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Homework Help: General relativity question - geodesics.

  1. Apr 27, 2010 #1
    I'm doing some revision for a General relativity exam, and came across this question:

    A Flat Earth space-time has co-ordinates (t, x, y, z), where z > 0, and a metric

    ds2 = ((1 + gz)2)dt2 − dx2 − dy2 − dz2

    where g is a positive constant.
    Write down the geodesic equations in this space-time.

    Hence, or otherwise, show that A Flat Earth physicist, stationary at a point with z = h, will measure a ‘gravitational acceleration’ of magnitude g(1 + gh)−1

    ---//---

    I've solved the geodesics, but can't get the 'show that' at the end:

    Using the Euler-Lagrange equation:

    [tex] \stackrel{d}{ds} \stackrel{\partial L}{\partial \dot{x^{\mu}}} - \stackrel{\partial L}{\partial x^{\mu}} =0[/tex]

    I get

    [tex]\ddot{z} = 2g(1+gz)\dot{t}^2[/tex]

    And

    [tex]\dot{t} = E/(1+gz)^2[/tex] ( where E is a constant )

    So

    [tex]\ddot{z} = g{E^2}/(1+gz)^3[/tex]

    which isn't right. Alternatively,

    [tex]\ddot{z}= \dot{t}^2 z'' + \ddot{t}z'[/tex] ( where ' is (d/dt) )

    it's stationary, so z' = 0 thus [tex]z''=\ddot{z}/\dot{t}^2=2g(1+gz)[/tex]

    which still isn't right. Any takers?
     
    Last edited: Apr 27, 2010
  2. jcsd
  3. Apr 27, 2010 #2

    Cyosis

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    I think the two in your geodesic equation shouldn't be there. You probably missed the 1/2 in the Christoffel symbol. Moreover you made a mistake in trying to find an expression for [itex](dt/ d\tau)^2[/itex]. Can you show us how you obtained your expression?
     
    Last edited: Apr 27, 2010
  4. Apr 27, 2010 #3
    Yes the 2 is wrong. My apologies.

    For the mu = 0 equation I have:

    [tex](d/ds)(2\dot{t}(1+gz)^2) =0[/tex]

    thus

    [tex] \dot{t} (1+gz)^2 = constant [/tex]

    also

    [tex]\ddot{z} = \dot{t}d\dot{z}/dt = \dot{t}(d/dt)(\dot{t}z') = \dot{t}(z'(d\dot{t})/dt +\dot{t}z'')=\dot{t}^2z''+\ddot{t}z'[/tex]
     
    Last edited: Apr 27, 2010
  5. Apr 27, 2010 #4
    Update: I've now solved it using the Christoffel symbols, but am still confused as to why the Euler-Lagrange method hasn't worked.


    This is wrong:
    [tex]\ddot{z} = -g{\dot{t}^2}(1+gz)[/tex] (1)

    It should read:

    [tex]\ddot{z} =-g{\dot{t}^2}/(1+gz)[/tex] (2)

    I got (1) direct from the EL equation

    [tex]-2\ddot{z} - {\dot{t}^2}d_{z}(1+gz)^2 = 0[/tex] (3)
     
    Last edited: Apr 27, 2010
  6. Apr 27, 2010 #5

    Cyosis

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    You're probably losing terms in places where you should be summing over the indices. Using the EL-equations I end up with [itex]\ddot{z}-g(1+gz)\dot{t}^2 =0 [/itex].

    Your second equation is wrong as well. Did you derive that one using the geodesic equation? Either way I just calculated it both ways and I get the same result, as it should be. I can't be of further assistance unless you show me your full derivation. Also I don't think you have gotten your [itex]\dot{t}[/itex] correct yet.
     
    Last edited: Apr 27, 2010
  7. Apr 27, 2010 #6
    Thanks. Given that's the case then [tex]z''=\ddot{z}/\dot{t}^2=g(1+gz)[/tex] can't be true (since this is the wrong answer).

    I can't see any thing wrong for [tex]\dot{t}[/tex]:

    whats wrong with:
    [tex]L = (1+gz)^2dt^2-dx^2-dy^2-dz^2)[/tex]

    therefore by EL:

    [tex](d/ds)(2\dot{t}(1+gz)^2) =0[/tex]

    so

    [tex] \dot{t} (1+gz)^2 = constant [/tex]
     
  8. Apr 27, 2010 #7

    Cyosis

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    Your Lagrangian is wrong, recall [itex]L=g_{\alpha \beta}\dot{x}^\alpha\dot{x}^\beta=(1+gz)^2 \dot{t}^2-\dot{x}^2-\dot{y}^2-\dot{z}^2[/itex].
     
    Last edited: Apr 27, 2010
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