Derivative of a delta function

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In the geometric algebra formulation of Maxwell’s equation (singular in GA), the Green’s function for the spacetime derivative ends up with terms like

\label{eqn:derivativeOfDeltaFunction:20}
\frac{d}{dr} \delta( -r/c + t – t’ ),

or
\label{eqn:derivativeOfDeltaFunction:40}
\frac{d}{dt} \delta( -r/c + t – t’ ),

where $$t’$$ is the integration variable of the test function that the delta function will be applied to. If these were derivatives with respect to the integration variable, then we could use the well known formula

\label{eqn:derivativeOfDeltaFunction:60}
\int_{-\infty}^\infty
\lr{ \frac{d}{dt’} \delta(t’) } \phi(t’) = -\phi'(0),

which follows by chain rule, and an assumption that $$\phi(t’)$$ is well behaved at the points at infinity. It’s not obvious to me that this can be applied to either of our delta function derivatives.

Let’s go back to square one, and figure out the meaning of these delta functions by their action on a test function. We wish to compute

\label{eqn:derivativeOfDeltaFunction:80}
\int_{-\infty}^\infty \frac{d}{du} \delta( a u + b – t’ ) f(t’) dt’.

Let’s start with a change of variables $$t” = a u + b – t’$$, for which we find

\label{eqn:derivativeOfDeltaFunction:100}
\begin{aligned}
t’ &= a u + b – t” \\
dt” &= – dt’ \\
\frac{d}{du} &= \frac{dt”}{du} \frac{d}{dt”} = a \frac{d}{dt”}.
\end{aligned}

Back substitution gives

\label{eqn:derivativeOfDeltaFunction:120}
\begin{aligned}
a \int_{\infty}^{-\infty} \lr{ \frac{d}{dt”} \delta( t” ) } f( a u + b – t” ) (-dt”)
&=
a \int_{-\infty}^{\infty} \lr{ \frac{d}{dt”} \delta( t” ) } f( a u + b – t” ) dt” \\
&=
\evalrange{a \delta(t”) f( a u + b – t”)}{-\infty}{\infty}

a \int_{-\infty}^{\infty} \delta( t” ) \frac{d}{dt”} f( a u + b – t” ) dt” \\
&=
– \evalbar{ a \frac{d}{dt”} f( a u + b – t” ) }{t” = 0} \\
&=
\evalbar{ a \frac{d}{ds} f( s ) }{s = a u + b}.
\end{aligned}

This shows that the action of the derivative of the delta function (with respect to a non-integration variable parameter $$u$$) is

\label{eqn:derivativeOfDeltaFunction:160}
\boxed{
\frac{d}{du} \delta( a u + b – t’ )
=
\evalbar{a \frac{d}{ds}}{s = a u + b}.
}

Line charge field and potential.

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When computing the most general solution of the electrostatic potential in a plane, Jackson [1] mentions that $$-2 \lambda_0 \ln \rho$$ is the well known potential for an infinite line charge (up to the unit specific factor). Checking that statement, since I didn’t recall what that potential was offhand, I encountered some inconsistencies and non-convergent integrals, and thought it was worthwhile to explore those a bit more carefully. This will be done here.

Using Gauss’s law.

For an infinite length line charge, we can find the radial field contribution using Gauss’s law, imagining a cylinder of length $$\Delta l$$ of radius $$\rho$$ surrounding this charge with the midpoint at the origin. Ignoring any non-radial field contribution, we have

\label{eqn:lineCharge:20}
\int_{-\Delta l/2}^{\Delta l/2} \ncap \cdot \BE (2 \pi \rho) dl = \frac{\lambda_0}{\epsilon_0} \Delta l,

or

\label{eqn:lineCharge:40}
\BE = \frac{\lambda_0}{2 \pi \epsilon_0} \frac{\rhocap}{\rho}.

Since

\label{eqn:lineCharge:60}
\frac{\rhocap}{\rho} = \spacegrad \ln \rho,

this means that the potential is

\label{eqn:lineCharge:80}
\phi = -\frac{2 \lambda_0}{4 \pi \epsilon_0} \ln \rho.

Finite line charge potential.

Let’s try both these calculations for a finite charge distribution. Gauss’s law looses its usefulness, but we can evaluate the integrals directly. For the electric field

\label{eqn:lineCharge:100}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int \frac{(\Bx – \Bx’)}{\Abs{\Bx – \Bx’}^3} dl’.

Using cylindrical coordinates with the field point $$\Bx = \rho \rhocap$$ for convience, the charge point $$\Bx’ = z’ \zcap$$, and a the charge distributed over $$[a,b]$$ this is

\label{eqn:lineCharge:120}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{(\rho \rhocap – z’ \zcap)}{\lr{\rho^2 + (z’)^2}^{3/2}} dz’.

When the charge is uniformly distributed around the origin $$[a,b] = b[-1,1]$$ the $$\zcap$$ component of this field is killed because the integrand is odd. This justifies ignoring such contributions in the Gaussing cylinder analysis above. The general solution to this integral is found to be

\label{eqn:lineCharge:140}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\evalrange{
\lr{
\frac{z’ \rhocap }{\rho \sqrt{ \rho^2 + (z’)^2 } }
+\frac{\zcap}{ \sqrt{ \rho^2 + (z’)^2 } }
}
}{a}{b},

or
\label{eqn:lineCharge:240}
\boxed{
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\lr{
\frac{\rhocap }{\rho}
\lr{
\frac{b}{\sqrt{ \rho^2 + b^2 } }
-\frac{a}{\sqrt{ \rho^2 + a^2 } }
}
+ \zcap
\lr{
\frac{1}{ \sqrt{ \rho^2 + b^2 } }
-\frac{1}{ \sqrt{ \rho^2 + a^2 } }
}
}.
}

When $$b = -a = \Delta l/2$$, this reduces to

\label{eqn:lineCharge:160}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\frac{\rhocap }{\rho}
\frac{\Delta l}{\sqrt{ \rho^2 + (\Delta l/2)^2 } },

which further reduces to \ref{eqn:lineCharge:40} when $$\Delta l \gg \rho$$.

Finite line charge potential. Wrong but illuminating.

Again, putting the field point at $$z’ = 0$$, we have

\label{eqn:lineCharge:180}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{dz’}{\lr{\rho^2 + (z’)^2}^{1/2}},

which integrates to
\label{eqn:lineCharge:260}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}.

With $$b = -a = \Delta l/2$$, this approaches

\label{eqn:lineCharge:200}
\phi
\approx
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ (\Delta l/2) }{ \rho^2/2\Abs{\Delta l/2}}
=
\frac{-2 \lambda_0}{4 \pi \epsilon_0 } \ln \rho
+
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \lr{ (\Delta l)^2/2 }.

Before $$\Delta l$$ is allowed to tend to infinity, this is identical (up to a difference in the reference potential) to \ref{eqn:lineCharge:80} found using Gauss’s law. It is, strictly speaking, singular when $$\Delta l \rightarrow \infty$$, so it does not seem right to infinity as a reference point for the potential.

There’s another weird thing about this result. Since this has no $$z$$ dependence, it is not obvious how we would recover the non-radial portion of the electric field from this potential using $$\BE = -\spacegrad \phi$$? Let’s calculate the elecric field from \ref{eqn:lineCharge:180} explicitly

\label{eqn:lineCharge:220}
\begin{aligned}
\BE
&=
-\frac{\lambda_0}{4 \pi \epsilon_0}
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 }
\PD{\rho}{}
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0}
\lr{
\inv{ b + \sqrt{ \rho^2 + b^2 }} \frac{ \rho }{\sqrt{ \rho^2 + b^2 }}
-\inv{ a + \sqrt{ \rho^2 + a^2 }} \frac{ \rho }{\sqrt{ \rho^2 + a^2 }}
} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ -b + \sqrt{ \rho^2 + b^2 }}{\sqrt{ \rho^2 + b^2 }}
-\frac{ -a + \sqrt{ \rho^2 + a^2 }}{\sqrt{ \rho^2 + a^2 }}
} \\
&=
\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ b }{\sqrt{ \rho^2 + b^2 }}
-\frac{ a }{\sqrt{ \rho^2 + a^2 }}
}.
\end{aligned}

This recovers the radial component of the field from \ref{eqn:lineCharge:240}, but where did the $$\zcap$$ component go? The required potential appears to be

\label{eqn:lineCharge:340}
\phi(\rho, z)
=
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}

\frac{z \lambda_0}{4 \pi \epsilon_0 }
\lr{ \frac{1}{\sqrt{\rho^2 + b^2}}
-\frac{1}{\sqrt{\rho^2 + a^2}}
}.

When computing the electric field $$\BE(\rho, \theta, z)$$, it was convienent to pick the coordinate system so that $$z = 0$$. Doing this with the potential gives the wrong answers. The reason for this appears to be that this kills the potential term that is linear in $$z$$ before taking its gradient, and we need that term to have the $$\zcap$$ field component that is expected for a charge distribution that is non-symmetric about the origin on the z-axis!

Finite line charge potential. Take II.

Let the point at which the potential is evaluated be

\label{eqn:lineCharge:360}
\Bx = \rho \rhocap + z \zcap,

and the charge point be
\label{eqn:lineCharge:380}
\Bx’ = z’ \zcap.

This gives

\label{eqn:lineCharge:400}
\begin{aligned}
\phi(\rho, z)
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_a^b \frac{dz’}{\Abs{\rho^2 + (z – z’)^2 }} \\
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_{a-z}^{b-z} \frac{du}{ \Abs{\rho^2 + u^2} } \\
&= \frac{\lambda_0}{4\pi \epsilon_0}
\evalrange{\ln \lr{ u + \sqrt{ \rho^2 + u^2 }}}{b-z}{a-z} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\ln \frac
{ b-z + \sqrt{ \rho^2 + (b-z)^2 }}
{ a-z + \sqrt{ \rho^2 + (a-z)^2 }}.
\end{aligned}

The limit of this potential $$a = -\Delta/2 \rightarrow -\infty, b = \Delta/2 \rightarrow \infty$$ doesn’t exist in any strict sense. If we are cavilier about the limits, as in \ref{eqn:lineCharge:200}, this can be evaluated as

\label{eqn:lineCharge:n}
\phi \approx
\frac{\lambda_0}{4\pi \epsilon_0} \lr{ -2 \ln \rho + \textrm{constant} }.

however, the constant ($$\ln \Delta^2/2$$) is infinite, so there isn’t really a good justification for using that constant as the potential reference point directly.

It seems that the “right” way to calculate the potential for the infinite distribution, is to

• Calculate the field from the potential.
• Take the PV limit of that field with the charge distribution extending to infinity.
• Compute the corresponding potential from this limiting value of the field.

Doing that doesn’t blow up. That field calculation, for the finite case, should include a $$\zcap$$ component. To verify, let’s take the respective derivatives

\label{eqn:lineCharge:420}
\begin{aligned}
-\PD{z}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ -1 + \frac{z – b}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ -1 + \frac{z – a}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ 1 + \frac{b – z}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ 1 + \frac{a – z}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
},
\end{aligned}

and

\label{eqn:lineCharge:440}
\begin{aligned}
-\PD{\rho}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ \frac{\rho}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ \frac{\rho}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rho \lr{
-(b-z) + \sqrt{ \rho^2 + (b-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (b-z)^2 } }

\frac{\rho \lr{
-(a-z) + \sqrt{ \rho^2 + (a-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (a-z)^2 } }
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0 \rho}
\lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
.
\end{aligned}

Putting the pieces together, the electric field is
\label{eqn:lineCharge:460}
\BE =
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rhocap}{\rho} \lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
+
\zcap \lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
}
}.

For has a PV limit of \ref{eqn:lineCharge:40} at $$z = 0$$, and also for the finite case, has the $$\zcap$$ field component that was obtained when the field was obtained by direct integration.

Conclusions

• We have to evaluate the potential at all points in space, not just on the axis that we evaluate the field on (should we choose to do so).
• In this case, we found that it was not directly meaningful to take the limit of a potential distribution. We can, however, compute the field from a potential for a finite charge distribution,
take the limit of that field, and then calculate the corresponding potential for the infinite distribution.

Is there a more robust mechanism that can be used to directly calculate the potential for an infinite charge distribution, instead of calculating the potential from the field of such an infinite distribution?

I think that were things go wrong is that the integral of \ref{eqn:lineCharge:180} does not apply to charge distributions that are not finite on the infinite range $$z \in [-\infty, \infty]$$. That solution was obtained by utilizing an all-space Green’s function, and the boundary term in that Green’s analysis was assumed to tend to zero. That isn’t the case when the charge distribution is $$\lambda_0 \delta( z )$$.

References

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

Jackson’s electrostatic self energy analysis

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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

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

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

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

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

\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}}. This sum over all unique pairs is somewhat unwieldy, so it can be adjusted by explicitly double counting with a corresponding divide by two $$\label{eqn:electrostaticJacksonSelfEnergy:80} W = \inv{2} \inv{4 \pi \epsilon} \sum_{1 \le i \ne j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}.$$ The point that causes the trouble later is the continuum equivalent to this relationship, which is $$\label{eqn:electrostaticJacksonSelfEnergy:100} W = \inv{8 \pi \epsilon} \int \frac{\rho(\Bx) \rho(\Bx')}{\Abs{\Bx - \Bx'}} d^3 \Bx d^3 \Bx',$$ or $$\label{eqn:electrostaticJacksonSelfEnergy:120} W = \inv{2} \int \rho(\Bx) \Phi(\Bx) d^3 \Bx.$$ 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 \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} 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 $$\label{eqn:electrostaticJacksonSelfEnergy:160} w = \frac{\epsilon}{2 } \BE^2.$$

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

\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},

so the energy density is
\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}.

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

\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}

for which we find

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

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

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

so the total interaction energy is
\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}

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

\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}

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
\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} } }.

In integral form this is
\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}

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

\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}}.

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

\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}

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

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

The total field energy can now be expressed as a sum of the self energies and the interaction energy
\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} }.

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.

Helmholtz theorem

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This is a problem from ece1228. I attempted solutions in a number of ways. One using Geometric Algebra, one devoid of that algebra, and then this method, which combined aspects of both. Of the three methods I tried to obtain this result, this is the most compact and elegant. It does however, require a fair bit of Geometric Algebra knowledge, including the Fundamental Theorem of Geometric Calculus, as detailed in [1], [3] and [2].

Question: Helmholtz theorem

Prove the first Helmholtz’s theorem, i.e. if vector $$\BM$$ is defined by its divergence

\label{eqn:helmholtzDerviationMultivector:20}
\spacegrad \cdot \BM = s

and its curl
\label{eqn:helmholtzDerviationMultivector:40}
\spacegrad \cross \BM = \BC

within a region and its normal component $$\BM_{\textrm{n}}$$ over the boundary, then $$\BM$$ is
uniquely specified.

The gradient of the vector $$\BM$$ can be written as a single even grade multivector

\label{eqn:helmholtzDerviationMultivector:60}
= \spacegrad \cdot \BM + I \spacegrad \cross \BM
= s + I \BC.

We will use this to attempt to discover the relation between the vector $$\BM$$ and its divergence and curl. We can express $$\BM$$ at the point of interest as a convolution with the delta function at all other points in space

\label{eqn:helmholtzDerviationMultivector:80}
\BM(\Bx) = \int_V dV’ \delta(\Bx – \Bx’) \BM(\Bx’).

The Laplacian representation of the delta function in \R{3} is

\label{eqn:helmholtzDerviationMultivector:100}
\delta(\Bx – \Bx’) = -\inv{4\pi} \spacegrad^2 \inv{\Abs{\Bx – \Bx’}},

so $$\BM$$ can be represented as the following convolution

\label{eqn:helmholtzDerviationMultivector:120}
\BM(\Bx) = -\inv{4\pi} \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’).

Using this relation and proceeding with a few applications of the chain rule, plus the fact that $$\spacegrad 1/\Abs{\Bx – \Bx’} = -\spacegrad’ 1/\Abs{\Bx – \Bx’}$$, we find

\label{eqn:helmholtzDerviationMultivector:720}
\begin{aligned}
-4 \pi \BM(\Bx)
&= \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’) \\
&= \gpgradeone{\int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’)} \\
&= -\gpgradeone{\int_V dV’ \spacegrad \lr{ \spacegrad’ \inv{\Abs{\Bx – \Bx’}}} \BM(\Bx’)} \\
\spacegrad’ \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
-\frac{\spacegrad’ \BM(\Bx’)}{\Abs{\Bx – \Bx’}}
} } \\
&=
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
\frac{s(\Bx’) + I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}
} \\
&=
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cdot \int_V dV’
\frac{I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}

By inserting a no-op grade selection operation in the second step, the trivector terms that would show up in subsequent steps are automatically filtered out. This leaves us with a boundary term dependent on the surface and the normal and tangential components of $$\BM$$. Added to that is a pair of volume integrals that provide the unique dependence of $$\BM$$ on its divergence and curl. When the surface is taken to infinity, which requires $$\Abs{\BM}/\Abs{\Bx – \Bx’} \rightarrow 0$$, then the dependence of $$\BM$$ on its divergence and curl is unique.

In order to express final result in traditional vector algebra form, a couple transformations are required. The first is that

\label{eqn:helmholtzDerviationMultivector:800}
\gpgradeone{ \Ba I \Bb } = I^2 \Ba \cross \Bb = -\Ba \cross \Bb.

For the grade selection in the boundary integral, note that

\label{eqn:helmholtzDerviationMultivector:740}
\begin{aligned}
&=
+
\gpgradeone{ \spacegrad (\ncap \wedge \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)
+
\gpgradeone{ \spacegrad I (\ncap \cross \BX) } \\
&=
\spacegrad (\ncap \cdot \BX)

\spacegrad \cross (\ncap \cross \BX).
\end{aligned}

These give

\label{eqn:helmholtzDerviationMultivector:721}
\boxed{
\begin{aligned}
\BM(\Bx)
&=
\spacegrad \inv{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}

\spacegrad \cross \inv{4\pi} \int_{\partial V} dA’ \ncap \cross \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}} \\
&-\spacegrad \inv{4\pi} \int_V dV’
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
+\spacegrad \cross \inv{4\pi} \int_V dV’
\frac{\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
}

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

[3] Garret Sobczyk and Omar Le’on S’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21:221–231, 2011. URL https://arxiv.org/abs/0809.4526.

Does the divergence and curl uniquely determine the vector?

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A problem posed in the ece1228 problem set was the following

Helmholtz theorem.

Prove the first Helmholtz’s theorem, i.e. if vector $$\BM$$ is defined by its divergence

\label{eqn:emtProblemSet1Problem5:20}
\spacegrad \cdot \BM = s

and its curl
\label{eqn:emtProblemSet1Problem5:40}
\spacegrad \cross \BM = \BC

within a region and its normal component $$\BM_{\textrm{n}}$$ over the boundary, then $$\BM$$ is uniquely specified.

Solution.

This problem screams for an attempt using Geometric Algebra techniques, since
the gradient of this vector can be written as a single even grade multivector

\label{eqn:emtProblemSet1Problem5AppendixGA:60}
\begin{aligned}
&= \spacegrad \cdot \BM + I \spacegrad \cross \BM \\
&= s + I \BC.
\end{aligned}

Observe that the Laplacian of $$\BM$$ is vector valued

\label{eqn:emtProblemSet1Problem5AppendixGA:400}
= \spacegrad s + I \spacegrad \BC.

This means that $$\spacegrad \BC$$ must be a bivector $$\spacegrad \BC = \spacegrad \wedge \BC$$, or that $$\BC$$ has zero divergence

\label{eqn:emtProblemSet1Problem5AppendixGA:420}
\spacegrad \cdot \BC = 0.

This required constraint on $$\BC$$ will show up in subsequent analysis. An equivalent problem to the one posed
is to show that the even grade multivector equation $$\spacegrad \BM = s + I \BC$$ has an inverse given the constraint
specified by \ref{eqn:emtProblemSet1Problem5AppendixGA:420}.

Inverting the gradient equation.

The Green’s function for the gradient can be found in [1], where it is used to generalize the Cauchy integral equations to higher dimensions.

\label{eqn:emtProblemSet1Problem5AppendixGA:80}
\begin{aligned}
G(\Bx ; \Bx’) &= \inv{4 \pi} \frac{ \Bx – \Bx’ }{\Abs{\Bx – \Bx’}^3} \\
\spacegrad \BG(\Bx, \Bx’) &= \spacegrad \cdot \BG(\Bx, \Bx’) = \delta(\Bx – \Bx’) = -\spacegrad’ \BG(\Bx, \Bx’).
\end{aligned}

The inversion equation is an application of the Fundamental Theorem of (Geometric) Calculus, with the gradient operating bidirectionally on the Green’s function and the vector function

\label{eqn:emtProblemSet1Problem5AppendixGA:100}
\begin{aligned}
\oint_{\partial V} G(\Bx, \Bx’) d^2 \Bx’ \BM(\Bx’)
&=
\int_V G(\Bx, \Bx’) d^3 \Bx \lrspacegrad’ \BM(\Bx’) \\
&=
\int_V d^3 \Bx (G(\Bx, \Bx’) \lspacegrad’) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) (\spacegrad’ \BM(\Bx’)) \\
&=
-\int_V d^3 \Bx \delta(\Bx – \By) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-I \BM(\Bx)
+
\inv{4 \pi} \int_V d^3 \Bx \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}

The integrals are in terms of the primed coordinates so that the end result is a function of $$\Bx$$. To rearrange for $$\BM$$, let $$d^3 \Bx’ = I dV’$$, and $$d^2 \Bx’ \ncap(\Bx’) = I dA’$$, then right multiply with the pseudoscalar $$I$$, noting that in \R{3} the pseudoscalar commutes with any grades

\label{eqn:emtProblemSet1Problem5AppendixGA:440}
\begin{aligned}
\BM(\Bx)
&=
I \oint_{\partial V} G(\Bx, \Bx’) I dA’ \ncap \BM(\Bx’)

I \inv{4 \pi} \int_V I dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-\oint_{\partial V} dA’ G(\Bx, \Bx’) \ncap \BM(\Bx’)
+
\inv{4 \pi} \int_V dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}

This can be decomposed into a vector and a trivector equation. Let $$\Br = \Bx – \Bx’ = r \rcap$$, and note that

\label{eqn:emtProblemSet1Problem5AppendixGA:500}
\begin{aligned}
\gpgradeone{ \rcap I \BC }
&=
\gpgradeone{ I \rcap \BC } \\
&=
I \rcap \wedge \BC \\
&=
-\rcap \cross \BC,
\end{aligned}

so this pair of equations can be written as

\label{eqn:emtProblemSet1Problem5AppendixGA:520}
\begin{aligned}
\BM(\Bx)
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) } \\
0
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\rcap}{r^2} \wedge \ncap \wedge \BM(\Bx’)
+
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}.
\end{aligned}

Consider the last integral in the pseudoscalar equation above. Since we expect no pseudoscalar components, this must be zero, or cancel perfectly. It’s not obvious that this is the case, but a transformation to a surface integral shows the constraints required for that to be the case. To do so note

\label{eqn:emtProblemSet1Problem5AppendixGA:540}
\begin{aligned}
\spacegrad \inv{\Bx – \Bx’}
&= -\spacegrad’ \inv{\Bx – \Bx’} \\
&=
-\frac{\Bx – \Bx’}{\Abs{\Bx – \Bx’}^3} \\
&= -\frac{\rcap}{r^2}.
\end{aligned}

Using this and the chain rule we have

\label{eqn:emtProblemSet1Problem5AppendixGA:560}
\begin{aligned}
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}
&=
\frac{I}{4 \pi} \int_V dV’ \lr{ \spacegrad’ \inv{ r } } \cdot \BC(\Bx’) \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r}

\frac{I}{4 \pi} \int_V dV’ \frac{ \spacegrad’ \cdot \BC(\Bx’) }{r} \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r} \\
&=
\frac{I}{4 \pi} \int_{\partial V} dA’ \ncap(\Bx’) \cdot \frac{\BC(\Bx’)}{r}.
\end{aligned}

The divergence of $$\BC$$ above was killed by recalling the constraint \ref{eqn:emtProblemSet1Problem5AppendixGA:420}. This means that we can rewrite entirely as surface integral and eventually reduced to a single triple product

\label{eqn:emtProblemSet1Problem5AppendixGA:580}
\begin{aligned}
0
&=
-\frac{I}{4 \pi} \oint_{\partial V} dA’ \lr{
\frac{\rcap}{r^2} \cdot (\ncap \cross \BM(\Bx’))
-\ncap \cdot \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\frac{\rcap}{r^2} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\lr{ \spacegrad’ \inv{r}} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\frac{\BM(\Bx’) \cross \ncap}{r}
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\frac{\BM(\Bx’) \cross \ncap}{r}.
\end{aligned}

Final results.

Assembling things back into a single multivector equation, the complete inversion integral for $$\BM$$ is

\label{eqn:emtProblemSet1Problem5AppendixGA:600}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\partial V} dA’
\lr{
\frac{\BM(\Bx’) \wedge \ncap}{r}
-\frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) }.

This shows that vector $$\BM$$ can be recovered uniquely from $$s, \BC$$ when $$\Abs{\BM}/r^2$$ vanishes on an infinite surface. If we restrict attention to a finite surface, we have to add to the fixed solution a specific solution that depends on the value of $$\BM$$ on that surface. The vector portion of that surface integrand contains

\label{eqn:emtProblemSet1Problem5AppendixGA:640}
\begin{aligned}
\gpgradeone{ \rcap \ncap \BM }
&=
\rcap (\ncap \cdot \BM )
+
\rcap \cdot (\ncap \wedge \BM ) \\
&=
\rcap (\ncap \cdot \BM )
+
(\rcap \cdot \ncap) \BM

(\rcap \cdot \BM ) \ncap.
\end{aligned}

The constraints required by a zero triple product $$\spacegrad’ \cdot (\BM(\Bx’) \cross \ncap(\Bx’))$$ are complicated on a such a general finite surface. Consider instead, for simplicity, the case of a spherical surface, which can be analyzed more easily. In that case the outward normal of the surface centred on the test charge point $$\Bx$$ is $$\ncap = -\rcap$$. The pseudoscalar integrand is not generally killed unless the divergence of its tangential component on this surface is zero. One way that this can occur is for $$\BM \cross \ncap = 0$$, so that $$-\gpgradeone{ \rcap \ncap \BM } = \BM = (\BM \cdot \ncap) \ncap = \BM_{\textrm{n}}$$.

This gives

\label{eqn:emtProblemSet1Problem5AppendixGA:620}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) +
\BC(\Bx’) \cross \frac{\rcap}{r^2} },

or, in terms of potential functions, which is arguably tidier

\label{eqn:emtProblemSet1Problem5AppendixGA:300}
\boxed{
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
-\spacegrad \int_V dV’ \frac{ s(\Bx’)}{ 4 \pi r }
+\spacegrad \cross \int_V dV’ \frac{ \BC(\Bx’) }{ 4 \pi r }.
}

Commentary

I attempted this problem in three different ways. My first approach (above) assembled the divergence and curl relations above into a single (Geometric Algebra) multivector gradient equation and applied the vector valued Green’s function for the gradient to invert that equation. That approach logically led from the differential equation for $$\BM$$ to the solution for $$\BM$$ in terms of $$s$$ and $$\BC$$. However, this strategy introduced some complexities that make me doubt the correctness of the associated boundary analysis.

Even if the details of the boundary handling in my multivector approach is not correct, I thought that approach was interesting enough to share.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.