Lagrangian - surface of sphere

Click For Summary
SUMMARY

The discussion focuses on deriving the motion of a free particle constrained to move on the surface of a sphere of fixed radius R using Lagrangian mechanics. The Lagrangian is defined as L = R^2 \dot{\theta}^2 + R^2 sin^2{\theta} \dot{\phi}^2. The Euler-Lagrange equations yield two key equations, leading to a complex integral that participants struggle to solve. The goal is to express the solution in the form θ(t) = arccos(√(1 - A^2/ω^2) cos(ωt + θ₀), which remains elusive for the contributors.

PREREQUISITES
  • Understanding of Lagrangian mechanics and the Euler-Lagrange equations
  • Familiarity with spherical coordinates and their applications in physics
  • Proficiency in solving differential equations and integrals
  • Knowledge of trigonometric identities and their manipulation
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equations in classical mechanics
  • Learn techniques for solving differential equations, particularly separable equations
  • Explore variable substitution methods in integral calculus
  • Investigate the physical implications of motion on a spherical surface
USEFUL FOR

Students and professionals in physics, particularly those focusing on classical mechanics, as well as mathematicians interested in differential equations and integrals related to motion on curved surfaces.

rolotomassi
Messages
52
Reaction score
0
I have a free particle moving on the surface of a sphere of fixed radius R. Gravity is ignored and m/2 is left out since its constant.
The lagrangian is L = R^2 \dot{\theta^2} + R^2 sin^2{\theta} \dot{\phi^2}

Using the Euler Lagrange equations I obtain
sin^2{\theta} \dot{\phi} = A = const \ (1) \\ \ddot{\theta} - sin{\theta}cos{\theta} \dot{\phi^2} = 0 \ (2)
and by substituting 1 into 2
\ddot{\theta} = A^2 cos{\theta}/sin^3{\theta}

By integrating w.r.t time and using the fact dt = d{\theta}/\dot{\theta} and that theta and its time derivative are treated as independent coordinates i get
\dot{\theta} + A^2/ 2 \dot{\theta} sin^2{\theta} \ = c_1
integrating w.r.t time again i get. [Using omega instead of theta dot now.]

\theta - A^2 cot{\theta}/\omega^2 \ = t c_1 + c_2

I can't see anything wrong but I am supposed to get this into the form

\theta(t) = arccos ( \sqrt{1 - A^2/\omega^2} cos(\omega t + \theta_0)

and I cant. If anyone can help I would appreciate it a lot. Thanks
 
Physics news on Phys.org
Your integration of the ##\ddot\theta## equation looks suspicious.
 
You're right it was. What I get now which I am pretty sure is correct is this horrible integral

\theta = \frac{1}{\omega} \int \sqrt[]{ c_1 - c_2/sin^2{\theta} } \ d \theta

Which I also have no idea how to solve :S
 
rolotomassi said:
You're right it was. What I get now which I am pretty sure is correct is this horrible integral

\theta = \frac{1}{\omega} \int \sqrt[]{ c_1 - c_2/sin^2{\theta} } \ d \theta

Which I also have no idea how to solve :S
You cannot have ##\theta## as the integration variable and the result!
 
wow I'm being stupid today.

If i separate the variables I get \int \frac{1}{ \sqrt[]{(c_1 - c_2/sin^2{\theta} )} }d \theta \ = \ \int dt
 
Indeed, so I suggest a change of variables in the left integral. (Doing this change of variables from the beginning significantly simplifies the differential equation. You might try it instead.)
 
I start again and get \dot \theta = c_1 \int \frac{u}{(1-u^2)^{3/2}} \frac{dt}{du}du But still abit stuck. If I sub in du/dt I keep getting a theta dot term. But I can't just assume u dot is independent of u and take it out the integral either. Brain is really hitting brick wall after brick wall with this
 
So start by solving the integral then.
 
Solving the integral IS the question. If i could solve the integral I wouldn't be asking for help on solving the integral.
 
  • #10
Well, so use the substitution in it. It is a standard integral.
 
  • #11
The integral is now \int d\theta (c_2 + \frac{c_1^2}{u^2 -1})^{-1/2}= \int dt As I find it. But i get this simplifying to \int du \frac{1}{u^2 - a^2} for the LHS where a^2 is a combination of the other constants. But this is a logarithm when integrating which doesn't make sense for the answer.
 
  • #12
\int 1/(c - 1/2sin^2x)^{1/2} dx is what I've got now
 
  • #13
You are not doing the arithmetics properly then. Please show your work.
 
  • #14
That was the correct integral bar a constant, which has now been solved. Thanks anyway
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 0 ·
Replies
0
Views
867
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
6
Views
2K