# Reflection behaviour of the radiation as a function of the used voltage

According all the answers to a question about What is the advantage of making an antenna resonant different lengths of the antenna rod could be tuned to resonance with the antenna generator.

You can introduce shift with additional components to get the antenna to look resistive again, and get the best signal power from it. But the gain will not be as in the ideal case. Source

In my understaning the tuning of antennas with different lengths to the same frequency leads to different voltages, with which the electrons in the rod get drived. Different voltages lead to different accelerations of these electrons and this leads to the emission of photons with different energy content.

I’m curious about the reflection behaviour of the radiation as a function of the used voltage. The reflection could be from the ionosphere as well as from buildings or other obstacles.

The energy of a photon is given by:

$$E={\frac {hc}{\lambda }}$$

where:

• $$E$$ is energy, in joules (though with the proper change in $$h$$, electron-volts are also frequently used)
• $$h$$ is the Planck constant, approximately 6.63×10−34 joule-seconds
• $$c$$ is the speed of light, 299792458 meters per second
• $$\lambda$$ is the wavelength, in meters

Changing the length of the antenna, the transmitter power, feedpoint impedance or transmitter power (and consequently feed voltage) does not change the photon energy. The only way to change the photon energy is to transmit at a different frequency.

Consequently, changing the antenna length will have no effect on how signals propagate, except to the extent that the new antenna may have changed efficiency or gain.

Radios are well described by classical electromagnetic theory, so there's not much reason to be thinking about photons or other quantum effects. In fact unless you are very familiar with quantum physics, thinking of radio operation in terms of photons is more likely to cultivate misconceptions than insight.

• Comments are not for extended discussion; this conversation has been moved to chat. – Mike Waters Jan 3 '19 at 4:17

I’m curious about the reflection behaviour of the radiation as a function of the used voltage. The reflection could be from the ionosphere as well as from buildings or other obstacles.

At reasonable power levels, the way in which the environment reflects (or refracts or absorbs) radio waves is mostly linear — it obeys the superposition principle. Saying that a radio wave has a higher voltage (between specified points) than another is the same as saying it has higher amplitude (or power). If you send 10 watts in some particular direction and observe 1 watt on a nearby receiver, then you can know that sending 20 watts will give you (almost) 2 watts back. It's linear.

Different voltages lead to different accelerations of these electrons and this leads to the emission of photons with different energy content.

This is not correct. I think you are thinking of optical phenomena where the photons are understood to be emitted from energy level transitions of individual atoms. Here, everything is bigger and slower. The radio-frequency photons come from the change in the overall electromagnetic field of the antenna, which is driven by the signal generated by the transmitter (also an electromagnetic wave, just one confined to a transmission line). The frequency, and therefore energy per photon, is set by the transmitter; it cannot be changed by the antenna.

https://en.wikipedia.org/wiki/Impedance_of_free_space

Once the single frequency RF signals leave the antenna, the photons all have the same energy content and the same ratio of electrical to magnetic field, approximately 377 ohms. In the far field, that ratio is unrelated to the voltage on the antenna. The energy content of all individual photons of the same frequency is identical, i.e. the individual photons from a QRP station have the same energy content as the individual photons from a QRO station on the same frequency. The total number of photons is what is different, not the individual photons.

Whether a conjugately matched antenna element is resonant or non-resonant, the voltage will be whatever it takes to cause the current to be the magnitude necessary for (the square of that current) times (the radiation resistance) to be the maximum available radiated power. That follows from the maximum power transfer theorem applied to RF signals.

Let's get some definitions straight. Assuming that we are feeding an antenna at a current maximum point, the feedpoint impedance can be thought of in two parts: The part that radiates the power is called the radiation resistance. The part that dissipates power is called the loss resistance and is made up of ground losses, I^2*R losses, dielectric losses, etc.

By definition, the radiation resistance radiates all of its power and dissipates zero power as heat. By definition, it is impossible for the radiation resistance to dissipate heat.

• Comments are not for extended discussion; this conversation has been moved to chat. – Kevin Reid AG6YO Jan 2 '19 at 23:19
• Gentlemen, as @KevinReidAG6YO asked, kindly post any continuation of that discussion in this chat room. – Mike Waters Jan 3 '19 at 17:23
• The reason why is here. Thanks! – Mike Waters Jan 3 '19 at 17:30