Proving Invariance of Lagrangian Equations with Coordinate Transformation?

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SUMMARY

This discussion centers on proving the invariance of Lagrangian equations under coordinate transformations. The participants emphasize the importance of establishing an invertible relationship between two sets of coordinates, denoted as \( q_{i} \) and \( q'_{i} \). They highlight the necessity of applying the chain rule to relate the transformed coordinates and their derivatives, ultimately leading to the conclusion that the Lagrangian remains invariant, expressed as \( L'(\mathbf{q'},\mathbf{\dot{q'}},t)=L(\mathbf{q},\mathbf{\dot{q}},t) \). The discussion concludes with a reminder of the fundamental calculus rules essential for these transformations.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with coordinate transformations
  • Proficiency in calculus, particularly the chain rule
  • Knowledge of the notation used in classical mechanics
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  • Study the derivation of Lagrangian equations in classical mechanics
  • Learn about coordinate transformations in Lagrangian mechanics
  • Explore the application of the chain rule in physics
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This discussion is beneficial for physics students, particularly those studying classical mechanics, as well as educators and researchers interested in the mathematical foundations of Lagrangian dynamics.

xman
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Hi, I'm hoping someone can give a little guidance, my taks is to prove that the Lagrangian equations are invariant under a change of coordinates.

So what I've done is said if we have a set of coordinates say
\left{ q_{i} \right\} , i = 1, \ldots ,N
where I'll assume this first set is adequate and complete. Since we want to show the tranformation of coordinates and the Lagrangian is invariant, I create another set of the form
q_{k} = \tilde{f}_{k} (q_{1}^{\ast}, \ldots , q_{N}^{\ast},t)
where we assume an invertible relationship defined by above. So the Lagragian becomes
L ( q_{i},t) = L ( \tilde{f}_{1} (q_{k}) , \ldots , \tilde{f}_{N} (q_{k}) ,t) \equiv L^{\ast}
Now the Lagragian equations are of course
\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_{i}} = \frac{\partial L}{\partial q_{i}}
So, I think we must also have
\frac{d}{dt} \frac{\partial L^{\star}}{\partial \dot{q}_{i}} = \frac{\partial L^{\ast}}{\partial q_{i}^{\ast}}
which leads me to my question. I am sure the last equation is correct just by the form, I'm failing to see how to relate the coordinates and the derivatives as they are transformed. Any ideas would be great thanks.
 
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What a concidence! I've been given the same problem for my classical mechanics assignments about a months ago, except the lecturer gave us the coordinate transformation:biggrin:

I think you've got the transformation right, but I prefer it written in this form

q'_{i}=q'_{i}(\mathbf{q},t), i=1,2,...,s, \mathhbf{q} \equiv {q_{j}, j=1,2,...,s}

Now, the lagrangian is invariant so

L'(\mathbf{q'},\mathbf{\dot{q'}},t)=L(\mathbf{q},\mathbf{\dot{q}},t)

You need to start with the lagrangian equation

\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_{i}} - \frac{\partial L}{\partial q_{i}}=0

Then apply the chain rule to transform the coordinate from q_{i} to q'_{i}.

It should be all right, although the last step is a bit tricky:devil:
 
Last edited:
thanks, i took your advice and got a nice relationship once i realized how to properply apply the chain rule. most of the time i guess it helps to remember the simple rules of calc.
 

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