Hello! I ask this question in this forum because I wish to be given a general explanation.(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# General question about E-L in revativity

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