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I need help to understand the following passage from Goldstein's book "Classical Mechanics. 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 $T$?

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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. $\endgroup$ – aaaaaa 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 $\endgroup$ – aaaaaa 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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