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Second order geodesic equation.

  1. Nov 30, 2013 #1
    Hello all,

    I have a geodesic equation from extremizing the action which is second order. I am curious as to what the significance is of having 2 independent geodesic equations is. Also I was wondering what the best way to deal with this is.
     
  2. jcsd
  3. Nov 30, 2013 #2

    WannabeNewton

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    What do you mean by two independent geodesic equations? Are you working in a 2 dimensional space with coordinates ##\{x^1,x^2\}## and have two geodesic equations, one for each coordinate?

    As an aside, usually the easiest way to deal with the geodesic equation is to not deal with it at all. What I mean by this is that if your space has obvious symmetries then just use first integrals of conserved quantities. It's the same thing as using conservation of energy instead of Newton's 2nd law for classical mechanics problems.
     
  4. Nov 30, 2013 #3
    Sorry I should have been more clear, I believe that I have two independent solutions to the geodesic equation for a single direction, but perhaps I am misinterpreting the result. The equation is written as

    [itex] \ddot{x}(\tau)^a = A_{ab}(\tau) x(\tau)^b [/itex]

    Here a,b are the two orthogonal directions to the wavefront of a pp-wave. Luckily I am in a system where [itex] A_{ab} [/itex] is an orthogonal matrix. This was derived using some symmetries and conversations with the Lagrangian. However I am having trouble interpreting what it means to have a second order geodesic equation (when we write down our geodesic equation in terms of Christoffel symbols it is always first order).
     
  5. Nov 30, 2013 #4

    WannabeNewton

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    Perhaps we're using different definitions of the geodesic equation but AFAIK it is always second order in ##\tau##: ##\ddot{x}^{\mu} = -\Gamma ^{\mu}_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}##.
     
  6. Nov 30, 2013 #5
    True but the equation above should only have one independent solution as best I can ascertain.. Suppose for simplicity we had a diagonal connection coefficient which is valued at 1 in some [itex] \bar{x} [/itex]direction, I don't see how that is different from writing [itex] \dot{x}= x^2 [/itex] (by substituting[itex] x = \dot{\bar{x}} [/itex]),this is only a first order equation... or am I missing something blatant?
     
  7. Dec 2, 2013 #6

    Bill_K

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    This is one of the standard tricks/methods for solving differential equations. Define p = dx/dt and hope that you can write a DE containing p alone. If so, it will be first order and you can solve it to get p(t). There will be one constant of integration.

    But then you still have to solve p = dx/dt to get x(t), and this will produce a second constant of integration. It's a second order DE, you haven't changed that, all you have done is to solve it in stages.
     
  8. Dec 2, 2013 #7
    Fair enough, I agree with what you are saying. My main question then is what do we do with these constants of integration? May we just arbitrarily set them equal to one?
     
  9. Dec 2, 2013 #8

    Bill_K

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    E.g. they can be used to specify the two initial conditions for the geodesic: the initial position and velocity.
     
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