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Beginner's Light Clock question

  1. Sep 3, 2009 #1
    I'm having a bit of trouble with the famous light clock example, where the photon bounces between 2 mirrors and the light clock is moving laterally across my view.

    The description is that I see the photon's path longer (spread out) and so it should take longer to travel between ticks.
    The answer about time slowing down for the moving clock (relative to me) makes sense.

    But, what if the clock is moving toward me rather than laterally, and we measure only the trip from the mirror closer to me to the further mirror.

    I see that path as shortened, not lengthened (because the clock is moving toward me), but according to SR the clock's time is still slowed down (moving clocks run slow).

    Shorter path + slower time = photon speed measured greater than c.

    This is a 'gut feeling' thing for me - I can follow the maths in the examples and sure, it all works.
    But I can't see the answer to the above scenario.

    Help?
     
  2. jcsd
  3. Sep 3, 2009 #2

    JesseM

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    What do you mean by "moving toward me"? As long as the direction the clock is moving is perpendicular to the axis between the two mirrors, the path is always longer as measured in your frame, regardless of the direction of motion. On the other hand, if you want to turn the light clock on its side so that the direction of motion is parallel to the axis between the mirrors, in this case you have to take into account the length contraction effect in SR, which will make the distance between the mirrors shorter in your frame than it is in the rest frame of the clock--if the distance is L in the clock frame, the distance will be only [tex]L*\sqrt{1 - v^2/c^2}[/tex] in your frame. Also, if you want to measure only half of the two-way trip, you also have to take into account the relativity of simultaneity, which tells us that if we had two stopwatches next to the mirrors that were synchronized in the clock rest frame and a distance L apart in the clock frame, then in the frame of an observer who sees the mirrors moving parallel to the axis between them, the two stopwatches will be out-of-sync, with the watch next to the rear mirror being ahead of the watch at the leading mirror by vL/c^2.
     
  4. Sep 3, 2009 #3
    Yes, the mirror is oriented so it has one mirror is in front of the other (according to me) i.e. turned on it's side.

    So the path the photon travels from the far mirror (from me) to the near mirror is shortened by the mirror's motion toward me and also by SR length contraction.

    Doesn't that mean that the speed of the photon would be measured greater than c?
    (I'm missing something, obviously).
     
  5. Sep 3, 2009 #4

    JesseM

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    Don't you mean from the near mirror to the far mirror? The path from the far mirror to the near one is lengthened by the mirror's motion towards you, since in your frame the near mirror is moving away from the point where the far mirror was when the photon bounced off it.
    I edited my post shortly before you responded--did you see the last part about the relativity of simultaneity? If you like I can give a numerical example to show how this effect combines with length contraction to ensure both frames measure the photon to move at c.
     
  6. Sep 3, 2009 #5
    Sorry I meant near mirror to far mirror, as you point out.

    So it's the relativity of simultaneity? I guess I have some more reading to do!

    Thanks for your help!
     
  7. Sep 3, 2009 #6

    JesseM

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    An example may help:

    Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case the relativistic gamma-factor (which determines the amount of length contraction and time dilation) is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.

    Now, when the front end of the moving ruler is lined up with the 100 light-seconds mark of my own ruler (with my own ruler at rest relative to me), and the back end is lined up with the 60 light-second mark on my ruler, I set up a light flash at the front end. Let's say at this moment the clock at the front of the moving ruler reads a time of 50 seconds, and since the clock at the back end is always ahead by 30 seconds in my frame, then in my frame the clock at the back must read 80 seconds at that moment. 25 seconds later in my frame, the back end will have moved (25 seconds)*(0.6c) = 15 light-seconds along my ruler, and since the back end was originally at 60 light seconds on my ruler, it will now be lined up with 60 + 15 = 75 light-seconds on my ruler. Since 25 seconds have passed, if the light beam is moving at c in my frame it must have moved 25 light-seconds in that time, so since it was emitted at the 100 light-second mark it will now be at the 100 - 25 = 75 light-second mark as well, just having caught up with the back end of the moving ruler.

    Since 25 seconds passed in my frame, this means 25/1.25 = 20 seconds have passed on the clocks at the front and back of the moving ruler due to time dilation. Since the clock at the back read 80 seconds when the flash was set off, it now reads 100 seconds as the light reaches it. And remember, the flash was set off at the front end when the clock there read 50 seconds, and the ruler is 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the front when the clock there read 50 seconds, and the light beam passed the clock at the back when its time read 100 seconds (a time of 100 - 50 = 50 seconds), so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.
     
  8. Sep 3, 2009 #7
    Interesting. The piece I was missing was the moving clocks being out-of-sync in the rest frame. So 25 seconds in the rest frame is 20 + 30 = 50 seconds in the moving frame, which allows the photon to be measured at speed c in the moving frame.

    This seems like it might be a bit recursive - but why are the clocks out-of-sync in the moving frame?
     
  9. Sep 3, 2009 #8

    JesseM

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    They're out of sync in frames where they're moving because of the Einstein synchronization convention, which says that clocks should be synchronized in their own frame using the assumption that light moves at the same speed in all directions in their frame. So, two clocks could be synchronized in their rest frame by setting off a flash at their midpoint, and setting them both to read the same time when the light from the flash hits them. Naturally in a frame where they're moving, the front clock is moving away from the point where the flash went off and the back clock is moving towards that point, so if you assume light moves at the same speed in both directions in this frame, the light will reach the back clock before the front clock, so if they both read the same time when the light hits them they must be out-of-sync.

    This is just one possible convention for defining what it means for clocks to be "synchronized", but it turns out that all the fundamental laws of physics we know of have a property known as "Lorentz-invariance", meaning the equations look the same when you write them in the coordinates of different inertial frames related by the Lorentz transformation; and different inertial frames will be related by the Lorentz transformation if each observer defines their own frame using readings on rulers and clocks at rest in their frame, with the clocks synchronized using Einstein's convention (if you used a different convention then the coordinate transformation relating different frames would be something other than the Lorentz transformation, and the equations for the laws of physics would not be invariant under this new transformation).
     
  10. Sep 3, 2009 #9
    I'll have to ponder that for a while.

    It seems a bit circular:
    The clocks are out-of-sync because the speed of light is constant in all frames, and the speed of light is constant in all frames because the clocks are out-of-sync.
     
  11. Sep 3, 2009 #10
    No, the clocks are out of sync because the speed of light is constant in all frames, and the speed of light is constant in all frames because it is an axiom of the theory.
     
  12. Sep 3, 2009 #11
    Ah, that's right. I think I'm starting to get it now.

    Thanks!
     
  13. Sep 3, 2009 #12

    JesseM

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    The Einstein synchronization convention only ensures that all frames measure the speed of light to be the same in all directions (so the speed of light is 'isotropic'), it doesn't ensure that they all measure the speed of light to have the same value of c; if there was no length contraction or time dilation, then you could still synchronize clocks in this way but different frames would disagree on the speed.

    Again, the Einstein synchronization convention is just that: a convention. You could choose a different method of synchronizing clocks and the one-way speed of light could be something other than c (although the two-way speed for a light ray to travel from a clock to a mirror and back to the clock would still be c, since this measurement requires only one clock and synchronization isn't an issue). As I said, the reason this is the most "natural" convention to use is that when inertial frames are defined using this convention, then all the fundamental laws of physics we know of will have the property of being written using identical equations in every inertial frame, which wouldn't be true if you used a different convention. It's an empirical observation that all the known fundamental laws of physics have this property of symmetry under the Lorentz transformation rather than under some other coordinate transformation (like the Galilei transformation used for inertial frames in Newtonian physics).
     
  14. Sep 3, 2009 #13
    Question:

    If we take the previous example and have the ruler moving away at 0.6c rather than toward us, how does the relativity of simultaneity treat the clocks? Is the front clock's (closest to us) time ahead of the back clock's time in this case?
     
  15. Sep 3, 2009 #14

    JesseM

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    It's always the clock that's trailing relative to the direction of motion that's ahead, but if you're just using "front clock" to mean the one that's closest to us rather than the one in front relative to the direction of motion, then yes, that one would be ahead.
     
  16. Sep 3, 2009 #15
    It's puzzled me for a long time that there seemed to be no 'directionality' in SR - for example 'moving clocks run slow' and nothing about what direction they're moving.

    The relativistic simultaneity piece solves the puzzle for me - that's where directionality got to.
     
  17. Sep 4, 2009 #16

    Saw

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    What could that "different method" for synch purposes that would not give off c as the 1-way speed of light?

    In terms of tick rate, it is said that the choice of the "ticker" does not alter anything. If the ticker of a the typical "light clock" is not light but a tennis ball, of course the instrument will be less precise and useful than an atomic clock, but it would suffer time dilation, as all physical processes.

    In terms of clock synch, I asked what would happen if I follow the Einstein procedure but use an elastic matter ball and got this answer:

    Slow clock transport as a synch method is said to render equivalent results as well.

    Have you thought of what would be the "different synch method" that would give "different results" = a one-way speed of light other than c?
     
  18. Sep 4, 2009 #17

    JesseM

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    Well, for example, you could set off a flash that was not at the midpoint of the two clocks, but rather twice as close to the one on the left as the one on the right, and set both clocks to show the same time when the light from the flash reached them. Or you could pick one "preferred" frame and synchronize all the clocks in that frame using the Einstein procedure, then other frames could synchronize their clocks so that their judgments of simultaneity would match those of the "preferred" frame.
     
  19. Sep 4, 2009 #18

    Saw

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    OK, understood. And could we also say that these methods would simply not work, since they're not "consistent"? I mean, the frames that synchronized their clocks to the preferred frame would soon lose this initial state of synchronicity with the preferred frame, due to time dilation. Since they would not know whether it's their clock or the "preferred" clock the one that slows down, they might be quite at a loss for finding a transformation between measurements... You said that their transformation would not be Lorentz invariant, which would certainly be a nuisance. But could we go even further and say that they would find no way for a transformation? Could we state that it's necessary that the observers get different results in the three legs of the puzzle (RS, TD and LC) so that they finally converge on solutions, i.e., possibility to make mutual transformations of measurements and agreement on events and thus on practical solutions to problems?
     
  20. Sep 4, 2009 #19

    JesseM

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    I was talking about simultaneity, not about whether they would assign the same time-coordinates to events as the preferred frame. They wouldn't, but they would still agree with the preferred frame about which pairs of events are simultaneous and which aren't; time dilation wouldn't cause a pair of comoving inertial clocks that were initially synchronized with one another in the preferred frame to become out-of-sync in the preferred frame.
    Why wouldn't they? Since there are no disagreements about simultaneity here, there can be no disagreements about which of two clocks is running slower. Everyone agrees that clocks in the preferred frame run faster than clocks in other frames.
    It's not meaningful to ask whether or not a coordinate transformation is Lorentz-invariant as far as I can tell...Lorentz invariance is a property of physics equations, if you have some equations describing dynamics in one Lorentz frame and then see how the equations transform into another Lorentz frame, Lorentz-invariance means the equations are unchanged in the second frame.
    You can always find a transformation, it's just a matter of looking at which clock-readings and ruler-markings in one physically-defined coordinate system would coincide with which clock-readings and ruler-marking in another. In the case of synchronizing clocks using a preferred frame, this gives the Mansouri/Sexl transformation provided here. The reason this is not a "natural" transformation to use is that the equations of the fundamental laws of physics that we know of would not remain unchanged under this transformation.
     
  21. Sep 4, 2009 #20

    Saw

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    Ok, that's clear. For the rest, I have to assimilate it, but it certainly looks reasonable.
     
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