I am having trouble understanding how the following statement (taken from some old notes) is true:
For a 2 dimensional space such that $$ds^2=\frac{1}{u^2}(-du^2+dv^2)$$ the timelike geodesics are given by $$u^2=v^2+av+b$$ where $a,b$ are constants.
When I see "geodesics" I jump to the Euler-Lagrange equations. They give me $$\frac{d}{d\lambda}(-2\frac{\dot u}{u^2})=(-\dot u^2+\dot v^2)(-\frac{2}{u^3})\\ \implies \frac{\ddot u}{u^2}-2\frac{\dot u^2}{u^3}=\frac{1}{u^3}(-\dot u^2+\dot v^2)\\ \implies u\ddot u-\dot u^2-\dot v^2=0$$ and $$\frac{d}{d\lambda}(2\frac{\dot v}{u^2})=0\\ \implies \dot v=cu^2$$ where $c$ is some constant.
Timelike implies $$\dot x^a\dot x_a=-1$$ where I have adopted the $(-+++)$ signature.
I can't for the life of me see how the statement results from these. Would someone mind explaining? Thanks.
Answer
I prefer to use Killing vectors and conservation laws to solve stuff like this, so let's analyze the problem using Killing vectors, and see if the results agree with your Euler-Lagrange equations.
Notice that the metric is invariant under translations of $v$. The associated killing vector is $\partial_v$ which in turn gives the following conserved quantity: $$ c_v = \dot x\cdot \partial_v = \frac{\dot v}{u^2} $$ This agrees precisely with your second Euler-Lagrange equation; so far so good. The timelike condition $\dot x\cdot\dot x = -1$ can be written in components as $$ \frac{-\dot u^2 + \dot v^2}{u^2} = -1 $$ Using the above conservation equation to eliminate $\dot v$ then gives a first order differential equation for $\dot u$ $$ \frac{-\dot u^2+c_v^2u^4}{u^2}=-1 $$ which simplifies to $$ \dot u^2 = c_v^2u^4 + u^2 $$ This is a first order, separable differential equation that can be solved by separation of variables and integration. Once you solve this for $u$, you can plug the solution back into the conservation equation $c_v = \dot v/u^2$ and solve this equation by integration as well. This yields the general solution to the system of differential equations, and then you can relate $u$ and $v$ in the way stated in the quote.
Warning; there may be simpler ways of showing what you want to show.
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