Prove the Langrangian is not unique

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Question:
If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations show by direct substitution that

http://qlx.is.quoracdn.net/main-74d090d14ee4fea0.png

also satisfies Lagrange's equations where F is any arbitrary but differentiable function of its arguments.
Attempt at a solution:
I'm not really sure how to solve this problem by direct substitution; I found a way to do it using the action integral but not direct substitution. Any clues or help on how to approach the problem?
 
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I got that L' satisfies that equation when the derivative for Lagrange's equation of F is 0.
 
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It's me said:
Question:
If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations show by direct substitution that

http://qlx.is.quoracdn.net/main-74d090d14ee4fea0.png

also satisfies Lagrange's equations where F is any arbitrary but differentiable function of its arguments.

Your question is wrong. The function [itex]F[/itex] can not depend on the velocity [itex]\dot{q}(t)[/itex]. If it does, then [itex]d F /d t[/itex] will depend (at least linearly) on the acceleration [itex]\ddot{q}(t)[/itex]. This in turns mean that the two Lagrangians are not equivalent to each other. So, try to solve the exercise for the function [itex]F(t) = F( q(t) , t )[/itex].

This is one part of well know theorem which states

A function [itex]\Lambda[/itex] of [itex]q(t)[/itex], [itex]\dot{q}(t)[/itex] and [itex]t[/itex] satisfies Lagrange’s equations identically (i.e., independent of the [itex]q_{a}(t)[/itex]) if, and only if, it is the total time derivative [itex]d F / dt[/itex] of some function [itex]F ( q(t) , t )[/itex].
 
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