Twin paradox (Pete and MacM)

Discussion in 'Physics & Math' started by Pete, Sep 6, 2004.

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  1. Paul T Registered Senior Member

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    Pete,

    Okay, let's try to apply Lorentz transformation for the exercise I gave to depict MacM's wrong view.

    Say, you are in a spaceship moving at v=0.6c from earth to moon. If earth-moon distance is 1.2c second, according to earth observer you need (1.2c second)/0.6c=2 seconds to reach moon. Now, from your point of view. You see earth-moon distance as 0.8*1.2c second = 0.96c second. According to MacM's interpretaion, you still need 2 second (instead of 1.6second suggested by standard SR) to complete the trip and therefore you will find that your velocity relative to earth (or moon) is (0.96c second)/2 = 0.48c. This cannot be right (Correct only based on MacM's relativity, which is the wrong version of relativity).Based on standard SR you will still see the relative velocity as (0.96c second)/1.6 = 0.6c! There is nothing to talk about on MacM's view on this matter, seriously, because his view is WRONG.​

    I believe you agree that for this case, the spaceship observer measures the time for the trip as 1.6 second and the earth-moon distance as 0.96c second, while earth observer measures them as 2 second and 1.2c second respectively.

    Applying LT for spaceship passing earth event gives us t=t'=0 and x=x'=0.

    Now, apply LT for t=2 seconds and x=0 that is an event occuring on earth at the time the spaceship arrives to the moon. This is the result:

    x' = 1.25*(0-0.6c*2) = 1.5c second
    t' = 1.25*(2-0) = 2.5 second​
    Next, apply LT for t=2 seconds and x=1.2c second that is an event occuring on the moon at the time the spaceship arrives to the moon. This is the result:

    x' = 1.25*(1.2c-0.6c*2) = 0
    t' = 1.25*(2-0.6*1.2) = 1.6 second​
    You can see that the correct result is obtained by applying LT for the spaceship arriving the moon event, which occurs at x'=0 at the same position as the spaceship passes earth event. The distance covered by the spaceship during the 1.6 second trip, according to the spaceship observer, is 0.6c*1.6 = 0.96c second.

    You can verify using other method that event at D occurs at t=0 in earth reference frame and event at E occurs at t=2sec in earth reference frame, say, by first synchronize clocks at D and E to ensure that they both tick simultaneously.
     
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  3. Pete It's not rocket surgery Moderator

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    Hi Paul,
    Yes, In Earth's frame, D and A are simultaneous, and C and E are simultaneous.

    But in the spaceship frame, D and A are not simultaneous, and neither are C and E.

    When the spaceship observer is timing the laser's transit, why would they time the interval between events that aren't simultaneous with the start and finish of the transit period?
     
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  5. Pete It's not rocket surgery Moderator

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    Yes, that's all correct.
    But, I have to confess that I don't see your point.
     
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  7. Pete It's not rocket surgery Moderator

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    Or perhaps I do...

    Have you considered how this situation is different to 2inquisitive's laser scenario?

    In your scenario, the end-points of the interval in question are in the same place in the rocket frame - hence, the elapsed time should be shortest in the rocket frame.

    In 2inquisitive's scenario, the end-points of the interval in question are in the same place in the Earth frame - hence, the elapsed time should be shortest in the Earth frame.
     
  8. MacM Registered Senior Member

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    Well, well, well. I have to correct myself. In the past I have referred to your posts as "Slanderous". I just looked up the definition in Webster and that is incorrect.

    WEBSTER:
    Slander: (1) The utterance in the presence of another person of a false statement or statements damaging to a third person's character or reputation: usually distinguished from Libel, which is written.

    Your posts are written therefore "Libelous". I have not entered the discussion regarding this scenario. Your assertions regarding what I think, what I believe and how I would address the issue are not only baseless but outright horseshit.

    Knock it off - asshole.
     
  9. Paul T Registered Senior Member

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    Pete,

    All we need are "In Earth's frame, D and A are simultaneous, and C and E are simultaneous", as for the fact they are not simulatneous in the spaceship's frame is not an issue.
     
  10. Paul T Registered Senior Member

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    If you look carefully (I am sure you already had), you would find that 2inquisitive's exercise is very similar to my suggested scenario. The similarity are on the times measured in spaceship's frame corresponding to the times measured by earth observer for the same set of events. They are 4sec or 1.0sec for 2inquisitive scenario and 2.5sec or 1.6sec for my scenario. The point come back to the question, which of them are correct. In my opinion, 1.0sec and 1.6sec, not 4sec and 2.5sec, that's the point I wanted to convey.
     
    Last edited: Nov 5, 2004
  11. Pete It's not rocket surgery Moderator

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    Hang on...
    Exactly what are we calculating?


    We're calculating the time that elapses on spaceship clocks during the laser's transit, but are we calculating that time in Earth's frame, or the spaceship's frame?

    From your statement, it seems that you are calculating it in Earth's frame, but that's not my understanding of the problem.
     
  12. Pete It's not rocket surgery Moderator

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    To recap, here's the scenario in the two frames:

    <img src="/attachment.php?attachmentid=3404&stc=1">

    <img src="/attachment.php?attachmentid=3406&stc=1">

    Look at the Spaceship frame diagram.

    Is the time for the laser transit the time between D and E, or the time between F and G?

    (Note that the diagrams are hand drawn, and there are some inconsistencies in the scaling)
     
  13. Pete It's not rocket surgery Moderator

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    I think the correct times are 4 seconds and 1.6 seconds.

    In your scenario, the end-points of the interval in question (ship leaves Earth, ship arrives Moon) are in the same place in the rocket frame - hence, the elapsed time should be shortest in the rocket frame.

    In 2inquisitive's scenario, the end-points of the interval in question (beam leaves Earth, beam arrives Earth) are in the same place in the Earth frame - hence, the elapsed time should be shortest in the Earth frame.
     
  14. Paul T Registered Senior Member

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    Pete,
    So, you did not see that 2inquisitive's scenario and mine are identical. Let's now make a little modification on my scenario.

    Firstly, move the origin from earth to a point exactly at the middle between earth and moon. Now, earth is located at x=-0.6c second and moon is at x=0.6c second.

    At t=0 (earth's time), the spaceship is at x=-0.6c second. Use LT to get t' and x':

    t' = 1.25*[0 - (0.6/c)*(-0.6c sec)] = 0.45 sec
    x' = 1.25*[-0.6c sec - 0.6c*0] = -0.75c sec​
    At t=2 sec (earth's time), the spaceship is at x=0.6c second. Use LT to get t' and x':

    t' = 1.25*[2 - (0.6/c)*(0.6c sec)] = 2.05 sec
    x' = 1.25*[0.6c sec - 0.6c*2sec] = -0.75c sec​
    The elapsed time remain 1.6 sec (2.05sec-0.45sec). It is apparent that changing the origin does not change the spaceship clock reading.

    Secondly, let's locate another point that is at rest relative to earth or moon(probably a satelite) at x=0 and y=-1c second. At t=0, this satelite fires a laser beam that will be reflected by the spaceship and return to the satelite. Total elapsed time between the firing and receiving back the laser is 2 sec according to earth's (also the satelite's) observer.

    Other than uses different figures, the set up of this modified exercise is now exactly resemble 2inquisitive's exercise. If for 2inquisitive's exercise, the correct elapsed time according to spaceship's observer is 4 second, in this modified exercise, shall be 2.5sec (1.25*2sec). This simply can't be true. I just moved the origin from earth to the middle of earth-moon and introduced (indeed for fancy purpose only) a flash of laser from a satelite that will be reflected by the spaceship. Such modification does not change the spaceship clock reading. What do you think?
     
  15. Pete It's not rocket surgery Moderator

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    Paul,
    I need to clear some things up about our common understanding...

    1) In your scenario it is clear that the spaceship clock runs slow in Earth's frame.

    Do you accept that the reverse also applies - that in exactly the same situation, the Earth clock also runs slow in the spaceship's frame?

    2) Do you agree with this statement:
    If two events occur in the same place in some reference frame, the time between those events will be longer in any other reference frame
     
    Last edited: Nov 6, 2004
  16. Pete It's not rocket surgery Moderator

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    Of course.

    Your implication is that the laser's transit time will be the same as the spaceship's transit time in both frames.

    I disagree.

    In the Earth/Moon/satellite frame:
    The firing of the laser is simultaneous with the spaceship departing Earth.
    The receiving back of the laser is simultaneous with the spaceship arriving on the Moon.
    Therefore, the transit time for the laser is the same as the transit time of the spaceship.

    In the spaceship frame:
    The firing of the laser occurs before the spaceship departs Earth.
    The receiving back of the laser occurs after the spaceship arrives on the Moon.
    Therefore, the transit time for the laser is longer than the transit time of the spaceship.



    The start and end points of the spaceship transit are equivalent to points D and E in the spacetime diagrams.
    The start and end times of the laser's transit are equivalent to points A and C in the spacetime diagrams.


    Clearly, (time E - time D) is less than (time C - time A) in the spaceship frame.
     
    Last edited: Nov 6, 2004
  17. Pete It's not rocket surgery Moderator

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    You haven't yet commented on my spacetime diagrams, nor on these statements:
    Why not?
    I'm happy to discuss them if you think I'm misled (I don't think I am, but I can't rule out the possibility).
     
    Last edited: Nov 6, 2004
  18. Paul T Registered Senior Member

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    Pete,

    More precisely, earth observer sees spaceship's clock runs slower and spaceship observer sees earth's clock runs slower. However, there are things to watch out. More on this later.

    Yes. The question is, on 2inquisitive's exercise, how do you consider the two events in one frame (such as laser beam leaves earth and the returning laser hits earth) corresponds to those detected in the other frame (spaceship)?

    My view

    From the earth observer's perspective, the spaceship must arrive to a point at x=-0.866c sec simultaneuosly with the laser firing event or else the ship would not arrive to the right location (at x=0) to reflect the laser beam. The ship then will arrive to x=0.866c sec simulteneously with the receiving back of the laser beam. If, for example, the ship emits light the moment it reaches at x=-0.866c sec and emits light again when the ship's clock passes 1 sec (in this case the ship reaches x=0.866c sec), the interval between the two light signals received by the earth observer should be 2 seconds.

    In the ship frame:
    1) at t=0 and x=-0.866c ====> x'=-2*0.866c sec
    2) at t=0 and x=0.866c ====> x'=2*0.866c sec​
    which tells us that the two points is separated by a distance of 4*0.866c sec. This same two points is separated by a distance of 2*0.866c sec in earth frame. So, in which frame the length is contracted relative to the other? It is the length in the spaceship frame; in this case two times of 2*0.866c in ship frame fit into earth's 2*0.866c. How much time required for the ship to move from x'=-2*0.866c sec to x'=2*0.866c sec? Not 4 seconds, but one second.

    Your view

    Now, let's see how it is based on your view. Forget about the laser beam, as it has no bearing on this issue other than its functionality as a somesort of clock (it only tells us that there are three events at x=0 at every one second interval in earth frame).

    At x=0 and t=0, the spaceship's origin is at x'=0 and t'=0. At this time (and indeed at all time after this), the spaceship is tailing 2*0.866c sec behind the origin (or 0.866c sec according to earth observer). The spaceship must arrive to the reflection point at x=0 and t=1 sec, which is at x'=-2*0.866c sec and t'=2 sec in the ship frame. Here the problem comes in. The ship moves at velocity 0.866c for 2 seconds, covering a distance of 2*0.866c second in the ship frame, while according to earth observer the ship only covers a distance of 0.866c sec.

    May be we are beating two different beasts, but the above (1 second elapsed time in ship frame equalt to 2 seconds on earth frame) is what I think important.
     
  19. Paul T Registered Senior Member

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    Take a look on my above post. Comment from me on your spacetime diagrams was indirectly given in that post.
     
  20. Pete It's not rocket surgery Moderator

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    OK, time for a reality check... I'm not convinced we're trying to answer the same problem.


    Here is the problem as I see it:

    "In the frame of the spaceship, how long is the time between when the laser is fired on Earth, and the time that the laser beam returns to Earth."


    Is that the problem you are addressing?
    Do you maintain that in the spaceship frame, the laser beam is in transit for one second?
     
  21. Paul T Registered Senior Member

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    The laser beam can be ignored but the situation remain the same; based on your view 2 seconds on earth corresponds to 4 seconds on the ship while based on my view it is 1 second.

    Consider just for the sake of guidance that there are two spaceship separated at a fixed distance of 4*0.866c second. Something moves at velocity 0.866c relative to these ships' frame need 4 seconds to travel from one ship to the other. On earth frame, these two ships are separated at a distance of 2*0.866c second. On the ship frame this 2*0.866c sec (earth frame) is covered in one second. Sound contradictory, but it is not. The ship observer sees the distance as 0.866c second only. The 4*0.866c second separation and the covered distance of 0.866c second do not contradict one to another because on the ships frame the front ship at x=0.866c sec and the back ship at x=-0.866c cannot happen simultenuously (they can in earth reference frame).

    Yes, I maintain that 2 seconds on earth (whether it is the transit time of laser beam or just mere clock ticking) corresponds to 1 second in the ship frame.
     
  22. Pete It's not rocket surgery Moderator

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    No, I can not ignore the laser. For me, it is what defines the time period in question.
    Without the laser, the problem I'm answering is undefined.

    Can you state the problem that you are answering without reference to the laser events?
     
  23. Paul T Registered Senior Member

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    Well, the laser might had hypnotized you.

    Please Register or Log in to view the hidden image!



    One more modification on 2inquisitive's exercise. Instead of laser beam just use normal light signal that propagate to all directions. The set up is as follow:

    There are three points (on earth reference frame), A, B and C on x-axis. A is at the origin (x=0) while B and C are respectively at x=0.866c sec and x=2*0.866c sec. Clock on those three points are set to tick simultenuosly and since they all in the same inertial reference frame, we expect them to tick at the same rate too. All those clocks trigger a light source in the respective location to emit light signal at t=0, t=1 sec and t=2 sec.

    A spaceship P passes A at constant velocity 0.866c (relative to A) when clock on A displays t=0 (so does the clock on B and C). I will just make the computation for event at B and Conly.

    At t=0:
    A: x=0 ====> x'=0 and t'=0​
    C: x=2*0.866c sec =====> x'=4*0.866c sec and t'=-3.0sec​

    At t=2 sec:
    A: x=0 ====> x'=-4*0.866c sec and t'=4 sec​
    C: x=2*0.866c sec =====> x'=0 and t'=1 sec​

    The above calculation show that the spaceship observer measures an elapsed times of 4 sec and 1 sec respectively for 2 second in A and C. You can see that your 4 seconds elapsed time result has nothing to do with that fancy laser beam.

    What that 4 seconds elapsed time really is? As I suggested in my earlier post, consider another spaceship Q that tailing behind P at the same constant velocity. On earth frame, P arrives to C and Q arrives to A simultenuously. However, on the ship frame, P arrives to C earlier than Q arrives to A. So, when Q arrives to A, P has already moved far ahead of C. Since P is the ship frame origin, "Q arrives to A" event occur at x'=-4*0.866c sec and t'=4 sec. This 4 seconds elapsed time therefore is not the ship clock elapsed time that correspond to the 2 second elepsed time on earth frame (the 1 second elapsed time is -- as it is the time at the same location in the ship frame).

    Hope this clarify my point.
     
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