Lagrangian - surface of sphere

[rolotomassi]

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

 

[Orodruin]

Your integration of the $\ddot\theta$ equation looks suspicious.

 

[rolotomassi]

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

 

[Orodruin]

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!

 

[rolotomassi]

wow I'm being stupid today.

If i separate the variables I get $\int \frac{1}{ \sqrt[]{(c_1 - c_2/sin^2{\theta} )} }d \theta \ = \ \int dt$

 

[Orodruin]

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.)

 

[rolotomassi]

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

 

[Orodruin]

So start by solving the integral then.

 

[rolotomassi]

Solving the integral IS the question. If i could solve the integral I wouldn't be asking for help on solving the integral.

 

[Orodruin]

Well, so use the substitution in it. It is a standard integral.

 

[rolotomassi]

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.

 

[rolotomassi]

$\int 1/(c - 1/2sin^2x)^{1/2} dx$ is what I've got now

 

[Orodruin]

You are not doing the arithmetics properly then. Please show your work.

 

[rolotomassi]

That was the correct integral bar a constant, which has now been solved. Thanks anyway