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

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For a free particle described by a Lagrangian L, if dL/dt=0, any twice differentiable function f(L) will yield the same equations of motion. The Euler-Lagrange equations can be modified by substituting L with f and applying the chain rule. This leads to a result where the equations of motion are multiplied by f'(L), which can be divided out since f'(L) is non-zero for a valid function. The conclusion is that f(L) indeed gives the same equations of motion as L. This demonstrates the equivalence of the two formulations in the context of free particle dynamics.
nikolafmf
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Homework Statement


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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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