# GR: find covariantly constant vector on a given curve

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

MY QUESTION:

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.

• 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. – Anthony Carapetis Dec 3 '17 at 6:08
• 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) – yourlazyphysicist Jan 5 '18 at 15:22