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

In summary, the conversation discusses how to compute geodesics using the Lagrangian method and the concept of canonical transformation. It is mentioned that there is no need to use the functional ##\int\sqrt{g_{ij}\dot x^i\dot x^j}dt## and instead, the geodesics can be determined using the functional ##\int g_{ij}\dot x^i\dot x^jdt## due to the energy integral and parametric invariance. The concept of canonical transformation is used to show that \dot q_x/L and \dot q_y/L are constants, which can be used to derive the equations for x(s) and y(s) in terms of arclength s
  • #1
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.

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  • #2
You have that [tex]
\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.[/tex] That tells you that [itex]\dot q_x/L[/itex] is a constant. Similarly [itex]\dot q_y/L[/itex] is constant, and since [itex](\dot q_x/L)^2 + (\dot q_y/L)^2 = 1[/itex] you can set [tex]
\frac{\partial L}{\partial \dot q_x} = \frac{\dot q_x}L = \cos \alpha,\qquad
\frac{\partial L}{\partial \dot q_y} = \frac{\dot q_y}L = \sin \alpha.[/tex] But now in terms of arclength [itex]s[/itex] you have [tex]
\frac{dx}{ds} = \frac{\dot q_x}{L} = \cos \alpha[/tex] so [itex]x(s) = x_0 + s\cos\alpha[/itex] and a similar result for [itex]dy/ds[/itex].
 
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  • #3
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.
 
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  • #4
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##
 
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