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Multiple clock thought experiment

  1. Sep 14, 2010 #1
    Consider a large number of clocks each within some apparatus programmed to accelerate away from, remain moving for a time, accelerate toward, remain moving for a time and then return to a point of origin. A similar clock simply remains at that point of origin. The accelerated clocks will reach relativistic speeds on their outward and inward journey. Each of the clocks both transmits its tick number and is capable of receiving and recording all the other clock's tick number.

    The experiment begins in deep space so that any outside gravitational effects can be ignored. The only difference between the clocks is the direction of travel. Each clock travels in a straight line, taking the same outbound and inbound path.

    When the clocks return to the point of origin, they have ticked fewer times than the clock that didn't move. Relative to the clock at the origin they have all accelerated and traveled at the same rate, so they should all have the same (lower) tick count.

    Relative to each other clock, though, the clock's velocities differ since they were sent in various directions. The acceleration effects should be identical since they all have the same general program. To each moving clock, each other moving clocks should appear to tick at a different rate.

    How then can the moving clocks be synchronized at the end of their journey? Do the differences in relative acceleration somehow compensate for the differences in velocity? Note that the clocks can spend any amount of time not accelerating, just moving at different relative velocities.

    Obviously I'm missing something basic here. Hopefully it's something simple enough to understand.
     
  2. jcsd
  3. Sep 14, 2010 #2
    Velocity plays no role, speed does. Since all the clocks , according to you, have had the same acceleration program over the same distance, they will have the same speed program. Since they had the same speed program, they will show the same elapsed time, so they are synchronized with each other (but not with the "stay at home" clock).
     
  4. Sep 14, 2010 #3
    Doesn't your answer assume, then, that there is an absolute measurement of speed? Only in relation to the stay at home clock are speed measurements the same.

    Clocks that go out in similar directions would move slower in relation to each other. Clocks that go out in differing directions would move faster in relation to each other. Clocks that go out in opposite directions would move fastest in relation to each other. The moving clocks are comparing their motion with other clocks, also in motion. Doesn't the direction of motion contribute to relative speed?
     
  5. Sep 14, 2010 #4
    No, it doesn't.

    The principle of relativity says that the result will be the same when judges from any other inertial frame,, so the "stay at home frame" is as good as any other frame for judging the results.


    True but irrelevant (see above).


    Not wrt the "stay at home" frame. Wrt that frame, all clocks move according to the same speed program as per your OP.
     
  6. Sep 14, 2010 #5

    JesseM

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    All frames always agree in their predictions about the times of two clocks when they meet at a single location. So, if in the rest frame of the central inertial clock we can see by symmetry that all the accelerating clocks should read the same time at the moment they reunite with the central clock, then this would be the prediction of other frames too, though the calculations would be more complicated since the clocks would have different functions for speed as a function of time in the other frames.
    The only speed relevant to calculating time dilation (and elapsed time) in a particular inertial frame is speed relative to that frame, not the speed of different clocks relative to one another (that relative speed would only be relevant if you were calculating things from the perspective of one of the clock's inertial rest frames, so the relative speed would be the same as the speed of the other clock in this frame).
     
  7. Sep 14, 2010 #6
    I'm sure that you're right and that the math will all work out in the end. Perhaps the simplest way of looking at it, from the stay at home clock, is the best.

    Still, I'd like to look at it from the point of view of some of the multiple clocks and just do the math for an example or two, but don't know all the proper equations.

    Here's a simpler and more definite example. Consider clocks W, X, Y, and Z. Clocks W, X, and Y are moving toward clock Z which is in our initial inertial frame of reference. In that frame of reference, at some point in time t, clocks W, X, and Y are one light second away from clock Z moving toward Z at 4/5 C. W and X are moving in opposite directions (towards each other.) Y is moving toward Z at a 45 degree angle from W and a 135 degree angle from X.

    Can all the frames reference agree on the synchronicity of the start time? By distance contraction, at the start time W, X and Y will be 3/5 of a light second away from Z, so they should be able to agree with Z (but not necessarily with each other) as to their start time. Is this actually the case?

    Do all the frames of reference agree on the relative approach angles? If not, how are they calculated?

    Since the clocks are moving at relativistic speeds, their velocities are not simply additive when they try to compare speeds with each other.
    So, I need to know, from each clocks frame of reference, how to 1) compute each other clock's observed speed and 2) compute their observed start time. If relative distance does not fall out from that, then I'd need that equation too. I'm sure that if I had the equations to do this that plugging in the numbers for any example would just work, but I'd like to just do it.
     
  8. Sep 14, 2010 #7

    JesseM

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    In general, if you know the position as a function of time x(t), y(t) and z(t) for a clock in one frame, you can find the position as a function of time in a different (primed) frame, i.e. x'(t'), y'(t') and z'(t') by using the Lorentz transformation equations (here written in a form which assumes the axes of the frames are aligned so that the primed frame's spatial origin is moving along the x-axis of the unprimed frame in the +x direction with speed v):

    x' = gamma*(x - vt)
    y' = y
    z' = z
    t' = gamma*(t - vx/c^2)

    x = gamma*(x' + vt')
    y = y'
    z = z'
    t = gamma*(t' + vx'/c^2)

    with gamma = 1/sqrt(1 - v^2/c^2)

    So for example if x = c^2 * t^2 for some period of time, you could substitute x=gamma*(x' + vt') t^2 = gamma^2*(t' + vx'/c^2)^2 and then solve for x' as a function of t'.

    However, the math is a lot simpler if you assume the accelerations are quasi-instantaneous, so that a clock moves away from the central clock at speed v1 for some time, then instantaneously accelerates so it's not moving back towards the central clock at speed v2, and it continues to move at that speed until it gets back to the central clock. In that case you just have to use the Lorentz transformation on the coordinates where the moving clock leaves the central clock, the coordinates where it accelerates to turn around, and the coordinates where it reunites with the central clock, and then in every frame the speed will be constant between each pair of events so you can just use (change in position)/(change in time) for a given pair to find the speed between that pair.
    No, because of the relativity of simultaneity, events at different spatial locations which happen simultaneously in some frame A must happen non-simultaneously in a different frame B (unless frame B is moving exactly perpendicular to the axis between the events in frame A).
    Different frames won't agree on all the angles, although if W and X were traveling along a straight rod at rest in frame Z, then other frames would agree that they are both traveling along a rod which is still straight in other frames (so the angle between them should still be 180 degrees in other frames). Calculating the angles would depend on the direction the other frame was moving relative to Z's rest frame, but you could do it using the Lorentz transformation above if you aligned the x axis of Z's rest frame to be in the same direction as the motion of the second frame's spatial origin.
     
  9. Sep 15, 2010 #8
    A good point, and unfortunately this particular form is mentioned too often to the detriment of the general form which is almost never referenced or used for educational reasons. In this particular topic the general form could be, although complicated, very instructive.

    Any good references where the general form is written down with a couple of good exercises? I would argue that "the beef" in SR is when we get multiple boosts or rotations.
     
  10. Sep 15, 2010 #9

    JesseM

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    The general transformation is given in matrix form in this section of the wikipedia article. I don't necessarily see why the x-axis form isn't sufficient for the question the OP is asking though.
     
  11. Sep 15, 2010 #10
    That is only the boosts right?

    Neither do I, but to make the calculations from one traveling clock to another one would obviously need a more general approach.
     
  12. Sep 15, 2010 #11

    JesseM

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    Yes, only boosts, if you want to consider coordinate transforms where the spacetime origins of the different frames don't coincide you have to use the Poincaré transformation discussed in this thread.
    But weren't the other clocks besides the central one moving non-inertially? Did the OP say anything specifically about wanting to calculate things from the perspective of frames where the other clocks were at rest?
     
  13. Sep 15, 2010 #12
    JesseM why are you so argumentative?

    I am simply saying that if you want to calculate from the traveling clock perspectives you need a more generic transformation.

    Please don't take everything so personal.
     
  14. Sep 15, 2010 #13

    JesseM

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    It seems like you're the one who's taking things personally, since you're suddenly making negative comments about my personal style/motivations. I'm just questioning whether the general form of the Lorentz transformation is relevant to the OP, since I didn't see anything there that was about "calculating from the traveling clock perspective" and in any case those clocks were moving non-inertially. It's not like I'm criticizing you personally or saying it was a terrible mistake to think it was relevant, no need to make a big deal out of such a trivial disagreement.
     
  15. Sep 19, 2010 #14
    Off to do the math. I presume OP is Original Premise. Since that's overly complex, let's get back to the Simplified Premise with just 4 clocks.

    One thing I'm not clear on. I think I need a time zero that is zero in multiple frames of reference, at least when calculated from clock Z's frame of reference. In the simplified premise, all three clocks are going to be at the same place (clock Z) in 5/4 seconds as read by clock Z, and some shorter time as read by each of the other clocks. In the SP, all of the clocks are moving in inertial frames of reference, so the moving clocks will all observe the shorter time elapse in clock Z.

    Can I use spacial closeness to determine time zero? That is, from clock Z's frame of reference, we make time zero the point at which the clock's converge. Since this will occur in 5/4 second and each of the other clocks are proceeding at 3/5 the rate of Z's clock, we know the corresponding clock time for W, X, and Y relative to Z (and Z relative to W, X, and Y) but nothing about the W, X, and Y clocks relative to each other.

    But they are all calculable from the equations above, as is the approach angle of Y. The basic assumption is that clocks that are very close to each other can instantaneously on their position in space and time. The secondary assumption is that I can correlate positions in instantaneous position in space and time in one frame of reference to those in multiple secondary frame of reference, and thereby correlate any desired frames of reference with each other. The key is that any measurement must be at the same point in space.

    Checkpoint one: From Z's frame of reference, at time -5/4, clocks W, X, and Y are one light second away from Z moving toward Z at 4/5 C. At time zero, W, X, Y and Z will be nearly at the same point in space and each of their clocks will read zero. This will also be the common point zero in space. From Z's frame of reference, clocks W, X and Y read 3/5 of -5/4, or -3/4. However, clocks W, X, and Y each measure their distance to Z as 3/5 of a light second. Since they are moving at 4/5C, they will reach Z in 3/4 second, at time zero. All this is as measured in frame Z. Correct so far?
     
  16. Sep 20, 2010 #15

    JesseM

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    Yes, the equations of the Lorentz transformation demand that all frames have their spatial origins (x=0, y=0, z=0) at the same position at a time-coordinate of 0, meaning the event (x=0, y=0, z=0, t=0) in one frame is the same as the event (x'=0, y'=0, z'=0, t'=0) in another frame. There is a more general transformation called the "Poincare transformation" that doesn't have this requirement, but to keep things simpler you will probably want to stick to the Lorentz transformation.
    Not sure what you mean by this. They will both agree they are very close to each other, but unless they are both at a position-coordinate of 0 and a time-coordinate of 0, they will in general disagree about the position and time coordinates assigned to the event of their being next to each other. That's the whole point of the Lorentz transformation, that it relates the coordinates on frame assigns to a given physical event (like the event of two clocks passing next to each other) to the coordinates a different frame assigns to the same physical event. You might take a look at the illustrations and diagrams I gave in this thread, showing how each frame can be imagined to assign coordinates to events using local measurements on a set of rulers and synchronized clocks. For example, see this image:

    MatchingClocks.gif

    The circled event shows two clocks passing right next to each other, and both frame A above and frame B below agree on the time each clock showed at the moment they passed next to each other. However, frame A assigns coordinates x=346.2 meters, t=1 microseconds to the moment these clocks passed, while frame B assigns coordinates x'=173.1 meters, t'=0 microseconds to the moment they passed.
    Agree with all of your numbers, but it's not quite right to say "all this is as measured in frame Z", since the sentences in bold refer to how things look in rest frames of the other clocks.

    Now, suppose we assume the W clock is moving along the x-axis of frame Z in the +x direction, and the x,y,z,t coordinates of frame Z are related to the x',y',z',t' coordinates of the W rest frame by the simple form of the Lorentz transformation:

    x' = gamma*(x - vt)
    y' = y
    z' = z
    t' = gamma*(t - vx/c^2)
    with gamma = 1/sqrt(1 - v^2/c^2)

    Then in this example, v=0.8 and gamma=1/0.6. So if in the Z frame, W is at position x=-1 light seconds at t=-1.25, and the W clock reads -3/4 = -0.75 at this time, then the transformation shows that the coordinates of this event in the W frame are:

    x' = (1/0.6)*(-1 + 0.8*1.25) = (1/0.6)*(-1 + 1) = 0
    t' = (1/0.6)*(-1.25 + 0.8*1) = -0.45/0.6 = -0.75

    So, you can see that in W's frame, W reads -0.75 at t'=-0.75 as you'd expect.
     
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