Sound means vibration of molecules and vibration produces electromagnetic waves. So, this means that sound produces electromagnetic waves directly.
Is this possible?
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asked Feb 13, 2020 at 10:04
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2$\begingroup$ Are you asking the question in general case of all types of sound? Or are you looking for a specific cases where you get significant EM waves? $\endgroup$– KF GaussFeb 13, 2020 at 17:14
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4$\begingroup$ In the case of liquid medium, there is a phenomenon called acoustic cavitation. The sudden collapse of cavitation bubbles can provoke local heating and an extreme raise of temperature (thousands of degrees) in a short period can lead to light emission from the bubbles. This is called sonoluminescence. $\endgroup$– VerktajFeb 14, 2020 at 1:06
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$\begingroup$ You can bounce lasers of reflecting surfaces at oblique angles and receive the motion and thus decode and hear the sound that way. $\endgroup$– FrankFeb 14, 2020 at 21:22
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$\begingroup$ There's something called an 'optical phonon' which has a vibration mode where opposite charges in a crystal move in contrary motion. I would start my online search with that topic and see where you can go. $\endgroup$– David ElmFeb 15, 2020 at 3:42
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$\begingroup$ Congratulations! You have discovered frictional heating, as the generated E-M waves are, except in peculiar and unusual cases, in the far infrared. E-M waves can only be generated in a quantized manner (see Planck) and thus the couplings between the acoustic frequencies and the vibrational modes of the material component atoms and molecules need to be determined. $\endgroup$– Pieter GeerkensFeb 15, 2020 at 10:10
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7 Answers
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A sound wave passing through a medium (e.g. air) indeed displaces molecules by a distance of a few nanometers. It seems reasonable that it should also displace the atoms, and thus electrons and protons in the process, which are charged particles and should radiate by Larmor's equation when undergoing acceleration.
Let us assume the sound frequency is on the order of kHz (which is in our audible range). Then molecules are accelerated by
$$a \approx \left(10^3\, \mathrm{Hz}\right)^2\times \left(10^{-9}\, \mathrm{m}\right) \approx 10^{-3}\, \mathrm{m/s}^2 $$
Then the predicted power of the radiation produced by Larmor's equation is ridiculously small $$P = \frac{2}{3}\frac{q^2 a^3}{c^3} \sim 10^{-73}\, \mathrm{W} $$ Even if one multiplies this by the number of molecules of air in a $\mathrm{m}^3$, $N\approx 10^{25}$, this would never be detectable. Therefore, this effect may well exist, but it is absolutely negligible in all respects.
N.b. My answer focused on direct effects of the acceleration of air molecules due to a sound wave. As other answers mention correctly, there are interesting secondary effects of (especially large-amplitude) sound waves involving EM radiation. Among these are sonoluminescence and heating of the air by sound dissipation leading to increased thermal radiation.
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10$\begingroup$ Not to mention most objects we interact with are electrically neutral $\endgroup$ Feb 13, 2020 at 13:01
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4$\begingroup$ @AaronStevens but they are composed of many charged particles, does the radiation cancel out somehow?... As a thought experiment I envisioned two charged regions of opposite charge, separated from each other. If I make them vibrate collectively I would expect that each would generate EM radiation in accordance to Larmor's equation, I don't see how they would cancel out even when considering the whole system is electrically neutral (because EM waves don't interact with each other in the classical limit). The point is, as long as particles vibrate individually, shouldn't they emit EM radiation? $\endgroup$– user137661Feb 13, 2020 at 15:38
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15$\begingroup$ @Aaoron Stevens, being overall electrically neutral is irrelevant. Antennas are overall neutral objects as well and radiate just fine. $\endgroup$– KF GaussFeb 13, 2020 at 17:09
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6$\begingroup$ @KFGauss: In an antenna, the positive charges are all stationary, while some of the negative charges move, so you have relative motion between the two, which is different from whole molecules moving. $\endgroup$ Feb 14, 2020 at 0:19
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3$\begingroup$ @thegreatemu : I wouldn't consider ultrasonic welding, a widespread industrial process, the equipment for which is COTS available at prosumer prices is an "edge case". $\endgroup$ Feb 15, 2020 at 16:05
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If it passes through a piezoelectric material, it may generate a measurable voltage and current, some fraction of which will be radiated. For most materials, however, there won't be a detectable effect.
answered Feb 13, 2020 at 19:57
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$\begingroup$ According to Wikipedia and Mahan, most materials are piezoelectric, so I guess the effect is measurable on most materials. $\endgroup$ Feb 15, 2020 at 9:00
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2$\begingroup$ A somewhat obvious example of this is a crystal microphone. $\endgroup$– TLWFeb 15, 2020 at 19:54
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$\begingroup$ Another example in non-piezoelectrics is the surface acoustic waves which achieves significant voltages because it lies at the surface where inversion is broken. See here: phys.org/news/2020-02-photons-stream-electrons.amp $\endgroup$– KF GaussFeb 16, 2020 at 14:43
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Physically vibrating atoms with a sound wave will technically accelerate the charges, but the radiation emitted will be very weak. However secondary effects from the sound wave are a different matter.
Indeed, sound waves in water can create light. This is called sonoluminescence. In the lab it's done with ultrasound passed through cavitation bubbles. It also happens naturally, with the pressure waves created by the rapid motion of the mantis shrimp's claw.
The exact mechanism is as yet unknown. One hypothesis is that the physical shock from the sound wave ionizes the particles of dissolved gases, which recombine and emit light.
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Motions of neutral atoms do not radiate
@Cyclone's answer provides very useful insight, but is ultimately irrelevant. As long as the medium that is vibrating is neutral, no energy is radiated. The EM waves emitted by the protons in the air interfere destructively with those emitted by the electrons and cancel out.
Put another way, the emitted radiation is proportional to the integral of the net current density. If positive and negative charges are moving the same, the net current density is 0.
In order to radiate, you would have to accelerate the atoms fast enough that they move faster than the electron cloud can catch up, causing a separation of the center of positive and negative charge (i.e. forming a dipole).
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1$\begingroup$ How exactly do you think mechanical waves can exist in materials made of neutral atoms? They're neither transmitted by strong or weak nuclear forces, nor by gravity. $\endgroup$– MSaltersFeb 14, 2020 at 13:11
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$\begingroup$ @MSalters I don't understand how your question is relevant. The OP's question refers to electromagnetic radiation, not electric forces. $\endgroup$ Feb 14, 2020 at 17:09
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$\begingroup$ So water, being a dipole, should radiate if exposed to a static electric field that causes the individual molecules to align? $\endgroup$– mic_eFeb 14, 2020 at 18:17
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3$\begingroup$ @mic_e When the field is first applied, yes! Before the field is turned on, the macroscopic body of water is neutral. After the field is turned on, it is a macroscopic dipole. In between those states, there was non-zero net current flow (individual molecules preferentially switching their orientation to align with the external field). Time varying current means radiation. It does not continue to radiate once the field is established though $\endgroup$ Feb 14, 2020 at 18:28
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If the vibrating material is charged, then electromagnetic radiation will be produced.
There is a type of microphone called the capacitor or condenser microphone, in which the microphone is a capacitor whose capacitance varies with the acoustic vibrations.
https://en.wikipedia.org/wiki/Microphone#Condenser_microphone
The capacitance can be detected either by measuring the impedance to an RF signal, or by applying a high voltage direct current bias and measuring the current flow in and out of the capacitor as its capacitance changes.
Normally the microphone is connected directly to the amplifier circuit, but it should be possible to couple it indirectly. In this case the electromagnetic radiation produced by acoustic vibration would be detectable.
If the vibrating material is magnetized, then electromagnetic radiation will be produced.
A typical electric guitar pickup consists of several magnets (usually one per string) surrounded by a coil of wire. The magnets induce magnetism in the string. The vibrations of the magnetized string are picked up by the coil of wire. Hence the electromagnic radiation produced by the string is detected.
However, in the general case where nothing is done to make the vibrating material an emitter of electromagnetic radiation, it will be difficult to detect (unless the vibration is so intense as to cause the vibrating material to heat up and emit radiation due to its increased temperature.)
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I am deeply surprised by the neglect of the energy balance. In the previous answers. All sound is dispersed within a body. The energy difference between the incoming sound and the outgoing sound turns into heat.
The increase in temperature of the body is accompanied by an increase of electromagnetic radiation. So your guess is right, the body is out of its thermal equilibrium and responsible for this this the sound.
answered Feb 14, 2020 at 10:32
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$\begingroup$ Not just the energy balance in particular, but the entire absence of any thermodynamic analysis in general. After Planck, E-M radiation can only be generated in a quantized manner, and the appropriate photon frequencies would have to be determined by analysis of the couplings between the acoustic and (potential) E-M modes. I suspect this is all in the infrared, explaining why materials heat up when acoustically forced - to deduce that we have discovered "friction". $\endgroup$ Feb 15, 2020 at 10:06
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The short answer is that some vibrations in materials can decay through photon emission, but that type is not excited by acoustic waves. Check out optical phonons.
