A slight variant on your fine answer...
A reference is Ramo et al, Fields and Waves in Communication Electronics, chapter 12.
First, reciprocity: $Z_{21}=Z_{12}$ tells you that (assuming a conjugate-matched load):
$$ g_{dt} A_{er} = g_{dr} A_{et}$$
For both transmitting (subscript t) and receiving (r) antennas, $g_d$ is the antenna directional gain.
$A_{er}$ is the effective area of the receiving antenna, defined as the ratio of useful power removed from the receiving antenna $W_r$ to average power density $P_{av}$ in the incoming radiation.
Thus the ratio $g_d/A_e$ is the same for both transmitting and receiving antennas.
For large aperture antennas, it can be shown that the maximum possible gain satisfies:
$$ \frac{(g_d)_{max}}{A_e} = \frac{4 \pi}{\lambda^2} $$
For other geometries, $A_e$ is defined to give the same result. For example, for a Hertzian dipole, with a maximum directivity of 1.5:
$$ (A_e)_{max} = \frac{\lambda^2}{4 \pi} (g_d)_{max} = \frac{3}{8 \pi} \lambda^2 $$
Anyway, for the problem at hand, as you deduced, the useful power removed from the receiving antenna is:
$$ W_r = P_{av} A_{er} \text{, with the power density } P_{av} = \frac{E_b^2}{2 Z_o} , Z_o=377 \text{ ohms} $$
(Here, electric field and voltage are sinusoids measured as peak values.)
With a conjugate-matched load with real part $R_L$, equating load power dissipated with power delivered gives for the receiving antenna's Thevenin equivalent source voltage $V_a$:
$$\frac{(V_a/2)^2}{2 R_L} = \frac{E_b^2}{2 Z_o} A_{er} $$
$$ V_a = 2 \sqrt{A_{er}} \sqrt{\frac{R_L}{Z_o}} \, E_b $$
Substituting for $A_{er}$ from the reciprocity relation, the maximum voltage $V_{a,max}$ is:
$$ V_{a,max} = \sqrt{\frac{(g_{dr})_{max}}{\pi }} \sqrt{\frac{R_L}{Z_o}} \,\, \lambda E_b $$
I'm cautious about the $\cos \psi$ factor because beam patterns differ for different antennas.
it seems the photons will have many different frequencies. And the exact same scenario stands for the acceleration of electrons in the other direction of the antenna.
That is exactly how an antenna works. The antenna generator is needed:
- to accelerate the electrons forth and back the rod to generate photons and
- to produce a carrier wave at the frequency to which the receiver is adapted.
The information is imprinted for example by frequency modulation or amplitude modulation.
So are the EM waves that are generated in this fashion, with a macroscopic antenna being fed an AC voltage signal, really just a wide variety of photon frequencies that are mimicking the AC signal's frequency only by landing in the corresponding succession at the receiving antenna?
That is exactly how the receiver works. A tiny amount of photons (in relation to the emitted amount of photons) hit the receiver, but again with the frequency of the wave carrier. And a lot of other EM radiation hits the rod too.
A filter has to let through the electrons, accelerated from the photons of the wave carrier frequency and after extract the information from the frequency or amplitude modulation.
Understanding this one has to clearly recognize the methodical difference between EM radiation (from IR/microwaves, visible light, UV, X- to gamma rays) and radio waves. The first in the most cases are thermal radiations, the last are modulated waves with photons from IR to X-rays. Dependent from the power the antenna could get warm or hot and it could dangerous to be near a broadcast antenna.
Best Answer
An antenna has an equivalent impedance both in transmit and in receive. If the load in receive mode is matched to that antenna impedance then there is no reflection from that load back out the antenna, the incident energy on the load will be completely absorbed. Note the emphasis on energy incident on the load not on the antenna. The actual reflectivity of an antenna will depend on the incoming direction of the wave, for example.
Whether the load is matched, therefore, just concerns the load and the way it is connected to the antenna ports via a transmission line, that is a matched load for one incident direction maybe mismatched for a different one. This problem does not show up in transmission mode because there is only one way and one transmission line mode used that the source energy may excite the antenna ports. (This is the main reason why antenna designers calculate the resulting antenna pattern in transmit mode instead of analyzing it in receive mode, by reciprocity the transmit and receive antenna patterns are the same.)
Following @JanLalisky perfectly correct answer I would like to add that the physical reason for the secondary emissions from a receive antenna is because the incident wave is usually emitted from a far away source so it is either spherical or planar. In either case it is not the type of EM field that a "perfectly matched" antenna would emit in its transmit mode. For example, around the antenna there is a reactive field (non-radiating) that would have be recreated by the incident wave so the antenna metal not reradiate parts of it in some unknown directions, a simple plane wave cannot do just that.