How to Derive the Theta Function for a Free Particle on a Spherical Surface?

[Matt atkinson]

Homework Statement

By finding the Lagrangian and using the metric:
$$\left(\begin{array}{cc}R^2&0\\0&R^2sin^2(\theta)\end{array}\right)$$
show that:
$$\theta (t)=arccos(\sqrt{1-\frac{A^2}{\omega^2}}cos(\omega t +\theta_o))$$

Homework Equations

The Attempt at a Solution

So I got the lagrangian to be $L=R^2 \dot{\theta^2} +R^2sin^2(\theta)\dot{\phi^2}$ and then used the E-L equation to find the equations of motion and the fact that $2R^2sin^2(\theta) \dot{\phi}=const=p$.
Using this and substituting into the equation i get for $\theta$ I get:
$$\frac{d}{dt}(2R^2\dot{\theta})=\frac{p^2}{2R^2}cot(\theta)csc^2(\theta)$$
which I then integrate using the substitution $dt=d\theta / \dot{\theta}$ to get:
$$\dot{\theta}=\frac{p}{2R^2}\sqrt{c-\frac{1}{2}sin^{-2}(\theta)}$$
Where c is the integration constant. Now if I separate variables to attempt to get a solution for $\theta$ i get:
$$\int _{\theta_o}^{\theta} \frac{d\theta}{\sqrt{c-\frac{1}{2}sin^{-2}}}=\frac{tp}{2R^2}$$
But i have absolutely no idea how to solve that integral. Please any pointers would be appreciated.

 

[Orodruin]

Make a variable substitution. The final form of the answer should give you some ideas.

See also: https://www.physicsforums.com/threads/lagrangian-surface-of-sphere.860009/

 

[Matt atkinson]

Ah okay, I did just try $u=cos(\theta)$ but it gives:
$$\int \frac{du}{\sqrt{c(1-u^2)-1/2}}$$
It didn't prove to be any easier to solve.
Also tired doing $u=cos(\theta)$ from the beginning just now as you suggested in the other post (although this could be the wrong substitution) and I must be doing something wrong because I get a complex square root on the LHS of the differential equation for $\dot{u}$.

 

[Orodruin]

Ah okay, I did just try $u=cos(\theta)$ but it gives:
$$\int \frac{du}{\sqrt{c(1-u^2)-1/2}}$$
It didn't prove to be any easier to solve.

This is a quite standard integral. It is of the form
$$
\int \frac{dx}{\sqrt{1 - x^2}}.
$$

 

[Matt atkinson]

Thankyou so much! I managed to get the answer now, i think it was just the fact i hadn't noticed that it was a standard integral.