Hello! I ask this question in this forum because I wish to be given a general explanation. Hope it is OK given this metric: [itex] ds^2= \frac {dt^2} {t^2}- \frac{dx^2} {t^2} [/itex] I wish to calculate the geodesics. [itex] S=\int{ \frac {d} {d\lambda} \sqrt{\frac {1} {t^2} \frac {dt^2} {d\lambda^2}- \frac{1} {t^2} \frac {dx^2} {d\lambda^2}}} [/itex] But [itex] \lambda [/itex] here, is any parameter I choose. Therefore, L does not depend on it (right? ) So now I want to use E-L equations: [itex] \frac {d}{dλ}\frac {∂L}{∂\dot t}=\frac {\partial L} {∂t}, \frac {d} {dλ}\frac {∂L} {∂\dot x} =\frac {\partial L} {∂x} [/itex] When [itex] \dot t [/itex] refers to [itex] \frac {dt} {d\lambda} [/itex] etc. But here I get confused: What does L depend on? I'll continue my solution: [itex]\frac {d} {dλ} 0.5 ({{\frac {{\dot t}^2} {t^2}-\frac {{\dot x}^2} {t^2}}})^{-0.5} 2 \frac {\dot t} {t^2}= \frac {\partial L} {\partial x} [/itex] As I stated, L does not depend on lambda, and so: [itex] \frac {d} {d\lambda} {[\frac {\dot t} {t^2}]}^2 = \frac {{\dot t}^2-{\dot x}^2} {t^3} [/itex] So how am I supposed to derive this? Do I add another dot? [itex] \frac {d} {d\lambda} \dot t = \ddot t [/itex] ? and do I refer to it as an operator? Meaning- [itex] \ddot t = {\dot t}^2? [/itex] How do I continue? These calculations on classical mechanics were so trivial to me- but for some reason I get lost in here... Thank you!
noamriemer, There are several different equivalent Lagrangians that may be used to calculate the geodesics. The easiest, I believe, is to use L = g_{μν} dx^{μ}/ds dx^{ν}/ds. This eliminates dealing with the square root. So for your case, L = t^{-2} (dt/ds)^{2} - t^{-2} (dx/ds)^{2}. You let t' = dt/ds and x' = dx/ds for short, and write the equations as, for example, d/ds(∂L/∂t') - ∂L/∂t = 0.