I need help to understand the following passage from Goldstein's book "Classical Mechanics", in which he writes that:
Consider first a generalized coordinate $q_j$ for which a change $dq_j$ represents the translation of the system as a whole in some given direction. [...] Then, clearly $q_j$ cannot appear in the kinetic energy of the system $T$, for velocities are not affected by a shift in the origin, therefore the partial derivative of $T$ with respect to $q_j$ must be zero.
This passage makes intuitive sense to me, but I am not seeing why this is true mathematically. If $r_i$ denote the position vectors, then the kinetic energy is given by
$$ T = \frac {1}{2} \sum_{k=1}^N m_k \dot{\mathbf{r}}_k \cdot \dot{\mathbf{r}}_k\,$$
where
$$\dot{\mathbf{r}}_k \cdot \dot{\mathbf{r}}_k = \sum_{i,j=1}^n \left(\frac{\partial \mathbf{r}_k}{\partial q_i}\cdot\frac{\partial \mathbf{r}_k}{\partial q_j}\right)\dot{q}_i\dot{q}_j + \sum_{i=1}^n \left(2\frac{\partial \mathbf{r}_k}{\partial q_i}\cdot\frac{\partial \mathbf{r}_k}{\partial t}\right) \dot{q}_i + \left(\frac{\partial \mathbf{r}_k}{\partial t}\cdot\frac{\partial \mathbf{r}_k}{\partial t}\right) \!$$
and since $\mathbf{r}_i$ is a function of the generalized coordinates $q_j$, it seems to me that that $q_j$ will appear in $T$ via the terms like $\frac {\partial \mathbf{r}_k}{\partial q_i}$ and $\frac {\partial \mathbf{r}_k}{\partial \dot q_i}$.
So, how come $q_j$ do not appear in $T$?
References:
- Goldstein, Classical Mechanics; Section 2.6, paragraph between eqs. (2.47) - (2.48).