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Repositioning clocks: continuing from the switch paradox

  1. Jul 31, 2010 #1
    In this hypothetical, an experiment is going to be conducted concerning the repositioning of clocks.

    two clocks are placed at the leading end of a vessel, that is 1 light second long, to an observer at rest relative to it.

    This vessel is shot into space from an observation station at 0.9C. As these clocks occupy essentially the same position, it can be safely assumed that these clocks in both the vessel frame and the observing station frame, that they are synchronised to each other enough that the difference is negligible between the frames.

    One of the clocks is going to be shot from the leading end to the trailing end at a velocity of
    -1/2.8 C according to the vessel or decelerate to a velocity of 0.8 to the observing station, where upon hitting the trailing end will return to the velocity of the vessel.

    Before this occurs predictions based on special relativity are conducted as to what the difference between the clocks will be in each frame.

    (Trailing - leading) in observing frame = duration of event in frame x (dilation of trailing - dilation of leading)


    (Trailing - leading) in vessel frame = duration of event in frame x (dilation of trailing - dilation of leading)

    special relativity suggests that any frame of reference can be treated as though it is an absolute frame, considering itself to have a velocity of 0. For relativity to hold when one translates the goings on in a particular frame to another, it must be consistent with what the frame would predict.

    for this situation it means, that the difference in clocks in the observing station frame converted into the vessel frame through the use of Lorentz transformations, should equal the difference in clocks in the vessel frame:

    difference in observing frame -simultaneity adjustment = difference in vessel frame

    or

    (((1-VS)/|S-V|)*(((1-((S-V)/(1-VS))^2)^0.5)-1))=(((1-V^2)^0.5/(V-S))*(((1-S^2)^0.5)-((1-V^2)^0.5))-V

    V: velocity of vessel / speed of light
    S: velocity of trailing clock / speed of light

    my calculations show in this instance
    (((1-VS)/|S-V|)*(((1-((S-V)/(1-VS))^2)^0.5)-1)) = -0.18466 seconds

    V-(((1-V^2)^0.5/(V-S))*(((1-S^2)^0.5)-((1-V^2)^0.5)) = -0.18466 seconds

    So it did work out in my calculations.
     
    Last edited: Aug 1, 2010
  2. jcsd
  3. Jul 31, 2010 #2
    This is impossible. You keep making the same mistake over and over.
     
  4. Jul 31, 2010 #3

    JesseM

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    OK. In the ship's frame, the ship's length is 1 light-second, so the clock will take a time of 1 light-second/0.866c = 1.1547 seconds to reach the back in the ship's frame. During this period of 1.1547, the clock was traveling at 0.866c so it was running slow by sqrt(1 - 0.866^2) = sqrt(1 - 0.75) = 0.5, meaning it only elapsed a time of 0.5*1.1547 = 0.57735 seconds. Meanwhile the front clock was at rest in this frame, so it elapsed the full time of 1.1547 seconds. So, by the time the back clock reaches the back and comes to rest again in the ship's frame, the clock at the back is behind the clock at the front by a time of 1.1547-0.57735=0.57735 seconds.

    In the observation tower frame, the ship is moving at 0.866c so its length is shrunk by a factor of 0.5, meaning its length is 0.5 light-seconds. So when one clock comes to rest in this frame and the back of the ship approaches it at 0.866c, the back of the ship will catch up to this clock in a time period of 0.5 light-seconds/0.866c=0.57735 seconds. During this time the clock was at rest so it ticked a full 0.57735 seconds. But during this time the clock at the front was moving at 0.866c so it was slowed down by a factor of 0.5, meaning it only ticked a time of 0.5*0.57735=0.288675 seconds. So, in the observation tower frame, by the time the clock reaches the back, it is ahead of the clock at the front by a time of 0.57735-0.288675=0.288675 seconds.
    In observing frame:

    duration = 0.57735 seconds
    dilation of trailing = 1
    dilation of leading = 0.5

    So, 0.57735*(1 - 0.5) = 0.288675, which is indeed what I got for (trailing-leading)
    In vessel frame:

    duration = 1.1547 seconds
    dilation of trailing = 0.5
    dilation of leading = 1

    So, 1.1547*(0.5 - 1) = -0.57735 seconds, which is again what I got for (trailing-leading)
    How can you convert a "difference" using the Lorentz transformation? The Lorentz transformation only transforms the coordinates of localized events from one frame to another.

    For example, let's assume both clocks read 0 at the moment they depart from one another. That means that the clock that goes to the back will read a time of 1.1547 seconds when it reaches the back wall--call that event A. Then since the clock is now 0.288675 ahead of the clock at the front in the observation tower frame, that must mean that if we let event B be the event of the front clock reading 1.1547-0.288675=0.866 seconds, then this event B is simultaneous with event A in the observation tower frame. So if we let A be at the origin of the observation tower frame, meaning A has coordinates x=0 and t=0, then B must also have coordinate t=0, and since the ship is 0.5 light seconds long in this frame B must have coordinate x=0.5.

    So in the observation tower frame, we have A (clock reaching back wall and reading 1.1547 seconds) at x=0, t=0 and B (clock at front wall reading 0.866 seconds) at x=0.5, t=0. If we use the Lorentz transformation with gamma=2 to find the coordinates of these events in the vessel frame, event A has coordinates x'=0, t'=0 while event B has coordinates:

    x' = gamma*(x-vt) = 2*(0.5 - 0.866*0) = 1
    t' = gamma*(t-vx/c^2) = 2*(0 - 0.866*0.5) = -0.866

    So in the vessel frame, at t'=-0.866 seconds, event B occurs, which is the event of the clock at the front reading 0.866 seconds. And the clock is at rest in this frame, so 0.866 seconds later at t'=0, the clock at the front reads 0.866 + 0.866 = 1.732 seconds. Call this event C. We know that event A also occurred at t'=0 in this frame, so event A is simultaneous with event C in the vessel frame. And remember that A was the event of the clock at the back reading 1.1547 seconds, so in this frame (trailing - leading) = 1.1547 - 1.732 = -0.5773 seconds. And that's just what I got for (trailing - leading) in the vessel frame by the earlier method of figuring out the velocity and time for the clock to go from front to back in the vessel frame! So you see, everything works out consistently.
    You need to explain your work, I have no idea what the basis for this complicated equation is.
     
  5. Jul 31, 2010 #4

    JesseM

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    No mistake there, he was just assuming the two clocks remained synchronized with one another as long as they were right next to each other at the front of the ship (and so following exactly the same worldline, and elapsing the same proper time to any point on that worldline), they do get out-of-sync as soon as one heads towards the back of the ship while the other remains at the front.
     
  6. Jul 31, 2010 #5
    In order for this to happen they need to be moved away from each other in the frame of the rocket. This means that they are no longer sychronized in the frame of the launcher.
     
  7. Jul 31, 2010 #6

    Fredrik

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  8. Jul 31, 2010 #7

    JesseM

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    Yeah, but I don't think he was saying they would stay synchronized once they were moved away from each other, in fact his whole analysis says they wouldn't. When he said "As these clocks occupy essentially the same position, it can be safely assumed that these clocks in both the vessel frame and the observing station frame, that they are synchronised to each other enough that the difference is negligible between the frames", I think he was just talking about the period of time when both clocks were together in the front of the rocket as it was shot away from the observation station, not about what happened after the two clocks began to move away from each other (the quoted paragraph was from before he mentioned moving them apart).
     
    Last edited: Aug 1, 2010
  9. Aug 1, 2010 #8
    JesseM has the understanding down of the situation

    "So, 0.57735*(1 - 0.5) = 0.288675, which is indeed what I got for (trailing-leading)" JesseM

    If 0.288675 seconds passed in the observer frame then half as much passed in the vessel

    then you need to adjust that based on the relativity of simultaneity equation which is simply V in as your dealing with a vessel that 1 light second long. Once this is done you will have the same answer as mine. You can also find if S was -(0.75)^0.5 instead of 0 congruency would emerge.
     
  10. Aug 1, 2010 #9
    i dilated time in a place i shouldn't have and the result is congruent. Least I get it now.
     
  11. Aug 1, 2010 #10

    JesseM

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    Not true, the time dilation equation only says that if two events happened at the same position but different times in the vessel frame (like different ticks of a clock at rest in the vessel frame), then the time between them in the vessel frame is half as much the time between these same events in the observer frame (i.e. more time passes between those two ticks of the vessel clock as seen in the observer frame, so the clock is running slow as seen by the observer). Likewise if two events happen at the same position and different times in the observer frame, the time in between them in the observer frame is half the time between them in the vessel frame. But 0.288675 isn't the time between two events at the same position in either frame. In fact it isn't really a time between a pair of events at all, instead 0.288675 is the amount the trailing clock is ahead of the leading one in the observing frame, so if at some moment the leading clock reads T seconds, then at the same moment the trailing clock reads T+0.288675 seconds.
    Relativity of simultaneity says that if two clocks are a distance L apart and synchronized in their mutual rest frame, then they are out-of-sync by vL/c^2 in the frame that sees them moving at v. But the clocks at the front and back of the ship aren't actually synchronized in the ship frame, so this wouldn't be the correct equation to tell you how out-of-sync they are in the observer frame.
    My math is correct, if it disagrees with your it's because yours is incorrect. If you disagree, you need to either explain your own reasoning in a more clear step-by-step manner (as I said, you just wrote down a giant complicated equation without explaining how you derived it), or point out where the flaw is in my own analysis.
     
  12. Aug 1, 2010 #11
    what i meant is that your answer is congruent as you didn't dilate time. This was bad math on my part.

    i'll be getting rid of the time dilation from the formula soon and then we will be on the same page.

    The parameters I used are big and my assumption that as you slow the velocity in which you reposition clocks, time dilation will become negligible, it is possible that the parmeters are just a sweet spot of congruency. My assumption could turn out to be similar to martingale strategy, as the assumption may only work if the velocity of clock repositioning is 0 in the vessel frame.

    We shall soon see, ill be editing the formula and using the parameters of s=0.8 v=0.9 in the first post.
     
  13. Aug 1, 2010 #12
    turns out the assumption was bad. But now at least im informed
     
  14. Aug 1, 2010 #13
    They are out of sync by [tex]\gamma(v)\frac{vL}{c^2}[/tex]
     
  15. Aug 1, 2010 #14
    L as defined above, is a distance in the clocks rest frame. Your equation would be correct applied to a length as measured in a relative frame , yes??
     
  16. Aug 1, 2010 #15
    no, L=proper length in the equation I posted.
     
  17. Aug 1, 2010 #16

    JesseM

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    No, if you have two clocks a distance L apart in their rest frame, and two events on their worldlines which occur simultaneously in their rest frame, then these same two events are separated by a time of [tex]\gamma(v)\frac{vL}{c^2}[/tex] in the observer's frame where the clocks are moving at speed v. But that's not telling us the amount they're out-of-sync in the observer's frame! To find that, we need to find two events on each worldline that are simultaneous in the observer's frame, and then find the time interval between them in the rest frame of the clocks (since the observer judges how out-of-sync the clocks are by looking at what each clock reads at a single moment in his own frame).

    In the observer's frame, the clocks are only separated by a distance of L/gamma(v), so suppose in the observer's frame the two simultaneous readings occur at these coordinates:

    x=0, t=0
    x=L/gamma, t=0

    In that case the time of the first event in the clock frame is t'=0, while the time of the second event is:

    t' = gamma*(0 - vL/gamma*c^2) = -vL/c^2

    So if the clocks are synchronized in their own rest frame, the difference in readings for these two events will be vL/c^2.
     
  18. Aug 1, 2010 #17
    It is very simple, really:

    [tex]t'=\gamma(v)(t-vx/c^2)[/tex]

    so:

    [tex]dt'=\gamma(v)(dt-vdx/c^2)[/tex]

    If the events are simultaneous in F, then [tex]dt=0[/tex] so:

    [tex]dt'=-\gamma(v)vdx/c^2[/tex]

    If the events are simultaneous in F' then [tex]dt'=0[/tex] and

    [tex]dt=vdx/c^2[/tex]

    I was quite clear in my answer that L=proper length, i.e. [tex]L=dx[/tex]
     
    Last edited: Aug 1, 2010
  19. Aug 1, 2010 #18

    JesseM

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    Yes, I agree with all of that. The point is that if you want to know how out-of-sync two clocks are in F' if the clocks are at rest and synchronized in F, and have a proper distance of dx in their rest frame, then you have to consider simultaneous readings in F' and figure out the time between them in F. So, the last equation tells you how out-of-sync the clocks are in F'.

    Likewise, if you had two clocks at rest and synchronized in F', and a proper distance dx' apart in F', then if you wanted to know how out-of-sync they are in F, you'd have to set dt=0, then you'd get dt'=vdx'/c^2. So regardless of what frame the clocks are at rest and synchronized in, if L is the proper distance between them in their own rest frame, it is always true that they are out-of-sync by vL/c^2 in the frame that sees them moving at speed v (in a direction parallel to the axis between them).
     
  20. Aug 1, 2010 #19
    Why would I do such a thing ? In F [tex]dt=0[/tex].


    I don't know why you persist in this error. I gave you the correct equation of desynchronization in F'.


    You can definitely do that but you are doing things backwards. Are you doing this because you are intent on justifying your initial formula? If this is what is driving you, it is easier to do it correctly:

    [tex]dx'=\gamma(dx-vdt)[/tex]
    [tex]dt'=\gamma(dt-vdx/c^2)[/tex]

    If [tex]dt=0[/tex] (events simultaneous in F) then:

    [tex]dx'=\gamma*dx[/tex]
    [tex]dt'=-\gamma*vdx/c^2=-vdx'/c^2[/tex]

    i.e. events are not simultaneous in F' and they are "desynchronized" (out of sync) by the amount [tex] dt'[/tex]
     
    Last edited: Aug 1, 2010
  21. Aug 1, 2010 #20

    JesseM

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    dt=0 for what two events? F' is the observing frame, if you want to figure out how out-of-sync the clocks are in F', you have to pick simultaneous readings in F'. For example, suppose the clocks are 10 light-seconds apart in their rest frame F, and synchronized in their rest frame. And suppose the clocks are moving at 0.8c in the frame of the observer F'. Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8? And that this is what it means for F' to say the two clocks are out-of-sync by 8 seconds in F'? So if event 1 is the leading clock reading T=0, and event 2 is the trailing clock reading T=8, then for these events dt=8 but dt'=0 (while dx=10 and dx'=6).
    I think you're just confused about what "desynchronization" means. Do you disagree that in my example above, we would say that in frame F' the two clocks are desynchronized by 8 seconds?
     
    Last edited: Aug 1, 2010
  22. Aug 1, 2010 #21
    Obviously, in F. The formulas are self-evident.

    [tex]dx=10[/tex] , [tex]v=0.8c[/tex], [tex]\gamma=1/0.6[/tex]

    [tex]dt'=\gamma vdx/c^2[/tex]

    Yes, I disagree. See above for the correct calculations. The correct anser is not 8s, as you claim but 8/0.6=13.33s
     
    Last edited: Aug 2, 2010
  23. Aug 1, 2010 #22

    JesseM

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    I didn't ask "what frame", I asked "what two events" (events are frame-independent). For example, I picked the two events "leading clock reads a time of T=0" and "trailing clock reads a time of T=8".
    That's not an answer to my question. I asked "Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?"

    Event 1: leading clock reading T=0
    Event 2: trailing clock reading T=8

    In frame F (where the clocks are at rest, synchronized, and have a separation of 10 light seconds) these events could have coordinates:

    Event 1: x=0 light-seconds, t=0 seconds
    Event 2: x=10 light-seconds, t=8 seconds

    Using the Lorentz transformation equation t'=gamma*(t - vx/c^2), you can see that in the observer's frame F' (where the clocks are moving at 0.8c) both of these events occur at the same time-coordinate t'=0. So, that's why I say "at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8." Please tell me if you agree or disagree with this statement, and if you disagree please tell me where you think my analysis above goes wrong.
     
  24. Aug 2, 2010 #23
    No, they don't. You are asked to find out the time separation between the two events, that are simultaneous in F (dt=0) as measured from frame F'. There is nothing entitling you to set t'=0.




    I don't understand why you have such a great difficulty in understanding the basic equations. Your analysis goes wrong in a simple attempt to calculate the time separation of two events that are simultaneous in frame F (dt=0) as measured in frame F' (where they are separated by [tex]dt'=\gamma v dx/c^2[/tex]). This is especially disturbing in the context of having showed you not only the formulas but also their derivation.
     
    Last edited: Aug 2, 2010
  25. Aug 2, 2010 #24

    JesseM

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    What part of my analysis is wrong? Do you disagree that if my two events have coordinates (x=0, t=0) and (x=10, t=8) in the F frame, then these events both have t'=0 in the F' frame?
    Your equation dt' = gamma*v*dx/c^2 is incorrect for those two events, because dt was nonzero for the events I picked. The full Lorentz transformation equation is

    dt' = gamma*(dt - v*dx/c^2)

    If dt is not zero, you can't reduce this to dt' = gamma*v*dx/c^2. For the two events I chose (the event of the leading clock reading T=0, and the event of the trailing clock reading T=8), dt was not zero, here dt=8.

    So, I ask again: "Would you then agree that at the same moment F' measures the leading clock reading T=0, he also measures the trailing clock to be reading T=8?" It's a simple yes-or-no question.
     
  26. Aug 2, 2010 #25
    ...which is precisely what I showed you earlier.

    ...but the discussion has ALWAYS been about simultaneous events in F. That is [tex]dt=0[/tex].

    But we are talking about [tex]dt=0[/tex]. I pointed out that your repeated dropping of [tex]\gamma[/tex] in the formula for dt' , for this PARTICULAR situation (i.e dt=0), is a mistake. Why are you now changing the goalposts to talk about [tex]dt>0[/tex]?
    BTW, even if you want to change the talk to [tex]dt>0[/tex]., your dropping of [tex]\gamma[/tex] is STILL a mistake.
     
    Last edited: Aug 2, 2010
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