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Question about the light clock

  1. Jul 23, 2004 #1
    Question about the "light clock"

    I'm trying to make sense of Brian Greene's explanations of SR in The Elegant Universe. For those who don't know it, he gives the example of a "light clock" that bounces a single photon between two mirrors and produces a tick after each round-trip. Another light clock moves past it. From the perspective of the stationary clock, the photon in the moving clock is tracing a diagonal pattern (moving forward while moving up and down), and therefore taking longer between ticks. Thus, from the stationary perspective, time is passing more slowly for the moving time clock.

    My question is, why would the light be constrained by the moving clock? Wouldn't it simply bounce out? It seems to me that the photon would hit the bottom head on, and by the time the new, upward photon reached the top, it would be hitting a section of mirror farther along, and so on, until it had escaped the light clock.

    This may be a minor point, I don't know, but it's making it very hard for me to grasp relativity. Visual explanations geared for us laypeople seem to hinge on light's being constrainable in a way that it's not.
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  3. Jul 23, 2004 #2
    From the stationary clock a human's psychological perception is proportionately distorted because it recieves information that is the expression of the clock (which changes proportionately with distance and relative speed of the perciever). For example, the photons are closer together at the source where they begin their voyage from the object, but proportionately spread out as they increase their distance from their origin.

    The expression of the moving clock to one who percieves it is proportionately different from the actual physical behavior of the moving clock. Is this distinguishment true and relevant to your example and question? If not, then maybe there is a field of viewer's aether affecting the moving clock. :biggrin:

    If it is true and relevant...

    The physical behavior of the clock in transit is not dictated by the person's perception who is viewing from a stationary position, therefore the photon in the moving clock has unaffected inertia.
  4. Jul 23, 2004 #3


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    One of the basic postulates of Special Relativity is that your frame of reference does not make any difference. So the clock must work for someone traveling with it the same as when they are stationary. Since the observer in the moving frame observes the clock working it must also work according to the stationary observer.
  5. Jul 23, 2004 #4
    For a description of the light clock please see
    One clock is giving an impulse to the photons in the light beam whereas the other clock isn't.

    That may be too difficult to visualize but I'm unable to concentrate hard enough to make that determination right now (back pain is blindingly painful at the moment). Does that description make any sense?

  6. Jul 23, 2004 #5
    I understand that a stationary observer's perception that an object is moving doesn't have any effect on the object, but that still doesn't solve the problem for me. Maybe my lack of comprehension goes deeper than I think.

    Let's look at it in terms of the earth. Our planet is moving (in several ways at once), but we don't perceive it as such in our daily life. Other bodies (Sun, Moon, etc.) appear to be moving around us, while our world appears stationary. When I jump up, I don't land in another country or get left behind in space, because I am part of the same system as the moving planet. The ground, the atmosphere, my surroundings and I are all moving at the same rate, which is why I can perceive myself as stationary when I am at rest.

    But what about light? If light is massless and constantly zips around at c, how can I trap it between two mirrors? Why is it part of the same system as our moving planet? It seems to me that if I set up a light clock on my table, the light would bounce out as the planet moved. The light would get left behind, not--of course--because it isn't fast enough, but because it doesn't share our velocity. Why isn't this the case?

    The relativity examples I've seen seem to work off the concept that the speed of light is constant, regardless of the velocity of the system in which it is operating (train moving along a track, light clock sliding perpendicular to the motion of the photon). I have no problem with that, but why should the velocity of the system matter at all? Why should it create different perceptions for different observers?
  7. Jul 23, 2004 #6


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    Here is one way to think about this.
    Imagine that at each bounce, a flash is triggered that sends light rays in all directions in space from that event. One of those light rays will strike the other mirror and trigger another such flash. Those other light rays are not timed and aimed correctly to hit the other mirror... they will escape from the clock.

    If you follow the sequence of light rays that successfully hit the mirrors, you'll find that they satisfy the law of reflection.
    Last edited: Jul 23, 2004
  8. Jul 23, 2004 #7


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    The problem is that you are still trying to think of things in terms of an absolute reference system. IOW, that there is some reference , call it "prime", by which movements can all be judged; The Earth has some velocity with respect to "prime", light travels at a constant speed with respect to "prime", etc. If something were motionless with respect to "prime", then it would be be in a state of absolute rest. This would make "prime" a prefered reference system. Relativity holds that no such animal exists; there is no such thing as a prefered reference system. "prime" does not exist.

    So when we say that the speed of light is constant, what are we saying it is constant realtive to?. Ourselves, or more correctly, whoever is measuring the speed of light. In a sense, every reference system is its own "prime".

    Example, you are sitting by a road way and a car zips by at .5c relative to you. The instant the car passes you, you turn on a flashlight pointing in the same direction as the car is moving, and the car turns on its headlights.

    What you see: Both the light from your flashlight and the light from the car's headlights run neck and neck at 300,000 km/sec away from you and races ahead of the car at 150,000 km/sec relative to it.

    What the driver of the car sees:

    Both lights still run neck and neck, but they rush at 300,000 km/sec away from him and travel at 450,000 km/sec relative to you.

    Both of you sees himself as "prime" and measures the speed of light as 300,000 km/sec relative to himself.

    The same happens with the light-clock. The light clock will always consider itself as stationary. Thus the light it shines on the mirror must return to it. it does not matter whether the light-clock can be seen as moving when measured from some other reference. If the light clock sees the light return, so must everyone else (anything else lead to major contradictions). For someone for which the clock is considered moving, this means that the light follows a diagonal path.
  9. Jul 23, 2004 #8
    Is that what would happen? It makes sense, but if that's the case, then the light clock would soon be flooded with light, yes? How could it continue to serve as a clock with all those photons bouncing around?
  10. Jul 23, 2004 #9


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    I'm a little confused by this last statement.
    The only impulse a mirror can apply on a photon is in a direction perpendicular to the mirror (i.e. parallel to the axis of the clock).

    In a transverse light clock (as shown in the url above), among all of the light rays emitted at the initial flash, the light ray that meets the other mirror is the one that has the same horizontal-component of velocity as the pair of mirrors. Upon reflection, only the vertical component (parallel to the axis of the clocks) changes. The mirror doesn't have to do anything to keep the light ray moving horizontally.
  11. Jul 23, 2004 #10
    So the driver sees the light traveling at 1.5c relative to me? It's just in relation to ourselves that the speed of light appears constant?

    And what if I drive in the opposite direction from him at .6c? Will each of us observe that we are traveling away from the other car at 1.1c? Not possible, right?
    Last edited by a moderator: Jul 23, 2004
  12. Jul 23, 2004 #11

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    Right. I think it's better to think of this speed as a "separation rate" since nothing is really traveling at this speed. From the driver's viewpoint you are moving 0.5c in one direction and the light is moving at 1.0c in the opposite direction. So the driver sees you and the light separate at the rate of 1.5c.
    Everyone measures light to move at the same speed c with respect to themselves.

    Let's say that you drive to the left at 0.6c with respect to the road while the other guy drives to the right at 0.5c. Someone standing on the road will measure you two to separate at 1.1c. But each of you will observe the other as traveling away at only around 0.85c.
  13. Jul 23, 2004 #12
    The light moves with the train, so for a person inside the train the light seems to move only vertically, but for a person outside the train the light also moves horizontally (because it moves with the train, which moves horizontally).

    This is just what happens, and what you were proposing just does not happen in this universe. It is not “really” clear why this is the case, we just try to make theories that can explain it. What you proposed just does not explain what happens. You should not ask why the universe does not behave the way you would expect it to behave, but try to learn to expect how it does behave :wink:
    The way in which this situation (which is just what happens in our universe) gives rise to the theory of relativity (time dilation) is as follows:

    Because of what was said above, the length of the path along which the light has traveled seems longer for the person outside the train than for the person inside the train. Yet, if both determine the speed of the light they get the same answer.
    - There are two events:
    (1) The light is at the lower mirror
    (2) The light is at the upper mirror
    - The distance between the two mirrors at these two events is different for the two observers.

    How can it be that the light moves a different distance between these two events for the two observers while they both determined that it moves at the same speed?
    Since speed is equal to the distance traveled divided by the amount of time the travel took, a solution to this problem is that less time passes between these two events for the observer that saw the light travel less distance.
  14. Jul 24, 2004 #13
    The light moves with the train, but no one knows why? Really?? It can't do so for long--if the light source stops producing light, the train will become dark, right? So for how long does the light travel with the train, and what is impelling the light to do so? This seems like a pretty important point to me--I may well be missing something basic (I'm no physicist, after all), but I'd really like to know what that something is . . .
  15. Jul 24, 2004 #14
    The light in the clock would just be moving between the mirrors and you would not see it. In order for you to see the light it has to enter your eyes. You see things because the light coming from it enters your eyes. In the light clock there is no source of the light, there is just one photon traveling between the mirrors forever it will never stop.

    If there is not some lamp in the train, which produces a lot of photons that reflect from the interior of the train into your eyes, it is dark in the train. This has no effect on the clock, it will simply count how many times the photon hits the mirror, whether it is dark in the train or not.
  16. Jul 24, 2004 #15


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    It seems like your difficulty is not with the implications of this thought experiment, but with the construction of the light clock itself.
    The light clock it obviously a modelized object. The reason why the light moves with the train is simply because it has been set it work inside the train. Inside the train is just moves vertically up and down, touching (say) the middle of the top and bottom plates everytime it bounces. This is surely possible. Suppose it would 'stay behind' when it moves and bounces out, then the observer on the ground would 'see' the photon move out the clock and the moving oberver would not. This disagreement is physical nonsense.

    Hey, but why worry about the light clock? We can do it differently. Say a light beam from the top of the train shoots down a photon vertically and measure the time it takes to reach the bottom. (not sure if that solves your problem though)

    If your problem is that earth, is rotating then you are right. Eventually after a few hours the earth will have rotated over some angle and (depending on your position on the earth) we will have been displaced from our position in space with the earth (because we are in contant with it and it sweeps us along by friction and the photon would leave the clock because the clock (presumably) is fixed to the train and the light is not. This is a practical objection and irrelevant in a though experiment where it's the theoretical implication is of importance.
  17. Jul 24, 2004 #16
    I guess it is helpful to look at descriptions like this more as rough examples of relativity rather than as proofs. I suppose if I want more thorough proof I need to spend a little more time with the math & physics books, eh? It just seems like the thought experiments I'm reading use things that wouldn't happen and draw really mind-blowing implications from them. Are the readers not supposed to come up with any of these objections? Are there any proofs of SR that are accessible to a non-physicist?
  18. Jul 25, 2004 #17


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    This thought experiment is not a 'proof' os SR. SR is based on two postulates, one if them is that the speed of light in vacuum is the same for all inertial observers, regardless of the motiuon of the source. This postulate comes from the experimental results of measuring the speed of loght by Michelson and Morley (although for Einstein, theoretical evidence was sufficient for this postulate).

    Time-dilation and Lorentz-contraction are logical consequences of these two postulates, but to make these phenomona quantitative we can use a thought experiment like the moving train with the light-clock.
    Therefore the thought experiment is no 'proof' of SR. (The justification has come from experiments in particle accelerators e.a.)

    Also, the laws of SR apply only in inertial reference frames; we only talk about relativistic effects for inertial observers moving with a constant velocity relative to each other. The train model is idealized in this way that
    we consider the ground observer to be in a inertial frame and the train in another one. The objection that this is not possible because the earth is NOT an inertial frame has nothing to do with the physical consequence of having time dilation between to observers moving with a constant velocity relative to each other.
  19. Jul 25, 2004 #18


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    Let me revise my original description.

    Suppose that two transverse light-clocks coincide at a starting event, call it O.
    Clock-A is at rest in our frame.
    Clock-B in moving with constant velocity to the right.

    At event-O, a flash of light sends light-rays in all spatial directions from that event.
    Among those rays:
    • the [upward] light-ray with zero horizontal-component of velocity is the light-ray that meets A's distant-mirror.
    • the [upward] light-ray with horizontal-component of velocity equal to that of clock-B is the light-ray that meets B's distant-mirror.
    • The other light rays do not meet any mirror and escape from the light clocks, continuing to travel until absorbed by some external object. (If you happen to see this flash with your own eyes, the flash appears briefly to you.)
    Upon reflection, the vertical-component of the velocity of each light-ray is [literally] reflected since the normal to the mirror is vertical. The horizontal-components are unchanged.
    It's the law of reflection.
    ... and so it goes for each subsequent reflection.

    From this point onward, only one light-ray is bouncing back and forth between the mirrors of a given light-clock. (Successive flashes sending light rays in all direction are not needed.)

    Why does that light-ray keep up with the mirrors? It keeps up because that successful light-ray has always had the same horizontal-component of velocity as that of the corresponding clock.
    It's the law of inertia.
    (Maybe it's useful to picture a photon bouncing between the mirrors. That successful photon is always on the segment joining its mirrors. A similar image is that of a ball dropped from the mast of an inertially moving ship.)

    If the moving transverse light-clock suddenly accelerates (e.g., slows down or stops), its light-ray will miss its mirror. To handle accelerations, one probably needs more than one light-clock... or something fancier. [For example, refer to R. F. Marzke and J. A. Wheeler, "Gravitation as geometry – I" in Gravitation and Relativity, HY Chiu and WF Hoffman, eds. (W. A. Benjamin, Inc., New York, 1964).]
    Last edited: Jul 26, 2004
  20. Jul 31, 2004 #19
    another viewpoint

    another way to picture this is to make a different kind of time clock. take the example from B.G.'s book Elegant Universe (excellent book). here is an 8-bit graphical depiction...
    first of all this is a thought experiment. we are assuming the following...
    1- there are no outside influences (other energy, gravity, etc)
    2- we can see this experiment with imaginary light and time
    the easiest way to envision this is to change it slightly

    | ~ |

    now the two VERTICAL mirrors are moving left to right as is the photon.
    so if one set of mirrors is moving faster than the other, the photon will have more distance to travel.
  21. Jul 31, 2004 #20


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    This is the "longitudinal light clock".
    Its treatment will have to incorporate length-contraction as well as time-dilation.
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