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- 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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- Thread starter lambdajitsu
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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

- 2

- 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

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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].

\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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##\int g_{ij}\dot x^i\dot x^jdt##

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