Generating Function for Lagrangian Invariant System

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The discussion revolves around finding the generating function G for a Lagrangian invariant system, specifically when the Lagrangian L(q, q̇) is invariant under the transformation q → q + K(q). The user attempts to derive G using Poisson brackets but encounters inconsistencies in their expressions for G. They initially assume a specific form for the Lagrangian, which leads to confusion about the relationship between the canonical momentum and the transformed coordinates. Suggestions are made to reconsider the connection between the canonical momentum and the transformation to resolve the discrepancies. The conversation emphasizes the importance of correctly applying the definitions and relationships in the context of Lagrangian mechanics.
Physgeek64
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Homework Statement


Given a system with a Lagrangian ##L(q,\dot{q})## and Hamiltonian ##H=H(q,p)## and that the Lagrangian is invariant under the transformation ##q \rightarrow q+ K(q) ## find the generating function, G.

Homework Equations

The Attempt at a Solution


##\delta q = \{ q,G \} = \frac{\partial G}{\partial p} ##
##\delta p = \{ p, G \} = -\frac{\partial G}{\partial q} ##

in this case ##\delta q =K(q)##
Hence
##G= pK + c_1(q) ##
Assuming the Lagrangian is of the form ##L=\frac{1}{2} m q^2 -V(q)##
then ##p=m\dot{q}##
and ##\delta p = m\dot{K} = mK' \dot{q}= pK'##

##\delta p = pK'=-\frac{\partial G}{\partial q} ##
## G= -pK +c_2(p)##

I know this is not right because i have two different expressions that can't be matched, but i can't think of another way to do this.

Many thanks
 
Last edited:
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Physgeek64 said:
Assuming the Lagrangian is of the form
What makes you think you can assume this?

Edit: Also note that { and } are LaTeX delimiters used for grouping. To actually get the brackets of the Poisson bracket you need to use \{ and \}.
 
Orodruin said:
What makes you think you can assume this?

Edit: Also note that { and } are LaTeX delimiters used for grouping. To actually get the brackets of the Poisson bracket you need to use \{ and \}.
Ahh okay then, so I guess i can't make that assumption. In which case i don't know how to proceed, any tips?

Thank you, i will change them :)
 
Consider how the canonical momentum ##P = \partial L/\partial \dot Q## connected to ##Q = q + K(q)## must change relative to ##p##. This should give you ##\delta p##.
 
Orodruin said:
Consider how the canonical momentum ##P = \partial L/\partial \dot Q## connected to ##Q = q + K(q)## must change relative to ##p##. This should give you ##\delta p##.

## \frac{\partial L}{\partial \dot{Q}} = \frac{\partial L}{\partial \dot{q}} \frac{\partial \dot{q}}{\partial \dot{Q}} =\frac{\partial L}{\partial \dot{q}} \frac{1}{1+\dot{K}'} = p \frac{1}{1+\dot{K}'} ##

?? is this along the right lines?

Many thanks
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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