Invariance of Lagrangian

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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
[tex] \left{ q_{i} \right\} , i = 1, \ldots ,N[/tex]
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
[tex] q_{k} = \tilde{f}_{k} (q_{1}^{\ast}, \ldots , q_{N}^{\ast},t) [/tex]
where we assume an invertible relationship defined by above. So the Lagragian becomes
[tex] L ( q_{i},t) = L ( \tilde{f}_{1} (q_{k}) , \ldots , \tilde{f}_{N} (q_{k}) ,t) \equiv L^{\ast} [/tex]
Now the Lagragian equations are of course
[tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_{i}} = \frac{\partial L}{\partial q_{i}} [/tex]
So, I think we must also have
[tex] \frac{d}{dt} \frac{\partial L^{\star}}{\partial \dot{q}_{i}} = \frac{\partial L^{\ast}}{\partial q_{i}^{\ast}} [/tex]
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.
 
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

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

Now, the lagrangian is invariant so

[tex]L'(\mathbf{q'},\mathbf{\dot{q'}},t)=L(\mathbf{q},\mathbf{\dot{q}},t)[/tex]

You need to start with the lagrangian equation

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

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

It should be all right, although the last step is a bit tricky:devil:
 
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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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