Absolute Reference Frame

Discussion in 'Physics & Math' started by Prosoothus, Mar 27, 2006.

  1. 2inquisitive The Devil is in the details Registered Senior Member

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    Yes, that can also be described as a Sagnac effect. The front mirror is moving away from the emission point while the photon is in flight, then the emission point is moving toward the photon reflected from the mirror while the photon is in flight after reflection. Same, but opposite motions, for the rear emission/mirror. Result is the two photons arrive back at the emission point simultaneously. See paper I linked to above.
     
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  3. DaleSpam TANSTAAFL Registered Senior Member

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    Hi 2inquisitive,

    I think I finally understood what you are talking about by a "linear Sagnac" effect. I still think it is a stretch, but more importantly I don't understand why you think that characterization is important. I assume that you think calling it a "linear Sagnac" effect represents some sort of challenge to SR, but I don't see it.

    -Dale
     
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  5. Montec Registered Senior Member

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    Can one apply SR from an object with a periodic or rotation motion to an object some distance away? Does SR allow for objects in different frames of reference to have rotation?

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  7. DaleSpam TANSTAAFL Registered Senior Member

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    Yes, here is an example I derived earlier in this thread:
    http://www.sciforums.com/showthread.php?p=1045127#post1045127

    -Dale
     
  8. 2inquisitive The Devil is in the details Registered Senior Member

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    My point in suggesting that the diagrams showing the Sagnac effect be looked at closely was to think of reference frames. Notice how the correct experimental results are met with the emitter in the rest frame, the receiver in the moving frame due to rotation. When the geostationary satellite is the emitter/reflecter, no rotational motion needs to be give to that frame, only to the Earth receiver that is moving due to Earth's rotation. Both the geostationary satellite and the Earth receiver are actually rotating when viewed by an observer external to the system. It is not necessary to show the emitter in motion in the diagrams because the emitted photon does not carry the forward motion of the emitter. The photon acts as if it were emitted from a stationary emitter, carrying none of the forward motion of the emitter. The receiver does rotate while the photon is in flight. It works the same way if the emitter is Earth station and the receiver is the geostationary satellite in that leg of the journey.
     
  9. DaleSpam TANSTAAFL Registered Senior Member

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    What is your point? The correct experimental results are indeed obtained in the emitter's rest frame. The correct experimental results are also obtained in the detector's rest frame and every other reference frame. This is a general principle and applies for any experiment and any reference frame that you might wish to use to describe the experiment. It's a consequence of the first postulate.

    -Dale
     
    Last edited: Jun 8, 2006
  10. Pete It's not rocket surgery Moderator

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    10,166
    But you maintain that the emitter is actually moving, do you not?
    In the chosen reference frame, the emitter has no forward motion.
    In a reference frame in which the emitter had forward motion, the path of the photon would be angled forward as well.

    What would the path of the photon look like in a reference frame in which the emitter was moving?

    The opposite conclusion is true:
    It is not necessary to show the emitter in motion in the diagrams because the emitted photon does carry the forward motion of the emitter.

    If the photon didn't carry the emitter's forward motion (which would imply the existence of an ether with some defined rest state), we would need to use the ether rest frame to find the photon's correct path.

    If we used the emitter's rest frame, the photon would be carried back by the motion of the ether.
     
  11. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    Dalespam,
    I put it in bold, Dale. Again: The photon does not retain any forward motion of the emitter.
    DaleSpam,
    Try it and see. If the ground detector is in the rest frame, the photon emitted from the moving reflector will hit on the wrong side of the detector.
     
  12. DaleSpam TANSTAAFL Registered Senior Member

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    Yes, that is the 2nd postulate.

    No. You did the math wrong if this is what you got. I have never gotten such a result.

    Could you please describe, in complete detail, the experiment you are proposing, the measurement you want to make, and which frames you think disagree? I am sure you either did not do the math or you did it wrong.

    -Dale
     
  13. DaleSpam TANSTAAFL Registered Senior Member

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    Huh? What are you talking about. The photon travels at c regardless of the velocity of the emitter (and regardless of the frame). That is the 2nd postulate. I agree with 2inquisitive on this point.

    -Dale
     
  14. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    Dale, the diagrams are posted at the site I gave the link to. Here again:
    http://www.timing.com/products/pubs/Airborn_ Platforms.pdf
    In the rest frame of the geostationary satellite: the photon is reflected straight toward the Earth's surface, which said surface is rotating. During the flight of the photon, the surface/detector move to the right assuming you are looking north from a frame at rest wrt the satellite. The photon hits the ground to the west (behind) the moving detector. This is what happens in actual experiments. Assume the same scenario, except you are at rest wrt the detector: if the photon retained the motion of the moving reflector, it would hit to the east (in front of) of the stationary detector. This frame is incorrect. If the photon did not retain any forward momentum of the moving reflector, the photon would travel directly to the detector. This frame is also incorrect. The photon hits behind (to the west) of the detector in reality. The only frame to give correct results is if a photon leaves a moving or stationary emitter/reflector with no forward momentum, then travels toward the detector, which moves while the photon is in flight. The rest frame of the emitter is the frame which reflects actual results.
     
  15. Pete It's not rocket surgery Moderator

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    Yes, but that's not what 2inq is talking about.

    He is saying that not only is the speed of the photon not affected by the motion of the emitter, but it's velocity is also not affected.

    He's correct if the photon velocity is parallel to the emitter's velocity, but not in any other case.
     
  16. Pete It's not rocket surgery Moderator

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    10,166
    Yes, but I'm fuzzy on exacty how.
     
  17. 2inquisitive The Devil is in the details Registered Senior Member

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    The photon's trajectory is perpendicular to the point where the photon was emitted. The emitter will keep travelling wrt this path. The photon's true seed and velocity are not effected as it is travelling a straight path, the motion of the emitter after the photon is emitted is of no consequence. The emitter can keep a constant velocity, stop, or turn while the photon is still in flight. The point in space at which the photon was emitted is all that matters to the photon. It achives instantaneous velocity and travels at 'c' in a straight line from that point in space, unaffected by the motions of the emitter.
     
  18. Pete It's not rocket surgery Moderator

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    10,166
    The motion of the emitter after emitting is, of course, of no consequence. The motion of the emitter at the instant of emitting is relevant.

    If it wasn't, then the result in the emitter's rest frame would differ... you'd need to know the emitter's "actual velocity" in order to determine the photon's path. But as you showed, we don't need to to that - we get the right results in a frame in which the emitter is at rest.
     
  19. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    No, you only need to draw your diagrams from the emitter's rest frame. The velocity of the emitter is of no consequence because the photon does not retain any 'velocity' of any kind of the emitter. The detector's velocity relative to the path of the photon does matter.
     
  20. Pete It's not rocket surgery Moderator

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    In the emitter's rest frame the velocity of the emitter is zero - so of course it is of no consequence.

    In the detector's rest frame, the velocity of the emitter is of consequence. Try drawing your diagrams in the detector's rest frame, and see what you discover.
     
  21. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    The detector has no rest frame unless you can identify an absolute rest frame. Think back to the aberration of starlight. The path of the starlight from the star is the rest frame. Any motion of the detector will result in aberration of the starlight. Bradley's version of aberration is the correct one, supported by modern experimental evidence. As I have said many times before, when the path of light is considered, it is a mistake to switch back and forth between 'rest frames'. The ray of light will seem to take a different path if the detector is assumed to be 'at rest' when it isn't. The path will be drawn as an incline when it is not in reality.
     
  22. Pete It's not rocket surgery Moderator

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    10,166
    :bugeye:
    That makes no sense.
    You don't seem to have a problem with the rest frame of the emitter, and it's not at rest in any absolute sense.
    Why is the rest frame of the detector so different?
     
    Last edited: Jun 8, 2006
  23. DaleSpam TANSTAAFL Registered Senior Member

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    Oh, I didn't notice that. That could explain why I wasn't understanding his scenario.

    -Dale
     

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