# D Alembert's Principle: Dependence of kinetic energy on generalized coordinates.

1. Jul 2, 2012

### puneeth9b

Hey!
I was reading Goldestein's book on classical mechanics and I came across this (Page 20 3rd Edition):

"Note that in a system of Cartesian coordinates the partial derivative of T with
respect to qj vanishes. Thus, speaking in the language of differential geometry,
this term arises from the curvature of the coordinates qj. In polar coordinates,
e.g., it is in the partial derivative of T with respect to an angle coordinate that the
centripetal acceleration teml appears."

Here T=Kinetic energy of the system
qj= the jth generalized coordinate.

I don't exactly understand how this works.
1.Why isn't it (dT/dq) zero in polar coordinates if it is zero in cartesian coordinates?
2.What if velocity was a function of coordinates? dT/dq can't possibly be zero even in cartesian coordinates then right?

2. Sep 4, 2012

### merickson45

I have the exact same question, no luck yet...

3. Sep 4, 2012

### vela

Staff Emeritus
I'm not sure why you'd expect $\partial T/\partial q_i$ to be zero in polar coordinates. The kinetic energy in polar coordinates is given by
$$T=\frac{1}{2}m\dot{r}^2+ \frac{1}{2}mr^2\dot{\theta}^2.$$ Clearly, $\partial T/\partial r = mr\dot{\theta}^2$ is not zero in general. Goldstein should have said it was the partial derivative with respect to the radial coordinate that gives rise to the centripetal acceleration term.

Regarding your second question, remember that the coordinates $q_i$ and velocities $\dot{q_i}$ are considered independent variables.