# A problem regarding to Lagrangian in Classical Mechanics

1. Dec 30, 2011

### iiternal

1. The problem statement, all variables and given/known data
I have a problem regarding to lagrangian.

If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations, show by direct substitution that

$L' = L + \frac{d F(q_1,...,q_n,t)}{d t}$

also satisfies Lagrange's equations where F is any ARBITRARY BUT DIFFERENTIABLE function of its arguments.

2. Relevant equations
Lagrange's equations:
$\frac{\partial L}{\partial q_i} - \frac{d}{d t}\frac{\partial L}{\dot{\partial q_i}} =0$

3. The attempt at a solution
Equivalently we have to find
$\frac{\partial F}{\partial q_i} - \frac{d}{d t}\frac{\partial F}{\partial \dot{q_i}} =0$
It is obvious that $\frac{\partial F}{\partial \dot{q_i}}=0$.
But how can I get $\frac{\partial F}{\partial q_i}=0$ ?

Thank you.

Last edited: Dec 30, 2011
2. Dec 30, 2011

### I like Serena

Welcome to PF, iiternal!

This looks like an exercise in multivariable differentiation.

You seem to have dropped a partial derivative there...

I'm afraid that since F is an arbitrary function of $q_i$, you won't get $\frac{\partial F}{\partial q_i}=0$.

I believe you have to substitute L' in Lagrange's equation and expand everything.
This means expanding dF/dt into partial derivatives.
Do you know how to do that?

3. Dec 30, 2011

### iiternal

Thank you very much! You are absolutely right!

let $G = \frac{dF}{dt} = \sum\frac{\partial F}{\partial q_i}\dot{q_i}+\frac{\partial F}{\partial t}$;
Substitute it into Lagrange's Equation
$\frac{\partial G}{\partial q_i} - \frac{d}{dt}\frac{\partial G}{\partial \dot{q_i}} = (\frac{\partial^2F}{\partial q_i^2}\dot{q} + \frac{\partial^2F}{\partial q_i\partial t}) - \frac{d}{dt}\frac{\partial F}{\partial q_i}$

I believe the last term can somehow cancel both the first two terms.
When I tried to expand the last term, I faced another problem
$\frac{d}{dt}\frac{\partial F}{\partial q_i} = \frac{\partial }{\partial q}\frac{\partial F}{\partial t} + ???$
How can I take derivative of the denominator of a differentiation?

4. Dec 30, 2011

### I like Serena

Good! :)

I'm afraid you should not use the index i everywhere and keep the summations.

$G = \frac{dF}{dt} = \sum\limits_j\frac{\partial F}{\partial q_j}\dot{q_j}+\frac{\partial F}{\partial t}$

Substitute it into Lagrange's Equation
$\frac{\partial G}{\partial q_i} - \frac{d}{dt}\frac{\partial G}{\partial \dot{q_i}} = \sum\limits_j ...$

And:
$\frac{d}{dt}\frac{\partial F}{\partial q_i} = \sum\limits_j {\partial^2 F \over \partial q_j\partial q_i} \dot q_j + {\partial^2 F \over \partial t\partial q_i}$

5. Dec 30, 2011

### iiternal

Great !!!
Thank you very much!
Happy New Year.