Comparing Generalised Momentum Calculations for Central Force Problems

[WWCY]

Homework Statement

I have an issue with understanding the idea of generalised momentum for the Lagrangian.

For a central force problem, the Lagrangian is given by,
$$L = \frac{1}{2}m(\dot{r} ^2 + p^2 \dot{\phi ^2}) - U(r)$$
with $r$ being radial distance.

The angular momentum is then,
$$P_{\phi} = \frac{\partial L}{\partial \dot{\phi}} = mr^2\dot{\phi}$$

Without the Lagrangian, I would have started with,
$$P_{\phi} = \vec{r} \times \vec{P}$$
$$|P_{\phi}| = rP\sin (\theta)$$
$$|P_{\phi}| = rmv \sin (\theta)$$
with $\theta$ being the angle between the two vectors, which I think can be reduced with the equation
$$v = r\dot{\phi}$$
and angular momentum is,
$$|P_{\phi}| = mr^2 \dot{\phi} \sin (\theta)$$
which varies significantly.

Could someone kindly
a) Explain the errors in my assumptions
b) Explain how I should have altered my "Newtonian" working in order to derive the same result as the Lagrangian method?

Thanks!

Homework Equations

The Attempt at a Solution

 

[mjc123]

v = rφ' is true only when the v and r vectors are perpendicular, as in circular motion. More generally, rφ' = v sinθ.

 

[WWCY]

v = rφ' is true only when the v and r vectors are perpendicular, as in circular motion. More generally, rφ' = v sinθ.

Hi, thanks for the response.

But this means that the particle in the Lagrangian problem has to be moving in a circular motion, which needn't be the case, no?

 

[mjc123]

No, the Lagrangian is general (as shown by the term in r'², which is 0 for circular motion.)

 

[WWCY]

No, the Lagrangian is general (as shown by the term in r'², which is 0 for circular motion.)

Do you mind explaining why $v\sin \theta = r \dot{\phi}$?

Thank you.

 

[mjc123]

Hope this helps

 

[WWCY]

View attachment 215433
Hope this helps

Yes it does, thanks!

 

[Dick]

View attachment 215433
Hope this helps

If you want to compare those two ways of calculating the momentum you need to express $\vec{r} \times \vec{P}$ completely in terms of $r$ and $\phi$ and their derivatives. Putting in the extra angle $\theta$ makes it hard to compare them.