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

• A
lambdajitsu
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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• lagrangian.png
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## Answers and Replies

Homework Helper
You have that $$\frac{d}{dt} \frac{\partial L}{\partial \dot q_x} = \frac{d}{dt} \frac{\dot q_x}{\sqrt{\dot q_x^2 + \dot q_y^2}} = \frac{d}{dt} \frac{\dot q_x}{L} = 0.$$ 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 $$\frac{\partial L}{\partial \dot q_x} = \frac{\dot q_x}L = \cos \alpha,\qquad \frac{\partial L}{\partial \dot q_y} = \frac{\dot q_y}L = \sin \alpha.$$ But now in terms of arclength $s$ you have $$\frac{dx}{ds} = \frac{\dot q_x}{L} = \cos \alpha$$ so $x(s) = x_0 + s\cos\alpha$ and a similar result for $dy/ds$.

Last edited:
• • NTuft, lambdajitsu and (deleted member)
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.

• NTuft
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