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Fitzgerald-Forentz contraction problem

  1. Feb 19, 2006 #1
    Dear All, Have a small problem:

    Two chaps attempt to demonstrate the phenomenon of Fitzgerald-Lorentz contraction. The two meet at the centre of a fast train which is L = 500m long and travelling at a speech of u = 0.7c to synchronize their watches.

    One sits at the front of the train while the other sits at the back. At a previously agreed time, both drop a bag of flour from the train so that it marks the ground beneath. When the train has stopped, they drive back to where they dropped the bags and measure the distance between the two marks. What distance do they measure? Have they actually demonstrated the Fitzgerald Lorentz contraction?


    *CONFUSED* Surely as they are in the same train, they are in the same inertial frame of reference? And hence, relativistic effects will not be relavent?
     
  2. jcsd
  3. Feb 19, 2006 #2

    lightgrav

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    same as each other? same as the ground?
    or same as they will be when they measure on the ground?

    Do you think they'll measure 500m? And NOT demonstrate contraction?
     
  4. Feb 19, 2006 #3

    nrqed

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    yes, they will have demonstrated the LF contraction. When they go measure the distances on the ground, it will be larger than 500 m.
    If you call the train the S` frame, this is a measurement corresponding to delta t'=0 and delta x'= 500 m. Plug that in the LF equations and find Delta x. That will be the value they will measure on the ground.
     
  5. Feb 19, 2006 #4
    LF equations? I'm working from the Lorentz equations. How do I derive them?

    Best Regards, James
     
  6. Feb 19, 2006 #5

    nrqed

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    Same equations.. LF= Lorentz-Fitzgerald, which is what you call Lorentz equations. Sorry.

    Btw, you are right that they are in the same frame, but they are effectively making a length measurement in the frame of the ground, so relativistic effects are relevant.

    Pat
     
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