Euler-Lagrangian Equations

1. Sep 24, 2006

UrbanXrisis

Euler-Lagrangian Equations

Let $$L=L(q_i(t), \dot{q}_i (t)$$ be a Lagrangian of a mechanical system, where $$q_i(t)$$ and $$\dot{q}_i (t)$$ are the short hand notations for $$q_1(t), q_2(t), . . . q_N(t)$$ and $$\dot{q}_1(t), \dot{q}_2(t), . . . \dot{q}_N(t)$$, respectively.

I need to prove that if $$L ' =L+\frac{d \phi}{dt}$$, then $$L=L(q_i(t), \dot{q}_i (t)$$ and $$L ' =L(q_i(t), \dot{q}_i (t)$$ give the same equations of motion (Euler-Lagrangian equations).

The Euler-Lagrangian equations is $$\frac{\partial L}{\partial x_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_i }= 0$$

I really have no idea where to begin. Any suggestions would be much appreciated.

Last edited: Sep 24, 2006
2. Sep 24, 2006

quasar987

More generally, the equations of motions are the same up to an added time dependant function in the lagrangian. The proof is as direct as can be... just write the E-L equations for L and L' (and perform the derivatives on L').

3. Sep 24, 2006

UrbanXrisis

$$\frac{\partial L}{\partial x_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_i }= 0$$

$$L ' =L+\frac{d \phi}{dt}$$

$$\frac{\partial L}{\partial x_i} + \frac{\partial }{\partial x_i} \frac{d \phi }{dt}- \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}_i } + \frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} \right)= 0$$

$$\frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} = 0$$

$$\frac{\partial L}{\partial x_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_i }= 0$$

something like that?

4. Sep 25, 2006

dextercioby

Does $\phi$ depend on the depend on the generalized coodinates & velocities...?

Daniel.

5. Sep 25, 2006

UrbanXrisis

I am told that $$\frac{d \phi }{dt}$$ is a "total time derivative"

6. Sep 25, 2006

dextercioby

As i know it, the trick is that $\phi=\phi\left(t,q^{i}\right)$.

Therefore

$$\frac{d\phi}{dt} =\frac{\partial \phi}{\partial t}+\frac{\partial \phi}{\partial q^{i}} \dot{q}^{i}$$.

Daniel.

7. Sep 25, 2006

quasar987

Dexter is probably right on this, because I checked the definition of "total time derivative" on wiki and it says "a derivative which takes indirect dependencies into account"*

This property of the equ. of motion you're trying to prove is mentioned in the "Landau & Lif****z" on mechanics in the first few pages of the book.

*http://en.wikipedia.org/wiki/Total_time_derivative

8. Sep 25, 2006

UrbanXrisis

frankly, i need some real help. My class has an awful professor and using a very mathematical book that I cant seem to grasp. Before I try to attack this problem, could someone explain to me what exactly the Euler-Lagrangian does? What does it solve for?

Also the notation $$L ' =L(q_i(t), \dot{q}_i (t)$$

Why is there a comma between q and q dot? What does the comma symbolize?

Also, could someone please explain $$\delta$$ notation?

What does it mean when $$\frac{dJ}{d \alpha} d \alpha = \delta J$$ ?

My whole class seems confused as to what the lectures are about, so I have taken it upon myself to learn the material by myself.

In class today, we were told that $$\phi=\phi\left(q_i,\dot{q}_i (t),t\right)$$

What is the significance of this statement dextercioby?

I am sorry for my ignorance, but I am willing to learn.

9. Sep 25, 2006

Hargoth

The Euler-Lagrange equations are the equations of motion for a system which motion is described by the generalized coordinates $q_i$. They are, so to say, for generalized coordinates what the Newtonian equations are for cartesian coordinates.

$L(q_i(t), \dot q_i(t) )$ simply means that L ist a function of both q_i and the total time-derivatives $\dot q_i$- of generalized velocities, so to say.

edit: The Euler-Lagrangian equations can be derived from the "priciple of least action", which you can take as an axiom instead of Newton's equations. From the formulas I think you did this in your course. For an introduction what it's all about I suggest Feynman's Lectures on physics, Volume 2, Chapter 19, after this f.e. Goldstein for a more detailed explanation.

Last edited: Sep 25, 2006
10. Sep 25, 2006

UrbanXrisis

what about $$\delta$$ notation? and what dextercioby wrote?

11. Sep 25, 2006

Hargoth

If $\Phi(q_i (t), \dot q_i(t),t )$ the total time derivative becomes
$\frac{d}{dt} \Phi = \sum \limits_i \frac{\partial \Phi}{\partial q_i} \dot q_i + \sum \limits_i \frac{\partial \Phi}{\partial \dot q_i} \ddot q_i + \frac{\partial \Phi}{\partial t}$ (dexter didn't write the sum out but used Einstein convention - sum is taken over i, since i is double).
$\delta$ means the "variation".

For example, $\delta \int \limits L(q_i, \dot q_i, t) dt =0$ means that the integral doesn't change when you change the "pathfunction" q_i over which is integrated a little (just as a function has a maximum or minimum and doesn't change "nearby" when the derivative is zero). This is the "Principle of least action", from which the Euler-Lagrange-equations can be derived.

Last edited: Sep 25, 2006
12. Sep 26, 2006

UrbanXrisis

was what I did here okay?

13. Sep 28, 2006

silimay

Hi....

I don't understand where you got $$\left( \frac{\partial L}{\partial \dot{x}_i } + \frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} \right)= 0$$. It is supposed to be equal to L right? How is it equal to L?

14. Sep 28, 2006

silimay

sorry, I just meant
$$\left( \frac{\partial L}{\partial \dot{x}_i } + \frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} \right)$$,
not the equal to zero part.

Last edited: Sep 28, 2006
15. Sep 29, 2006

dextercioby

It doesn't work if $\phi$ depends on the generalized velocities.

Daniel.