## Reciprocity theorem in Geometric Algebra

The reciprocity theorem involves a Poynting like antisymmetric difference of the following form

\label{eqn:reciprocityTheoremGA:20}
\BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)}.

This smells like something that can probably be related to a combined electromagnetic field multivectors in some sort of structured fashion. Guessing that this is related to the antisymmetic sum of two electromagnetic field multivectors turns out to be correct. Let

\label{eqn:reciprocityTheoremGA:60}
F^{(a)} = \BE^{(a)} + I c \BB^{(a)}

\label{eqn:reciprocityTheoremGA:80}
F^{(b)} = \BE^{(b)} + I c \BB^{(b)}.

Now form the antisymmetic sum

\label{eqn:reciprocityTheoremGA:100}
\begin{aligned}
\inv{2} \lr{ F^{(a)} F^{(b)} – F^{(b)} F^{(a)} }
&=
\inv{2} \lr{\BE^{(a)} + I c \BB^{(a)}}
\lr{\BE^{(b)} + I c \BB^{(b)}} \\
&-
\inv{2} \lr{\BE^{(b)} + I c \BB^{(b)}}
\lr{\BE^{(a)} + I c \BB^{(a)}} \\
&=
\inv{2} \lr{ \BE^{(a)} \BE^{(b)} -\BE^{(b)} \BE^{(a)} }
+ \frac{I c}{2} \lr{ \BE^{(a)} \BB^{(b)} – \BB^{(b)} \BE^{(a)} }\\&
+ \frac{I c}{2} \lr{ \BB^{(a)} \BE^{(b)} – \BE^{(b)} \BB^{(a)} }
+ \frac{c^2}{2} \lr{ \BB^{(b)} \BB^{(a)} – \BB^{(a)} \BB^{(b)} } \\
&=
\BE^{(a)} \wedge \BE^{(b)} + c^2 \lr{ \BB^{(b)} \wedge \BB^{(a)} }
+ I c \lr{
\BE^{(a)} \wedge \BB^{(b)}
+
\BB^{(a)} \wedge \BE^{(b)}
} \\
&=
I \BE^{(a)} \cross \BE^{(b)} + c^2 I \lr{ \BB^{(b)} \cross \BB^{(a)} }

c \lr{
\BE^{(a)} \cross \BB^{(b)}
+
\BB^{(a)} \cross \BE^{(b)}
}
\end{aligned}

This has two components, the first is a bivector (pseudoscalar times vector) that includes all the non-mixed products, and the second is a vector that includes all the mixed terms. We can therefore write the antisymmetic difference of the reciprocity theorem by extracting just the grade one terms of the antisymmetric sum of the combined electromagnetic field

\label{eqn:reciprocityTheoremGA:120}
\BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)}
=
-\frac{1}{2 c \mu_0} \gpgradeone{ \lr{ F^{(a)} F^{(b)} – F^{(b)} F^{(a)} } }.

Observing that the antisymmetrization used in the reciprocity theorem is only one portion of the larger electromagnetic field antisymmetrization, introduces two new questions

1. How would the reciprocity theorem be derived directly in terms of $$F^{(a)} F^{(b)} – F^{(b)} F^{(a)}$$?
2. What is the significance of the other portion of this antisymmetrization $$\BE^{(a)} \cross \BE^{(b)} – c^2 \mu_0^2 \lr{ \BH^{(a)} \cross \BH^{(b)} }$$ ?

… more to come.

## Reciprocity theorem: background

The class slides presented a derivation of the reciprocity theorem, a theorem that contained the integral of

\label{eqn:reciprocityTheorem:360}
\int \lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot d\BS = \cdots

over a surface, where the RHS was a volume integral involving the fields and (electric and magnetic) current sources.
The idea was to consider two different source loading configurations of the same system, and to show that the fields and sources in the two configurations can be related.

To derive the result in question, a simple way to start is to look at the divergence of the difference of cross products above. This will require the phasor form of the two cross product Maxwell’s equations

\label{eqn:reciprocityTheorem:100}
\spacegrad \cross \BE = – (\BM + j \omega \mu_0 \BH) % \BM^{(a)} + j \omega \mu_0 \BH^{(a)}

\label{eqn:reciprocityTheorem:120}
\spacegrad \cross \BH = \BJ + j \omega \epsilon_0 \BE, % \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)}

so the divergence is

\label{eqn:reciprocityTheorem:380}
\begin{aligned}
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
&=
\BH^{(b)} \cdot \lr{ \spacegrad \cross \BE^{(a)} } -\BE^{(a)} \cdot \lr{ \spacegrad \cross \BH^{(b)} } \\
&-\BH^{(a)} \cdot \lr{ \spacegrad \cross \BE^{(b)} } +\BE^{(b)} \cdot \lr{ \spacegrad \cross \BH^{(a)} } \\
&=
-\BH^{(b)} \cdot \lr{ \BM^{(a)} + j \omega \mu_0 \BH^{(a)} } -\BE^{(a)} \cdot \lr{ \BJ^{(b)} + j \omega \epsilon_0 \BE^{(b)} } \\
&+\BH^{(a)} \cdot \lr{ \BM^{(b)} + j \omega \mu_0 \BH^{(b)} } +\BE^{(b)} \cdot \lr{ \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)} }.
\end{aligned}

The non-source terms cancel, leaving

\label{eqn:reciprocityTheorem:440}
\boxed{
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
=
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}

Should we be suprised to have a relation of this form? Probably not, given that the energy momentum relationship between the fields and currents of a single source has the form

\label{eqn:reciprocityTheorem:n}
\PD{t}{}\frac{\epsilon_0}{2} \left(\BE^2 + c^2 \BB^2\right) + \spacegrad \cdot \inv{\mu_0}(\BE \cross \BB) = -\BE \cdot \BJ.

(this is without magnetic sources).

This suggests that the reciprocity theorem can be expressed more generally in terms of the energy-momentum tensor.

## far field integral form

Employing the divergence theorem over a sphere the identity above takes the form

\label{eqn:reciprocityTheorem:480}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
=
\int_V \lr{
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV

In the far field, the cross products are strictly radial. That surface integral can be written as

\label{eqn:reciprocityTheorem:500}
\begin{aligned}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cross \lr{ \rcap \cross \BE^{(b)}} – \BE^{(b)} \cross \lr{ \rcap \cross \BE^{(a)}} } \cdot \rcap dS \\
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cdot \BE^{(b)} – \BE^{(b)} \cdot \BE^{(a)}
}
dS \\
&= 0
\end{aligned}

The above expansions used \ref{eqn:reciprocityTheorem:540} to expand the terms of the form

\label{eqn:reciprocityTheorem:560}
\lr{ \BA \cross \lr{ \rcap \cross \BC } } \cdot \rcap
= \BA \cdot \BC -\lr{ \BA \cdot \rcap } \lr{ \BC \cdot \rcap },

in which only the first dot product survives due to the transverse nature of the fields.

So in the far field we have a direct relation between the fields and sources of two source configurations of the same system of the form

\label{eqn:reciprocityTheorem:580}
\boxed{
\int_V \lr{
\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV
=
\int_V \lr{
\BH^{(b)} \cdot \BM^{(a)} +\BE^{(a)} \cdot \BJ^{(b)}
}
dV
}

## Application to antenna

This is the underlying reason that we are able to pose the problem of what an antenna can recieve, in terms of what the antenna can transmit.

More on that to come.

## Identities

Lemma: Divergence of a cross product.

\label{thm:polarizationReview:400}
\spacegrad \cdot \lr{ \BA \cross \BB } =

Proof.

\label{eqn:reciprocityTheorem:420}
\begin{aligned}
\spacegrad \cdot \lr{ \BA \cross \BB }
&=
\partial_a \epsilon_{a b c} A_b B_c \\
&=
\epsilon_{a b c} (\partial_a A_b )B_c

\epsilon_{b a c} A_b (\partial_a B_c) \\
&=
\end{aligned}

Lemma: Triple cross product dotted
\label{thm:polarizationReview:520}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }
\end{aligned}

Proof.

\label{eqn:reciprocityTheorem:540}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
&=
\epsilon_{a b c} A_b \epsilon_{r s c } B_r C_s D_a \\
&=
\delta_{[a b]}^{r s}
A_b B_r C_s D_a \\
&=
A_s B_r C_s D_r
-A_r B_r C_s D_s \\
&=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }.
\end{aligned}