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triple cross product

Transverse gauge

November 16, 2016 math and physics play , , , , , , , , , , , , , , , , ,

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Jackson [1] has an interesting presentation of the transverse gauge. I’d like to walk through the details of this, but first want to translate the preliminaries to SI units (if I had the 3rd edition I’d not have to do this translation step).

Gauge freedom

The starting point is noting that \spacegrad \cdot \BB = 0 the magnetic field can be expressed as a curl

\begin{equation}\label{eqn:transverseGauge:20} \BB = \spacegrad \cross \BA. \end{equation}

Faraday’s law now takes the form
\begin{equation}\label{eqn:transverseGauge:40} \begin{aligned} 0 &= \spacegrad \cross \BE + \PD{t}{\BB} \\ &= \spacegrad \cross \BE + \PD{t}{} \lr{ \spacegrad \cross \BA } \\ &= \spacegrad \cross \lr{ \BE + \PD{t}{\BA} }. \end{aligned} \end{equation}

Because this curl is zero, the interior sum can be expressed as a gradient

\begin{equation}\label{eqn:transverseGauge:60} \BE + \PD{t}{\BA} \equiv -\spacegrad \Phi. \end{equation}

This can now be substituted into the remaining two Maxwell’s equations.

\begin{equation}\label{eqn:transverseGauge:80} \begin{aligned} \spacegrad \cdot \BD &= \rho_v \\ \spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\ \end{aligned} \end{equation}

For Gauss’s law, in simple media, we have

\begin{equation}\label{eqn:transverseGauge:140} \begin{aligned} \rho_v &= \epsilon \spacegrad \cdot \BE \\ &= \epsilon \spacegrad \cdot \lr{ -\spacegrad \Phi – \PD{t}{\BA} } \end{aligned} \end{equation}

For simple media again, the Ampere-Maxwell equation is

\begin{equation}\label{eqn:transverseGauge:100} \inv{\mu} \spacegrad \cross \lr{ \spacegrad \cross \BA } = \BJ + \epsilon \PD{t}{} \lr{ -\spacegrad \Phi – \PD{t}{\BA} }. \end{equation}

Expanding \spacegrad \cross \lr{ \spacegrad \cross \BA } = -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } gives
\begin{equation}\label{eqn:transverseGauge:120} -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } + \epsilon \mu \PDSq{t}{\BA} = \mu \BJ – \epsilon \mu \spacegrad \PD{t}{\Phi}. \end{equation}

Maxwell’s equations are now reduced to
\begin{equation}\label{eqn:transverseGauge:180} \boxed{ \begin{aligned} \spacegrad^2 \BA – \spacegrad \lr{ \spacegrad \cdot \BA + \epsilon \mu \PD{t}{\Phi}} – \epsilon \mu \PDSq{t}{\BA} &= -\mu \BJ \\ \spacegrad^2 \Phi + \PD{t}{\spacegrad \cdot \BA} &= -\frac{\rho_v }{\epsilon}. \end{aligned} } \end{equation}

There are two obvious constraints that we can impose
\begin{equation}\label{eqn:transverseGauge:200} \spacegrad \cdot \BA – \epsilon \mu \PD{t}{\Phi} = 0, \end{equation}

or
\begin{equation}\label{eqn:transverseGauge:220} \spacegrad \cdot \BA = 0. \end{equation}

The first constraint is the Lorentz gauge, which I’ve played with previously. It happens to be really nice in a relativistic context since, in vacuum with a four-vector potential A = (\Phi/c, \BA) , that is a requirement that the four-divergence of the four-potential vanishes ( \partial_\mu A^\mu = 0 ).

Transverse gauge

Jackson identifies the latter constraint as the transverse gauge, which I’m less familiar with. With this gauge selection, we have

\begin{equation}\label{eqn:transverseGauge:260} \spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \epsilon\mu \spacegrad \PD{t}{\Phi} \end{equation}
\begin{equation}\label{eqn:transverseGauge:280} \spacegrad^2 \Phi = -\frac{\rho_v }{\epsilon}. \end{equation}

What’s not obvious is the fact that the irrotational (zero curl) contribution due to \Phi in \ref{eqn:transverseGauge:260} cancels the corresponding irrotational term from the current. Jackson uses a transverse and longitudinal decomposition of the current, related to the Helmholtz theorem to allude to this.

That decomposition follows from expanding \spacegrad^2 J/R in two ways using the delta function -4 \pi \delta(\Bx – \Bx’) = \spacegrad^2 1/R representation, as well as directly

\begin{equation}\label{eqn:transverseGauge:300} \begin{aligned} – 4 \pi \BJ(\Bx) &= \int \spacegrad^2 \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\ &= \spacegrad \int \spacegrad \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ + \spacegrad \cdot \int \spacegrad \wedge \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\ &= -\spacegrad \int \BJ(\Bx’) \cdot \spacegrad’ \inv{\Abs{\Bx – \Bx’}} d^3 x’ + \spacegrad \cdot \lr{ \spacegrad \wedge \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ } \\ &= -\spacegrad \int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ +\spacegrad \int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ – \spacegrad \cross \lr{ \spacegrad \cross \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ } \end{aligned} \end{equation}

The first term can be converted to a surface integral

\begin{equation}\label{eqn:transverseGauge:320} -\spacegrad \int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ = -\spacegrad \int d\BA’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}}, \end{equation}

so provided the currents are either localized or \Abs{\BJ}/R \rightarrow 0 on an infinite sphere, we can make the identification

\begin{equation}\label{eqn:transverseGauge:340} \BJ(\Bx) = -\spacegrad \inv{4 \pi} \int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ + \spacegrad \cross \spacegrad \cross \inv{4 \pi} \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \equiv \BJ_l + \BJ_t, \end{equation}

where \spacegrad \cross \BJ_l = 0 (irrotational, or longitudinal), whereas \spacegrad \cdot \BJ_t = 0 (solenoidal or transverse). The irrotational property is clear from inspection, and the transverse property can be verified readily

\begin{equation}\label{eqn:transverseGauge:360} \begin{aligned} \spacegrad \cdot \lr{ \spacegrad \cross \lr{ \spacegrad \cross \BX } } &= -\spacegrad \cdot \lr{ \spacegrad \cdot \lr{ \spacegrad \wedge \BX } } \\ &= -\spacegrad \cdot \lr{ \spacegrad^2 \BX – \spacegrad \lr{ \spacegrad \cdot \BX } } \\ &= -\spacegrad \cdot \lr{\spacegrad^2 \BX} + \spacegrad^2 \lr{ \spacegrad \cdot \BX } \\ &= 0. \end{aligned} \end{equation}

Since

\begin{equation}\label{eqn:transverseGauge:380} \Phi(\Bx, t) = \inv{4 \pi \epsilon} \int \frac{\rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’, \end{equation}

we have

\begin{equation}\label{eqn:transverseGauge:400} \begin{aligned} \spacegrad \PD{t}{\Phi} &= \inv{4 \pi \epsilon} \spacegrad \int \frac{\partial_t \rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’ \\ &= \inv{4 \pi \epsilon} \spacegrad \int \frac{-\spacegrad’ \cdot \BJ}{\Abs{\Bx – \Bx’}} d^3 x’ \\ &= \frac{\BJ_l}{\epsilon}. \end{aligned} \end{equation}

This means that the Ampere-Maxwell equation takes the form

\begin{equation}\label{eqn:transverseGauge:420} \spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \mu \BJ_l = -\mu \BJ_t. \end{equation}

This justifies the transverse in the label transverse gauge.

References

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

Magnetic moment for a localized magnetostatic current

October 13, 2016 math and physics play , , , , , , , , ,

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Motivation.

I was once again reading my Jackson [2]. This time I found that his presentation of magnetic moment didn’t really make sense to me. Here’s my own pass through it, filling in a number of details. As I did last time, I’ll also translate into SI units as I go.

Vector potential.

The Biot-Savart expression for the magnetic field can be factored into a curl expression using the usual tricks

\begin{equation}\label{eqn:magneticMomentJackson:20} \begin{aligned} \BB &= \frac{\mu_0}{4\pi} \int \frac{\BJ(\Bx’) \cross (\Bx – \Bx’)}{\Abs{\Bx – \Bx’}^3} d^3 x’ \\ &= -\frac{\mu_0}{4\pi} \int \BJ(\Bx’) \cross \spacegrad \inv{\Abs{\Bx – \Bx’}} d^3 x’ \\ &= \frac{\mu_0}{4\pi} \spacegrad \cross \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’, \end{aligned} \end{equation}

so the vector potential, through its curl, defines the magnetic field \BB = \spacegrad \cross \BA is given by

\begin{equation}\label{eqn:magneticMomentJackson:40} \BA(\Bx) = \frac{\mu_0}{4 \pi} \int \frac{J(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’. \end{equation}

If the current source is localized (zero outside of some finite region), then there will always be a region for which \Abs{\Bx} \gg \Abs{\Bx’} , so the denominator yields to Taylor expansion

\begin{equation}\label{eqn:magneticMomentJackson:60} \begin{aligned} \inv{\Abs{\Bx – \Bx’}} &= \inv{\Abs{\Bx}} \lr{1 + \frac{\Abs{\Bx’}^2}{\Abs{\Bx}^2} – 2 \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^2} }^{-1/2} \\ &\approx \inv{\Abs{\Bx}} \lr{ 1 + \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^2} } \\ &= \inv{\Abs{\Bx}} + \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^3}. \end{aligned} \end{equation}

so the vector potential, far enough away from the current source is
\begin{equation}\label{eqn:magneticMomentJackson:80} \BA(\Bx) = \frac{\mu_0}{4 \pi} \int \frac{J(\Bx’)}{\Abs{\Bx}} d^3 x’ +\frac{\mu_0}{4 \pi} \int \frac{(\Bx \cdot \Bx’)J(\Bx’)}{\Abs{\Bx}^3} d^3 x’. \end{equation}

Jackson uses a sneaky trick to show that the first integral is killed for a localized source. That trick appears to be based on evaluating the following divergence

\begin{equation}\label{eqn:magneticMomentJackson:100} \begin{aligned} \spacegrad \cdot (\BJ(\Bx) x_i) &= (\spacegrad \cdot \BJ) x_i + (\spacegrad x_i) \cdot \BJ \\ &= (\Be_k \partial_k x_i) \cdot\BJ \\ &= \delta_{ki} J_k \\ &= J_i. \end{aligned} \end{equation}

Note that this made use of the fact that \spacegrad \cdot \BJ = 0 for magnetostatics. This provides a way to rewrite the current density as a divergence

\begin{equation}\label{eqn:magneticMomentJackson:120} \begin{aligned} \int \frac{J(\Bx’)}{\Abs{\Bx}} d^3 x’ &= \Be_i \int \frac{\spacegrad’ \cdot (x_i’ \BJ(\Bx’))}{\Abs{\Bx}} d^3 x’ \\ &= \frac{\Be_i}{\Abs{\Bx}} \int \spacegrad’ \cdot (x_i’ \BJ(\Bx’)) d^3 x’ \\ &= \frac{1}{\Abs{\Bx}} \oint \Bx’ (d\Ba \cdot \BJ(\Bx’)). \end{aligned} \end{equation}

When \BJ is localized, this is zero provided we pick the integration surface for the volume outside of that localization region.

It is now desired to rewrite \int \Bx \cdot \Bx’ \BJ as a triple cross product since the dot product of such a triple cross product has exactly this term in it

\begin{equation}\label{eqn:magneticMomentJackson:140} \begin{aligned} – \Bx \cross \int \Bx’ \cross \BJ &= \int (\Bx \cdot \Bx’) \BJ – \int (\Bx \cdot \BJ) \Bx’ \\ &= \int (\Bx \cdot \Bx’) \BJ – \Be_k x_i \int J_i x_k’, \end{aligned} \end{equation}

so
\begin{equation}\label{eqn:magneticMomentJackson:160} \int (\Bx \cdot \Bx’) \BJ = – \Bx \cross \int \Bx’ \cross \BJ + \Be_k x_i \int J_i x_k’. \end{equation}

To get of this second term, the next sneaky trick is to consider the following divergence

\begin{equation}\label{eqn:magneticMomentJackson:180} \begin{aligned} \oint d\Ba’ \cdot (\BJ(\Bx’) x_i’ x_j’) &= \int dV’ \spacegrad’ \cdot (\BJ(\Bx’) x_i’ x_j’) \\ &= \int dV’ (\spacegrad’ \cdot \BJ) + \int dV’ \BJ \cdot \spacegrad’ (x_i’ x_j’) \\ &= \int dV’ J_k \cdot \lr{ x_i’ \partial_k x_j’ + x_j’ \partial_k x_i’ } \\ &= \int dV’ \lr{J_k x_i’ \delta_{kj} + J_k x_j’ \delta_{ki}} \\ &= \int dV’ \lr{J_j x_i’ + J_i x_j’}. \end{aligned} \end{equation}

The surface integral is once again zero, which means that we have an antisymmetric relationship in integrals of the form

\begin{equation}\label{eqn:magneticMomentJackson:200} \int J_j x_i’ = -\int J_i x_j’. \end{equation}

Now we can use the tensor algebra trick of writing y = (y + y)/2 ,

\begin{equation}\label{eqn:magneticMomentJackson:220} \begin{aligned} \int (\Bx \cdot \Bx’) \BJ &= – \Bx \cross \int \Bx’ \cross \BJ + \Be_k x_i \int J_i x_k’ \\ &= – \Bx \cross \int \Bx’ \cross \BJ + \inv{2} \Be_k x_i \int \lr{ J_i x_k’ + J_i x_k’ } \\ &= – \Bx \cross \int \Bx’ \cross \BJ + \inv{2} \Be_k x_i \int \lr{ J_i x_k’ – J_k x_i’ } \\ &= – \Bx \cross \int \Bx’ \cross \BJ + \inv{2} \Be_k x_i \int (\BJ \cross \Bx’)_j \epsilon_{ikj} \\ &= – \Bx \cross \int \Bx’ \cross \BJ – \inv{2} \epsilon_{kij} \Be_k x_i \int (\BJ \cross \Bx’)_j \\ &= – \Bx \cross \int \Bx’ \cross \BJ – \inv{2} \Bx \cross \int \BJ \cross \Bx’ \\ &= – \Bx \cross \int \Bx’ \cross \BJ + \inv{2} \Bx \cross \int \Bx’ \cross \BJ \\ &= -\inv{2} \Bx \cross \int \Bx’ \cross \BJ, \end{aligned} \end{equation}

so

\begin{equation}\label{eqn:magneticMomentJackson:240} \BA(\Bx) \approx \frac{\mu_0}{4 \pi \Abs{\Bx}^3} \lr{ -\frac{\Bx}{2} } \int \Bx’ \cross \BJ(\Bx’) d^3 x’. \end{equation}

Letting

\begin{equation}\label{eqn:magneticMomentJackson:260} \boxed{ \Bm = \inv{2} \int \Bx’ \cross \BJ(\Bx’) d^3 x’, } \end{equation}

the far field approximation of the vector potential is
\begin{equation}\label{eqn:magneticMomentJackson:280} \boxed{ \BA(\Bx) = \frac{\mu_0}{4 \pi} \frac{\Bm \cross \Bx}{\Abs{\Bx}^3}. } \end{equation}

Note that when the current is restricted to an infintisimally thin loop, the magnetic moment reduces to

\begin{equation}\label{eqn:magneticMomentJackson:300} \Bm(\Bx) = \frac{I}{2} \int \Bx \cross d\Bl’. \end{equation}

Refering to [1] (pr. 1.60), this can be seen to be I times the “vector-area” integral.

References

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

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