Wikipedia:Reference desk/Archives/Science/2022 October 4

= October 4 =

Planck relation and radio wave
It seems to me that Planck's relationship (E=hv) cannot apply to radio waves, because for example, for a short wave transmitter from 1.6Mhz to 26Mhz we will have the same radiated power. So what is the difference between a radio wave and light, for these two types of electromagnetic radiations ? Malypaet (talk) 21:36, 4 October 2022 (UTC)
 * There is no qualitative difference between light and short waves. Planck's relation implies that the energy in a short-wave transmission at (say) 10MHz will be emitted and received in multiples of 6.62607×10-27 J.  This quantity is so fantastically small that the power sent or received appears to be continuously variable.catslash (talk) 23:19, 4 October 2022 (UTC)
 * Would that mean that your quanta of energy, which is also called a photon, at equal emission power would be less when the frequency increases?
 * Outside the flow of electrons is the same in the antenna and their number of acceleration / deceleration is proportional to the frequency, how do you explain that? Malypaet (talk) 07:45, 5 October 2022 (UTC)
 * No, the energy per photon goes up with frequency and down with wave length. But you can still emit the same overall amount of energy by emitting different numbers of photons. We don't really need to explain this in classical terms, because it is am observed fact. --Stephan Schulz (talk) 07:58, 5 October 2022 (UTC)
 * Indeed, the photoelectric effect is an experimental "refutation" of classical electromagnetism. "Refutation" is a strong word - see "all models are wrong". Some famous guy got the 1921 Nobel prize for that. Tigraan Click here for my talk page ("private" contact) 11:31, 5 October 2022 (UTC)
 * The earlier responses are totally valid; the wave can actually be described using a model based on individual photons.
 * At the same time, it is an uncommon circumstance when we encounter a radio-frequency wave of such low energy that its quantized nature is immediately apparent. In other words, the kind of radio wave that you would probably find coming out of an HF emitter would be characterized as having "lots" of photons - that is to say, such electromagnetic waves satisfy the "continuum assumption" or the "macroscopic asymptote" of a quantum-mechanical description.  We rarely encounter a radio wave with "one single photon in it."  (Contrast that to other electromagnetic phenomena: when we work with visible light, or waves of similar frequency, we often count individual, discrete photons.  In fact, we've built machines that can count individual photons for over a century.  We just don't usually design or build those kinds of machines to count individual radio-frequency-photons.  That kind of thing just wouldn't be as useful for doing your day-job!
 * The energy in the electromagnetic wave is the sum of the energy of each photon. Two waves at different frequencies would have a different amount of energy per photon - that's the Planck relation E = hν.  Therefore, if two waves have equal energy, but different energy-per-photon, these waves must have a different number of photons.  (Quantum mechanics!  It's weird!)  In reasonable scenarios, both waves have such a large number of photons that we aren't concerned with the discrete count of individual photons: we can meaningfully assume that "there are a lot of 'em."
 * What in the world would a single, individual, discrete radio frequency photon look like? Well, that's a fun thought experiment.  At face value, "that's a dumb question."  Your eye sees visible light photons.  You can't "see" a radio-frequency photon.
 * But if you want to stretch your cognitive processes a little bit, have a bit of a read about NLK, a radio research facility and/or "totally operational actual telecommunications facility that isn't actually the only thing standing between our flimsy civilization and total nuclear annihilation", and it's designed to operate at very low radio frequency. The wavelengths should be 5 or 8 miles long; one individual photon emitted by this radio receiver would be kind of like "the size of a large mountain valley."  So if we put just enough energy on that wire - let's say we hooked up a battery on one side of the wire, and stretched it over a valley ten miles long, and grounded the other side of the wire - and then we put a tenth of a nano-volt on one side, and turned that on and off twenty thousand times in a second...   we should get like, one photon coming out of that wire, emitted (where?!) in all directions, kinda sorta, as a kind of wibble-wobble of the electric field, just below the detection threshold - (after all, you'd need an equally-large photomultiplier tube to even detect this wibble-wobble of one single photon, and something about Heisenberg's uncertainty principle tells us that we don't even know which direction it got emitted in until we detect it) - so most of the time you'd measure nothing - ... which is great if you're counting on the non-receipt of a single one-way signal to decide not to launch the missiles ...  ( Besides, this is all madness!  And who's paying for this nonsense...?'' )
 * On the other hand, if we're being really rigorous, we have to stomp on the notion and say "photons don't work like that." So we can rest easy.
 * Nimur (talk) 19:26, 5 October 2022 (UTC)
 * Exactly, about the photoelectric effect here is what I read on your wikipedia link:
 * "electrons are dislodged only when the light exceeds a certain frequency".
 * Replace "light" with "electromagnetic radiation" and explain to me how the photons of radio waves manage to dislodge electrons from their atoms in a receiving antenna ? Malypaet (talk) 20:35, 5 October 2022 (UTC)
 * Most radio-frequency antennas are made of metal, and function by virtue of its conductivity. The conduction electrons in metal are not bound to individual atoms, but flow freely through the material.  In any case, the electrons in a receiving antenna only serve to intercept the incoming photons and scatter them into the receiver. catslash (talk) 21:58, 5 October 2022 (UTC)
 * Indeed. The photoelectric effect is concerned with getting electrons out of the metal (where you need to reach the binding energy), normal antenna operation is concerned with the movement of electrons inside the antenna. Metals are shiny because their electrons move freely within the crystal structure, so they can react to any wavelength. Technically, the individual energy states of the atoms merge into an electronic band structure, allowing electrons in the band to have a range of energies. --Stephan Schulz (talk) 05:34, 6 October 2022 (UTC)


 * For a radio wave where individial photons did matter, consider the signal send by the Huygens probe as it decended through Titan's atmosphere, as received by some pretty big radio dishes on earth. They only got (I think, but can't find the transmitter power right away) on the order of a few hundred photons per second. Now do some Doppler measurement and VLBI with that. PiusImpavidus (talk) 09:52, 6 October 2022 (UTC)