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I don't understand photon clocks

  1. Nov 25, 2011 #1
    The depiction of time dilation that is typically used to describe the situation,
    shows the stationary photon clock, with mirrors one above the other in the observer's frame and the photon "bounces" back and forth between them.

    In the moving frame the mirror pair moves as a unit let's say to the right and the photon bounces at an angle to these mirrors...and from this geometry the longer length traveled by the photon (always traveling at c) is the "cause" of time dilation.

    My problems are:
    why does the photon bounce at an angle to the mirrors?
    the motion (angled trajectory) of the photon should not be dependent on the mirror's motion to the right, only to the angle of the mirror's surface.
    So I think that depiction is not correct and some change in the angle of the mirrors is required.
    OR is the photon shown as a bouncing point not accurate and should be shown as a circular wave front propagation thus intercepting the displaced mirrors at greater distance?

    Thank you for your help.
     
  2. jcsd
  3. Nov 25, 2011 #2

    ghwellsjr

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    It's not the motion of the mirrors that makes the photon follow a zig-zag path, it would do the same thing if you just had two very long stationary mirrors and you started the photon going on an angled path, don't you agree?
     
  4. Nov 26, 2011 #3
    Is the photon emitted on an angled path to the mirrors? By definition, I believe not. Then what causes the change to an angled path?
     
  5. Nov 26, 2011 #4

    Dale

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    Maxwell's equations hold in all frames, so the angle of incidence is equal to the angle of reflection in all frames. It bounces out at an angle becomes it bounces in at the same angle.

    Correct. The angle of incidence is equal to the angle of reflection regardless of the speed of the mirror.

    Not in this case, although in general the angle of the mirrors is frame-variant.

    Either way works for the light clock.
     
  6. Nov 26, 2011 #5

    Dale

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    Yes, it is emitted on an angled path. Path directions are frame-variant.

    What definition are you refering to here?
     
  7. Nov 26, 2011 #6

    jtbell

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    Suppose you are standing on a railway flatcar and throw a ball straight upward (from your point of view). If the flatcar is moving, someone standing on the ground sees the ball going up at an angle. The photon in the moving light-clock moves at an angle for the same reason.
     
  8. Nov 26, 2011 #7

    ghwellsjr

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    Let's pretend you have a photon gun with some sort of a barrel that guides the photon along the path you want it to take, in other words, it let's you aim the photon. This barrel will not be angled but pointing straight up. As the photon travels up the barrel, it has to travel at an angle as viewed from the stationary frame because the barrel is moving to the right during the time that the photon is moving up the barrel. If it doesn't move on an angle, it will hit the left side of the barrel and not make it out. Doesn't this make sense to you?
     
  9. Nov 26, 2011 #8

    Janus

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    Try looking at it from the other direction. The clock is "stationary" and you are moving past it. The photons for the clock go straight up and down, but from your perspective, relative to you, they follow a zig-zag pattern.

    The thing is, according to the principle of Relativity, there is no difference between the clock moving past you and you moving past the clock. There is no absolute frame against which you can use to say which of you is "really" moving and it actually meaningless to talk about who is "really" moving. All you can know is that have have a relative motion with respect to each other. So when we talk about a "stationary" clock and a "moving" clock, we just mean stationary or moving relative to some convenient frame of reference (usually ourselves) and not in any absolute sense.

    So if you also have a light clock, you would see it tick straight up and down and see the other clock's photons travel in a zig-zag, while someone with the other light clock would see your clock's photons follow a zig zag pattern and see his clock's tick straight up and down.
     
  10. Nov 26, 2011 #9
    I thought laws of physics are preserved in every refernece frame and between reference frames.....So angle of incidence does not equal angle of reflection?
    when both reference frames were traveling at the same speed the photon bounces perpendicular to perpendicular mirrors and when one reference frame moves to the right at constant velocity with repsect to the other, the observed photon start to reflect at an angle to the perpendicular mirrors? What am I missing.
     
  11. Nov 26, 2011 #10

    Dale

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    Angle of incidence does equal angle of reflection in every frame. Otherwise Maxwell's equations would not hold in every frame and it would be possible to determine an absolute rest frame by measuring angles of incidence and reflection.

    I am not certain but I suspect that you are missing the fact that the angle of incidence is frame variant.
     
  12. Nov 26, 2011 #11

    ghwellsjr

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    Maybe you're thinking that both reference frames and the mirrors at rest in them start out with zero velocity between them and then one of them starts to move to the right and you're wondering how a photon that was launched in the vertical direction at the stationary mirror would then hit the same mirror after it started moving while the photon was still in transit. Is that what you're thinking?
     
  13. Nov 26, 2011 #12
    Yes, that's one thing I am thinking. What actually happens in that case. If initialy aimed at the mirror. it always hits the displaced mirror or does it miss the displaced mirror?

    Also, what initially causes the photon to become angled to the mirror/emitter? i didn't think that moving the emitter/mirror to the right could effect the speed and/or direction of the emitted light.

    The only response above, that I resonate with is "either way works for the light clock" i.e. that the photon could be thought of as a spherically propogating wavefront (circular in 2D) and that wavefront contacts the right displaced top mirror at a greater distance, than it would, a mirror directly above the bottom mirror. But I still would like to understand the emitted photon description of the clock.
     
    Last edited by a moderator: Nov 26, 2011
  14. Nov 26, 2011 #13
    My advice is to go with the spherical waveform. I doubt that a pulsed photon clock could be constructed in real life.
     
  15. Nov 26, 2011 #14
    I'm getting the feeling that my original thought, that the zigzaging photon is not a good depiction and the spherical (2D circular) photon depiction is better, may be acceptable.
    Thank you for your responses.
    Burt
     
  16. Nov 26, 2011 #15
    Originally Posted by PatricPowers "My advice is to go with the spherical waveform. I doubt that a pulsed photon clock could be constructed in real life".

    Ok, so I'm getting some feeling that my original thought that the typical zigzag bouncing photon is not a good depiction and a circular wavefront depiction maybe more accurate.

    Thank you all for your responses.
    Burt
     
  17. Nov 27, 2011 #16

    Dale

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    Both are fine. You should be able to understand both. If you do not understand the diagonal line, you should not try to "sweep it under the rug" but really confront the source of the confusion.

    Let's, for a minute, consider a rubber-ball clock rather than a light clock (in the absence of gravity and with all v<<c). The clock is constructed with a spring-loaded rubber-ball launcher that launches the ball vertically at a fixed speed, it travels a fixed distance and hits a mark on the horizontal ceiling, bounces back, hits a mark on the horizontal floor right above the launcher, bounces up, ...

    Now, for a minute, consider the same rubber clock moving inertially to the right at a speed equal to the lanuch speed. Immediately upon exiting the vertical launcher tube, which direction is the rubber ball travelling? Why is it travelling in that direction?
     
  18. Nov 27, 2011 #17

    jtbell

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    To elaborate on Dale's setup, imagine the rubber-ball launcher to be a box with a spring inside it, on the bottom, and a hole at the top which ensures that the ball comes out moving vertically in the launcher's rest frame. Sometimes the spring pushes the ball off-center so it starts off at an angle and thereby misses the hole and fails to leave the launcher.
     
  19. Nov 27, 2011 #18

    ghwellsjr

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    In this case, where the mirrors are accelerated after the photon is emitted, the photon will not hit the mirror, but that's why we insist on inertial frames of reference when discussing a light clock. We have to assume that the mirrors have always been at a constant speed. So you need to assume that even though we might start out thinking about the mirrors as they first move away from each other, we really are considering the moving mirrors to have always been in motion and we're picking up the scene as they get close together.
    You shouldn't give up on the single-photon idea. If a circular wavefront makes sense to you, why can't you think of it made up of individual photons, all traveling in a straight line at the same speed but at different angles so that the sum total of all of them looks like a circular wavefront? But then if you just remove all the photons except the one that hits the mirror, you will have the solution for the single-photon idea. And then go back and read post #7 and see if that now makes more sense to you.
     
  20. Nov 27, 2011 #19
    So what I am hearing:
    1). A moving emitter can effect the direction of the emitted photon.
    2). The moving frame's velocity can add to the light's velocity.The light will then have a velocity in the emitted direction of c and a velocity in the right direction of the moving frame? but the reultant velocity is still c?
    3). A moving flashlight has an angled beam in the observers frame?
    4). In an accelerating frame, other effects take place (in addition?) to the velocity effect aka gravitational lensing?
     
  21. Nov 27, 2011 #20
    An easy way to think about it is look at it as you are moving and the emitter is stationary. Then that you are stationary and the emitter is moving. This is one situation, just looking it from different positions.
     
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