It is written in the Goldstein's Classical Mechanics text that $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}=\sum_k \frac{\partial^2{r_i}}{\partial {q_j}\partial{q_k}}\dot q_k+\frac{\partial^2{r_i}}{\partial {q_j}\partial t},\tag{1.50b}$$ where $$\dot r_i=\frac{\mathrm d}{\mathrm dt}r_i=\sum_k\frac{\partial{}r_i}{\partial{q_k}}\dot q_k+\frac{\partial {r_i}}{\partial t}.\tag{1.46}$$ But it seems to me that there is another term in $\frac{\partial {\dot r_i}} {\partial {q_j}}$ because of product rule which is $$\sum_k \frac{\partial{r_i}}{\partial{q_k}}\frac{\partial{\dot q_k}}{\partial{q_j}},$$ which I think is equal to $$\frac{\partial{r_i}}{\partial{q_j}}\frac{\partial{\dot q_j}}{\partial{q_j}}$$ since $q_j$'s are independent among themselves.

Then how come $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}~?\tag{1.50b}$$ Does $$\frac{\partial{\dot q_j}}{\partial{q_j}} = 0\ ?$$


In the Lagrangian formalism position and velocity are considered as independent variables, so indeed $\frac{\partial \dot{q}_j}{\partial q_j} = 0$. See Calculus of variations -- how does it make sense to vary the position and the velocity independently?

  • $\begingroup$ But why is $\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}$ ? Is it simply an interchange of derivatives? $\endgroup$ – Anjan May 4 '19 at 9:38
  • $\begingroup$ Yes, you can exchange those derivatives. Consider $f(g(x,t),t)$. Then $\dot{f} = \frac{\partial f}{\partial g} \dot{g} + \frac{\partial f}{\partial t}$ and $\frac{\partial \dot{f}}{\partial g} = \frac{\partial^2 f}{\partial g^2} \dot{g} + \frac{\partial^2 f}{\partial g \partial t} = \frac{d}{dt} \frac{\partial f}{\partial g}$, where we used $\frac{\partial \dot{g}}{\partial g} = 0$. $\endgroup$ – jkb1603 May 14 '19 at 19:04

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