Why is there a sin^2(theta) factor in the diatomic molecule Hamiltonian?

[Andurien]

Homework Statement

I have to find the hamiltonian for a diatomic molecule, where the molecule can only rotate and translate and we supose that potencial energy doesn't change.

Homework Equations

The Attempt at a Solution

Okey so I used Spherical coordinate system such as the kinetic energy of the molecule is
T=\frac{1}{2}(m\dot{r}^2+I(\dot{\phi}^2+I(\dot{θ})^2)=L

To find the Hamiltonian I've considered:

P_r=m\dot{r} \Rightarrow \dot{r}=\frac{P_r}{m}

And so on for the other momentum so the final solution for the hamiltonian is:

H=\frac{1}{2}(\frac{P_r^2}{m}+\frac{P_\phi^2}{I}+ \frac{P_θ^2}{I})

BUT the correct solution given by my professor is:
H=\frac{1}{2}(\frac{P_r^2}{m}+\frac{P_\phi^2}{I \sin^2\theta}+ \frac{P_θ^2}{I})

So I don't know why is there a sin^2\theta factor.

 

[vela]

Okey so I used Spherical coordinate system such as the kinetic energy of the molecule is
T=\frac{1}{2}(m\dot{r}^2+I(\dot{\phi}^2+I(\dot{θ})^2)=L

The problem lies in where you started. How'd you come up with this?

 

[Andurien]

Yep I already solved the problem. You were right Vela i started wrong. I divided the Hamiltonian between the lagrangian for the translation kinetic energy and the rotation kinetic energy so:
T_t=\frac{1}{2}M(\dot{x}^2+\dot{y}^2+\dot{z}^2)= \frac{1}{2M} (P_x^2+P_y^2+P_z^2)

And then the rotation one:
T_{rot}=\frac{1}{2}Mr^2[\dot{\theta^2}+\dot{\phi}^2\sin^2(\theta)]

And then just like before i get the hamiltonian.

Thank you for the response and sorry for my english ;)