SR conundrum.

Discussion in 'Pseudoscience' started by Zeno, Nov 4, 2013.

  1. Zeno Registered Senior Member

    Messages:
    242
    Situation from M1 frame of reference...
    Code:
     
                                  A2(.8c---->)
                      A1          M1          B1
                   (4:00)                   (4:00)
    
    Situation from the A2 frame of reference...
    Code:
             
                                  A2
     (<-----.8c)      A1          M1          B1
                   (4:00)                   (4:15)
    
    M1 is at the midpoint of A1-B1. The time is flashed from both A1 and B1 simultaneously in the A1-M1-B1 frame, (say 4:00 for example), and reaches M1 in the middle simultaneously just as A2 reaches him. Now then, A2 also sees the flashes simultaneously, but according to him the flash from B1 occurred before the flash from A1. So, the clock at B1 is running ahead of the clock at A1 with 4:00 occurring at B1 before it occurred at A1. When A2 reaches B1, he looks at the time on the clock at B1. Now then, both A2 and B1 will agree on the time displayed on B1's clock, but they will disagree on the time at A1.

    We have the following two situations:
    according to B1:
    Code:
     
                                              A2(.8c---->)(The clock at A1 reads 4:45)
                      A1          M1          B1
                    (5:00)                   (5:00) 
    
    according to A2:
    Code:
     
                                                                                 A2(The clock at A1 reads 5:00)
                      A1          M1          B1
    (<-----.8c)     (5:00)                  (5:15) 
    
    This is impossible because it requires A2 to be located in two different places simultaneously.
     
    Last edited: Nov 4, 2013
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  3. arfa brane call me arf Valued Senior Member

    Messages:
    7,832
    What mechanism is used to ensure that A1 and B1 signal M1 simultaneously? Or do A1 and B1 just broadcast their local times continuously?
     
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  5. Zeno Registered Senior Member

    Messages:
    242
    I'm not sure of the relevance of the question to the opening post. There really isn't any difference if the time is flashed or shown continuously.
     
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  7. arfa brane call me arf Valued Senior Member

    Messages:
    7,832
    I suppose not, as long as A1, M1 and B1 are in the same rest frame.

    However this:
    . . . is contradictory. If A2 sees simultaneous flashes then A2 cannot see one flash occur before the other (in which case the flashes are not simultaneous).
    Your conclusion seems to have the same problem.

    Or can you explain how A2 sees the clock at A1 is 4:45 (15 minutes slower than the clock at B1), and simultaneously sees the time at A1 is 5:00 (0 minutes slow)?
     
  8. Janus58 Valued Senior Member

    Messages:
    2,385
    How is A2 located at two places simultaneously? For any observer (A1, M1, B1, A2), A2 is never is more than one spot at a time. Your conclusion comes from improperly mixing frames. Just because A2 is next to B1 when both agree that B1 reads 5:00, and that according to B1 A1 reads 5:00, it does not mean If A2 is not next to B1 when A1 reads 5:00 according to A2,that A2 is in two different places at the same time. Simultaneity cannot be applied between two different frames like that because simultaneity is frame dependent.

    To prove that A2 would have to be in two places at once, you would have to show that this is the case according to one single frame, and not by trying to mash together simultaneity from two different frames.
     
  9. Zeno Registered Senior Member

    Messages:
    242
    When A2 arrives at M1 at the same time both flashes of light reach M1, A2 must also see both flashes of light simultaneously, but in order for the speed of light to be constant relative to all observers, the light emitted from B1 must have occurred before the light emitted from A1 according to A2. Therefore, according to A2, the clock at B1 is running ahead of the clock at A1, but for M1 the two clocks always simultaneously display the same time.
    When A2 arrives at B1, the clock (according to A2) at A1 reads 4:45, only some time later (according to A2) after A2 has passed by B1, does the clock at A1 read 5:00 (according to A2). According to B1, A1 is at B1 when the clock at A1 reads 5:00. So in answer to the question "Where is A2 when the clock at A1 reads 5:00?" A2 is both at B1 and away from B1.
    When the clock at A1 reads 5:00, we have A2 at two different places according to both frames of reference. I'm just following SR to its logical conclusion.
     
  10. arfa brane call me arf Valued Senior Member

    Messages:
    7,832
    If you get a result like that, you've made a mistake.
    You're confusing frames, and that's the cause of the error.

    And no, I can't be arsed trying to explain where you made the error.
     
  11. Zeno Registered Senior Member

    Messages:
    242
    When the clock at A1 reads 5:00, A2 says "I've already passed by B1".
    When the clock at A1 reads 5:00, B1 says "A2 is here right next to me."
    Can you please elaborate on how that is "confusing frames"?
     
  12. Janus58 Valued Senior Member

    Messages:
    2,385
    Because A2 and B2 have different rest frames. Each frame determines what is simultaneous in its own frame and that is the only way simultaneity can be defined. In the rest frame of B1 the event of A2 passing him and A1 reading 5:00 is simultaneous, while in the rest frame of A1, they are not.

    You are "confusing frames" when you try to treat "when the at A1 reads 5:00" as if it has some absolute meaning.

    When the clock at A1 reads 5:00 in the rest frame of A2, A2 says,"Ive already passed by B1"
    When the clock at A1 reads 5:00 in the rest frame of A1 and B1 , B1 says, "A2 is here right next to me."

    This is not impossible nor a conundrum because simultaneity is frame dependent. that is the whole point of the Relativity of Simultaneity. I'm sorry that you cannot see this and that you keep wanting to apply some universal meaning to simultaneity that doesn't exist.
     
  13. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    According to B1, the events in the first diagram are simultaneous:
    A1 reads 5:00
    B1 reads 5:00
    A2 is next to B1

    According to A2, the events in the second diagram are simultaneous:
    A1 reads 5:00
    B1 reads 5:15
    A2 is not next to B1

    But there is no reference frame in which the events "A2 is next to B1" and "A2 is not next to B1" are simultaneous.


    (Optional - We could define the "A2 is not next to B1" event better by putting a clock on A2, or by adding another clock C1 at rest with A1,B1,M1 that A2 passes when A1 reads 5:00 in A2's rest frame)
     
  14. Zeno Registered Senior Member

    Messages:
    242
    Yes. I believe it has absolute meaning.

    So, this isn't a problem? Let's say that the clock at A1 flashes its time when it reads 5:00. So when A2 reaches B1, A1 flashes its time. Then, after A2 has passed by B1, A1 flashes its time again. So, we have two separate groups of photons flying through space for the clock at A1 reading 5:00, even though the clock only reached 5:00 once. This is obviously garbage and nonsense.
     
  15. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    No, every reference frame agrees that A1 only flashes 5:00 once.
     
  16. Zeno Registered Senior Member

    Messages:
    242
    Yes. In every reference frame clock A1 flashes 5:00 only once. Once for the A1-B1 frame (at that moment the clock at A1 reads 4:45 according to A2, so it doesn't yet flash). Then the clock flashes once for the A2 frame once it finally reaches 5:00. These events are not simultaneous so we have two groups of photons flying through space.
     
  17. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Does anyone see the flash twice?
     
  18. Zeno Registered Senior Member

    Messages:
    242
    If observers B1 and A2 both see the clock at A1 showing 5:00 only once, one can ask "How's that possible?".
    If observers B1 and A2 both see the clock at A1 showing 5:00 twice, one can ask "How's that possible?"
     
  19. arfa brane call me arf Valued Senior Member

    Messages:
    7,832
    Lets also assume that for A1, this event occurs exactly once, at 5:00 for A1, M1 and B1, since these are all in the same rest frame.
    So A1 flashes two different times, or otherwise it flashes the same time twice (in which case the two flashes are the same flash).
    See, what you're doing is saying that A1 can flash its time at 5:00, then "after A2 has passed B1", i.e. after 5:00 for A1, it does so again. You've inserted an absolute: A2 passes B1, but forgotten that this is relative to each observer.
    You need to explain how a clock can advance past 5:00 after flashing that time to somewhere, then flash 5:00 again. Is the clock at A1 not working?
    Well, if you assume A1 can flash 5:00 at two different times, yes it is.
     
  20. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    It flashes once. They each see it once. What's the problem?
    Do you think that's what happens?
     
  21. Zeno Registered Senior Member

    Messages:
    242
    Because the A1-M1-B1 frame is moving it is length contracted.
    Here is a view with frame of A2 at rest. Observer B2 is co-located with observer A1 when A2 is co-located with B1.
    Code:
                        B2                   A2
      <----(0.8c)       A1                   B1
                      (4:45)               (5:00)
    
    Now, here is a view with the A1-B1 frame at rest.
    Code:
           C2                 B2                   A2----->(0.8c)
           A1                                      B1
         (5:00)                                  (5:00)
    
    So here the A2-B2-C2 frame is length contracted and A1 is co-located with observer C2 when A2 is co-located with B1.
    A1 passes by B2 when the clock reads 4:45 and passes by C2 when it reads 5:00. All is well and good.
    However, when viewed from the A2 frame at rest...
    Code:
                        B2                   A2
      <----(0.8c)       A1                   B1
                      (4:45)               (5:00)
    
    becomes
    Code:
                   C2             B2             A2
      <----(0.8c)  A1                   B1
                 (5:00)               (5:15)
    
    Viewed from the A1-B1 at rest...
    Code:
           C2                 B2                 A2----->(0.8c)
           A1                                    B1
         (5:00)                                (5:00)
    
    So when the clock at A1 reads 5:00, A2 is co-located with B1 and not co-located with B1.
    To me, that looks like A2 is in two places at the same time.
    Or look at this...
    View from the A2 frame at rest...
    Code:
                        B2                   A2
      <----(0.8c)       A1                   B1
                      (4:45)               (5:00)
    
    View from the A1-B1 frame at rest...
    Code:
           C2                 B2                 A2----->(0.8c)
           A1                                    B1
         (5:00)                                (5:00)
    
    How can A2 be co-located with B1 when B2 is co-located with A1 and then again A2 is co-located with B1
    when C2 is co-located with A1? Is A1 simultaneously in two different places at B2 and C2?
    While A1 is in transit from B2 to C2 which direction should B1 look in order to look at A2?
    Should he look towards A1 or away from A1?
     
    Last edited: Nov 9, 2013
  22. arfa brane call me arf Valued Senior Member

    Messages:
    7,832
    The A1-M1-B1 frame is moving relative to what?
    Ok, the A2-B2-C2 frame is length contracted relative to the A1-M1-B1 frame.
    Where you say "the clock reads 4:45", which observers see this?
    Again, which observers see A1's clock reading 5:00?

    You're still confusing frames, or assuming that everyone sees the same time on A1's clock.
    Or something.
     
  23. Zeno Registered Senior Member

    Messages:
    242
    Both observers A1 and B2.
    Observers A1 and C2.
    Yes, I'm assuming that everyone sees the same time on A1's clock since it would be impossible for say B2
    to arrive at A1, both look at the same clock located at A1 and disagree on the time.
     

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