Unperturbed propagation of a particle?

Discussion in 'Physics & Math' started by al onestone, Jul 8, 2013.

  1. al onestone Registered Senior Member

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    What is the explanation given when one questions how a particle propagates without loss? For example, a photon stays at its velocity which is determined by the index of refraction of the medium that it traverses. This could be explained as simply being due to its inertia. But, let's say the photon is travelling through a vaccuum at c, then it hits a glass plate and "slows down" while traversing the plate, but then as it exits the plate it "speeds up" to its original velocity c. Assuming we are in a lab with a fixed frame of reference, how do we explain this unperturbed propagation?
     
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  3. RJBeery Natural Philosopher Valued Senior Member

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    My response would be that the photon never loses velocity; it is absorbed and re-emitted by electrons in the refractive medium, and maintains its original velocity of c between electrons. The time spent during these processes determines the refractive index. This means that the change in momentum of the medium itself upon its initial absorption event is counter-balanced by the change in momentum (in the opposite direction) by the final emission event.
     
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  5. al onestone Registered Senior Member

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    This is a good explanation. If the photon is constantly absorbed and re-emitted then it gives no energy to the medium via radiation pressure, and it also explains why the photon is emitted at a velocity which is greater than the "percieved velocity" inside the medium. I wonder if there is a way to back this up? Possibly the electron density in the medium (along any straight line within the medium) could be shown to be in direct proportion to the index of refraction.

    However, this model fails when considering that the absorption is a measurement which would reduce the position vector of the photon to a single eigenvalue and destroy its state. But if it is re-emitted in the initial position state then it could be possible. But then again, each absorption's collapse is irreversible, therefore any "path information" concerning the history of the photon would also be erased. And we know from interferometry that the insert of a plate does not destroy path information, but measurements do. Therefore the propagation of a photon through a plate cannot be modelled with measurements like absorption.
     
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  7. RJBeery Natural Philosopher Valued Senior Member

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    The model does not necessarily fail. We simply differentiate between transparent media and opaque/translucent media. In the former, pathing information (the original velocity vector) is propagated, maintained and unknowable; in the latter, it is destroyed and effectively "measured". We could consider the two types of photon absorption/re-emission as two separate phenomena.
     
  8. al onestone Registered Senior Member

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    I would say you're going out on a limb as concerns

    because measurement is a well defined process in QM, but what other explanation do we have? In order to describe propagation through a medium we require an alternative, and your model of absorption/emission is helpful. I will definitely keep this in mind.

    Anyone else have an alternative view?
     
  9. przyk squishy Valued Senior Member

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    Absorption in itself is not a measurement. It is simply a type of interaction that would ultimately be described by a relevant quantum field theory such as quantum electrodynamics.


    Actually it's arguably the least well defined part of any mainstream physical theory. It's simply not defined what a measurement is or objectively under what conditions state collapse occurs. It's not even clear that we actually need the measurement/state collapse postulate at all (there are interpretations of quantum physics that deny wavefunction collapse entirely and try to view measurement as interaction that results in an observer becoming entangled with the system they're observing).
     
  10. al onestone Registered Senior Member

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    przyk

    My take on this argument is that we need an irrational process (collapse) to take place in measurement or else interference would not work. There would be no other explanation as to why interference necessitates a lack of path knowledge, indistinguishability. If measurement was rational, it would be reversible, and then you would be able to get back the original state of the measured system which would give you the path knowledge back. Measurement permanently erases path information because it is irreducibly irrational.

    The only way out of this is to have measurement as rational yet not reversible. This could be the case but there is no such proof.

    And remember, this whole "God doesn't play dice" thing is supported by people who have an inheirent hatred of the irrational. They've argued hidden variables. They've argued about a multiverse. They've argued interpretations

    But they would happily through all these interpretations under the bus for a different one, as long as it gives a complete rational account of collapse.

    Why don't we just accept the irrational statistical interpretation that the experiments tell us quantum mechanics is? It reeks of audacity to try and "tell" the laws of physics. Let the laws of physics tell us.
     
  11. RJBeery Natural Philosopher Valued Senior Member

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    Yes, I'm throwing in my personal views here. I think there is a physical distinction between an absorption/emission process by electrons in a transparent medium vs the same process in an translucent one; the case of photons reflecting off of a surface may either be a third type of physical interaction or a special case of the latter one. The first preserves the photon's original path, the second effectively randomizes it, and the third preserves the angle of incidence. The latter two cases would yield a net transfer of momentum to the medium.
     
  12. RJBeery Natural Philosopher Valued Senior Member

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    I feel that this is what most of Physics has done; "accept" the irrationality of things after decades of trying to makes sense of them. I'm simply too naive to give up quite yet.

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  13. przyk squishy Valued Senior Member

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    What's an "irrational" process? Do you mean random?


    It has already been explained to you that this "lack of path knowledge" rule is not part of the axiomatic formulation of quantum theory and is best thought of as a rule of thumb. To the extent it is true, it can readily be derived from the rest of quantum theory. In particular it can be understood in terms of entanglement/decoherence.


    I don't see the issue. Decoherence is always reversible in theory, but would typically be practically (e.g. thermodynamically) irreversible.


    If by "irrational" you mean "randomness", then no, not all interpretations of quantum physics are motivated by trying to restore determinism.
     
  14. przyk squishy Valued Senior Member

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    I don't think that's a fair assessment, given there's a fairly substantial segment of the quantum information community working on quantum foundations.

    The problem is that there's little to go on, since there's no clear contradiction in quantum physics indicating that something or other will definitely have to go. You could take the view that the problems with quantum physics were largely solved in the 1950s by Hugh Everett or David Bohm. On the other hand, it could just as well be that quantum physics is wrong and will eventually need to be replaced. But if quantum physics breaks down on some experimental scale far beyond what's technologically accessible at the moment it could be a very long time before we get the experimental clues we need to know how to proceed.
     
  15. Farsight

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    I know I'm late to the party, but maybe I can volunteer something useful.

    That it's a wave propagating through space. It isn't some billiard-ball thing that you can speed up or slow down.

    That isn't quite true, the photon still travels at c through glass. RJ said it get absorbed and remitted, and as you said, that isn't quite right either. Maybe the best way to say it simply is to say its path isn't dead straight any more.

    You have to take care with this. A photon conveys inertia in that a radiating body loses mass and the absorbing body gains it, and the photon does have an "inertial mass", but inertia is usually associated with rest mass which doesn't apply for a photon.

    We just say the photon had to zigzag a bit when going through the glass. But again, don't think of it as some kind of billiard-ball, think of it as a wave, something like this. The top half shows your typical sinusoidal electromagnetic wave, the bottom half shows the integral, which is the four-potential. See the wiki derivation from electromagnetic theory: where you can read this: "the curl operator on one side of these equations results in first-order spatial derivatives of the wave solution, while the time-derivative on the other side of the equations, which gives the other field, is first order in time". You take the spatial derivative of potential for the E waveform, it's the "slope of the hump". For the B waveform which is usually drawn orthogonal to E, you take the time derivative, it's the "rate of change of slope". Also note that the photon is wherever the grid lines are curved, it is not some localized point-particle. Think seismic waves rather than billard balls to get the general idea. Also see In praise of weakness.

    If I might go out on a limb, I'd say wavefunction collapse isn't something mysterious, but instead is related to the optical Fourier transform, see Steve Lehar's webpage at http://sharp.bu.edu/~slehar/fourier/fourier.html but note that it seems to be down at the moment.
     
  16. RJBeery Natural Philosopher Valued Senior Member

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    I know our views on this differ, but I'll reiterate my objections to yours here. If the photon's "apparent" velocity is reduced simply because it's bouncing around in the glass then what would be the mechanism to preserve it's original direction of travel? Why wouldn't all materials become translucent?
     
  17. Tach Banned Banned

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    So, Duffield, you are back to doing physics by picking random images off the web, right?
     
  18. brucep Valued Senior Member

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    It's not a 'fair assessment' under any circumstance. The fair assessment would need to include the huge experimental literature which supports the predictions of QM. Berry's assessment is the type I associate with intellectually dishonest cranks. You really think we need more technology to conclude the predictions of QM are valid? What we need further technology to test are predictions for a quantum universe at energies associated with the Planck era and the center of black holes. Probably not going to get to test many predictions associated with such extreme environments.
     
  19. brucep Valued Senior Member

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    You really don't understand what you seem to think is irrational. Typical nonsense from you.
     
  20. wellwisher Banned Banned

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    Photons are in two references at the same time. They travel at C, but can display inertial reference parameters that do not coincide with the length and times predictions of SR for something moving at C. Something moving at C, according to SR, should look like a point in our reference, yet photons moving at C, can have a range of finite expressions called wavelength, that depart from point.

    As an analogy, if you had a wave tank with a wave of one meter in our reference, and caused the tank to move at C, the wavelength would appear to shrink to zero in our reference. Photons don't do this, but will display a range of expressions greater than zero, at C, from gamma to large radio waves. Photons have one leg in C, and the other leg in finite reference, with the two legs connected.

    The C-state of the photon is connected to the particle state and this is maintained at C. What changes are the finite leg or waves components when photons interact with matter. Atoms are defined by wave functions, instead of particle functions, such that only the finite leg wave aspect of the photon changes, with the particle ignoring this and maintaining C.

    The reason this is so is, C is the ground state of the universe. It is the only reference that is the same in all references. It is the zero point from which all finite wave functions in space-time emanate.

    Picture it this way. Say we have a number of devices all stopped in our reference. We give some velocity, some rotation and some accelerations. There is one ground state that is the same for all. But say we were on each of the devices and don't know there is a ground state but we see each moving relative to each other, we may use our own reference the ground state; earth centric. Photons maintain the original reference leg and another for finite reference since mass can't go C.
     
  21. brucep Valued Senior Member

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    It's a bad analogy from somebody who likes to make up physics that is already well understood. A little less bullshit from you. Please.
     
  22. ananymousse Banned Banned

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    Our physical bodies possibly are not able to recognize the phase change through certain or most materials.
     
  23. Tach Banned Banned

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    It is not clear why you keep posting nonsense.
     

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