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An other train paradox.

  1. Jan 23, 2010 #1
    Dear forum,

    when I first read about special relativity I came up with this paradox.

    I went to the teacher who gave me the booklet but he couldn't answer my question.

    I'm quite confident that this forum is the place to answer my question:

    Does time goes slower or faster inside the train?

    www.trainparadox.tk

    grtz,

    S.

    ps: you can't react on the site itself but this forum will do just fine of course...
     
  2. jcsd
  3. Jan 24, 2010 #2

    HallsofIvy

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    That's a meaningless question. One of the basic concepts of "relativity" is that any statement or question about properties of an even must include "relative to ..."
    From a frame of reference motionless relative to the earth and moving relative to the train, time goes slower in the train. From a frame of reference motionless relative to the train and moving relative to the earth, time has not changed in the train but time is goes slower outside the train.

    Also you state at your site:
    "If you speed is s1 and the speed of the train is s2 you move at speed s1 + s2 when you are stepping in the same direction and s1 - s2 (or s2 - s1) in the other case.
    Einstein stated that this is not the case for the speed of light."

    That is not true. The same formula applies to any speeds: if your speed is s1 (relative to the train) and you are on a train with speed s2 (relative to the ground) then your speed, relative to the ground, is
    [tex]\frac{s1+ s2}{1+ \frac{s1s2}{c^2}}[/tex]
    And if either s1 or s2 is equal to c, the result will be just c.
     
    Last edited by a moderator: Jan 24, 2010
  4. Jan 24, 2010 #3

    Doc Al

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    Staff: Mentor

    In addition to what HallsofIvy points out regarding addition of velocities, your analysis ignores both length contraction and--more importantly--the relativity of simultaneity. If you want more detailed comments, post your analysis here.
     
  5. Jan 30, 2010 #4
    First of all:
    congrats about the truly professional way of welcoming me here...
    right to the case and stuff,
    I'm quite impressed.

    Secondly:
    I must admit that I'm relatively new to relativity, especially special relativity so try to bear with me...

    And now the analysis:

    I gave the article on wikipedia about the ladder paradox an attempt, but I'm not sure I get it quite well, so once again: bare with me.

    As I stated on the site (I hope you don't expect me to repeat the whole page here, so I quote the site):
    So my best guess until now is that the relativity of simultaneity will do the trick...

    I quickly glanced at the formulas of relativity of simultaneity and in my example I would have to substitute t2, v and x by something complicated so I'm going to do the math later on when I find the courage for it...

    Sorry for the useless post.

    S.
     
  6. Jan 31, 2010 #5
    First, I am going to be unprofessional and say welcome to PF matdoya :smile:

    Second, you are right that "relativity of simultaneity" will do the trick, but it is not complicated. Basically, what appears simultaneous to one observer will not appear simultaneous to another observer with relative motion. The equation for relativity of simultaneity is simply -Lv/c^2 where L is the length of the train as measured by an obvserver onboard the train. For example if the train has a rest length of 1.0 lightsecond and a velocity relative to the track of 0.8c then when the clocks on the train are synchronised according to Einstein's method, the clock at the rear of the train will be ahead of the clock at the front of the train by 0.8 seconds. If a light signal is sent from the front of the train when the front clock is reading zero seconds, the rear clock will already have a head start of 0.8 seconds showing on it at the moment the light signal started out. When the light signal reaches the rear, the rear clock will have advanced by 0.2 seconds and will be reading 1 second when the signal arrives. On the other hand when a signal starts out from the rear of the train at time zero according to the rear clock the front clock will be reading -0.8 seconds and the front clock advances by 1.8 seconds in the time it takes the light signal from the back to catch up with the front and the front clock will also be reading 1 second when the signal arrives. In both cases the amount the clocks advance during the signal travel times, is less than elapsed time measured by observers at rest with the track. Given that at a relative velocity of 0.8c the time dilation factor is 1.66666 and length contraction factor is 0.6, you should have enough information to work out the ellapsed times measured by the track observers.
     
  7. Feb 5, 2010 #6
    tsss ;)

    Yeah,
    well,
    I kinda hold on to the idea until I can do the whole calculation and the whole equation is solved on my own special way, so expect either the aha-post or, I hope, a calculation that needs further investigation...

    Anyway, I know already know that an important element was left out,
    chances are that the site will be off-line soon...

    best regards,

    Sander
     
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