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Relativity Paradox

  1. Jun 28, 2004 #1
    According to the dictates of Special Relativity, the rate of a clock is unaffected by motion of the frame at which it is at rest. But a puzzle arises when one considers the reciprocity of the frames in a case where a precedent determination of the temporal interval is given for each frame.

    Here there are two trains, each 5 light years in length from beginning to end
    - they are approaching each other on parallel tracks at velocity approximately c.

    E* ---------------train 2---->B* .............................. S* frame
    S frame................................B <--------train 1---------E

    B is the beginning of the train at rest in the S frame and E is its end,
    B* is the beginning of the train at rest in the S* frame and E* is its end.

    E and B represent clocks that are synchronized in S,
    E* and B* represent clocks that are synchronized in S*

    As beginning points BB* pass each other, each takes note of the others time -
    we will assume this to be “0" for convenience.
    This is the position shown in the Fig 1. At this time, the operator of train 1 conducts an experiment that produces a muon M, and the operator of train 2 does a similar experiment and produces a muon M* Both muons have a lifetime of 2 μsec in
    their own frame and M remains at rest with respect to clock B and M* remains at rest with respect to B*

    If frame S is considered at rest, the time required for B* to reach E will be 5 years as measured by clocks B and E.
    In the S* frame, the muon M* will be at rest relative to B* and if the relative
    velocity is sufficiently close to c, will not yet have decayed when B* reaches E.
    So, B* clock will read only 2 μsec when the S clock has logged 5 years.

    Now consider the S* frame at rest. The time that has passed in S* as measured by the synchronized clocks B* and E* when B reaches E* should be 5 years (not 2 usec) since the proper distance in S* is 5 light years. But muon M will
    only have aged 2 usec in its rest frame S - so clock B should read 2 usec, not 5 years.

    Question: When B* is opposite E and B is opposite E*
    Does clock B read 2 μsec or 5 years? Does clock B* read 2 μsec or 5 years?
    Last edited: Jun 28, 2004
  2. jcsd
  3. Jun 28, 2004 #2

    Doc Al

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    Just a quick comment
    Realize that "B* is opposite E" and "B is opposite E*" are two distinct events. Neither frame observes these events happening simultaneously.
  4. Jun 28, 2004 #3
    These are two distinct events in both frames. From frame S the other ship (S*) seems to be Lorentz contracted and - if their relative speed is almost c - it is possible that just after B and B* have coincided E* reaches B too, in less than 2 microseconds. The scenario is the same from S* where E reaches B* just after B and B* have met. But, of course, it takes B* (or B) 5 years to reach E (or E*) in the frame from which B* (or B) seems to be moving.
  5. Jun 28, 2004 #4
    So if the 5 ly length appears contracted in the other frame, we can still expect the muon to decay in 2 usec as measured by its own local clock. This corresponds to about 600 meters. Each muon will interpret the 5 ly length it has traveled in the other frame (to reach the end of the other train) as 600 meters - does this resolve anything?
  6. Jun 28, 2004 #5

    Doc Al

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    what is v?

    I don't understand the point of introducing the muons. They are clocks, just like any other. The 600 meters? That's the distance light travels in 2 usec? What's that got to do with anything? I guess each observer will see the other train travel about 600 meters in the time it takes his muon to decay. Of course, since you haven't specified the actual speed of the trains, we can't tell what the time dilation & length contraction would be. That "5 ly" train may contract down to 0.5 meters, for all we know. :smile:

    What paradox are you trying to resolve? I'm not seeing one.

    Another problem with your setup:
    The devil is in the details. "approximately" c is vastly different than equal to c. The relativistic effects depend on the exact speed v (v < c).
  7. Jun 29, 2004 #6
    Doc AL: Muon decay is a familiar scenario - and most agree that the decay time in the frame of the muon is independent of the velocity of the frame in which the muon resides. While it is not necessary to introduce muons, it sets the stage in terms of a physical event that has a preset time in each frame (2 usec). If the lifetime of the muon is 2 usec in its own frame, and it decays at the instant it reaches the 5 ly mark as measured by two clocks in another frame, the relative velocity can be computed - but just exactly how close this is to c is immaterial to the conceptual question posed.
  8. Jun 29, 2004 #7

    Doc Al

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    If the speeds are just right, when point E* just passes observer B, observer B's muon will have just decayed. And vice versa: When point E just passes observer B*, observer B*'s muon will have just decayed. Sounds good to me. What's the question?
  9. Jun 29, 2004 #8
    Let's say I'm located at point B on the ship denoted by S. If the relative velocity of the two ships is high enough it is possible that my muon (M) will have just decayed when the end of the other ship (E*) passes me. This means that the Lorentz contracted length of S* equals the relative speed times the lifetime of my muon in my own frame. This whole story is interpreted by an other obsrever sitting on ship S* at point B* as my clock ticking slower compared to his one and so my muon still being "alive" even when me (point B) reaches the end of his ship (E*) (which takes me almost 5 years according to his clock if the relative speed is almost c). In other words the time dilation is so huge that my clock is ticking slow enough to show only 2 microseconds (lifetime of my muon in my own frame) when I reach E* in 5 years time in his frame.

    I hope that was clear enough. There is no paradox here, it's just how relativity works.
  10. Jun 30, 2004 #9
    The question(s) posed is: 1) When B* reaches E, what does each clock read?
    2) When B reaches E*, what does each clock read?
  11. Jun 30, 2004 #10

    Doc Al

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    B* clock reads 2 usec; E clock reads 5 years.
    B clock reads 2 usec; E* clock reads 5 years.

    Realize that the relative speed is extremely close to c.
  12. Jun 30, 2004 #11
    But Doc - B* is synchronized with E* - both read the same in the S* frame. How can
    B* read two usec in answer to question 1 and 5 years in answer to question 2?

    Likewise, E has be sync(ed) to B. E and B should read the same in the S frame at all times. How can B read 2 usec in answer to question 2 and 5 years in answer to question 1?
  13. Jun 30, 2004 #12

    Doc Al

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    two non-simultaneous events

    You seem to still be thinking that the two events:
    (1) B* opposite E
    (2) B opposite E*​
    happen at the same time. They don't. Read my first response in this thread (post #2) and DanielK's as well.
  14. Jun 30, 2004 #13


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    Here maybe this will help. it shows the sequence of events according to both frames.


    Note that while E and B are in sync according the S frame they are not according to the S* frame, and the same is true for E* and B* according to the S frame.
  15. Jun 30, 2004 #14
    Hi janus - good animation graphics - let me see - If I follow your lesson, it is essentially the same as Doc's in that, in the moving frame, the beginning and end will have different temporal times consequent to their displacement (as observed from the other frame), and according to SR the B* and E* clock will not appear in sync when viewed from the non-moving S frame and E and B will not be in sync when viewed from S*. But the premise is that B* and E* are in sych in S* when S is considered the rest frame, and B and E are in sync when S* is considered the rest frame. In other words, measurements are made locally (proper times) at the events of concurrence. At root is the reality of time dilation - same old issue - if its apparent, then the riddle is resolved.
  16. Jul 2, 2004 #15
    There is still nothing to be resolved. B* and E* are in sync in S*, but not in S, B and E are in sync in S, but not in S*. This is because events that happen simultaneously in their rest frame don't happen simultaneously in another frame which is moving relative to the previous one. Just remember Einstein's famous thought experiment with a moving train and two flashes of light.

    Time dilation and Lorentz contraction are apparent phenomena in the sense that different observers in differently moving frames measure different times and spatial locations for the same events, but not apparent in the sense that these effects are not measurable at all. The equivalence of inertial frames requires these effects to be reciprocal, but many people regard this reciprocity as an evidence that time dilation and Lorentz contraction are not real. It is enough to think about muons reaching the ground from a hight of 70-80 km though their lifetime would be too short to do it, if relativity didn't come and explain this by time dilation (or L. contraction, as seen from the frame of a muon).
  17. Jul 4, 2004 #16
    DaneilK - Relativists dodge in and out of the reality/apparency issue of time retardation and length contraction to evade paradoxical results - in fact there are several separate and inconsistent schools of thought among the advocates of the special theory. Different viewpoints evoke different interpretations to rationalize a particular explication to avoid inconsistent results that might indicate the need for some theoretic overhaul. There are a lot of bright people who have trouble with SR, and its not because they don't understand it. The problem is not with what the LT predict, it is in the transformational consequences. With regard to the present thread, you might take a look at Ling Jun Wangs treatment of the same problem labeled as the 4-J experiment:


    Also take a look at Selleri's approach to these issues using "Inertial Transforms"

  18. Jul 6, 2004 #17
    I think that every relativist agree that your two-ship-paradox is essentially nothing more than a modern version of Einstein's classic thought experiment with the trains. It involves nothing more than the basic concepts of relativity with the usual LT equations. If it still looks paradoxical then it is because this situation doesn't resemble our everyday life at all and causes such paradox-looking effects. Unfortunately, many people attribute the shocking and unusual results of relativity to its paradoxical theoretical structure. Not to mention that they forget that the word "paradox" means "apparent contradiction" and not "contradiction". That is it is explained by the theory even if the explanation is a bit strange.
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