Is Gravity Faster than Light?

Discussion in 'General Science & Technology' started by Bowser, Jan 18, 2018.

  1. Write4U Valued Senior Member

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    The speed of light is a constant. It can slow down travelling through a medium but gravity is not a medium, it is a warp in spacetime. Gravitational pull can bend light but that does not mean light slows down.

    I believe someone else made a prior post that passing through a gravity field does not slow anything down. The net momentum is preserved. I'm not sure if I posited this correctly, but that was my understanding. In fact, for massive objects gravity may well be used for a sling shot effect, actually using gravity to speed up an object, except light of course.

    Black Holes may be different, but the lensing effect (caused by gravity) does not interfere with the speed of light, IMO..
     
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  3. Write4U Valued Senior Member

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    But I thought that as soon as the photon hits the plate its wave function collapses.
    So how can you arrive at a conclusion that there are residual interference fringes if the original wave function has collapsed altogether?

    Are you proposing that the wave function persists after it has collapsed?

    If there is any sign of residual wave function, that would tend to support Bohm's Pilot Wave, which continues independent of the collapse of a particle's own wave function.
     
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  5. RainbowSingularity Valued Senior Member

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    Thanks, i am quite interestedin this concept.

    soo if the speedof light is not effected by gravitational lensing... is time altered ?
    thus is there a different time other than the basic light speed measurement relative to distance and visual anotation of the observer ?
    i.e
    observer looking through gravitational lensing at a planet...
    sees things 1000 years ago
    without gravitational lensing sees things 1200 years ago on the same planet ?
     
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  7. Write4U Valued Senior Member

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    I prefer to do my own research, it makes for better discussions...

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    https://www.quora.com/What-is-smaller-a-quark-the-Higgs-boson-or-a-photon

    I see no problem with post #61 in that respect. In context of the actual posit ;
    More Energy = smaller size, More Mass = larger size. That's wrong?

    As to post #62, please explain where it is wrong and why.....

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    Please note that I already qualified it as speculative.
    But instead of providing a correct answer, you just repeat my original qualifier. Not very helpful to me or others.

    Do you know the answer? Then enlighten me, please......

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    Last edited: Mar 1, 2018
  8. Q-reeus Valued Senior Member

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    Some confusion there. As soon as a photon hits the (absorbing) screen aka 'plate' it is annihilated i.e. ceases to exist. It's position on the screen is partially random, in keeping with the probabilistic nature of QM, but is partially ordered in keeping with the wave interference picture. Hence the observed one-photon-at-a-time speckle pattern that starts out looking completely random, but over time reveals the interference fringes expected of a classical wave picture.
    In the Feynman path integral picture, a photon takes all possible paths, through both slits, and thence to the screen. But only paths that have a phase difference close to an integer multiple of the photon wavelength will significantly reinforce and consequently have a high probability of corresponding to a screen hit.
     
  9. Write4U Valued Senior Member

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    And what happens after the entire wave function has collapsed when the photon hits the plate.?
    You mentioned "residual" wave function. Of what?
     
  10. Q-reeus Valued Senior Member

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    Many probably most physicists maintain a photon does not possess a wavefunction since as a massless 'particle' there is no position operator like as applies to massive particles e.g. electron. See last main para p2 this article, whose authors take a contrary view: https://arxiv.org/abs/quant-ph/0604169

    At any rate once the photon is annihilated there is obviously nothing left to have any possible wavefunction. The energy-momentum of said photon has been transferred to whatever in the screen absorbed it. Could be an expelled electron, or a chemical change e.g. silver halide molecule decomposing to pure silver etc. - depending on the particular nature of detection screen.
     
  11. Write4U Valued Senior Member

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    That article assumes a wave function associated with photons. If it did not, it would answer to Bohm's Pilot Wave model.
    Ok, I agree with that (if the wave function is associated with the photon).
    Now we enter secondary or even tertiary conditions as well as changing the chemical composition of the receiving medium. That's changing the experiment and the way you stated it, sounds speculative.
    Any evidence of such phenomena?
     
  12. Q-reeus Valued Senior Member

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    How is it 'changing the experiment'? What restrictions are you imposing on the extent of the experiment? Have a careful read through:
    https://www.quora.com/In-the-double...ding-is-the-wave-state-maintained-Why?share=1
     
  13. Write4U Valued Senior Member

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    Consistency to assure reliable results.
    I'll need some time to absorb all that. But my intuition tells me that the more complicated a theory (experiment) becomes, the more complicated the proof becomes. The inclusion of mind/brain functions, introduces a completely subjective aspect.

    Moreover I already introduced an example of a room filled with photons, but that requires a constant emitter. Turn the emitter (light) off and the room instantly becomes dark. i.e. all wave functions collapse instantly.

    I wonder why that poster did not replace the room windows with narrow slits. IMO, that might tell us more about the double slit expariment.

    But we are talking about the behavior of single photons. Thus using an example of a (any) lit room does not tell us anything about the double slit experiment.
     
  14. Q-reeus Valued Senior Member

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    There were several different answers including one where a wavefunction is never part of the picture. None where a lit room is invoked iirc. It was the first one I thought more relevant to your outlook. I suggest thinking about that explanation some more. It may lift at least some of your continued confusion.
     
  15. exchemist Valued Senior Member

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    This is not what happens. When the emitter is turned off, the last photons travel towards the walls as usual and are duly absorbed, as Q-reeus explained earlier by whatever it is they hit and are absorbed by. There is no instantaneous "collapse". The absorption process is well described by quantum mechanics. Both the travel of the photon and the absorption process take a finite time to occur.
     
  16. Write4U Valued Senior Member

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    This makes no sense to me. Is this assuming that when an obervation device is in an off-state it still causes decoherence? Does that mean that the observation device in off-state is somehow still projecting something that will cause decoherence ?​

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    The right side of the picture depicts a photon detector which is "attached' to the slits by leads and thus draws energy from the photon as it passes the slit, clearly removing its wave function, but replacing it with an electromagnetic "thread", which is a different state altogether.
    Seems to me;
    Even if you turned photon detector "off", but leave the leadst connected with the slits, the configuration will still produce a field which affects the energy of the photon and possibly decoherence. I should like to know what happens when you remove the leads from the slits altogether. Will the detector still interfere with the photon?

    Thus, IMO, it is not the observation device itself which is causal to the wave collapse. Its the method, the attached leads, which results in the collapse at the slits.

    Can they not use some gas which will show the interference pattern without needing a connection to the slits and thus avoid adding (another field) to the configuration of the slits?

    Perhaps this might also reveal what happens in a large area such as a room.
     
    Last edited: Mar 1, 2018
  17. Write4U Valued Senior Member

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    OK, I can visualize that.
    If we were to cover all walls with black photographic plates, would the walls change color to white or show a decoherent scattering of white dots?
    How many photons would it take? If we used a very small light source or limit the time of emision, would the walls turn gray?

    I am drawing on my experience with B/W photography. Intensity (energy) of light and exposure time both affect the results when developing a negative which determines the black and white shading on the photograph.
    I assume this problem of negative to positive has been eliminated by our digital cameras, so that extra step can be ignored.
     
    Last edited: Mar 1, 2018
  18. exchemist Valued Senior Member

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    These are remarkably silly questions for someone claiming to be familiar with photography. I can't be bothered to answer.
     
  19. Write4U Valued Senior Member

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    But earlier, I asked a serious question and received no answer either.......

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    Anyone?
     
  20. Q-reeus Valued Senior Member

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    Can't really follow the logic there but will remark that assuming the gas strongly interacts with a photon initially enroute to the screen (photon absorption then random emission of another photon by a gas molecule), the screen is then irrelevant. No interference pattern could be expected.
     
  21. Write4U Valued Senior Member

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    I was thinking of a medium (gas) that does not interfere with the photon and it's interference patterns can actualy be demonstrated and can be measured without interfering with the photon itself.
     
  22. Q-reeus Valued Senior Member

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    Can't envisage how that would work. Many double slit experiments have been carried out - in normal air filled lab settings. The air molecules only weakly interact via tiny collective dipole-moment fluctuations. That amount to no more than a very small departure from unity relative permittivity of a perfect vacuum. Since there is no inhomogeneous bias in such permittivity shift, the net effect is zero i.e. experiment might as well be done in perfect vacuum.
    And btw this is getting right away from OP topic - agreed?
     
  23. Write4U Valued Senior Member

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    Thanks for that explanation and your patience...

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    And I agree, but to me it is valuable info about the nature of light (photons), which is part of the OP question.

    Next comes Gravity.......

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