This Wikipedia entry tells me that the Thevenin equivalent circuit for an arbitrary receiving antenna on which an electric field $E_b$ is incident is a voltage source $V_a$ in series with an impedance $R_a + j X_a$ where (I have re-arranged the terms a bit to frame my question...) $$ V_a = E_b\; \frac{\cos {\psi}}{\sqrt{\pi Z_0}} \; \left( \lambda \sqrt{R_a G_a} \right) $$ ... given that $G_a$ is the directive gain of the antenna in the directive gain of the antenna in the direction of incidence, and $\psi$ is the angle by which the electric field is 'misaligned' with the antenna.
The article does mention that this is derived from reciprocity, from which I assume that there should be some reasoning beginning with the Rayleigh-Carson theorem: $$ \iiint_{V_1} \vec{J}_1 \cdot \vec{E}_2 \;dV = \iiint_{V_2} \vec{J}_2 \cdot \vec{E}_1 \;dV $$ I am trying to understand how I can apply this, and in fact how I can approach an arbitrary antenna structure in general (I do understand how a dipole and a loop can be analyzed)
Unfortunately the article itself doesn't point out any sources where this relationship is derived, so I was wondering if anyone could point me to any textbook or paper where this derivation may be found?
My motivation is something like this -- the relation mentioned in the Wikipedia article is actually for a sinusoidal input -- and the frequency determines $\lambda$, $R_a$, and $G_a$ in the expression (and $X_a$ in the equivalent circuit). I am trying to understand if any insight can be obtained about the equivalent voltage source $V\left(t\right)$ given an arbitrary $E\left(t\right)$ -- maybe, for example, as a differential or integral equation? The $X_a$ can be replaced by frequency independent $C_a$ and $L_a$ in series -- and for the voltage source I would integrate over $\lambda$ -- but I don't know how to deal with $R_a\left(\lambda\right)$ (which is ideally only the radiation resistance) and $G_a\left(\lambda\right)$ for arbitrary antenna geometries. So I was hoping that the derivation would offer me some clues...
Update OK, so it seems that I went on the wrong track here -- it is actually quite easy. I am answering my own question below.