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  1. Relevant equations

I have computed the christoffel symbols via comparing the Euler-Lagrange equations to the form expected from geodesic equation.

geodesic equation: $\ddot{x^a}+\Gamma^a_{bc}\dot{x^b}\dot{x^c}=0$

covariantly constant equation: $ V^a \nabla_a W^b = V^a (\partial_a W^b) + V^a \Gamma^b_{ac} W^c= 0 $ 1 where $V^a $ is the tangent vector to the geodesic.

I have computed the christoffel symbols as:

$\Gamma^{x}_{tt}=\frac{-1}{2x^2}$ and $ \Gamma^{t}_{tx}=\frac{-1}{2x}$

  1. The attempt at a solution

From the information given $x^u=(t,1) \implies V^u=(1,0)=\delta^u_t $

Therrefore 1 non-zero equations reduces to:

$ \nabla_t W^b = (\partial_t W^b) + V^t \Gamma^b_{tc} W^c= 0 $

Using the christoffel symbols non-zerro equations further reduce to:

$\partial_t W^t - \frac{1}{2x}W^x=0$

and $ (\partial_t W^x) -\frac{W^t}{x^2}= 0 $


so it is at this point that I am stuck. the only way I can see to proceed is to differentiate either one of the equations again wrt $t$ to get a second-order equation and then substitute in the other equation. However to then solve completely we would need 2 boundary conditions, but are only given one.

Many thanks in advance

  • $\begingroup$ Since you're on the curve $x=1$, this is a $2 \times 2$ system of linear first-order ODE with constant coefficients, which hopefully you can solve. For example you could write it as a matrix ODE and solve it using a matrix exponential. $\endgroup$ – Anthony Carapetis Dec 3 '17 at 6:08
  • $\begingroup$ ok many thanks I think I have it now. $\endgroup$ – yourlazyphysicist Jan 5 '18 at 15:16
  • $\begingroup$ ok many thanks I think I have it now. so I have $w=w_0e^{\vec{A}t} $ where $A=(0,1/2) $ top row and $A=(1,0)$ bottom row, (apologies I don't know how to do a matrix with LaTeX ). so now my question is how to simplify $e^{\vec{A}t} $ . I have the result: if $A$ can be written as: $A=PDP^{-1}$ then $e^{\vec{A}t}=Pe^{Dt}P^{-1} $ (which I can't find a proof for ) anyway so if I find the eigenvalues and eigenvectors I can use this to finally simplify and so then the eigenvectors become absorbed into the constant in front with this result? (assuming there are linealry independent eigenvectors) $\endgroup$ – yourlazyphysicist Jan 5 '18 at 15:22
  • $\begingroup$ in the case the eigenvectors do not happen to be linearly independent with m of them for a m x m matrix , how would I simplify the exponent of the matrix? $\endgroup$ – yourlazyphysicist Jan 5 '18 at 15:23

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