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Acceleration in Born rigidity

  1. Aug 3, 2009 #1
    Suppose I am standing in an inertial reference frame and watching two point masses A and B accelerate according to the rules of the Born rigidity. The distance between the masses is L in an inertial frame in which A and B are simultaneously at rest. Mass A is 'in front' of mass B. Mass B has a fixed constant acceleration aB. What is the expression for the acceleration aA of mass A?
  2. jcsd
  3. Aug 3, 2009 #2


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    If they move in a Born rigid way while one of them is doing constant proper acceleration, the other one is too. The rear has a higher proper acceleration than the front. (Otherwise the distance between them in the original rest frame wouldn't decrease). Check out the Wikipedia page for Rindler coordinates. The world lines of A and B are two different hyperbolic arcs in the first picture. See also this thread, in particular DrGreg's posts.
  4. Aug 3, 2009 #3


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    Assuming the accelerations are proper accelerations (not coordinate acceleration, otherwise it wouldn't be Born rigid motion)

    [tex] L = \frac{c^2}{a_A} - \frac{c^2}{a_B} [/tex]​
  5. Aug 3, 2009 #4
    I prefer to state that as aA=aB/(1+aBL/c2).
    Taking Newtons formula for gravity F=GMm/r2 for mA and mB and assuming mass A is above mass B in a planet M gravity field, or
    L=rA-rB, you can get the same type of expression for a planet gravity field: aA=aB/(1+L/rB)2

    The Equivalence Principle says these two should be equal? Am I wrong?
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