A Lagrangian for straight line in XY-plane (dependent on time)

[lambdajitsu]

TL;DR Summary

Really stuck on this one. Any help (even a slight nudge in the right direction) would be greatly appreciated. Seems I am hitting a fundamental barrier in the Lagrangian formulation.

https://dst-public.s3-us-west-2.amazonaws.com/lagrangian.png

 

[pasmith]

You have that <br /> \frac{d}{dt} \frac{\partial L}{\partial \dot q_x} = \frac{d}{dt} \frac{\dot q_x}{\sqrt{\dot q_x^2 + \dot q_y^2}} = \frac{d}{dt} \frac{\dot q_x}{L} = 0. That tells you that \dot q_x/L is a constant. Similarly \dot q_y/L is constant, and since (\dot q_x/L)^2 + (\dot q_y/L)^2 = 1 you can set <br /> \frac{\partial L}{\partial \dot q_x} = \frac{\dot q_x}L = \cos \alpha,\qquad <br /> \frac{\partial L}{\partial \dot q_y} = \frac{\dot q_y}L = \sin \alpha. But now in terms of arclength s you have <br /> \frac{dx}{ds} = \frac{\dot q_x}{L} = \cos \alpha so x(s) = x_0 + s\cos\alpha and a similar result for dy/ds.

 

[lambdajitsu]

Is your solution above an example of a canonical transformation? I admit I don't quite get it yet, but I'm getting closer. Thank you very much for your help, btw.

 

[wrobel]

Actually, there is no need to use functional $\int\sqrt{g_{ij}\dot x^i\dot x^j}dt$ to compute geodesics. Due to the energy integral $g_{ij}\dot x^i\dot x^j=const$ and parametric invariance , the geodesics can equivalently be determined from the functional
$\int g_{ij}\dot x^i\dot x^jdt$