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Rindler metric and proper acceleration.

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that observers whose world-lines are the curves along which only
    ξ varies undergo constant proper acceleration with invariant magnitude
    ημν(duμ/dτ )(duν/dτ ) = 1/χ2.

    2. Relevant equations

    ds2 = −a2χ2dξ2 + dχ2 + dy2 + dz2
    xμ = {ξ, χ, y, z}
    t = χsinh(aξ)
    x =χcosh(aξ)

    3. The attempt at a solution

    The trajectory of this particle in the rest frame would be:
    xκ={χsinh(aξ),χcosh(aξ),0,0} as a function of ξ.

    dxκ/ dξ = {aχcosh(aξ), aχsinh(aξ),0,0} Velocity of xκ.

    Since I want proper time, I would expand dxκ/ dξ into (dxκ/ dξ)*(dξ/dτ)
    So I need to define dξ/dτ. For this I would use the metric:
    letting ds^2= -dτ^2
    dτ/dξ= aχ
    γ= dξ/dτ = 1/aχ
    The acceleration is then: γ^2{a^2χsinh(aξ),a^2χcosh(aξ),0,0} = d2xk/dτ^2
    ημν(duμ/dτ )(duν/dτ ) = γ^4(-1*a^4χ^2sinh^2(aξ) + 1*a^4χ^2cosh^2(aξ))
    Which cancels out to: 1/χ^2. I just wanted to make sure that although the math works out, my reasoning is okay.
    Where I took ds^2= -dτ^2 my first thought was that dτ^2= (a^2)(χ^2)(dξ)^2 - (V(ξ))^2(dL)^2. I realized that dropping the velocity term would allow me to get the right answer.. so it must = 0. Since the velocity itself is not actually zero I would be inclined to guess that dividing the equation by (dξ)^2 would leave (dL)^2/(dξ)^2 and that must equal zero?
     
  2. jcsd
  3. Apr 10, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Since only ξ changes, that means dχ = dy = dz = 0. I think that's why the "velocity term" is zero. That is,

    ds2 = −a2χ22 + dχ2+ dy2 + dz2 = −a2χ22.

    So, dτ2 = a2χ22
     
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