Seasons
This is a forum or general chit-chat, small talk, a "hey, how ya doing?" and such. Or hell, get crazy deep on something. Whatever you like.
Posts 3,998 - 4,010 of 6,170
Now, let me define “propagating-1” as: “being somewhere where it wasn’t a little while ago.” Then it is true that the wave function of the entire ocean doesn’t propagate-1, since the ocean is (let us assume) already wavy everywhere. Likewise, in Quantum Mechanics, there are wave functions employed which have a non-zero amplitude everywhere. For example, the one-dimensional function e^i(kx-wt), often taken to be the wave function of a single free particle, has non-zero amplitude everywhere. So it doesn’t propagate-1.
But now imagine an infinite, flat ocean, with an infinite number of individual crests, all of the same amplitude, parallel and heading in the same direction at a certain speed. The entire system can be said to be “propagating,” though it is not propagating-1. It is ‘propagating’ in the same direction as the individual crests. A surfer would find that he can only surf on one side of the crests. In the case of e^i(kx-wt), the momentum of the ‘particle’, if it is not zero, will have the positive direction if k is positive, otherwise the negative direction. The sign of k tells us which way the individual crests are going. So when I say that a wave function “propagates-2”, I mean that it is composed of individual crests, each of which propagates-1. The propagation-2 of the wave function is the totality of propagation-1’s of all the crests. So the wave function of all ocean waves, everywhere, does not propagte-1, but it does propagate-2. Likewise for e^i(kx-wt). I think that Penrose would grant that non-trivial wave functions propagate-2.
[I’ve written the word “propagation” so often that it’s beginning to sound like a pregnant Mozart character.]
Now, let us consider a third sort of example. Let’s say we are in a soundproof room, and someone strikes a tuning fork and lets it ring. Suppose there is no other sound in the room. The sound quickly fills the room, it is everywhere at once, so it cannot propagate-1. It does propagate-2, however: each individual crest starts at the fork, spreads outward as a sphere until it his a wall, and then (let’s say the walls are perfect absorbers) stops. But when the tuning fork was first struck, the sound spread out from the fork to the rest of the room. That is, the region of non-zero amplitude (loudness) spread out from the fork to the rest of the room. I will call this sort of thing “propagation-3”. More generally, I would say that a wave function propagates-3 if and only if there is some point at which its amplitude changes. I think Penrose would grant that there is such a thing as propagation-3 of the wave function (Psi) in Quantum Mechanics.
Another example: imagine a police car with its siren on going down a road. The amplitude of the sound wave function dies away as it gets far from the car. Let’s say that the sound issues from the siren at 10 decibels. Then we can say that the region in which the sound has a loudness of 6 decibels or more travels down the road, i.e., the sound propagates-3.
OK, now when I said that Psi propagates, I usually meant it in the sense of propogates-2 or propagates-3. Penrose’s argument, though, is an argument against propagation-1. I will try to be explicit in the future, since there may be a source of misunderstanding here, and I humbly request that you do the same.
In particular, the two-slit experiment is very similar to the tuning fork example; in fact, if you use a tuning fork as the source of waves, and have a double slit, you will get the same sort of interference pattern. But let’s say we are using light; let’s say we have a flashlight aimed at the slits. Initially, let’s say, the flashlight is off. Psi now has amplitude zero everywhere (for sound, amplitude is loudness; for light, amplitude is brightness, for Psi, it is just “amplitude”). No scintillae appear on the screen. Now we turn on the flashlight; Psi now propagates-3 from the bulb of the flashlight, to the barrier, through the slits, and (interfering with itself in the process) onto the screen. When it arrives, we begin to see scintillae. There is a tiny delay between turning on the flashlight and the appearance of scintillae, since light travels at a finite velocity. If we turn the flashlight off, then (after a tiny delay) scintillae cease to appear. If there were no propagation-3, why is there a delay in the appearance of the scintillae when the flashlight is turned on, and another delay in their disappearance when it is turned off? For that matter, wouldn’t we see scintillae regardless of whether the flashlight was turned on or off? If Psi (the wave function) doesn’t go through the slits (but not the rest of the barrier), why would the presence of the slits have any effect on the distribution of scintillae on the screen?
OK, let’s say we turn the flashlight on and leave it on for awhile, at the same amplitude. Then after a moment a kind of equilibrium is reached; the amplitude of Psi remains the same everywhere. Thus there is no more propagation-3. There is, however, still propagation-2. Because the wave-function is going from the flashlight to the screen, the diffraction and interference happen on the side of the barrier opposite to the flashlight. Light diffracts when it passes through a slit. Well, perhaps you or Penrose will come up with an entirely different explanation of the interference patterns; but then you will have refuted quantum Mechanics, rather than explaining it.
Posts 3,998 - 4,010 of 6,170
Now, let me define “propagating-1” as: “being somewhere where it wasn’t a little while ago.” Then it is true that the wave function of the entire ocean doesn’t propagate-1, since the ocean is (let us assume) already wavy everywhere. Likewise, in Quantum Mechanics, there are wave functions employed which have a non-zero amplitude everywhere. For example, the one-dimensional function e^i(kx-wt), often taken to be the wave function of a single free particle, has non-zero amplitude everywhere. So it doesn’t propagate-1.
But now imagine an infinite, flat ocean, with an infinite number of individual crests, all of the same amplitude, parallel and heading in the same direction at a certain speed. The entire system can be said to be “propagating,” though it is not propagating-1. It is ‘propagating’ in the same direction as the individual crests. A surfer would find that he can only surf on one side of the crests. In the case of e^i(kx-wt), the momentum of the ‘particle’, if it is not zero, will have the positive direction if k is positive, otherwise the negative direction. The sign of k tells us which way the individual crests are going. So when I say that a wave function “propagates-2”, I mean that it is composed of individual crests, each of which propagates-1. The propagation-2 of the wave function is the totality of propagation-1’s of all the crests. So the wave function of all ocean waves, everywhere, does not propagte-1, but it does propagate-2. Likewise for e^i(kx-wt). I think that Penrose would grant that non-trivial wave functions propagate-2.
[I’ve written the word “propagation” so often that it’s beginning to sound like a pregnant Mozart character.]
Now, let us consider a third sort of example. Let’s say we are in a soundproof room, and someone strikes a tuning fork and lets it ring. Suppose there is no other sound in the room. The sound quickly fills the room, it is everywhere at once, so it cannot propagate-1. It does propagate-2, however: each individual crest starts at the fork, spreads outward as a sphere until it his a wall, and then (let’s say the walls are perfect absorbers) stops. But when the tuning fork was first struck, the sound spread out from the fork to the rest of the room. That is, the region of non-zero amplitude (loudness) spread out from the fork to the rest of the room. I will call this sort of thing “propagation-3”. More generally, I would say that a wave function propagates-3 if and only if there is some point at which its amplitude changes. I think Penrose would grant that there is such a thing as propagation-3 of the wave function (Psi) in Quantum Mechanics.
Another example: imagine a police car with its siren on going down a road. The amplitude of the sound wave function dies away as it gets far from the car. Let’s say that the sound issues from the siren at 10 decibels. Then we can say that the region in which the sound has a loudness of 6 decibels or more travels down the road, i.e., the sound propagates-3.
OK, now when I said that Psi propagates, I usually meant it in the sense of propogates-2 or propagates-3. Penrose’s argument, though, is an argument against propagation-1. I will try to be explicit in the future, since there may be a source of misunderstanding here, and I humbly request that you do the same.
In particular, the two-slit experiment is very similar to the tuning fork example; in fact, if you use a tuning fork as the source of waves, and have a double slit, you will get the same sort of interference pattern. But let’s say we are using light; let’s say we have a flashlight aimed at the slits. Initially, let’s say, the flashlight is off. Psi now has amplitude zero everywhere (for sound, amplitude is loudness; for light, amplitude is brightness, for Psi, it is just “amplitude”). No scintillae appear on the screen. Now we turn on the flashlight; Psi now propagates-3 from the bulb of the flashlight, to the barrier, through the slits, and (interfering with itself in the process) onto the screen. When it arrives, we begin to see scintillae. There is a tiny delay between turning on the flashlight and the appearance of scintillae, since light travels at a finite velocity. If we turn the flashlight off, then (after a tiny delay) scintillae cease to appear. If there were no propagation-3, why is there a delay in the appearance of the scintillae when the flashlight is turned on, and another delay in their disappearance when it is turned off? For that matter, wouldn’t we see scintillae regardless of whether the flashlight was turned on or off? If Psi (the wave function) doesn’t go through the slits (but not the rest of the barrier), why would the presence of the slits have any effect on the distribution of scintillae on the screen?
OK, let’s say we turn the flashlight on and leave it on for awhile, at the same amplitude. Then after a moment a kind of equilibrium is reached; the amplitude of Psi remains the same everywhere. Thus there is no more propagation-3. There is, however, still propagation-2. Because the wave-function is going from the flashlight to the screen, the diffraction and interference happen on the side of the barrier opposite to the flashlight. Light diffracts when it passes through a slit. Well, perhaps you or Penrose will come up with an entirely different explanation of the interference patterns; but then you will have refuted quantum Mechanics, rather than explaining it.
To all:
Is anyone other than Psimagus and I interested in these QM posts? I don't want to take up huge amount of Seasons space if it's not of general interest. I've seen responses from other people in the past, so I've assumed that there is boader interest, but don't hesitate to say so if you feel crowded out.
Is anyone other than Psimagus and I interested in these QM posts? I don't want to take up huge amount of Seasons space if it's not of general interest. I've seen responses from other people in the past, so I've assumed that there is boader interest, but don't hesitate to say so if you feel crowded out.
Psimagus (3991):
[Classical Electromagnetic Theory is] Not wrong, just incomplete.
No, wrong. Max Planck invented his eponymous constant because Classical Electromagnetic Theory gave the wrong predictions for blackbody radiation. CET also said that electrons orbiting the nucleus of an atom should constantly radiate energy, and hence fall into the nucleus. CET gave the wrong predictions for the Photoelectric Effect. CET implies that all energy in the electromagnetic field would eventually drift into the high frequencies. Any example of tunnelling is a refutation of CET.
No, wrong. Max Planck invented his eponymous constant because Classical Electromagnetic Theory gave the wrong predictions for blackbody radiation. CET also said that electrons orbiting the nucleus of an atom should constantly radiate energy, and hence fall into the nucleus. CET gave the wrong predictions for the Photoelectric Effect. CET implies that all energy in the electromagnetic field would eventually drift into the high frequencies. Any example of tunnelling is a refutation of CET.
Prob123:
I actually know one guy who uses the term, "testosterone poisoning." But only one. He might use it to explain why, for example, so many more young men die in car crashes than young women.
I'm sorry if my remark contributed to some negative stereotype of females. I didn't intend it to, but that's no excuse.
I actually know one guy who uses the term, "testosterone poisoning." But only one. He might use it to explain why, for example, so many more young men die in car crashes than young women.
I'm sorry if my remark contributed to some negative stereotype of females. I didn't intend it to, but that's no excuse.
Irina, I think anyone who doesn't want to read up on the QM stuff will just learn to read around. It's a perfectly legitimate topic for Seasons. I do find it interesting even if I can only follow about half of it. My Physics studies stopped at the end of Year 11 when I decided Literature was more my thing.
Thanks, Prob123. Please do not hesitate to ask questions, make comments, etc.. I want to learn how to express myself clearly.
Thanks, Corvin. Perhaps I can express myself in a more accessible way. That's what I'm trying to do when I compare Psi to ocean waves, operatic soprano sounds, and the like.
Irina,
In everyday speech, a ‘wave’ tends to be an entity with a single crest, like a single ocean wave.
Indeed, this is mere colloquial usage, and no physics, classical or otherwise.
A surfer can only surf on one ‘wave’ at a time. But in Physics, a wave (or wave function) can have multiple crests.
MUST have multiple crests, and troughs.
For example, a sound wave produced by a voice or instrument is a whole series of crests and troughs; if an oboist (or a bagpiper) plays A-440, crests are going past your ear 440 times a second. Thus, the motion of the surface of the entire ocean can be regarded, in Physics, as the motion of a single ‘wave’ or ‘wave function.’
If it were all a single wave, that might be the case. But it is, in fact, a lot of different waves interfering with each other. There is, I would suggest, no feasible model that can regard the surface of even a relatively small area of the ocean as a single wave - there are currents and tides pulling in different directions all the time. The surf simultaneously breaks in diametrically opposed directions on either side of an island, after all.
In order to avoid ambiguity, I suggest that we use the term, “crest” for the everyday notion of ‘wave’, and ‘wave function’ for the more general notion.
I had assumed that was what we were doing, since these are the proper terms.
Or at least that we were calling a wave a "wave" and a crest a "crest", in accordance with the accepted scientific terminology.
Indeed, this is mere colloquial usage, and no physics, classical or otherwise.
MUST have multiple crests, and troughs.
If it were all a single wave, that might be the case. But it is, in fact, a lot of different waves interfering with each other. There is, I would suggest, no feasible model that can regard the surface of even a relatively small area of the ocean as a single wave - there are currents and tides pulling in different directions all the time. The surf simultaneously breaks in diametrically opposed directions on either side of an island, after all.
I had assumed that was what we were doing, since these are the proper terms.
Or at least that we were calling a wave a "wave" and a crest a "crest", in accordance with the accepted scientific terminology.
Irina,
Now, let me define “propagating-1” as: “being somewhere where it wasn’t a little while ago.” Then it is true that the wave function of the entire ocean doesn’t propagate-1, since the ocean is (let us assume) already wavy everywhere. Likewise, in Quantum Mechanics, there are wave functions employed which have a non-zero amplitude everywhere. For example, the one-dimensional function e^i(kx-wt), often taken to be the wave function of a single free particle, has non-zero amplitude everywhere. So it doesn’t propagate-1.
I hesitate to flatly contradict you, but I have to say that my understanding of Maxwell's e^i(kx-wt) has always been that it in fact describes propagating waves that do inevitably have an amplitude of zero at the points between each crest and trough, and trough and crest. A sinewave's amplitude cannot pass from a positive "crest" value to a negative "trough" value without passing through zero. Or are you proposing some sort of sawtooth wave? That would be a novelty!
I fear you are mixing this up with Psi again - Psi has a non-zero amplitude everywhere, precisely because it has no troughs (and thus has only fluctuations of positive values,) and is continuous. (And is only debatably a wave at all.)
If you have been dabbling with Hawking Radiation (I'm second guessing the source of confusion here - please ignore this if you haven't,) the zero-value is merely delayed by the acceleration exerted on the infalling particle by the black hole. It's effectively stuck with a crest that's stretching as fast as the particle is falling, until the black hole eventually evaporates in a few trillion years. Then the next trough will appear, the wave having been redshifted to within a gnat's crotchet of its life. But it's still a wave, and still with a trough to come, necessitating a point of zero amplitude at the transition from crest to trough. Just as there had been between the previous trough and the crest that stretched. This is the only scenario I have ever come across where a photon experiences an extended period of non-zero amplitude, but it is not non-zero "everywhere".
I confess - I don't know where to even begin with "being somewhere where it wasn’t a little while ago" so I won't, if you don't mind.
Propagation is propagation - it is not a massively complex concept (alright, it has some potentially abstruse ramifications, but the basic principle is unproblematic.)
There's a handy set of animations @http://id.mind.net/~zona/mstm/physics/waves/propagation/huygens1.html and a good overview of the principles @ http://www.fas.org/man/dod-101/navy/docs/es310/propagat/Propagat.htm that I would recommend.
And (I hesitate to mention, lest the multicoloured, 3-dimensional undulations should overexcite you, but it's too good a toy to keep to myself,) a most entertaining java applet @
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35
which demonstrates the electric and magnetic components of the wave excellently.
I hesitate to flatly contradict you, but I have to say that my understanding of Maxwell's e^i(kx-wt) has always been that it in fact describes propagating waves that do inevitably have an amplitude of zero at the points between each crest and trough, and trough and crest. A sinewave's amplitude cannot pass from a positive "crest" value to a negative "trough" value without passing through zero. Or are you proposing some sort of sawtooth wave? That would be a novelty!
I fear you are mixing this up with Psi again - Psi has a non-zero amplitude everywhere, precisely because it has no troughs (and thus has only fluctuations of positive values,) and is continuous. (And is only debatably a wave at all.)
If you have been dabbling with Hawking Radiation (I'm second guessing the source of confusion here - please ignore this if you haven't,) the zero-value is merely delayed by the acceleration exerted on the infalling particle by the black hole. It's effectively stuck with a crest that's stretching as fast as the particle is falling, until the black hole eventually evaporates in a few trillion years. Then the next trough will appear, the wave having been redshifted to within a gnat's crotchet of its life. But it's still a wave, and still with a trough to come, necessitating a point of zero amplitude at the transition from crest to trough. Just as there had been between the previous trough and the crest that stretched. This is the only scenario I have ever come across where a photon experiences an extended period of non-zero amplitude, but it is not non-zero "everywhere".
I confess - I don't know where to even begin with "
Propagation is propagation - it is not a massively complex concept (alright, it has some potentially abstruse ramifications, but the basic principle is unproblematic.)
There's a handy set of animations @
And (I hesitate to mention, lest the multicoloured, 3-dimensional undulations should overexcite you, but it's too good a toy to keep to myself,) a most entertaining java applet @
which demonstrates the electric and magnetic components of the wave excellently.
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