Coordinate transformation of lagrangian

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SUMMARY

The discussion focuses on the coordinate transformation of the Lagrangian in classical mechanics, specifically how to derive the relationship between time ratios along different paths for a particle with homogeneous potential energy. The key transformation equations are r' = ar and t' = a^(1-k/2)t, leading to the derived ratio t'/t = (l'/l)^(1-k/2). The user seeks clarification on the derivation process, particularly how to correctly apply these transformations to solve related problems in their textbook.

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nlake27
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Hey all,
According to my physics textbook, if the potential energy of a particle is a homogeneous function of the spatial coordinate r, one can transform r by some factor a and t by some factor b=a^(1-.5k) such that the Lagrangian of the particle is multiplied by a^k. I understand all of this, but I do not understand how one gets from here to the result given: t'/t=(l'/l)^(1-k/2); t'/t being the ratio of the times along the paths l and l'.

My textbook has some problems that require (I assume) similar logic ("Find the ratio of the times in the same path for particles having the same mass but potential energies differing by a constant factor").
The book solution is t'/t=(U/U')^(1/2); I tried solving it by making the "constant factor" a^k so that l=a*l', which eventually gives me an answer that is off by a factor of 1/a. If someone could explain the logic that leads to the derivation of the equation in bold, I'm pretty sure I can find the error in my own calculations.
 
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Well, you've got two coordinate transformations [itex]r'=ar[/itex] and [itex]t'=a^{(1-k/2)}t[/itex]. Now suppose we let l and l' be the paths traversed by the particle in times t and t' respectively. Then the first equation yields [itex]a=l'/l[/itex] which can be substituted into the second equation to give [tex]t'=\left(\frac{l'}{l}\right)^{(1-k/2)}t[/tex].
 

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