm \), and \( \spacegrad 1/r^n = -n \Bx/r^{n+2} \), this reduces to

\begin{equation}\label{eqn:magneticFieldFromMoment:60}
\begin{aligned}
\BB
&=
\frac{\mu_0}{4 \pi}
\lr{
– \frac{\Bm}{r^3}
+ 3 \frac{(\Bm \cdot \Bx) \Bx}{r^5} %
+ 3 \Bm \inv{r^3}
-3 \Bm \frac{\Bx}{r^5} \cdot \Bx
} \\
&=
\frac{\mu_0}{4 \pi}
\frac{3 (\Bm \cdot \ncap) \ncap -\Bm}{r^3},
\end{aligned}
\end{equation}

which is the desired result.

Magnetostatic force and torque

October 18, 2016 math and physics play No comments , , , , , , ,

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In Jackson [1], the following equations for the vector potential, magnetostatic force and torque are derived

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:20}
\Bm = \inv{2} \int \Bx’ \cross \BJ(\Bx’) d^3 x’
\end{equation}
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:40}
\BF = \spacegrad( \Bm \cdot \BB ),
\end{equation}
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:60}
\BN = \Bm \cross \BB,
\end{equation}

where \( \BB \) is an applied external magnetic field and \( \Bm \) is the magnetic dipole for the current in question. These results (and a similar one derived earlier for the vector potential \( \BA \)) all follow from
an analysis of localized current densities \( \BJ \), evaluated far enough away from the current sources.

For the force and torque, the starting point for the force is one that had me puzzled a bit. Namely

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:80}
\BF = \int \BJ(\Bx) \cross \BB(\Bx) d^3 x
\end{equation}

This is clearly the continuum generalization of the point particle Lorentz force equation, which for \( \BE = 0 \) is:

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:100}
\BF = q \Bv \cross \BB
\end{equation}

For the point particle, this is the force on the particle when it is in the external field \( BB \). i.e. this is the force at the position of the particle. My question is what does it mean to sum all the forces on the charge distribution over all space.
How can a force be applied over all, as opposed to a force applied at a single point, or against a surface?

In the special case of a localized current density, this makes some sense. Considering the other half of the force equation \( \BF = \ddt{}\int \rho_m \Bv dV \), where \( \rho_m \) here is mass density of the charged particles making up the continuous current distribution. The other half of this \( \BF = m\Ba \) equation is also an average phenomena, so we have an average of sorts on both the field contribution to the force equation and the mass contribution to the force equation. There is probably a centre-of-mass and centre-of-current density interpretation that would make a bit more sense of this continuum force description.

It’s kind of funny how you can work through all the detailed mathematical steps in a book like Jackson, but then go right back to the beginning and say “Hey, what does that even mean”?

Force

Moving on from the pondering of the meaning of the equation being manipulated, let’s do the easy part, the derivation of the results that Jackson comes up with.

Writing out \ref{eqn:magnetostaticsJacksonNotesForceAndTorque:80} in coordinates

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:320}
\BF = \epsilon_{ijk} \Be_i \int J_j B_k d^3 x.
\end{equation}

To first order, a slowly varying (external) magnetic field can be expanded around a point of interest

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:120}
\BB(\Bx) = \BB(\Bx_0) + \lr{ \Bx – \Bx_0 } \cdot \spacegrad \BB,
\end{equation}

where the directional derivative is evaluated at the point \( \Bx_0 \) after the gradient operation. Setting the origin at this point \( \Bx_0 \) gives

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:340}
\begin{aligned}
\BF
&= \epsilon_{ijk} \Be_i
\lr{
\int J_j(\Bx’) B_k(0) d^3 x’
+
\int J_j(\Bx’) (\Bx’ \cdot \spacegrad) B_k(0) d^3 x’
} \\
&=
\epsilon_{ijk} \Be_i
\Bk_0 \int J_j(\Bx’) d^3 x’
+
\epsilon_{ijk} \Be_i
\int J_j(\Bx’) (\Bx’ \cdot \spacegrad) B_k(0) d^3 x’.
\end{aligned}
\end{equation}

We found

earlier
that the first integral can be written as a divergence

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:140}
\int J_j(\Bx’) d^3 x’
=
\int \spacegrad’ \cdot \lr{ \BJ(\Bx’) x_j’ } dV’,
\end{equation}

which is zero when the integration surface is outside of the current localization region. We also found

that

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:160}
\int (\Bx \cdot \Bx’) \BJ
= -\inv{2} \Bx \cross \int \Bx’ \cross \BJ = \Bm \cross \Bx.
\end{equation}

so
\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:180}
\begin{aligned}
\int (\spacegrad B_k(0) \cdot \Bx’) J_j
&= -\inv{2} \lr{ \spacegrad B_k(0) \cross \int \Bx’ \cross \BJ}_j \\
&= \lr{ \Bm \cross (\spacegrad B_k(0)) }_j.
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:200}
\begin{aligned}
\BF
&= \epsilon_{ijk} \Be_i \lr{ \Bm \cross (\spacegrad B_k(0)) }_j \\
&= \epsilon_{ijk} \Be_i \lr{ \Bm \cross \spacegrad }_j B_k(0) \\
&= (\Bm \cross \spacegrad) \cross \BB(0) \\
&= -\BB(0) \cross (\Bm \cross \lspacegrad) \\
&= (\BB(0) \cdot \Bm) \lspacegrad – (\BB \cdot \lspacegrad) \Bm \\
&= \spacegrad (\BB(0) \cdot \Bm) – \Bm (\spacegrad \cdot \BB(0)).
\end{aligned}
\end{equation}

The second term is killed by the magnetic Gauss’s law, leaving to first order

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:220}
\BF = \spacegrad \lr{\Bm \cdot \BB}.
\end{equation}

Torque

For the torque we have a similar quandary at the starting point. About what point is a continuum torque integral of the following form

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:240}
\BN = \int \Bx’ \cross (\BJ(\Bx’) \cross \BB(\Bx’)) d^3 x’?
\end{equation}

Ignoring that detail again, assuming the answer has something to do with the centre of mass and parallel axis theorem, we can proceed with a constant approximation of the magnetic field

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:260}
\begin{aligned}
\BN
&= \int \Bx’ \cross (\BJ(\Bx’) \cross \BB(0)) d^3 x’ \\
&=
-\int (\Bx’ \cdot \BJ(\Bx’)) \BB(0) d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’ \\
&=
-\BB(0) \int (\Bx’ \cdot \BJ(\Bx’)) d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’.
\end{aligned}
\end{equation}

Jackson’s trick for killing the first integral is to transform it into a divergence by evaluating

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:280}
\begin{aligned}
\spacegrad \cdot \lr{ \BJ \Abs{\Bx}^2 }
&=
(\spacegrad \cdot \BJ) \Abs{\Bx}^2
+
\BJ \cdot \spacegrad \Abs{\Bx}^2 \\
&=
\BJ \cdot \Be_i \partial_i x_m x_m \\
&=
2 \BJ \cdot \Be_i \delta_{im} x_m \\
&=
2 \BJ \cdot \Bx,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:300}
\begin{aligned}
\BN
&=
-\inv{2} \BB(0) \int \spacegrad’ \cdot \lr{ \BJ(\Bx’) \Abs{\Bx’}^2 } d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’ \\
&=
-\inv{2} \BB(0) \oint \Bn \cdot \lr{ \BJ(\Bx’) \Abs{\Bx’}^2 } d^3 x’
+\int (\Bx’ \cdot \BB(0)) \BJ(\Bx’) d^3 x’.
\end{aligned}
\end{equation}

Again, the localized current density assumption kills the surface integral. The second integral can be evaluated with \ref{eqn:magnetostaticsJacksonNotesForceAndTorque:160}, so to first order we have

\begin{equation}\label{eqn:magnetostaticsJacksonNotesForceAndTorque:360}
\BN
=
\Bm \cross \BB.
\end{equation}

References

[1] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

Green’s function inversion of the magnetostatic equation

September 27, 2016 math and physics play No comments , , , , , , , , ,

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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.