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The geodesic equation

  1. Aug 22, 2013 #1
    I understand why the geodesic equation works in flat space. It just basically gives a set of differential equations to solve for a path as a function of a single variable s where the output is the coordinates of whichever parameterization of the space you are using. But the derivation I know and understand relies on the existence of a straight line in your space. Which isn't true for curved spaces. I assume the equation must be true also for curved spaces as it underpins the entire derivation of the Riemann curvature tensor, so if it was only valid for flat spaces then the RCT would be a bit pointless. But how do you prove this? I think you can prove it with the euler lagrange equation and the metric tensor field for the space, but last time I attempted this it went horribly wrong. So does anyone know any more intuitive ways of seeing that the equation should hold true for curved spaces as well as flat ones?
     
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  3. Aug 22, 2013 #2

    PeterDonis

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    Do you have a reference for this derivation? The geodesic equation *defines* what a "straight line" is, so I'm not sure I see how you can assume the existence of straight lines in deriving the equation.
     
  4. Aug 22, 2013 #3

    mfb

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    Locally, space always looks like a Minkowski space, curvature is relevant for the global structure only.
     
  5. Aug 22, 2013 #4

    WannabeNewton

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    Exactly as Peter stated, ##\nabla_\xi \xi = 0## is the definition of a geodesic with tangent field ##\xi##. The equation ##\ddot{x}^{\mu} + \Gamma ^{\mu}_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta} = 0## is simply the local coordinate form of ##\nabla_\xi \xi = 0##. In abstract index notation, this is ##\xi^a\nabla_a\xi^b = 0## so intuitively, as you can see easily from the notation, we are basically taking ##\nabla_a \xi^b## and projecting it onto the curve in the direction of ##\xi^a##; the statement then is essentially that from the perspective of someone riding along the curve, the tangent vector remains constant.
     
    Last edited: Aug 22, 2013
  6. Aug 22, 2013 #5
    Surely the definition of the geodesic is the path of shortest length between two points which remains in the space?
    Derivation - So take an arbitrary coordinate system to describe our D dimensional space, call this X. Then take the Cartesian coordinate system, denote this C. If we have a path through this space we can describe it in both coordinate systems. Each x coordinate will be a function of s and equivalently each Cartesian coordinate will be a function of s, where s is a parameterization for the curve. Now for the curve to be a straight line the second derivative of all the Cartesian coordinates as a function of s must be zero. d2c/ds2 =0 for all c coordinates. So this is a system of ODEs which can be solved for the equations which describe the path. Now lets try and find the equivalent system for the x coordinates. we know dc(I)/ds = dc(I)/dx(u) dx(u)/ds, by the chain rule so just differentiate again wrt s using the product rule 0=d2c(I)/dx(u)dx(v) dx(u)/ds dx(v)/ds + dc(I)/dx(u) d2x(u)/ds2. So dc(I)/dx(u) d2x(u)/ds2 = d2c(I)/dx(k)dx(v) dx(k)/ds dx(v)/ds . Contract both sides with dc(I)/dx(j) and you get g(uj)d2x(u)/ds2 =dc(I)/dx(j) d2c(I)/dx(k)dx(v) dx(k)/ds dx(v)/ds. inverse the matrix g(ui) and you get d2x(u)/ds2 = -Gamma(u; k,v) dx(k)/ds dx(v)/ds. This system of ODEs can then be solved to get a path relative to the x coordinate system that s a straight line.
     
  7. Aug 22, 2013 #6

    WannabeNewton

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    Consider a smooth manifold ##M##. If we endow ##M## with a metric ##g_{ab}## then there is a unique (torsionless) derivative operator ##\nabla_a## such that ##\nabla_a g_{bc} = 0##. The definition of a ##\nabla_a## induced geodesic is, as stated above, a curve ##\gamma## with tangent field ##\xi^a## such that ##\xi^a\nabla_a \xi^b = 0##. It just so happens that for ##\nabla_a##, ##\gamma## also happens to be a curve which locally extremizes arc-length (or, equivalently, energy) where arc-length is defined as ##l = \int (g_{ab} \xi^a \xi^b)^{1/2}d\lambda##. This is not true if we instead took an arbitrary derivative operator ##\tilde{\nabla}_a## that had no association with a metric tensor ##g_{ab}## (in such a case arc-length has no meaning); these much more general geodesics are called affine geodesics and they are, as already mentioned, defined by ##\xi^a \tilde{\nabla}_a \xi^b =0##. What you are referring to are specifically geodesics induced by that unique derivative operator ##\nabla_a## associated with a metric ##g_{ab}##.
     
    Last edited: Aug 22, 2013
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