## Lagrangian

1. The problem statement, all variables and given/known data
I am trying to prove that Lagrangian L is not uniquely defined, but only up to a time derivative of a function:
$$\frac{d\Lambda}{dt}, \Lambda(\vec{q}, t)$$

So

$$L > L+\frac{d\Lambda}{dt} = L+\frac{\partial \Lambda}{\partial q}~\dot{q}+\frac{\partial \Lambda}{\partial t}$$

But when I put it in the E-L eqns they definitely aren't as before.

Where have I gone wrong?

2. Relevant equations

3. The attempt at a solution
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Show us what you got when you tried to crank out the Euler-Lagrange equations.
 Alright: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\frac{\partial L}{\partial q}$$ $$\frac{d}{dt}\frac{\partial}{\partial \dot{q}}(L+\frac{\partial \Lambda}{\partial q}~\dot{q}+\frac{\partial \Lambda}{\partial t})=\frac{\partial}{\partial q}(L+\frac{\partial \Lambda}{\partial q}~\dot{q}+\frac{\partial \Lambda}{\partial t})$$ $$\frac{d}{dt}(\frac{\partial L}{\partial q}+\frac{\partial \Lambda}{\partial q})=\frac{\partial L}{\partial q}+\frac{\partial^2 \Lambda}{\partial q^2}~\dot{q}+\frac{\partial^2 \Lambda}{\partial q \partial t}$$

Recognitions:
Gold Member
Homework Help
Staff Emeritus

## Lagrangian

Now calculate what

$$\frac{d}{dt}\left(\frac{\partial \Lambda}{\partial q}(q,t)\right)$$

is equal to.

(Do you only have one coordinate, or should you have qi's?)
 Well I can't, so I am asking for help.
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Don't be intimidated by the notation. The partial of Λ with respect to q is just another function of q and t. You find the total time derivative of it the same way you found the total time derivative of Λ(q,t).
 I see. Thank you a lot.