Post by: lenadorap on 25/11/2011 03:56:07
Post by: Soul Surfer on 25/11/2011 08:32:32
Post by: JP on 25/11/2011 14:14:39
Post by: Soul Surfer on 25/11/2011 18:45:40
Post by: imatfaal on 25/11/2011 19:08:41
Post by: simplified on 26/11/2011 18:52:26
Post by: yor_on on 26/11/2011 19:34:30
You won't see a photon anywhere, except in its release of 'energy' dying, as I understands it. But as we also seem to assume that waves can 'reflect' elastically, changing 'momentum' (direction) but not 'kinetic energy'?
It most definitely should be a duality.
Post by: simplified on 27/11/2011 13:30:46
You mean in its annihilation?No. I can know way and speed. And at all maybe photon is small tornado in gravitational field of universe. [:P]
You won't see a photon anywhere, except in its release of 'energy' dying, as I understands it. But as we also seem to assume that waves can 'reflect' elastically, changing 'momentum' (direction) but not 'kinetic energy'?
It most definitely should be a duality.
Post by: yor_on on 27/11/2011 21:23:13
Post by: simplified on 29/11/2011 14:39:59
I don't agree there Simplified, you can deduce a path, but not 'know' it. It's enough comparing it to a balls path to see the difference there. And all deducing involves statistical approaches, as far as I know. You can only measure a photon once, in its annihilation. That you can define them as being 'identical' of property and then assume that by measuring a large amount coming from a same source, at different positions, do not give you a photons path, as various 'slit experiment' also can show you, depending on interpretation.I think you confuse different waves.
Photon has quantum wave.Such wave is wave in a narrow corridor of travel.
Post by: yor_on on 29/11/2011 23:03:40
As for if a photon is a 'quantum wave'? Don't know, but I doubt it. A photon is not a wave, it is a quanta of energy.
Post by: simplified on 30/11/2011 14:50:40
Then quanta of energy of electron is photon. Does it travel at 'c'?
Post by: yor_on on 30/11/2011 22:43:54
Post by: simplified on 01/12/2011 13:33:00
yes, if you are talking about a 'photon', not a electron.Can electron have quanta of energy? What is quanta of energy?
Post by: JP on 01/12/2011 16:31:49
Photons are, by definition, the smallest quanta of energy you can get from an electromagnetic field of a given frequency. If I shine a green laser at you, the green photons that make it up are the tiniest possible piece of energy you can pick out of that beam. Because they are defined as energy quanta, they do not have a simple description in terms of how they spread out over space, so you can't properly draw a photon wave traveling through space.
Post by: simplified on 19/12/2011 14:57:39
Quantum states are often defined in terms of quantized values: physical quantities such as energy, momentum, position, angular momentum, and so on that take on only discrete values. You can't generalize to all particles about what values get quantized, but for each type of particle and situation there are rules.Can you properly draw a corridor of travel of photon in space?
Photons are, by definition, the smallest quanta of energy you can get from an electromagnetic field of a given frequency. If I shine a green laser at you, the green photons that make it up are the tiniest possible piece of energy you can pick out of that beam. Because they are defined as energy quanta, they do not have a simple description in terms of how they spread out over space, so you can't properly draw a photon wave traveling through space.
Post by: JP on 19/12/2011 15:02:57
Post by: simplified on 19/12/2011 15:38:05
No. Photons don't follow classical paths through space.Then I don't understand why shadows of objects can be predicted? :D
Post by: yor_on on 19/12/2011 15:56:35
Post by: JP on 19/12/2011 16:48:26
No. Photons don't follow classical paths through space.Then I don't understand why shadows of objects can be predicted? :D
You can predict where billions of photons will go, or where a photon interacting with matter (a photon in an optical fiber, for example) will go. But you can't predict the path taken by a true photon (meaning one that is described as the smallest packet of energy of an electromagnetic field).
Post by: simplified on 19/12/2011 16:54:10
There are different approaches Simplified. Some do want to see it as one path, but that isn't how the experiments I know of describe it. But all agree in that the source (laser) and the sink (detector) are connected, and they also, like Feynman's interference or otherwise, leave only one path existing as the defined outcome. And it's those 'final' paths that then defines the shadow you cast.Then your astronomers can predict an eclipse only after this event. ;)
Post by: JP on 19/12/2011 17:06:33
There are different approaches Simplified. Some do want to see it as one path, but that isn't how the experiments I know of describe it. But all agree in that the source (laser) and the sink (detector) are connected, and they also, like Feynman's interference or otherwise, leave only one path existing as the defined outcome. And it's those 'final' paths that then defines the shadow you cast.Then your astronomers can predict an eclipse only after this event. ;)
Simplified, there is no problem with making predictions if intensity by averaging over enormous numbers of photons.
If the sun released only one photon, they couldn't predict the eclipse.
Post by: simplified on 20/12/2011 16:15:52
You try to prove undefiniteness by another undefiniteness. Let know a finding of a concrete source (we should know a solar point which will radiate the one photon).There are different approaches Simplified. Some do want to see it as one path, but that isn't how the experiments I know of describe it. But all agree in that the source (laser) and the sink (detector) are connected, and they also, like Feynman's interference or otherwise, leave only one path existing as the defined outcome. And it's those 'final' paths that then defines the shadow you cast.Then your astronomers can predict an eclipse only after this event. ;)
Simplified, there is no problem with making predictions if intensity by averaging over enormous numbers of photons.
If the sun released only one photon, they couldn't predict the eclipse.
I can not agree with you because don't know the experiments,which prove you are right. :-'(
Post by: imatfaal on 21/12/2011 17:21:56
i. we can and do predict eclipses (you cannot find fault with our ability to do that)
ii. using the sun as a source for a single photon is a bonkers idea (you are the person raising the idea that we must be able to reduce the sun to a singular source - it isn't, and we don't try to make it so)
Post by: simplified on 21/12/2011 19:39:54
SimplifiedI thought Yor_on and JP were serious. :I
i. we can and do predict eclipses (you cannot find fault with our ability to do that)
ii. using the sun as a source for a single photon is a bonkers idea (you are the person raising the idea that we must be able to reduce the sun to a singular source - it isn't, and we don't try to make it so)
Post by: Bill S on 29/12/2011 16:39:42
Quote
I thought Yor_on and JP were serious.
You lost me there, Simplified, just when I thought I'd kept a handle on the thread. :(
Post by: Bill S on 29/12/2011 18:09:08
Quote from: JP
Time doesn't actually stand still for a photon according to our theories.
Could we have some more information about these theories, please.
Post by: imatfaal on 30/12/2011 10:29:55
Quote from: JPTime doesn't actually stand still for a photon according to our theories.
Could we have some more information about these theories, please.
The time standing still is an extrapolation of SR - but it is not really a correct one.
SR uses the formula
For a massive object as it approaches c the bottom of the fraction will approach zero - and time dilation will become more and more pronounced when viewed from a local FoR in relative rest. It might seem for an outside observer that time has completely stopped as an object reaches c - but SR is quite clear on the fact that a frame of reference of the photon or any other massless particle which must be travelling at c is not defined (you need to be able to define an inertial frame where the photon is at rest and that is a nono), you cannot just extrapolate that equation.
Post by: simplified on 30/12/2011 15:08:10
You were lost earlier. They tried to prove probable inability of the moon to block a way of a visible photon(length of wave has meaning) even by central part . :PQuoteI thought Yor_on and JP were serious.
You lost me there, Simplified, just when I thought I'd kept a handle on the thread. :(
Post by: JP on 30/12/2011 15:50:16
Post by: simplified on 02/01/2012 18:29:17
I suggest you reread what we were describing, simplified. You keep thinking of photons as "little bullets," which they aren't.If we do not know height of photon wave then what creates an eclipse? Diameter of the Moon (parallel to a way of a photon) or diameter of the Moon (perpendicular to way of photon)? :-\
Post by: JP on 02/01/2012 18:37:09
Its weird and nonintuitive, but that's quantum mechanics.
Post by: simplified on 02/01/2012 18:58:58
Post by: JP on 02/01/2012 21:25:47
The simple question is too complex(difficult) for the backward science. :D
Nope. But if you aren't willing to put effort into understanding it, it's not worth trying to explain it further! *<;o)
Post by: Phractality on 02/01/2012 23:59:22
If photons travel at the speed of light, a clock in a photon shall be stand-still. If it is stand-still photon shall not change or undistructable. How this could be explained?As a moving clock passes a succession of observer's clocks (which are stationary and synchronized in the observers reference frame) the time difference between the moving clock and each successive observer's clock will change. The moving clock's time will advance more slowly than the time on successive observer's clocks.
The closer a clock's relative speed is to the speed of light, relative to the observer's reference frame, the slower that clock appears to be compared to the succession of observer's clocks that it passes. A clock is a mechanism made of matter. Matter cannot move as fast as light. Even if a clock is moving at .99999999999999999999999999999999999999 c relative to an observer, a photon will still pass that clock at the speed of light in the clock's reference frame. It doesn't even matter which dirction the photon is moving in the clock's reference frame, the speed of the photon is always the same. So it makes no sense to talk about a clock moving along with a photon. The speed of the photon is always the same relative to any clock.
Post by: LetoII on 18/01/2012 04:09:09