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Temporal distortion of a moving train

  1. Nov 5, 2006 #1
    Hi i have a question about how time is distorted along the length of a train moving at near light speeds.

    Will the front carriage be farther foward in time than the rear carriage from stationary frame of reference? or the otherway round?
    Assume that the front and rear sections had clocks that were synchronized when the train was stationary and the train took time to accelerate to near light speeds.

    If the clocks were synchronized after the train accelerated will the front still be farther ahead in time than the rear section relative to the stationary frame? or the otherway round?

    Dave
     
  2. jcsd
  3. Nov 5, 2006 #2
    Good questions.

    Foreward clocks always lag aftward clocks, so at any instant in the stationary embankment frame, the moving engine clock will lag the moving caboose clock by gamma(-vx/c^2). This is the temporal offset portion of the time transform ... T=gamma(t-vx/c^2). The -vx/c^2 relates to Minkowski's frame rotation. Gamma is related to the required Fitzgerald contraction.

    It doesn't matter when the train's clocks are synchronized. Clocks in sync in the train frame before acceleration, remain in sync per the train passengers even during train accelrational periods. So as long as the clocks are synchronized in the train at some point before the embankment observer inspects them on driveby, the moving engine clock will lag the moving caboose clock by gamma(-vx/c^2) units of time.

    pess
     
  4. Nov 5, 2006 #3
    Geoffrey A Landis disagrees

    Geofrrey A Landis a NASA scientist disagreed tho, do a google search for him if u want his contact details and find out who he is but here is what he wrote to me:

    Here's a quick thought experiment.

    Suppose your "train" consists of two super-incredible rockets, one
    sitting next to kilometer marker zero, one sitting next to kilometer
    marker zero. To start with, nobody's moving, so everybody's clock is
    synchronized, and there's no ambiguity about what we mean by time t=0.

    So, at time t=0, the front rocket (piloted by Francis) and the back
    rocket (piloted by Buck) both turn on their rockets. They accelerate
    at 30 quadrillion gravities for one nanosecond. This gets them up to
    eighty some percent of the speed of light.

    Now, suppose their clocks were still synchronized in the accelerated
    frame. In that case, since they were always synchronized in their
    own frame, and they both were accelerating the same, then in that
    frame they would still be exactly one kilometer apart.

    Francis takes a photo out her window, and it shows the 1 kilometer
    marker (since, despite their acceleration, she moved only a foot or
    so in the nanosecond of acceleration). Buck takes a photo out his
    window, and it shows the zero kilometer marker. They compare notes,
    and Francis radios to Buck "Lorentz contraction is false! The two
    kilometer markers are moving past our one-kilometer long train at
    relativistic speed, but they are exactly one kilometer apart in our
    frame, instead of Lorentz contracted!

    Meanwhile, somebody on the ground standing at kilometer marker zero
    takes a photo, and somebody else standing at kilometer marker one
    takes a photo, both at one nanosecond after ignition. They then say
    "Lorentz contraction is false! The ships are one kilometer apart in
    their own frame, and the photos prove that they're one kilometer
    apart in our frame, too! They're not Lorentz contracted at all!"

    Paradox? No. The assumption that the clocks in the front ship and
    the back ship are still synchronized is false. And the assumption
    that the two ships are rigorously the same distance apart (in their
    own frame) is false, since the clocks aren't synchronized.

    Clocks synchronized before acceleration are not synchronized after
    acceleration (assuming that they are at different positions x, and
    the acceleration is in the x direction-- y and z are unaffected).

    --
    Dr. Geoffrey A. Landis
    Ronald E. McNair-NASA Visiting Professor of Astronautics at MIT
    http://mit.edu/aeroastro/www/people/landis/landis.html
     
  5. Nov 5, 2006 #4
    part 2 of Landis' reply

    Hi,

    Right. Once they start moving, their clocks lose simultaneity, and
    when the clocks begin to differ, the front and back no longer agree
    on things like acceleration.

    The ship in back sees the clock on the ship in front as running
    fast. The ship in front sees the clock on the ship in back as
    running slow. The clocks at the front and back are no longer
    synchronized in anybody's reference frame.

    The classic way to analyze this problem is assuming that each ship
    emits a pulse at regular intervals, and look at when the pulses
    arrive at a central (fixed) station. Any special relativity book
    with a discussion of the twin paradox should do this analysis.

    Oddly, last week a friend forwarded me this, which is a different
    discussion of the same thing. I'm not sure I really like his
    analysis, but he gets to the same result.

    Bell's Spaceship Paradox
    http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html

    --
    Dr. Geoffrey A. Landis
    Ronald E. McNair-NASA Visiting Professor of Astronautics at MIT
    http://mit.edu/aeroastro/www/people/landis/landis.html
     
  6. Nov 5, 2006 #5

    Doc Al

    User Avatar

    Staff: Mentor

    First point: I hope you realize that "farther forward in time" depends on who is doing the observing.

    Second point: The only way you can maintain synchronization of the front and rear clocks with respect to the station frame as the train accelerates is by accelerating the front and rear of the train uniformly with respect to the station. That will introduce forces that will rip the train apart. (On the other hand, if the train accelerates uniformly with respect to its own instantaneous frames, then the clocks will maintain synchronization--according to train observers.)

    As pess5 explained, if the clocks are synchronized in the train frame they will be unsynchronized in the station frame, the clock in front will lag the clock in the rear--according to station observers.
     
  7. Nov 5, 2006 #6

    when u say the front clock will lag the rear, do u mean the front clock will read behind the rear clock?
     
  8. Nov 5, 2006 #7
    Why - if the train is accelerated, the effect is the same as a G field, the clock at the lower potential will be viewed as being in a lower gravitational potential - and therefore from the perspective of the front clock, will be seen to run slower; the forward clock will be seen to run faster from the standpoint of the caboose clock. Or did I misinterpret the problem?
     
  9. Nov 5, 2006 #8

    Doc Al

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    Staff: Mentor

    That's right. In the train frame, the clocks are synchronized and both strike 3:00pm at the same time. But in the station frame, when the rear clock strikes 3:00pm the front clock reads 2:55pm (for example).
     
  10. Nov 5, 2006 #9
    Hi Doc Al,

    I agree in that the train would be ripped apart.

    However, it seems to me that clocks synchronized in the train frame can never remain in sync per the station observer, given they are in motion. Are you saying that the clocks remain in sync, but that they slow down identically as the train accelrates?

    pess
     
    Last edited: Nov 5, 2006
  11. Nov 6, 2006 #10

    DrGreg

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    Science Advisor
    Gold Member

    If the train accelerates such that its length, measured by a train observer, remains constant, then the clocks will not be synchronised, neither to train observers nor station observers.

    See posts #3 and #4 in this thread.
     
  12. Nov 6, 2006 #11

    Doc Al

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    Staff: Mentor

    That's what makes sense to me. Do you disagree? Since--according to the track observers--the clocks undergo the same acceleration for the same time, on what basis would they have different readings? (Again: according to the track frame.)

    Let me know if I'm missing something.
     
  13. Nov 6, 2006 #12

    Doc Al

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    Staff: Mentor


    I don't see how you get that from posts #3 and #4. In those posts, the two rockets are given equal "bursts" with respect to the track. That is not an example of the acceleration being uniform with respect to the "train". Despite the claim that "To start with, nobody's moving, so everybody's clock is synchronized, and there's no ambiguity about what we mean by time t=0.", that's only true for the first instant. As soon as they start moving, the rockets no longer agree that they are undergoing equal accelerations. (Instead of a single nanosecond burst, it may be helpful to think of it as a series of 1000 picosecond bursts.)

    By the way, I highly recommend Bell's paper "How to teach special relativity", in which he describes--and properly analyzes--the infamous Bell spaceship paradox. :wink:
     
    Last edited: Nov 6, 2006
  14. Nov 6, 2006 #13
    Doc Al,

    Interesting, because I had never given it much thought before about a scenario such as this one.

    I just checked it by inspecting Minkowski worldlines wrt another scenario. The clocks must drop out of sync per the station observers even if the train stretched in such a way to always maintain the same length per the station observer. In fact, the worldlines suggest that the de-synchronization becomes worse, and not better ... that is, they are further out of sync than had the train not stretched at all. The rotation of the frames doesn't change in your scenario, since the train is v per the station whether the train maintain's its original length or not as it accelerates. Although the rotation of frames doesn't change, the clock separations do change since the train lengthens. Hence, the clocks must become more desynchronized than had the train not stretched.

    And as DrGreg mentioned, it seems the train clocks must also de-synchronize since they are moving wrt each other if the train stretches, per the train passgengers.

    Does this sound reasonable to you?

    pess
     
    Last edited: Nov 6, 2006
  15. Nov 6, 2006 #14

    Doc Al

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    Staff: Mentor

    Why would the clock separations change with respect to the station? At all times both clocks move with the same speed--according to the station observers--so they remain at the same separation. (Of course, in their own frame the clock separation does increase and the clocks lose synchronization.)


    See my response to DrGreg above. Certainly the train clocks, if accelerated uniformly with respect to the station, will lose synchronization in their own frame.
    Not yet! :wink:
     
  16. Nov 6, 2006 #15
    During acceleration,

    will the train's rear clock appear to tick faster than the train's front clock from stationary frame of reference ?

    im asking because i don't see how else the train's rear clock would read farther into the future than the train front clock. Either the rear clock would appear to speed up or the front clock appear to slow down or both.

    2nd question: if yes to above, will the train's rear clock tick faster than stationary clock from stationary frame of reference during acceleration?

    regarding the latest debate , are u guys taking into account that the train is made up of atoms that are electrostatically attracted to each other and prevent a trully uniform acceleration like that of 2 rockets 1 in front of each other that have no electrostatic forces between them to interfere with their simultaneous accelerations? im assuming absence of any electrostatic forces between the 2 rockets means the space between rockets do not get lorenz contracted from station frame, but each individual rocket body is lorenz contracted?
     
    Last edited: Nov 6, 2006
  17. Nov 6, 2006 #16
    You're right, the clocks do not change in separation per the station. A poor choice of wording on my behalf. Let me word it differently here then ...

    The rotation of frames is the same no matter if the train accelerates normally whereby the train contracts in length by 1/gamma, versus the train accelerating in a way to maintain the train length per the station. The degree of rotation between frames is a function of v alone, and drives the de-synchronization of the moving clocks. So given the same frame rotation either way, a larger x means greater desynchronization ...

    Since v is the same per either accelerational method, then the desynchronization must become more pronounced if the train maintains its length per the station during train acceleration. This because the separation of moving clocks (x) is greater than it would be have been had the train accelerated normally with clock separation becoming contracted by 1/gamma...

    t'=gamma(t-vx/c^2)​

    Since v & c are the same no matter if the moving train maintains its length as it accelerates, as opposed to the train accelerating normally whereby the train contracts, then the only variable is x...

    Since x is larger per the station if the moving train maintains its length as it accelerates (as opposed to the train accelerating normally whereby the moving train contracts), then the desynchronization must be larger than had the station seen the moving train contract in length.

    pess
     
  18. Nov 6, 2006 #17
    davidong3000,

    The scernario is impossible of course. I was assuming a train made of stretchable rubber, for example. Actually, I am envisioning a continuous string of clocks which would (say) be sitting in train windows facing outward so the station observer could see them ... then just assume the train vanished but clocks remain. So a string of self powered moving clocks that could emulate where they would be in a rubber train had it stretched.

    pess
     
  19. Nov 6, 2006 #18
    how can the train accelerate and at the same time remain uncontracted? surely the atoms would each lorenz contract if an external force were to manually stretch the train during acceleration? if this happens wouldn't the train "try" to contract it self electrostatically between atoms? (from stationary frame)

    im assuming that during normal acceleration, rear atoms are electrostatically attracted to atoms infront of it that are farther in the future and thus farther infront (had more accelerating time) causing rear atoms to speed up electrostatically to catch up while foward atoms slow down a little as the forces of atoms behind them pulling them back are farther behind both in time and space.
     
    Last edited: Nov 6, 2006
  20. Nov 6, 2006 #19
    ok above msg posted or edited out of sync of other above ones lol but correct any mistakes in it please if any.
     
    Last edited: Nov 6, 2006
  21. Nov 6, 2006 #20
    repeating my question from prev page...

    During acceleration,

    will the train's rear clock appear to tick faster than the train's front clock from stationary frame of reference ?

    im asking because i don't see how else the train's rear clock would read farther into the future than the train front clock. Either the rear clock would appear to speed up or the front clock appear to slow down or both.

    2nd question: if yes to above, will the train's rear clock tick faster than stationary clock from stationary frame of reference during acceleration?
     
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