How do different Lagrangians affect the equations of motion?

[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.

 

[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').

 

[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?

 

[dextercioby]

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

Daniel.

 

[UrbanXrisis]

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

 

[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.

 

[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

 

[UrbanXrisis]

frankly, i need some real help. My class has an awful professor and using a very mathematical book that I can't 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)

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.

What is the significance of this statement dextercioby?

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

 

[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.

 

[UrbanXrisis]

what about \delta notation? and what dextercioby wrote?

 

[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.

 

[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?

was what I did here okay?

 

[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?

 

[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.

 

[dextercioby]

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

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

Daniel.