3 ms·

PS: so in electromagnetic radiation, shorter wavelengths carry more energy, whereas in sound, longer wavelengths (bass) carry more energy. (Right?) I'm not phys

PS: so in electromagnetic radiation, shorter wavelengths carry more energy, whereas in sound, longer wavelengths (bass) carry more energy. (Right?) I'm not physics-savvy enough why these effects are opposite; is there an explanation for this?

jessaustin 12y ago
Sound energy is unrelated to audio frequency. Intuitively, it's possible to have quiet, deep sounds and loud, high-pitched sounds.
- schoen 12y ago
  I think you're right (audio frequency doesn't appear in the formulas I found for sound energy), but I don't think the analogy fully demonstrates why, since it's also possible to have bright or dim light of a particular color! The issue about the energy of electromagnetic radiation depending on frequency is, if I understand correctly, a quantum issue. https://en.wikipedia.org/wiki/Photoelectric_effect https://en.wikipedia.org/wiki/Photoelectric_effect That is, quantum theory explains why there is such a thing as nonionizing electromagnetic radiation.
- jwmerrill 12y ago
  It's also possible to have faint blue light, and bright red light. Sound energy is quantized too, but for audible frequencies, hf is small compared to kT, so quantum effects are swamped by thermal effects.
pdonis 12y ago
"More energy" can mean two different things. In the case of sound, only one of those meanings--the amplitude of the wave--is relevant: soft sounds have lower amplitude and loud sounds have higher amplitude. The frequency of the wave, as jessaustin pointed out, is independent of how soft or loud it is, and for sound, there isn't any useful sense in which any frequency carries "more energy" than any other frequency, given that the amplitudes are the same. With regard to light, however, there is a second possible meaning for "more energy": not a higher amplitude of the classical wave, but more energy contained in a single quantum of the light (i.e., a single photon). Light of higher frequency (toward the blue/ultraviolet end of the spectrum) has more energy per quantum; light of lower frequency (toward the red/infrared/microwave end of the spectrum) has less. This sense of "more energy" is still independent of wave amplitude: wave amplitude corresponds to number of photons (more precisely, the number of photons is proportional to the amplitude squared). So you can have faint light (few photons) of high frequency (more energy per photon); or you can have bright light (lots of photons) of low frequency (less energy per photon). The total energy contained in the light depends on both of these factors; but the total energy in the light is not the crucial factor involved (see below). It turns out that, when you are looking at the possible effects of radiation on the body, the energy per photon is the most important factor, because that is what determines what kind of interactions the radiation can have with the molecules that make up your body. UV, X-rays, and gamma rays have enough energy per photon to break apart the chemical bonds that hold together things like proteins in your cells; that's what makes them potentially cancer-causing. Microwaves, OTOH, don't have enough energy per photon to break chemical bonds; all they can do is make the molecules, like proteins, vibrate more rapidly, which just means heating them up.
- thinkling 12y ago
  Thanks.
drostie 12y ago
It's worth understanding that the light which you see is the result of many photons -- though it's usually estimated that maybe, if our rod/cone cells were a little more sensitive, they might have been able to perceive individuals. So actual energy/mass transfer comes from a lot of particles each offering their own individual kicks. So the overall intensity is not just (energy per kick) but really (number of kicks) * (energy per kick). The number of kicks counts. Phonons, for sound, actually are quantized in the exact same way as for light -- just replacing the speed of light c with the speed of sound in the medium v. (Both arguments are based on analogies to a harmonic oscillator problem which every quantum mechanics course works out; basically any vibration in the bottom of a potential well X can be dealt with by approximating that potential well with a parabola, which leads to particles, we could call them X-ons, of one form or another. Quantum mechanics makes all particles act like waves but also makes all waves quantize into particles.) So, long-story-short, the physics is the same, and long wavelength phonons also carry less energy per kick than shorter ones, just like photons. That's not the difference, and the E = h f equation is still the same. But for phonons it's more about numbers. The biggest sound particles you can hear correspond to 20,000 Hz frequencies. The smallest light particles you can see correspond to 700 nm wavelengths or 10^15 Hz frequencies. So photons are more than 50 billion times more energetic than phonons, as a rule. This changes a lot of features about them. For example, every degree of freedom in a system has some average energy due to just random energy-sharing, which is known colloquially as the "temperature" of the system. If you work out how many 20kHz phonons there are simply due to a room being at room temperature, it works out to about 300 million. Any detector which registers 20kHz phonons, including the hairs in your ear, have 300 million of those phonons in them just from random thermal excitation. On the other hand, the red light receptors in your eye have on average (300 million) / (50 billion) = 0.006 photons in them on average, meaning that if you look at a thousand of those photon-absorbers you'll see on average only six of them which are excited randomly just by thermal excitations. So we can, at room temperature, detect individual photons, where we can't detect even ten thousand phonons at once without it getting hopelessly buried in thermal noise. (But our eyes simply do not fire the neurons when a light-sensitive cell only sees one photon.)

jessaustin 12y ago

Sound energy is unrelated to audio frequency. Intuitively, it's possible to have quiet, deep sounds and loud, high-pitched sounds.

schoen 12y ago
I think you're right (audio frequency doesn't appear in the formulas I found for sound energy), but I don't think the analogy fully demonstrates why, since it's also possible to have bright or dim light of a particular color! The issue about the energy of electromagnetic radiation depending on frequency is, if I understand correctly, a quantum issue. https://en.wikipedia.org/wiki/Photoelectric_effect https://en.wikipedia.org/wiki/Photoelectric_effect That is, quantum theory explains why there is such a thing as nonionizing electromagnetic radiation.
jwmerrill 12y ago
It's also possible to have faint blue light, and bright red light. Sound energy is quantized too, but for audible frequencies, hf is small compared to kT, so quantum effects are swamped by thermal effects.

pdonis 12y ago

"More energy" can mean two different things. In the case of sound, only one of those meanings--the amplitude of the wave--is relevant: soft sounds have lower amplitude and loud sounds have higher amplitude. The frequency of the wave, as jessaustin pointed out, is independent of how soft or loud it is, and for sound, there isn't any useful sense in which any frequency carries "more energy" than any other frequency, given that the amplitudes are the same. With regard to light, however, there is a second possible meaning for "more energy": not a higher amplitude of the classical wave, but more energy contained in a single quantum of the light (i.e., a single photon). Light of higher frequency (toward the blue/ultraviolet end of the spectrum) has more energy per quantum; light of lower frequency (toward the red/infrared/microwave end of the spectrum) has less. This sense of "more energy" is still independent of wave amplitude: wave amplitude corresponds to number of photons (more precisely, the number of photons is proportional to the amplitude squared). So you can have faint light (few photons) of high frequency (more energy per photon); or you can have bright light (lots of photons) of low frequency (less energy per photon). The total energy contained in the light depends on both of these factors; but the total energy in the light is not the crucial factor involved (see below). It turns out that, when you are looking at the possible effects of radiation on the body, the energy per photon is the most important factor, because that is what determines what kind of interactions the radiation can have with the molecules that make up your body. UV, X-rays, and gamma rays have enough energy per photon to break apart the chemical bonds that hold together things like proteins in your cells; that's what makes them potentially cancer-causing. Microwaves, OTOH, don't have enough energy per photon to break chemical bonds; all they can do is make the molecules, like proteins, vibrate more rapidly, which just means heating them up.

thinkling 12y ago
Thanks.

drostie 12y ago

It's worth understanding that the light which you see is the result of many photons -- though it's usually estimated that maybe, if our rod/cone cells were a little more sensitive, they might have been able to perceive individuals. So actual energy/mass transfer comes from a lot of particles each offering their own individual kicks. So the overall intensity is not just (energy per kick) but really (number of kicks) * (energy per kick). The number of kicks counts. Phonons, for sound, actually are quantized in the exact same way as for light -- just replacing the speed of light c with the speed of sound in the medium v. (Both arguments are based on analogies to a harmonic oscillator problem which every quantum mechanics course works out; basically any vibration in the bottom of a potential well X can be dealt with by approximating that potential well with a parabola, which leads to particles, we could call them X-ons, of one form or another. Quantum mechanics makes all particles act like waves but also makes all waves quantize into particles.) So, long-story-short, the physics is the same, and long wavelength phonons also carry less energy per kick than shorter ones, just like photons. That's not the difference, and the E = h f equation is still the same. But for phonons it's more about numbers. The biggest sound particles you can hear correspond to 20,000 Hz frequencies. The smallest light particles you can see correspond to 700 nm wavelengths or 10^15 Hz frequencies. So photons are more than 50 billion times more energetic than phonons, as a rule. This changes a lot of features about them. For example, every degree of freedom in a system has some average energy due to just random energy-sharing, which is known colloquially as the "temperature" of the system. If you work out how many 20kHz phonons there are simply due to a room being at room temperature, it works out to about 300 million. Any detector which registers 20kHz phonons, including the hairs in your ear, have 300 million of those phonons in them just from random thermal excitation. On the other hand, the red light receptors in your eye have on average (300 million) / (50 billion) = 0.006 photons in them on average, meaning that if you look at a thousand of those photon-absorbers you'll see on average only six of them which are excited randomly just by thermal excitations. So we can, at room temperature, detect individual photons, where we can't detect even ten thousand phonons at once without it getting hopelessly buried in thermal noise. (But our eyes simply do not fire the neurons when a light-sensitive cell only sees one photon.)