A problem regarding to Lagrangian in Classical Mechanics

[iiternal]

Homework Statement

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.

Homework Equations

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

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.

 

[I like Serena]

Welcome to PF, iiternal!

This looks like an exercise in multivariable differentiation.

Let's start with your relevant equation.
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?

 

[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?

Welcome to PF, iiternal!

This looks like an exercise in multivariable differentiation.

Let's start with your relevant equation.
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?

 

[I like Serena]

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?

Good! :)

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

Your equations should read:
$G = \frac{dF}{dt} = \sum\limits_j\frac{\partial F}{\partial q_j}\dot{q_j}+\frac{\partial F}{\partial t}$

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

 

[iiternal]

Great !
Thank you very much!
Happy New Year.

Good! :)

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

Your equations should read:
$G = \frac{dF}{dt} = \sum\limits_j\frac{\partial F}{\partial q_j}\dot{q_j}+\frac{\partial F}{\partial t}$

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