I (desperately) need help with the following:
What is the null geodesic for the space time $$ds^2=-x^2 dt^2 +dx^2?$$
I don't know how to transform a metric into a geodesic…! There is no need to start from the Lagrangian. I know that $$0=g_{ij}V^iV^j$$ where $V^i={dx^i\over d\lambda}$
where $\lambda$ is some parameter. But I don't know what that parameter is nor how to find the geodesic.
Many thanks!! Please help!
Best Answer
As you mentioned a null geodesic implies:
$$g_{\mu \nu} U^{\mu} U^{\nu}=\left (\frac{ds}{d \lambda} \right )^2=0$$
where $\lambda$ is some affine parameter. If you take $\lambda=t$, then this implies:
$$-x^2+\left ( \frac{dx}{dt} \right )^2=0$$
So $~dx/dt=\pm x$ is a null geodesic. (Take a second time derivative to get the actual geodesic.)