Help with example from goldstein (lagrangian)

In summary: I think you are misunderstanding what the zdot^2 term is. It is just the z derivative of the kinetic energy.
  • #1
cahill8
31
0

Homework Statement


From pages 124-125 in edition 3.

This is about the restricted three body problem (m3 << m1,m2)

[PLAIN]http://img718.imageshack.us/img718/7012/3bdy.jpg


Homework Equations


L = T-V

Euler-Lagrange equations

The Attempt at a Solution



I'm interested in m3, the negligible mass object.
Latex isn't showing up in the preview so note that by xdot I mean d/dt(x), abs = absolute value, and r1,r2,r3 are vectors.

L = 1/2 m3 (xdot^2+ydot^2) - V

my first question is: is the potential simply this?
[PLAIN]http://img408.imageshack.us/img408/9834/potential.jpg

and I'm calling this V(r,theta,t)

changing to polar coordinates (x=rcos(theta), y=rsin(theta)) and simplifying gives

[PLAIN]http://img11.imageshack.us/img11/6784/14361215.jpg

This is where the example starts, except the potential is not stated, it is just given as V(r,theta,t)
the vectors in my potential depend on r, theta and t so I guess that is right?

Next in the example, the coordinate system is made to rotate at the same frequency as m3 so that m1 and m2 appear stationary.
theta' = theta + wt (I'm using w for omega)
theta = theta' - wt
thetadot = theta'dot - w
so I replace thetadot above in the equation for L and I eventually get
[PLAIN]http://img198.imageshack.us/img198/2047/lchanged1.jpg

now my main confusion comes. In the example, as well as making the coordinate system rotate, they say switch to cylindrical coordinates and state that this becomes

[PLAIN]http://img685.imageshack.us/img685/421/lchanged2.jpg

questions:
The t dependency in the potential changes into a z dependency?
z seems a strange choice as it is a two-dimensional problem?

The only reason I can see for this is to get the lagrangian in the form L = (rho,theta,z,rhodot,thetadot,zdot)

The book does not say what the transformations were except for theta. Just that it was a cylindrical transform using rho, theta and z

comparing the two equations (the one I derived and the one given) the transformation must have been

rho = r (since x and y are already in polar coordinates x=rcos(theta)=rhocos(theta))
theta = theta' - wt
z = t
and then they added the zdot^2 to the kinetic energy term which is fine since it is 0 anyway.

but can you simply make a change like this? z = t doesn't feel right.
So I'm stuck up to this point.

The book then says to find the euler-lagrange equations and look for solutions where rhodot=zdot=thetadot=0 and these will be the five lagrange points.

So the first thing I need is a complete lagrangian with a defined and transformed potential in order to get the derivatives for the euler-lagrange equations.

Hopefully someone can help me out, thanks
 
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  • #2
cahill8 said:
my first question is: is the potential simply this?
[PLAIN]http://img408.imageshack.us/img408/9834/potential.jpg

and I'm calling this V(r,theta,t)

Shouldn't there be another term in your potential...namely the interaction between [itex]m_1[/itex] and [itex]m_2[/itex]?

Also, are you treating [itex]x_1[/itex], [itex]x_2[/itex], [itex]r_1[/itex], [itex]r_2[/itex] and [itex]r_3[/itex] as distances or vectors? If you are treating them as distances, then [itex]x_1\neq r_2-r_3[/itex] and [itex]x_2\neq r_1-r_3[/itex]. If you are treating them as vectors, then your potential (which is always a scalar!) involves only the magnitudes of [itex]\textbf{x}_1[/itex], [itex]\textbf{x}_2[/itex] and [itex]\textbf{r}_2-\textbf{r}_1[/itex] (from your missing 3rd term).
 
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  • #3
Thanks for that, those are some pretty big mistakes. They are vectors as that's how they are in the book.

About the 3rd term in the potential, I agree it would be needed if it was the potential was for the system but as far as I understand I'm only interested in the potentials that act upon m3? Just as I have excluded the kinetic energies of m2 and m1, I exclude the potential that acts between them? Let me know if I'm wrong.

I'll add the third term for now though as I can always change it later

[PLAIN]http://img693.imageshack.us/img693/9287/newvandl.jpg

So I'm assuming this is right apart from whether the 3rd term in the potential should be included or not.

Then the following transformation is applied to make to coordinate system rotate at the same frequency.

[PLAIN]http://img534.imageshack.us/img534/1777/rotatep.jpg

Then thetadot in the equation for L is changed and expanded giving
[PLAIN]http://img198.imageshack.us/img198/2047/lchanged1.jpg

Then the book says switch to cylindrical coordinates I'll quote here:
we can write the Lagrangian in terms of the rotating system by using theta' = theta + wt as the transformation to the rotating frame. Thus, the Lagrangian in the rotating coordinates can be written in terms of the cylindrical coordinates, rho, theta and z, with rho being the distance from the center of mass and theta the counterclockwise angle from the line joining the two masses
Thus:
[PLAIN]http://img685.imageshack.us/img685/421/lchanged2.jpg

Which is exactly the same, just changing r to rho and t to z?
In the standard cylindrical coordinates, z = z. So why does the t vanish and become a z in the book? z is not just another variable for the time either as they add in the zdot^2 term in the kinetic energy implying they mean the kinetic energy in the z direction. So how did my time dependent potential turn into a z dependent potential? Especially since it is a two-dimensional problem
 
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  • #4
I tried a different approach not using vectors, is this valid?

[PLAIN]http://img59.imageshack.us/img59/9499/45953816.jpg

[PLAIN]http://img411.imageshack.us/img411/7677/42701744.jpg

[PLAIN]http://img265.imageshack.us/img265/7100/33037853.jpg

and I then get the Euler-Lagrange equations for r and theta. When differentiating w.r.t r, am I correct in thinking that r1 and r2 and constants and the only thing I have to differentiate is r?
 
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  • #5
in advance.

As a scientist, it is important to approach problems in a systematic and logical manner. In this case, the first step would be to carefully review the given information and equations from the textbook, making sure to understand the concepts and notations used. It may also be helpful to consult other resources or seek clarification from a professor or fellow students.

Based on the information provided, it seems that the main confusion lies in the transformation to cylindrical coordinates and the resulting change in the potential. It may be helpful to carefully go through the steps of the transformation and consider the implications of each variable being transformed.

Regarding the choice of z as a dependent variable in the potential, it is possible that this is simply a convention used in this particular problem. However, it is important to make sure that this choice makes sense in the context of the problem and the given equations.

Once a complete and transformed Lagrangian has been derived, the next step would be to use the Euler-Lagrange equations to find the Lagrange points. This involves setting the derivatives of the Lagrangian with respect to the dependent variables equal to zero and solving for the values of the independent variables. It may also be helpful to check the solutions against known solutions or simulations of the problem.

In conclusion, as a scientist, it is important to carefully and methodically approach problems, seeking clarification and consulting additional resources when needed. With persistence and a thorough understanding of the concepts and equations involved, the solution to this problem can be successfully obtained.
 

1. What is the significance of Goldstein's Lagrangian in physics?

The Lagrangian, also known as the Lagrangian function, is a fundamental concept in classical mechanics that describes the dynamics of a physical system. Goldstein's Lagrangian is a specific formulation of the Lagrangian that is widely used in the study of mechanics and electromagnetism.

2. How does Goldstein's Lagrangian differ from other formulations of the Lagrangian?

Goldstein's Lagrangian takes into account the constraints and symmetries of a physical system, while other formulations may not. This allows for a more accurate and comprehensive description of the system's dynamics.

3. Can you provide an example of how Goldstein's Lagrangian is applied in physics?

One example is in the study of rigid body dynamics. Goldstein's Lagrangian is used to describe the motion of a rigid body, taking into account its rotational and translational kinetic energies.

4. How does Goldstein's Lagrangian relate to Hamilton's principle?

Goldstein's Lagrangian is based on Hamilton's principle, which states that the true path of a physical system is the one that minimizes the action integral. Goldstein's formulation of the Lagrangian allows for the application of Hamilton's principle to a wider range of physical systems.

5. Are there any limitations to using Goldstein's Lagrangian?

Goldstein's Lagrangian is a powerful tool in classical mechanics, but it may not be applicable in all cases. It is most commonly used in systems with a finite number of degrees of freedom and in systems without dissipative forces.

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