# Interpreting generalised momenta

• Jan 15th 2012, 06:44 AM
supaman5
Interpreting generalised momenta
In my notes we are using polar coordinates, so r=(xcosy,xsiny) implying r(dot)^2=x(dot)^2 + (x^2)(y^2)

so L= 0.5m(x(dot)) + (x^2)(y^2)) - U(x,y) (we aren't given exactly what the potential is)

I understand that the generalised momentum of x= mx(dot)
.................................................. ...........of y= m(x^2)y(dot)

but then it says the gen. mom. of x can be interpreted as the radial component of the linear momentum, and the gen. mom. of y is angular momentum. I understand that the angular momentum can be checked using the definition, but is there any way you could notice this without checking? (i.e. is it something about the formulas for gen. mom. of y that means it is the angular momentum)?

Sorry for the essay
• Jan 15th 2012, 08:01 AM
ILikeSerena
Re: Interpreting generalised momenta
Welcome to MHF, supaman5! :)

There are a couple of typos in your lagrangian, but you did get the proper generalized momentums.

The generalized momentum for a generalized coordinate q is defined as $p_q = {\partial \mathcal{L} \over \partial {\dot q}}$.

If ${\partial \mathcal{L} \over \partial q} = 0$ is zero, the langrangian equation tells us that the generalized momentum is conserved (its time derivative is zero).
This is often the case for angular coordinates, meaning both the kinetic energy and the potential energy often do not depend on the angle.

For a coordinate that is angular, the generalized momentum corresponds exactly to the angular momentum.
• Jan 15th 2012, 09:01 AM
supaman5
Re: Interpreting generalised momenta
Cheers, angular coordinate implies angular momentum... seems so obvious now.