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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:

  1. Goldstein, Classical Mechanics; Section 2.6, paragraph between eqs. (2.47) - (2.48).
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  • $\begingroup$ $q_j$ don't appear in $T$, but $\partial q_j$ and $d q_j$ (in form of $\dot{q_j}$) do. $\endgroup$ – aaaaa says reinstate Monica Feb 23 '17 at 21:41
  • $\begingroup$ Why do they not appear? If $r_i = r_i(q_1,q_2,\dots,...)$, why not $\frac {\partial \mathbf{r}_k}{\partial q_i} = \frac {\partial \mathbf{r}_k}{\partial q_i} (q_1,q_2,...)$? $\endgroup$ – nan Feb 23 '17 at 21:46
  • $\begingroup$ "does not appear" in this context means that $\frac{\partial{T}}{\partial q_j}=0$, and I think you can see that's to be true since nowhere you see $q_j$ explicitly $\endgroup$ – aaaaa says reinstate Monica Feb 23 '17 at 22:12
  • $\begingroup$ @aaaa How do you know that $q_j$ would not appear explictily when you compute $\partial r_i \partial q_j$? If $f(x) = x$ and $g(x) = f(x) \dot x$, then of course I do not see explicilty $x$ in $g$, but it will be explicit if I write out what $g$ is $\endgroup$ – nan Feb 23 '17 at 22:37
  • $\begingroup$ Of course in general the kinetic energy may depend on $q_i$. It is $T=\sum_{ij} a(q)_{ij}\dot q_i\dot q_j$, where $a(q)_{ij}$ are functions of the generalized coordinates. However, Goldstein is considering the particular case where the change in the coordinates corresponds to a translation of the system. $\endgroup$ – Diracology Feb 24 '17 at 0:11
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$\frac {\partial \mathbf{r}_k}{\partial q_i}$ and $\frac {\partial \mathbf{r}_k}{\partial \dot q_i}$ are base vectors at $\mathbf{r}_k$. Hence the dot products vanish.

Thus kinetic energy does not contain any $q_j$ term.

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  • $\begingroup$ If $r_i = r_i(q_1,q_2,\dots,...)$, why not $\frac {\partial \mathbf{r}_k}{\partial q_i} = \frac {\partial \mathbf{r}_k}{\partial q_i} (q_1,q_2,...)$? $\endgroup$ – nan Feb 23 '17 at 21:46
  • $\begingroup$ Is it so? Or do you mean $${d {r_k}}= {\partial {r_k}\over\partial{q_i}}d {q_i}$$ with summation over the repeated index? $\endgroup$ – Sayontön Vöttacharjo Feb 23 '17 at 21:52
  • $\begingroup$ All I am saying is that since $r_i$ is a function of the variables $q_i$s, its partial derivative with respect to one of those variables is a function of the variables. $\endgroup$ – nan Feb 23 '17 at 21:55
  • $\begingroup$ Okey. Let's see. $${\vec{r}}=x\hat i + y\hat j + z\hat k=\vec{r}(x, y, z)$$ Then by your argument $${\partial{\vec {r}}\over\partial{x}}=\hat i$$ Is $\hat i$ a function of $(x, y, z)$? $\endgroup$ – Sayontön Vöttacharjo Feb 23 '17 at 22:00
  • $\begingroup$ I understand your example. But, if I have generalized coordinates $q_j$ and $r_i = r_i(q_1,q_2,...)$, how do I know that $q_j$ do not appear in a partial derivative of $r_i$? $\endgroup$ – nan Feb 23 '17 at 22:11
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OP is right that the kinetic term $T(q,\dot{q},t)$ in general depends on the generalized coordinates $q^j$. The quoted paragraph is indeed not one of Goldstein's best formulations. However note that the whole section 2.6 is devoted to symmetry and conservation laws. In the quoted paragraph Goldstein essentially wants to assume translational symmetry $$q^j ~\to~ q^j+ \epsilon^j$$ of the kinetic term $T(q,\dot{q},t)$, i.e. that $$\frac{\partial T}{\partial q^j}~=~0.$$

References:

  1. Goldstein, Classical Mechanics; Section 2.6, paragraph between eqs. (2.47) - (2.48).
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