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.