Replacing Lagrangian L with function f(L) for free particle

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SUMMARY

The discussion centers on the relationship between the Lagrangian L for a free particle and a function f(L) in the context of the Euler-Lagrange equations. It is established that if dL/dt=0, any twice differentiable function f(L) will yield the same equations of motion as L. The key step involves substituting f into the Euler-Lagrange equations and applying the chain rule, leading to the conclusion that the resulting equations are equivalent, provided f'(L) is non-zero and can be divided out.

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nikolafmf
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Homework Statement


[/B]
If L is Lagrangian for a (system of) free particle(s) and dL/dt=0, show that any twice differentiable function f(L) gives the same equations of motions.

Homework Equations


[/B]
Euler-Lagrange equations.

The Attempt at a Solution



Well, after some calculation, I get [itеx] $\frac{d}{dt}\frac{\partial f}{\partial \dot{r}}-\frac{\partial f}{\partial r}=0$ [/itеx].

Can I conclude from this that f(L) gives the same equations of motion? If not, what should I do?
 
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Well, in my Latex the command worked as should do. I don't know why in my previous message the equation didn't show up. :(
 
Substitute the lagrangian with f in the euler-lagrange equations. Then use chainrule.
 
exclamationmarkX10 said:
Substitute the lagrangian with f in the euler-lagrange equations. Then use chainrule.

Thank you for your suggestion. I already did that and got zero as a result. What should I conclude from that?
 
nikolafmf said:
Thank you for your suggestion. I already did that and got zero as a result. What should I conclude from that?

After that, you should get the same equations of motion except they are multiplied by f\prime(L). You have to then argue that you can divide out the f\prime(L).
 
Last edited:

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