How does an antenna capture the energy of EM radiation?

Question

I am having trouble reconciling the descriptions I have read of how antennas work to recieve and transmit radiation.

My simple understanding is that in recieveing the oscilating electric field of an EM wave causes the electrons in the antenna wire to oscilate producting an AC current.

On the other hand in transmiting and AC current is aplied to the wire and the osciations of electrons in the wire create an oscilating electric field which leaves the antenna as an EM wave.

If these two processes are happening simultaniously (ie EM wave incuces AC current which creates EM wave) how can an antenna "keep hold of" any of the energy it recieves as EM radiation?

Answer 1

A charge in an external field moves indeed, but its own radiative losses are small, practically negligeable. Thus, the external wave is mostly "absorbed" by the receiver antenna.

Answer 2

I imagine it like a resonant circuit in which the antenna itself plays the role of a capacitor. When the electrons are moved to one side of the antenna, the capacitor is charged. The wire in between or the readout, are the inductance and the resistance of this circuit.

The electrons radiate, if they get accelerated, so mainly when they change direction. How much energy is captured and radiated away, depends strongly on the frequency and geometry of the antenna. When designing an antenna, these effects have to be considered as the reflections can potentially also interfere destructively with the signal or with other reflections of the environment.

Answer 3

To understand the working of an antenna, consider the electromagnetic wave propagation, rather than the electromagnetic field oscillation: A perturbation in the electromagnetic field propagates with light speed in vacuum (this approximately holds also for air), but with considerable lower speed in condensed (solid or liquid) materials, such as dielectrics or metals. The wavefront in contact with a metal (e.g. a wire) or with a dielectric (e.g. a slab) is thus slowed down at the contact edge, and bent towards the metal or the dielectric, causing more of the wave to enter into the metal or into the dielectric. This is the antenna effect. As an empirical fact, a metel surface collects, in addition to the energy directly falling onto it, as well the energy in a margin of 0.15 wavelengths around its borders. E.g. a dipole antenna of a radio station operating at 7 MHz frequency (40 m wavelength) consists of 2 quarter-wave wires of 10 m lenght each. The energy-collecting surface of this antenna is thus 2 x 10 x (0.15+0.15) x 40 = 240 $m^2$ (the wire diameter being negligible). Part of the collected energy can be retrieved from the wire as an electric oscillation; another part is radiated back into space again.

Answer 4

Capture of EM energy from work of external EM wave on charges in an antenna is not a 100% efficiency process. Oscillating charges in the antenna will produce secondary EM waves (wave due to antenna), so some energy goes away from the antenna (a kind of reflection). But presence of this reflection does not mean that all work of external EM field on the charges in the antenna gets reflected. Some part of it goes into the metal and increases internal energy of the metal (antenna gets a little hotter), some other part is funnelled along the wires into an amplifier or a receiver - in general, a load. How much is this useful part, depends also on characteristics of the load.