Timelike geodesic curves for two-dimensional metric

[Fisherlam]

Homework Statement

For the two-dimensional metric $ds^2 = [dx^2 + c^2dt^2] /(\alpha t^{-2})$, with $\alpha$ being a constant of appropriate dimensions, show that $$\frac{dx/dt}{\sqrt{1-(dx/dt)^2}}$$ is constant and hence, or otherwise, find all timelike geodesic curves.

Relevant Equations

$$L=g_{ab}\dot{x}^a\dot{x}^b $$ $$\frac{\partial L}{\partial x}=\frac{\partial }{\partial u}\left(\frac{\partial L}{\partial \dot{x}}\right) $$

Using EL equation, $$L=\left(\frac{t^2}{\alpha}\dot{x}^2-\frac{c^2t^2}{\alpha}\dot{t}^2\right)^{0.5} \Longrightarrow \mathrm{constant} =\left(\dot{x}^2 -c^2 \dot{t}^2\right)^{-0.5} \left(\frac{t^2}{\alpha}\right)^{0.5} \dot{x}$$.

Get another equation from the metric: $$ds^2=-\frac{c^2t^2}\alpha dt^2+\frac{t^2}\alpha dx^2=c^2d\tau^2\quad\Longrightarrow\quad-\frac{c^2t^2}\alpha t^2+\frac{t^2}\alpha\dot{x}^2=c^2\quad\Longrightarrow\quad\frac{t^2}\alpha=\frac{c^2}{\dot{x}^2-c^2\dot{t}^2}$$

Substitution and set $c=1$: $$\mathrm{constant}=\left(\dot{x}^2-c^2\dot{t}^2\right)^{-0.5}\left(\frac{t^2}\alpha\right)^{0.5}\dot{x}=\frac{c\dot{x}}{\dot{x}^2-c^2\dot{t}^2}=\frac{\dot{x}}{\dot{x}^2-\dot{t}^2}=\cdots?$$

I think I am close but clearly missing something...

 

[DAntanov]

You might take this SageMath code for geodesics in a Schwarzschild metric and rewrite the code to fit your simpler 2D metric. I highly recommend learning/using SageMath for anything Euler-Lagrange related.

 

[NoMoniker]

I've done some digging and if this comes from the Cambridge Part II Physics 2021 past paper, then this is actually a typo. They actually wanted you to find the first conserved quantity which you've obtained from the E-L equations.