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An array of lightclocks

  1. Apr 6, 2009 #1
    I asked this question in an earlier thread, but I have thought a bit more about it and would like to add to it, hopefully focussing more successfully on the bit the bothers me.

    First some preparation: At first, I was thinking of two light clocks set in motion relative to an observer A but at rest with respect to observer B. One light clock was parallel with that motion and I call that the "longitudinal light clock". The other was perpendicular to the motion and I call that the "transverse light clock".

    I am fully aware that there are simultaneity issues so "right" and "wrong" are not necessarily hard and fast terms.

    As clarification, I see the "half clock" as being set up so that the "tick" is collocated with B.

    The centiclock will tick one hundred times for one tick of a "normal" light clock, so long as both are at rest with respect to each other.

    The initiation is based on Einstein synchronisation. The initiation of the centiclock and the longitudinal light clock will be simultaneous for B, but not for A. For the sake of the exercise, I am making the assumption that B is at the end of the longitudinal light clock which is closest to A and the centiclock is at the other end of the longitudinal light clock so that, according to A, the longitudinal light clock initiates before the centiclock.

    I can break up the inertial journeys like this:

    • initiation photon leaves centre of longitudinal clock towards "tick mirror" - a quicker journey according to A (because the tick mirror is moving towards the start point)
    • initiation photon leaves centre of longitudinal clock towards "tock mirror" and "centiclock" - a slower journey according to A (because the tock mirror is moving away from the start point)
    • clock photon leaves longitudinal light clock's tick mirror towards tock mirror - a slower journey according to A.
    According to A, the sequence of the longitudinal light clock is this: longer journey from tick to tock mirror, shorter journey from tock to tick mirror.

    According to A, the period that centiclock is in operation (between the time the initiation photon hits the centiclock and the time the longitudinal light clock photon hits the tock mirror) is determined by the lengths of the journeys. The journey the longitudinal light clock photon takes between tick and tock mirror is twice that of the initiation photon from start point to centiclock.

    I think they do.

    The centiclock stops emitting photons when the longitudinal light clock's photon hits the tock mirror, according to both A and B.

    According to A, the centiclock starts emitting photons after the longitudinal light clock's photon starts its journey from the tick mirror.

    So, according to A, there are three phases:

    • Longitudinal light clock in operation (photon on way from tick mirror to tock mirror), centiclock not yet in operation
    • Longitudinal light clock in operation (photon on way from tick mirror to tock mirror), centiclock in operation
    • Longitudinal light clock in operation (photon on way back from tock mirror to tick mirror), centiclock no longer in operation
    According to A, the phases during which longitudinal light clock's photon is on the way from tick mirror to tock mirror, when combined, are greater than the phase during which that photon is on the way back from tock mirror to tick mirror.

    According to A, the phases during which the centiclock is not in operation, when combined, equal the phase during which the centiclock is operation.

    I really think I have answered my own question, but confirmation would be great.

    cheers,

    neopolitan
     
    Last edited by a moderator: Apr 24, 2017
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  3. Apr 7, 2009 #2

    JesseM

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    Agree with everything up to this, but here I need to clarify--are you saying that "the period that centiclock is in operation" is half the time for the photon to travel from the tick mirror to the tock mirror, or are you just saying that the time for the initiation photon to travel from the center to the tock mirror is half the time for the longitudinal photon to travel from the tick mirror to the tock mirror, and we can "determine" the period of the centiclock's operation from this? If the latter I agree, if the former I disagree. Suppose the distance between the tick and tock mirrors is 16 light-seconds in A's frame, and that the mirrors are moving at 0.6c. In this case, when the photon is moving from the tick mirror to the tock mirror, the distance is shrinking at 0.4c in A's frame, and when the photon is moving from tock mirror to tick mirror, the distance is shrinking at 1.6c. So, it'll take 16/0.4 = 40 seconds for the photon to go from tick mirror to tock mirror, and 16/1.6=10 seconds to go from tock mirror back to tick mirror. This means that the time for one of the two initiation photons to go from the center to the tock mirror (where it starts the centiclock) is 20 seconds, and the time for the other of the two initiation photons to go from the center to the tick mirror is 5 seconds. So, suppose at t=0 in A's frame the initiation photons are sent out. At t=5 one initiation photon hits the tick mirror, and then we have a photon traveling from the tick mirror to the tock mirror, which takes an additional 40 seconds, so it'll reach the tock mirror (and stop the centiclock) at t=45 seconds. Meanwhile the other initiation photon hits the tock mirror at t=20 seconds, with the centiclock starting at that moment. So, the time for the photon to go from the tick mirror to the tock mirror is 45-5 = 40 seconds, while the time between the centiclock starting and stopping is 45-20 = 25 seconds.
    Yes, this works. You can see in my above example that the first phase (light traveling from tick to tock, centiclock not in operation) lasts from t=5 to t=20, or 15 seconds; the second phase (centiclock in operation) lasts from t=20 to t=45, or 25 seconds; and the third phase (photon traveling back from tock to tick, centiclock no longer in operation) lasts 10 seconds, so the total time the centiclock is not in operation is 25 seconds just like the total time it is in operation. This makes sense, since we can see by analyzing things in the B frame that if the centiclock is repeatedly turning on simultaneously with the longitudinal photon hitting the tick mirror (according to B's definition of simultaneity) and turning off again each time the longitudinal photon hits the tock mirror, then any inertial observer should spend equal amounts of time seeing the centiclock turned on and seeing it turned off.
     
  4. Apr 7, 2009 #3
    Amazingly, it was the latter, and we both agree :smile:

    Initially, I thought the solution would be more complicated. In reality, explaining the "apparatus" was the more difficult task.

    As a further step, I imagine that (save paralax error) if we had two centiclocks, both aimed at A, both located at the tock mirrors - one at a longitudinal tock mirror, the other at a transverse tock mirror - and the clocks were synchronised in their shared rest frame, then two series of 50 photons would be received together by A (because those photons would be received together by B, if aimed appropriately).

    From your figures, B would see a tick, then 16 seconds later see photons from the centiclocks in lockstep, with 50 arriving over 16 seconds until there's another tick.

    A, on the other hand, would see a tick, then 25 seconds later see photons from the centiclocks in lockstep, with 50 arriving over 25 seconds until there's another tick.

    I'm pretty certain that is right. If both centiclocks were aimed at B, B would certainly receive all photons together and if there was a mirror at 45 degrees to deflect the transverse photon towards A, then A would get all photons together too.

    cheers,

    neopolitan
     
  5. Apr 7, 2009 #4

    JesseM

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    In my figures the distance between the tick and tock mirrors of the longitudinal clock was 16 light-seconds in A's frame, so this figure would include length contraction, meaning the distance would actually be 20 light-seconds in B's frame where the clock is at rest. We can assume that the distance between the tick and tock mirrors of the "transverse light clocks" is also 20 light-seconds. So if B is located next to the two tick mirrors (we're assuming both tick mirrors are next to each other, right?), it would be 20 seconds later that B sees photons from the centiclocks, and they would continue for 20 seconds.
    If you're talking about when A would see light from different events such as the centiclocks starting and stopping, as opposed to just the time-coordinates these events would happen at in A's frame (which was what I was calculating), then you have to figure out where the mirrors and centiclocks are in relation to A as a function of time. For example, suppose A is at x=0,y=0, and at t=0 the two tick mirrors are also at this same position receiving their initiation photons, with the transverse centiclock being at x=0,y=20 at this moment and the longitudinal centiclock being at x=16,y=0 at this moment, both moving at 0.6c in the +x direction in A's frame. In this case the transverse light clock's centiclock has x(t) = 0.6c*t and y(t) = 20, while the longitudinal light clock's centiclock has x(t) = 0.6c*t + 16 and y(t) = 0. The transverse centiclock starts emitting photons at t=0 and stops at t=25; at t=0 it is at a distance of 20 l.s. from A, at t=25 it is at a distance of [tex]\sqrt{15^2 + 20^2}[/tex] = 25 l.s. from A. So, A begins to receive photons from the transverse centiclock at t=20 seconds, and stops receiving them at t=50 seconds. Meanwhile, I found in the previous post that the longitudinal light clock's centiclock starts emitting photons 15 seconds after the initiation photon hits the tick mirror in A's frame, and stops 40 seconds after the initiation photon hits the tick mirror. At t=15 seconds the longitudinal centiclock is at a distance of 0.6*15 + 16 = 25 light-seconds from A, and at t=40 seconds the longitudinal centiclock is at a distance of 0.6*40 + 16 = 40 light-seconds from A. So, A begins to receive photons from this centiclock at t = 15 + 25 = 40 seconds, and stops receiving photons from this centiclock at t = 40 + 40 = 80 seconds.
     
  6. Apr 7, 2009 #5
    I didn't read closely enough.

    I'm happy so long as:

    the length of the longtiduinal clock in B's frame = 1/2 tick-tick period times speed of light (ie 20 light seconds = 0.5 * 40 seconds * c)

    and

    according to A, the tick and tock mirrors of the longitudinal clock (in motion) are closer together (ie 16 light seconds)

    and

    according to A, the ticks - when received at A - are further apart (ie 50 seconds)

    I think it is better consider that the photons from the centiclock travel to B first (yes B and both tick mirrors of the longer clocks are collocated). Otherwise there will be a delay between arrival of photons at A from each of the centiclocks, due with their different positions.

    Even if we don't do that if we aim the centiclock's photons directly at A, I find it highly counterintuitive (not necessarily wrong) that depending on the orientation of your light clock, you will see different periods of centiclock photon emission (highlighted in red).

    If we channel all photons past B, then they should arrive in lockstep, according to B.

    The photons from the longitudinal light clock will then pass by B (assuming some sort of reusable photon here, which can be "seen" without being absorbed) towards A. The photons from the transverse light clock will be reflected off a 45 degree mirror towards A.

    In this case, the photons will arrive at A in lockstep - yes?

    But what you are saying is, if the photons don't go past B, they don't have the journey along a longer path, including a leg moving away from A the effect of which would have be a dilation of the period between the arrivals of each photon at A. Because they don't take that longer path, they arrive over a shorter period. Correct?

    I wonder if you missed the words "(save paralax error)" in my post. I was intellectually eliminating parallax error along with the second "l" in parallax. If that is unacceptable, I guess an arrangement could be made back where A is located, so that photons don't travel at an angle (ie mirrors, set at the same distance apart as the transverse tick and tock mirrors, channel photons to A). There will be delays due to the extra distances travelled, but we should be talking about two centiclocks travelling at the same speed, both emitting photons for equal periods (with different timings of when the emissions start and stop).

    B should certainly see 50 photons over 20s ending in a tick, then 20s of nothing, then 50 photons over 20s, etc - from both clocks.

    I thought we worked out that the period that the centiclock associated with the longitudinal light clock would be transmitting would be half the tick to tick period (independent of frame).

    The geometry of the transverse lightclock indicates to me that the associated centiclock would transmit photons for half the tick to tick period (independent of frame).

    The two standard light clocks, longitudinal and transverse, will both run slower (according to A) - but at the same rate.

    In that case, logically, A should see photons from both centiclocks arrive over the same period (so long as parallax error is eliminated). I repeat, timings may differ, but not the emission durations.

    How does this reconcile with your figures (which I am not saying are wrong)?

    cheers,

    neopolitan
     
  7. Apr 7, 2009 #6

    JesseM

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    Yeah, all of these should be true.
    I was just assuming that each time the centiclock flashes up it radiates light spherically in all directions, so no matter where an observer is positioned they'll see light that traveled in a straight line from the event of the flash to their eyes at the moment they enter the future light cone of each flash. Are you suggesting that the light is aimed like a laser at B, and when B receives the light he then redirects the light towards A? This would mean that A doesn't actually receive a signal from that clock lighting up until well after he's entered the future light cone of the event of the clock lighting up.
    Think of it this way: the Doppler effect shows that the observed period of a given clock depends on its direction of motion relative to you (minimized if the clock is traveling straight at you, maximized if it's traveling in a straight line away from you), and in the case of the transverse centiclock, it is neither traveling straight towards A nor straight away from him.
    Yes, if the photons from the transverse clock are deflected in this way rather than traveling to A directly, then since they arrive in lockstep for B they'll also arrive in lockstep for A.
    I don't think the negative fact that they aren't diverted in the specific way you suggest really counts as a full explanation for why the photons from the transverse clock arrive over a shorter period--a positive explanation might be one like the one I suggested earlier about the Doppler effect, since it's the longitudinal clock that's moving straight away from A and that's the direction where observed time is stretched the most (note that in A's frame the actual coordinate time-interval that both centiclocks are in operation is the same, 25 seconds).
    What do you mean by "parallax error" in this context? The wikipedia article seems to say the term has something to do with errors in estimating distance using parallax, but I don't think that's what you're talking about.
    But that's the coordinate time that the centiclock is transmitting, not the time that a given observer actually sees the centiclock in operation--the two are different due to the Doppler effect (which is just a consequence of the fact that light signals from different ticks of a moving clock have different distances to travel to reach your eyes). For example, in A's frame the longitudinal centiclock started up at t=15 and ended at t=40 (a 25-second time interval), but A first saw a flash from the centiclock at t=40 and saw the last flash at t=80 (a 40-second time interval). And although the centiclock was in operation for 25 seconds in A's frame, in the centiclock's own frame it was only in operation 20 seconds, the difference due to time dilation. The fact that the clock has a frequency of 50 ticks/20 seconds in its own frame but is seen to have a frequency of 50 ticks/40 seconds (half the previous frequency) by A is exactly what you'd predict if you used the relativistic Doppler shift formula for an object moving straight away at 0.6c: [tex]\sqrt{\frac{1 - 0.6}{1 + 0.6}} = \sqrt{0.25}[/tex] = 0.5.
    Yes, if you're talking about the time-interval in your frame that the transverse centiclock is transmitting photons (25 seconds in A's frame), but no if you're talking about the time that you'll actually see the clock transmitting photons to you (30 seconds in A's frame).
    Again, you have to distinguish between the time interval between events in a given observer's frame and the time interval between the same observer seeing light from each event, in general they are not the same.
     
  8. Apr 7, 2009 #7
    I put together a diagram, since I am a visual sort of guy.

    The parallax error I meant was due to the offset of the tock mirror of the transverse light clock. If you consider A a baseline, rather than a point, then the whole clock moves away from the baseline at a constant rate (the minimum distance between a point on the light clock and the nearest point on the baseline will increase at a constant rate). Because we usually consider A as a point, there won't be such a constant rate - eventually, when the distance between A and the transverse light clock is sufficiently great, there will be effectively no parallax error.

    Alternatively, you can use the arrangement in the diagram, which overcomes this "parallax error".

    Note that there is a question in the bottom right had corner of the diagram.

    I think that ??? will tend towards 50s - at first it won't be, because of "parallax error", but as separation -> infinity, ??? -> 50s.

    Is that right?

    cheers,

    neopolitan
     

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    Last edited: Apr 8, 2009
  9. Apr 8, 2009 #8

    JesseM

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    By "parallax error" do you mean the fact that the tock mirror of the transverse light clock is not moving in a straight line away from A, so that its velocity away from A is changing, and thus if we allow the centiclock to turn on and off repeatedly the time that A sees the centiclock being on for will be changing too? If that's what you're getting at I agree, for example I found above that if the transverse centiclock turned on at t=0 and off at t=25, then A would see it turn on at t=20 and see it turn off at t=50, a visual period of 25 + 5 = 30 seconds; but if the centiclock turned on again at t=50 and off again at t=75, then since the distance at t=50 is [tex]\sqrt{0.6^2*50^2 + 20^2}[/tex] = 36.055513, and the distance at t=75 is [tex]\sqrt{0.6^2*75^2 + 20^2}[/tex] = 49.244289, that means A will see the visual period as 25 + (49.244289 - 36.055513) = 25 + 13.18878 = 38.18878 seconds, which is somewhat larger. More generally, if the centiclock starts at time T and stops at time T+25, then the visual period will be [tex]25 + \sqrt{0.6^2*(T+25)^2 + 20^2} - \sqrt{0.6^2*T^2 + 20^2}[/tex] = [tex]25 + \sqrt{0.36*(T^2 + 50T + 625) + 400} - \sqrt{0.36*T^2 + 400}[/tex] = [tex]25 + \sqrt{0.36T^2 + 18T + 225 + 400} - \sqrt{0.36T^2 + 400}[/tex] = [tex]25 + \sqrt{0.36T^2 + 18T + 625} - \sqrt{0.36T^2 + 400}[/tex], and plugging this expression into the limit calculator and taking the limit as T approaches infinity (type 'inf' into the 'compute at' box) gives an answer of 40, or 25 + 15 seconds. This is the same as the visual period of the longitudinal clock, which makes sense since the further away the transverse centiclock gets, the closer it gets to just moving radially outward at 0.6c just like the longitudinal centiclock, so in this limit you should be able to use the same Doppler shift formula. But yes, this means that if you have set the period of the longitudinal centiclock such that A receives signals from it at a rate of 1 per second, then in the limit as the distance of the transverse centiclock becomes large, A will also receive signals from it at a rate arbitrarily close to 1 per second.
     
    Last edited: Apr 8, 2009
  10. Apr 8, 2009 #9
    So it seems we are in agreement.

    I have one further question.

    There is no doppler consideration applied to length contraction, is there?

    cheers,

    neopolitan
     
  11. Apr 8, 2009 #10

    JesseM

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    "Length contraction" refers to coordinate lengths rather than apparent visual lengths, so no. But just as the doppler effect causes the visual time that an observer sees certain processes to last to be different from how long those processes "really" last in the observer's inertial rest frame, so there is a visual effect which can cause how long a moving object looks to an observer (as it passes next to another object at rest in this frame and their ends briefly appear to line up, for example) to be different from how long it "really" is in the observer's rest frame--this is known as "Penrose-Terrell rotation", or sometimes just Terrell rotation, and it was discussed on this thread, for example.
     
  12. Apr 8, 2009 #11
    The wikipedia entry http://en.wikipedia.org/wiki/Relativistic_Doppler_effect" [Broken] is interesting.

    Read after "However, due to the relativistic time dilation" (3rd paragraph).

    cheers,

    neopolitan
     
    Last edited by a moderator: May 4, 2017
  13. Apr 8, 2009 #12
    For anyone who is interested as to why I find the previously posted link interesting, the composite authors of that article effectively draw upon temporal contraction. Probably there are other better derivations of the relativistic doppler effect which don't do so.

    I asked about a doppler consideration for length contraction because I was wondering if time dilation is the combined effect of doppler effect and this temporal contraction. I fiddled with the figures and at first blush it doesn't seem to.

    I'd prefer to stick with the temporal analogue for length contraction explanation discussed in an earlier thread ("bump"). I imagine that this is what is being used in the wikipedia article.

    cheers,

    neopolitan
     
  14. Apr 8, 2009 #13

    JesseM

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    I would say that in that instance they're using what I called the "reversed time dilation equation", i.e. the regular time dilation equation with both sides divided by gamma. They were first calculating the time t between wavefronts hitting the observer in the frame of the source, so this is the time between two events that are not colocated in this frame; then they were using this to find the time [tex]t_0[/tex] between the same two events in the frame of the observer, where they both happen at the same position since the observer is at rest. The "regular" time dilation equation is:

    (time between events in frame where they are not colocated) = (time between events in frame where they are colocated) * gamma

    Which in these terms would be:

    [tex]t = t_0 * \gamma[/tex]

    So, just divide both sides by gamma and you have the equation in the wikipedia article:

    [tex]t_0 = \frac{t}{\gamma}[/tex]

    In contrast, if they were using the "temporal analogue for length contraction" they would be calculating the time [tex]t_0[/tex] in the observer's frame between surfaces of simultaneity in the source's frame which have a separation of t in the source's frame--I don't think it makes much sense to interpret it that way.
     
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