Month: October 2016

Vector Area

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

[Click here for a PDF of this post with nicer formatting]

One of the results of this problem is required for a later one on magnetic moments that I’d like to do.

Question: Vector Area. ([1] pr. 1.61)

The integral

\begin{equation}\label{eqn:vectorAreaGriffiths:20}
\Ba = \int_S d\Ba,
\end{equation}

is sometimes called the vector area of the surface \( S \).

(a)

Find the vector area of a hemispherical bowl of radius \( R \).

(b)

Show that \( \Ba = 0 \) for any closed surface.

(c)

Show that \( \Ba \) is the same for all surfaces sharing the same boundary.

(d)

Show that
\begin{equation}\label{eqn:vectorAreaGriffiths:40}
\Ba = \inv{2} \oint \Br \cross d\Bl,
\end{equation}

where the integral is around the boundary line.

(e)

Show that
\begin{equation}\label{eqn:vectorAreaGriffiths:60}
\oint \lr{ \Bc \cdot \Br } d\Bl = \Ba \cross \Bc.
\end{equation}

Answer

(a)

\begin{equation}\label{eqn:vectorAreaGriffiths:80}
\begin{aligned}
\Ba
&=
\int_{0}^{\pi/2} R^2 \sin\theta d\theta \int_0^{2\pi} d\phi
\lr{ \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta } \\
&=
R^2 \int_{0}^{\pi/2} d\theta \int_0^{2\pi} d\phi
\lr{ \sin^2\theta \cos\phi, \sin^2\theta \sin\phi, \sin\theta\cos\theta } \\
&=
2 \pi R^2 \int_{0}^{\pi/2} d\theta \Be_3
\sin\theta\cos\theta \\
&=
\pi R^2
\Be_3
\int_{0}^{\pi/2} d\theta
\sin(2 \theta) \\
&=
\pi R^2
\Be_3
\evalrange{\lr{\frac{-\cos(2 \theta)}{2}}}{0}{\pi/2} \\
&=
\pi R^2
\Be_3
\lr{ 1 – (-1) }/2 \\
&=
\pi R^2
\Be_3.
\end{aligned}
\end{equation}

(b)

As hinted in the original problem description, this follows from

\begin{equation}\label{eqn:vectorAreaGriffiths:100}
\int dV \spacegrad T = \oint T d\Ba,
\end{equation}

simply by setting \( T = 1 \).

(c)

Suppose that two surfaces sharing a boundary are parameterized by vectors \( \Bx(u, v), \Bx(a,b) \) respectively. The area integral with the first parameterization is

\begin{equation}\label{eqn:vectorAreaGriffiths:120}
\begin{aligned}
\Ba
&= \int \PD{u}{\Bx} \cross \PD{v}{\Bx} du dv \\
&= \epsilon_{ijk} \Be_i \int \PD{u}{x_j} \PD{v}{x_k} du dv \\
&=
\epsilon_{ijk} \Be_i \int
\lr{
\PD{a}{x_j}
\PD{u}{a}
+
\PD{b}{x_j}
\PD{u}{b}
}
\lr{
\PD{a}{x_k}
\PD{v}{a}
+
\PD{b}{x_k}
\PD{v}{b}
}
du dv \\
&=
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{a}{x_j}
\PD{u}{a}
\PD{a}{x_k}
\PD{v}{a}
+
\PD{b}{x_j}
\PD{u}{b}
\PD{b}{x_k}
\PD{v}{b}
+
\PD{b}{x_j}
\PD{u}{b}
\PD{a}{x_k}
\PD{v}{a}
+
\PD{a}{x_j}
\PD{u}{a}
\PD{b}{x_k}
\PD{v}{b}
} \\
&=
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{a}{x_j}
\PD{a}{x_k}
\PD{u}{a}
\PD{v}{a}
+
\PD{b}{x_j}
\PD{b}{x_k}
\PD{u}{b}
\PD{v}{b}
}
+
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{b}{x_j}
\PD{a}{x_k}
\PD{u}{b}
\PD{v}{a}

\PD{a}{x_k}
\PD{b}{x_j}
\PD{u}{a}
\PD{v}{b}
}.
\end{aligned}
\end{equation}

In the last step a \( j,k \) index swap was performed for the last term of the second integral. The first integral is zero, since the integrand is symmetric in \( j,k \). This leaves
\begin{equation}\label{eqn:vectorAreaGriffiths:140}
\begin{aligned}
\Ba
&=
\epsilon_{ijk} \Be_i \int
du dv
\lr{
\PD{b}{x_j}
\PD{a}{x_k}
\PD{u}{b}
\PD{v}{a}

\PD{a}{x_k}
\PD{b}{x_j}
\PD{u}{a}
\PD{v}{b}
} \\
&=
\epsilon_{ijk} \Be_i \int
\PD{b}{x_j}
\PD{a}{x_k}
\lr{
\PD{u}{b}
\PD{v}{a}

\PD{u}{a}
\PD{v}{b}
}
du dv \\
&=
\epsilon_{ijk} \Be_i \int
\PD{b}{x_j}
\PD{a}{x_k}
\frac{\partial(b,a)}{\partial(u,v)} du dv \\
&=
-\int
\PD{b}{\Bx} \cross \PD{a}{\Bx} da db \\
&=
\int
\PD{a}{\Bx} \cross \PD{b}{\Bx} da db.
\end{aligned}
\end{equation}

However, this is the area integral with the second parameterization, proving that the area-integral for any given boundary is independant of the surface.

(d)

Having proven that the area-integral for a given boundary is independent of the surface that it is evaluated on, the result follows by illustration as hinted in the full problem description. Draw a “cone”, tracing a vector \( \Bx’ \) from the origin to the position line element, and divide that cone up into infinitesimal slices as sketched in fig. 1.

conevectorareafig1

Fig 1. Cone subtended by loop

The area of each of these triangular slices is

\begin{equation}\label{eqn:vectorAreaGriffiths:160}
\inv{2} \Bx’ \cross d\Bl’.
\end{equation}

Summing those triangles proves the result.

(e)

As hinted in the problem, this follows from

\begin{equation}\label{eqn:vectorAreaGriffiths:180}
\int \spacegrad T \cross d\Ba = -\oint T d\Bl.
\end{equation}

Set \( T = \Bc \cdot \Br \), for which

\begin{equation}\label{eqn:vectorAreaGriffiths:240}
\begin{aligned}
\spacegrad T
&= \Be_k \partial_k c_m x_m \\
&= \Be_k c_m \delta_{km} \\
&= \Be_k c_k \\
&= \Bc,
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:vectorAreaGriffiths:200}
\begin{aligned}
(\spacegrad T) \cross d\Ba
&=
\int \Bc \cross d\Ba \\
&=
\Bc \cross \int d\Ba \\
&=
\Bc \cross \Ba.
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:vectorAreaGriffiths:220}
\Bc \cross \Ba = -\oint (\Bc \cdot \Br) d\Bl,
\end{equation}

or
\begin{equation}\label{eqn:vectorAreaGriffiths:260}
\oint (\Bc \cdot \Br) d\Bl
=
\Ba \cross \Bc.
\end{equation}

References

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

Magnetic moment for a localized magnetostatic current

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

[Click here for a PDF of this post with nicer formatting]

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.

Corollaries to Stokes and Divergence theorems

October 12, 2016 math and physics play , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

In [1] a few problems are set to prove some variations of Stokes theorem. He gives some cool tricks to prove each one using just the classic 3D Stokes and divergence theorems. We can also do them directly from the more general Stokes theorem \( \int d^k \Bx \cdot (\spacegrad \wedge F) = \oint d^{k-1} \Bx \cdot F \).

Question: Stokes theorem on scalar function. ([1] pr. 1.60a)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:20}
\int \spacegrad T dV = \oint T d\Ba.
\end{equation}

Answer

The direct way to prove this is to apply Stokes theorem

\begin{equation}\label{eqn:stokesCorollariesGriffiths:80}
\int d^3 \Bx \cdot (\spacegrad \wedge T) = \oint d^2 \Bx \cdot T
\end{equation}

Here \( d^3 \Bx = d\Bx_1 \wedge d\Bx_2 \wedge d\Bx_3 \), a pseudoscalar (trivector) volume element, and the wedge and dot products take their most general meanings. For \(k\)-blade \( F \), and \(k’\)-blade \( F’ \), that is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:100}
\begin{aligned}
F \wedge F’ &= \gpgrade{F F’}{k+k’} \\
F \cdot F’ &= \gpgrade{F F’}{\Abs{k-k’}}
\end{aligned}
\end{equation}

With \( d^3\Bx = I dV \), and \( d^2 \Bx = I \ncap dA = I d\Ba \), we have

\begin{equation}\label{eqn:stokesCorollariesGriffiths:120}
\int I dV \spacegrad T = \oint I d\Ba T.
\end{equation}

Cancelling the factors of \( I \) proves the result.

Griffith’s trick to do this was to let \( \Bv = \Bc T \), where \( \Bc \) is a constant. For this, the divergence theorem integral is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:160}
\begin{aligned}
\int dV \spacegrad \cdot (\Bc T)
&=
\int dV \Bc \cdot \spacegrad T \\
&=
\Bc \cdot \int dV \spacegrad T \\
&=
\oint d\Ba \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Ba T.
\end{aligned}
\end{equation}

This is true for any constant \( \Bc \), so is also true for the unit vectors. This allows for summing projections in each of the unit directions

\begin{equation}\label{eqn:stokesCorollariesGriffiths:180}
\begin{aligned}
\int dV \spacegrad T
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad T } \\
&=
\sum \Be_k \lr{ \Be_k \cdot \oint d\Ba T } \\
&=
\oint d\Ba T.
\end{aligned}
\end{equation}

Question: ([1] pr. 1.60b)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:40}
\int \spacegrad \cross \Bv dV = -\oint \Bv \cross d\Ba.
\end{equation}

Answer

This also follows directly from the general Stokes theorem

\begin{equation}\label{eqn:stokesCorollariesGriffiths:200}
\int d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv } = \oint d^2 \Bx \cdot \Bv
\end{equation}

The volume integrand is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:220}
\begin{aligned}
d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv }
&=
\gpgradeone{ I dV I \spacegrad \cross \Bv } \\
&=
-dV \spacegrad \cross \Bv,
\end{aligned}
\end{equation}

and the surface integrand is
\begin{equation}\label{eqn:stokesCorollariesGriffiths:240}
\begin{aligned}
d^2 \Bx \cdot \Bv
&=
\gpgradeone{ I d\Ba \Bv } \\
&=
\gpgradeone{ I (d\Ba \wedge \Bv) } \\
&=
I^2 (d\Ba \cross \Bv) \\
&=
-d\Ba \cross \Bv \\
&=
\Bv \cross d\Ba.
\end{aligned}
\end{equation}

Plugging these into \ref{eqn:stokesCorollariesGriffiths:200} proves the result.

Griffiths trick for the same is to apply the divergence theorem to \( \Bv \cross \Bc \). Such a volume integral is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:260}
\begin{aligned}
\int dV \spacegrad \cdot (\Bv \cross \Bc)
&=
\int dV \Bc \cdot (\spacegrad \cross \Bv) \\
&=
\Bc \cdot \int dV \spacegrad \cross \Bv.
\end{aligned}
\end{equation}

This must equal
\begin{equation}\label{eqn:stokesCorollariesGriffiths:280}
\begin{aligned}
\oint d\Ba \cdot (\Bv \cross \Bc)
&=
\Bc \cdot \oint d\Ba \cross \Bv \\
&=
-\Bc \cdot \oint \Bv \cross d\Ba
\end{aligned}
\end{equation}

Again, assembling projections, we have
\begin{equation}\label{eqn:stokesCorollariesGriffiths:300}
\begin{aligned}
\int dV \spacegrad \cross \Bv
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad \cross \Bv } \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint \Bv \cross d\Ba } \\
&=
-\oint \Bv \cross d\Ba.
\end{aligned}
\end{equation}

Question: ([1] pr. 1.60e)

Prove
\begin{equation}\label{eqn:stokesCorollariesGriffiths:60}
\int \spacegrad T \cross d\Ba = -\oint T d\Bl.
\end{equation}

Answer

This one follows from
\begin{equation}\label{eqn:stokesCorollariesGriffiths:320}
\int d^2 \Bx \cdot \lr{ \spacegrad \wedge T } = \oint d^1 \Bx \cdot T.
\end{equation}

The surface integrand can be written
\begin{equation}\label{eqn:stokesCorollariesGriffiths:340}
\begin{aligned}
d^2 \Bx \cdot \lr{ \spacegrad \wedge T }
&=
\gpgradeone{ I d\Ba \spacegrad T } \\
&=
I (d\Ba \wedge \spacegrad T ) \\
&=
I^2 ( d\Ba \cross \spacegrad T ) \\
&=
-d\Ba \cross \spacegrad T.
\end{aligned}
\end{equation}

The line integrand is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:360}
d^1 \Bx \cdot T = d^1 \Bx T.
\end{equation}

Given a two parameter representation of the surface area element \( d^2 \Bx = d\Bx_1 \wedge d\Bx_2 \), the line element representation is
\begin{equation}\label{eqn:stokesCorollariesGriffiths:380}
\begin{aligned}
d^1 \Bx
&= (\Bx_1 \wedge d\Bx_2) \cdot \Bx^1 + (d\Bx_1 \wedge \Bx_2) \cdot \Bx^2 \\
&= -d\Bx_2 + d\Bx_1,
\end{aligned}
\end{equation}

giving

\begin{equation}\label{eqn:stokesCorollariesGriffiths:400}
\begin{aligned}
-\int d\Ba \cross \spacegrad T
&=
\int
-\evalbar{\lr{ \PD{u_2}{\Bx} T }}{\Delta u_1} du_2
+\evalbar{\lr{ \PD{u_1}{\Bx} T }}{\Delta u_2} du_1 \\
&=
-\oint d\Bl T,
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:stokesCorollariesGriffiths:420}
\int \spacegrad T \cross d\Ba
=
-\oint d\Bl T.
\end{equation}

Griffiths trick for the same is to use \( \Bv = \Bc T \) for constant \( \Bc \) in (the usual 3D) Stokes’ theorem. That is

\begin{equation}\label{eqn:stokesCorollariesGriffiths:440}
\begin{aligned}
\int d\Ba \cdot (\spacegrad \cross (\Bc T))
&=
\Bc \cdot \int d\Ba \cross \spacegrad T \\
&=
-\Bc \cdot \int \spacegrad T \cross d\Ba \\
&=
\oint d\Bl \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Bl T.
\end{aligned}
\end{equation}

Again assembling projections we have
\begin{equation}\label{eqn:stokesCorollariesGriffiths:460}
\begin{aligned}
\int \spacegrad T \cross d\Ba
&=
\sum \Be_k \lr{ \Be_k \cdot \int \spacegrad T \cross d\Ba} \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint d\Bl T } \\
&=
-\oint d\Bl T.
\end{aligned}
\end{equation}

References

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

Jackson’s electrostatic self energy analysis

October 10, 2016 math and physics play , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

Motivation

I was reading my Jackson [1], which characteristically had the statement “the […] integral can easily be shown to have the value \( 4 \pi \)”, in a discussion of electrostatic energy and self energy. After a few attempts and a couple of pages of calculations, I figured out how this can be easily shown.

Context

Let me walk through the context that leads to the “easy” integral, and then the evaluation of that integral. Unlike my older copy of Jackson, I’ll do this in SI units.

The starting point is a statement that the work done (potential energy) of one charge \( q_i \) in a set of \( n \) charges, where that charge is brought to its position \( \Bx_i \) from infinity, is

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:20}
W_i = q_i \Phi(\Bx_i),
\end{equation}

where the potential energy due to the rest of the charge configuration is

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:40}
\Phi(\Bx_i) = \inv{4 \pi \epsilon} \sum_{i \ne j} \frac{q_j}{\Abs{\Bx_i – \Bx_j}}.
\end{equation}

This means that the total potential energy, making sure not to double count, to move all the charges in from infinity is

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:60}
W = \inv{4 \pi \epsilon} \sum_{1 \le i < j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}. \end{equation} This sum over all unique pairs is somewhat unwieldy, so it can be adjusted by explicitly double counting with a corresponding divide by two \begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:80} W = \inv{2} \inv{4 \pi \epsilon} \sum_{1 \le i \ne j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}. \end{equation} The point that causes the trouble later is the continuum equivalent to this relationship, which is \begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:100} W = \inv{8 \pi \epsilon} \int \frac{\rho(\Bx) \rho(\Bx')}{\Abs{\Bx - \Bx'}} d^3 \Bx d^3 \Bx', \end{equation} or \begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:120} W = \inv{2} \int \rho(\Bx) \Phi(\Bx) d^3 \Bx. \end{equation} There's a subtlety here that is often passed over. When the charge densities represent point charges \( \rho(\Bx) = q \delta^3(\Bx - \Bx') \) are located at, notice that this integral equivalent is evaluated over all space, including the spaces that the charges that the charges are located at. Ignoring that subtlety, this potential energy can be expressed in terms of the electric field, and then integrated by parts \begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:140} \begin{aligned} W &= \inv{2 } \int (\spacegrad \cdot (\epsilon \BE)) \Phi(\Bx) d^3 \Bx \\ &= \frac{\epsilon}{2 } \int \lr{ \spacegrad \cdot (\BE \Phi) - (\spacegrad \Phi) \cdot \BE } d^3 \Bx \\ &= \frac{\epsilon}{2 } \oint dA \ncap \cdot (\BE \Phi) + \frac{\epsilon}{2 } \int \BE \cdot \BE d^3 \Bx. \end{aligned} \end{equation} The presumption is that \( \BE \Phi \) falls off as the bounds of the integration volume tends to infinity. That leaves us with an energy density proportional to the square of the field \begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:160} w = \frac{\epsilon}{2 } \BE^2. \end{equation}

Inconsistency

It’s here that Jackson points out the inconsistency between \ref{eqn:electrostaticJacksonSelfEnergy:160} and the original
discrete analogue \ref{eqn:electrostaticJacksonSelfEnergy:80} that this was based on. The energy density is positive definite, whereas the discrete potential energy can be negative if there is a difference in the sign of the charges.

Here Jackson uses a two particle charge distribution to help resolve this conundrum. For a superposition \( \BE = \BE_1 + \BE_2 \), we have

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:180}
\BE
=
\inv{4 \pi \epsilon} \frac{q_1 (\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3}
+ \inv{4 \pi \epsilon} \frac{q_2 (\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3},
\end{equation}

so the energy density is
\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:200}
w =
\frac{1}{32 \pi^2 \epsilon} \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+
2 \frac{q_1 q_2}{32 \pi^2 \epsilon}
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3}.
\end{equation}

The discrete potential had only an interaction energy, whereas the potential from this squared field has an interaction energy plus two self energy terms. Those two strictly positive self energy terms are what forces this field energy positive, independent of the sign of the interaction energy density. Jackson makes a change of variables of the form

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:220}
\begin{aligned}
\Brho &= (\Bx – \Bx_1)/R \\
R &= \Abs{\Bx_1 – \Bx_2} \\
\ncap &= (\Bx_1 – \Bx_2)/R,
\end{aligned}
\end{equation}

for which we find

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:240}
\Bx = \Bx_1 + R \Brho,
\end{equation}

so
\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:260}
\Bx – \Bx_2 =
\Bx_1 – \Bx_2 + R \Brho
R (\ncap + \Brho),
\end{equation}

and
\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:280}
d^3 \Bx = R^3 d^3 \Brho,
\end{equation}

so the total interaction energy is
\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:300}
\begin{aligned}
W_{\textrm{int}}
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int d^3 \Bx
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int R^3 d^3 \Brho
\frac{ R \Brho }{ R^3 \Abs{\Brho}^3 } \cdot
\frac{R (\ncap + \Brho)}{R^3 \Abs{\ncap + \Brho}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon R}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}.
\end{aligned}
\end{equation}

Evaluating this integral is what Jackson calls easy. The technique required is to express the integrand in terms of gradients in the \( \Brho \) coordinate system

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:320}
\begin{aligned}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
&=
\int d^3 \Brho
\lr{ – \spacegrad_\Brho \inv{\Abs{\Brho}} }
\cdot
\lr{ – \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} } \\
&=
\int d^3 \Brho
\lr{ \spacegrad_\Brho \inv{\Abs{\Brho}} }
\cdot
\lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }.
\end{aligned}
\end{equation}

I found it somewhat non-trivial to find the exact form of the chain rule that is required to simplify this integral, but after some trial and error, figured it out by working backwards from
\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:340}
\spacegrad_\Brho^2 \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
=
\spacegrad_\Brho \cdot \lr{ \inv{\Abs{\Brho}} \spacegrad_\Brho \inv{ \Abs{\ncap + \Brho} } }
+
\spacegrad_\Brho \cdot \lr{ \inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho \inv{ \Abs{\Brho} } }.
\end{equation}

In integral form this is
\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:360}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\int d^3 \Brho’
\spacegrad_{\Brho’} \cdot \lr{ \inv{\Abs{\Brho’ – \ncap}} \spacegrad_{\Brho’} \inv{ \Abs{\Brho’} } }
+
\int d^3 \Brho
\spacegrad_\Brho \cdot \lr{ \inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho \inv{ \Abs{\Brho} } } \\
&=
\int d^3 \Brho’
\lr{ \spacegrad_{\Brho’} \inv{\Abs{\Brho’ – \ncap} } \cdot \spacegrad_{\Brho’} \inv{ \Abs{\Brho’} } }
+
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \spacegrad_{\Brho’}^2 \inv{ \Abs{\Brho’} } \\
&+
\int d^3 \Brho
\lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}}} \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} }
+
\int d^3 \Brho
\inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho^2 \inv{ \Abs{\Brho} } \\
&=
2 \int d^3 \Brho
\lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}}} \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} } \\
&- 4 \pi
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \delta^3(\Brho’)
– 4 \pi
\int d^3 \Brho
\inv{\Abs{\Brho + \ncap}} \delta^3(\Brho) \\
&=
2 \int d^3 \Brho
\lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}}} \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} }
– 8 \pi.
\end{aligned}
\end{equation}

This used the Laplacian representation of the delta function \( \delta^3(\Bx) = -(1/4\pi) \spacegrad^2 (1/\Abs{\Bx}) \). Back-substitution gives

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:380}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi
+
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}.
\end{equation}

We can argue that this last integral tends to zero, since

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:400}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\oint dA’ \ncap’ \cdot \lr{
\lr{ \spacegrad_\Brho \inv{ \Abs{\Brho}} } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }
} \\
&=
-\oint dA’ \ncap’ \cdot \lr{
\frac{ \Brho } {\inv{ \Abs{\Brho}}^3 } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \frac{ (\Brho + \ncap) }{ \Abs{\ncap + \Brho}^3 }
} \\
&=
-\oint dA’ \inv{\Abs{\Brho} \Abs{\Brho + \ncap}}
\lr{
\frac{ \ncap’ \cdot \Brho }{
{\Abs{\Brho}}^2 }
+\frac{ \ncap’ \cdot (\Brho + \ncap) }{
{\Abs{\Brho + \ncap}}^2 }
}.
\end{aligned}
\end{equation}

The integrand in this surface integral is of \( O(1/\rho^3) \) so tends to zero on an infinite surface in the \( \Brho \) coordinate system. This completes the “easy” integral, leaving

\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:420}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi.
\end{equation}

The total field energy can now be expressed as a sum of the self energies and the interaction energy
\begin{equation}\label{eqn:electrostaticJacksonSelfEnergy:440}
W =
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+ \inv{ 4 \pi \epsilon}
\frac{q_1 q_2}{\Abs{\Bx_1 – \Bx_2} }.
\end{equation}

The interaction energy is exactly the potential energies for the two particles, the this total energy in the field is biased in the positive direction by the pair of self energies. It is interesting that the energy obtained from integrating the field energy density contains such self energy terms, but I don’t know exactly what to make of them at this point in time.

References

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

Electric and magnetic fields at an interface

October 9, 2016 math and physics play , , ,

[Click here for a PDF of this post with nicer formatting]

As pointed out in [1] the fields at an interface that is not a perfect conductor on either side are related by

\begin{equation}\label{eqn:fieldsAtInterface:20}
\begin{aligned}
\ncap \cdot \lr{ \BD_2 – \BD_1 } &= \rho_{es} \\
\ncap \cross \lr{ \BE_2 – \BE_1 } &= -\BM_s \\
\ncap \cdot \lr{ \BB_2 – \BB_1 } &= \rho_{ms} \\
\ncap \cross \lr{ \BH_2 – \BH_1 } &= \BJ_s.
\end{aligned}
\end{equation}

Given the fields in medium 1, assuming that boths sets of media are linear, we can use these relationships to determine the fields in the other medium.

\begin{equation}\label{eqn:fieldsAtInterface:40}
\begin{aligned}
\ncap \cdot \BE_2 &= \inv{\epsilon_2} \lr{ \epsilon_1 \ncap \cdot \BE_1 + \rho_{es} } \\
\ncap \wedge \BE_2 &= \ncap \wedge \BE_1 -I \BM_s \\
\ncap \cdot \BB_2 &= \ncap \cdot \BB_1 + \rho_{ms} \\
\ncap \wedge \BB_2 &= \mu_2 \lr{ \inv{\mu_1} \ncap \wedge \BB_1 + I \BJ_s}.
\end{aligned}
\end{equation}

Now the fields in interface 2 can be obtained by adding the normal and tangential projections. For the electric field

\begin{equation}\label{eqn:fieldsAtInterface:60}
\begin{aligned}
\BE_2
&=
\ncap (\ncap \cdot \BE_2 )
+ \ncap \cdot (\ncap \wedge \BE_2) \\
&=
\inv{\epsilon_2} \ncap \lr{ \epsilon_1 \ncap \cdot \BE_1 + \rho_{es} }
+
\ncap \cdot (\ncap \wedge \BE_1 -I \BM_s).
\end{aligned}
\end{equation}

Note that this manipulation can also be done without Geometric Algebra by writing \( \BE_2 = \ncap (\ncap \cdot \BE_2 ) – \ncap \cross (\ncap \cross \BE_2) \)).
Expanding \( \ncap \cdot (\ncap \wedge \BE_1) = \BE_1 – \ncap (\ncap \cdot \BE_1) \), and \( \ncap \cdot (I \BM_s) = -\ncap \cross \BM_s \), that is

\begin{equation}\label{eqn:fieldsAtInterface:80}
\boxed{
\BE_2
=
\BE_1
+ \ncap (\ncap \cdot \BE_1) \lr{ \frac{\epsilon_1}{\epsilon_2} – 1 }
+ \frac{\rho_{es}}{\epsilon_2}
+ \ncap \cross \BM_s.
}
\end{equation}

For the magnetic field

\begin{equation}\label{eqn:fieldsAtInterface:100}
\begin{aligned}
\BB_2
&=
\ncap (\ncap \cdot \BB_2 )
+
\ncap \cdot (\ncap \wedge \BB_2) \\
&=
\ncap \lr{ \ncap \cdot \BB_1 + \rho_{ms} }
+
\mu_2 \ncap \cdot \lr{ \lr{ \inv{\mu_1} \ncap \wedge \BB_1 + I \BJ_s} },
\end{aligned}
\end{equation}

which is

\begin{equation}\label{eqn:fieldsAtInterface:120}
\boxed{
\BB_2
=
\frac{\mu_2}{\mu_1} \BB_1
+
\ncap (\ncap \cdot \BB_1) \lr{ 1 – \frac{\mu_2}{\mu_1} }
+ \ncap \rho_{ms}
– \ncap \cross \BJ_s.
}
\end{equation}

These are kind of pretty results, having none of the explicit angle dependence that we see in the Fresnel relationships. In this analysis, it is assumed there is only a transmitted component of the ray in question, and no reflected component. Can we do a purely vectoral treatment of the Fresnel equations along these same lines?

References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

%d bloggers like this: