# Generalized Coordinates Property for a System of Particles

I"m looking at "Principles of Dynamics: Second Edition" by Donald T Greenwood. I'm trying to figure out how he obtains Eq. (6-64)

$$\frac{\partial\dot x_j}{\partial\dot q_i} = \frac{\partial x_j}{\partial q_i}\tag{6-64}$$

from Eq (6-54)

$$\dot x_j = \sum_{i=1}^n \frac{\partial x_j}{\partial q_i}\dot q_i + \frac{\partial x_j}{\partial t}.\tag{6-54}$$

where the transformation equations from a set of $$3N$$ Cartesian coordinates to a set of $$n$$ generalized coordinates are of the form given by Eq (6-1)

$$x_1=f_1(q_1,q_2,...,q_n,t)$$ $$x_2=f_2(q_1,q_2,...,q_n,t)$$ $$\vdots$$ $$x_{3N}=f_{3N}(q_1,q_2,...,q_n,t).\tag{6-1}$$

When I differentiate Eq (6-54) with respect to $$q_i$$ I get second derivatives and I have no idea how the term $$\frac{\partial^2 x_j}{\partial t\partial \dot q_i}$$ is dealt with. Any insight appreciated.

Greenwood is asking how does $$\dot x_j$$ change if we vary $$\dot q_i$$ at some specified time $$t$$ and point $$q_i$$ (or equivalantly a specified point $$x_i$$ as there is a - perhaps time dependent - 1-1 relation between the $$q$$'s and $$x$$'s). At that specified point and time all the quantities on the RHS of your second equation, with the exception of the $$\dot q$$'s are to be treated as constants. The answer then, is exactly your first equation.
• Ahhhhh, I think I see now. Just to clarify, $\frac{\partial\dot q_i}{\partial \dot q_j}=0$, because $q_i$ is independent of $q_j$ for $i \ne j$? – eball Jun 14 '19 at 17:14
• Yes. You can vary the $\dot \q_i$'s independently of one another. – mike stone Jun 14 '19 at 21:40