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

  • #1
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
pasmith
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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].
 
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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.
 
  • #4
wrobel
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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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