Why $\frac{d}{dt}r_{a}\nabla_{a}U_{ab}+\frac{d}{dt}r_{b}\nabla_{b}U_{ba}=\frac{d}{dt}U_{ab}?$

Question

In classical mechanics for two mass particles $a$,$b$ we assume the symmetric potential arising from $F_{ab}$ and $F_{ab}$ given by $$U_{ab}(r)=-\int^{r}_{r_{0}}F_{ab}(r')dr'$$ and $$U_{ba}(r)=-\int^{r}_{r_{0}}F_{ba}(r')dr'$$

The book mechanics by Florian Scheck gives $$\frac{d}{dt}r_{a}\nabla_{a}U_{ab}+\frac{d}{dt}r_{b}\nabla_{b}U_{ba}=\frac{d}{dt}U_{ab}$$ because $$[\frac{d}{dt}r_{a}\nabla_{a}+\frac{d}{dt}r_{b}\nabla_{b}]U_{ab}=\frac{d}{dt}U_{ab}$$

I am confused how we get the summation form. My derivation goes as follows: Notice $F_{ab}=-\nabla_{b} U_{ab}$. Thus we should have $$\frac{d}{dt}U_{ab}=\frac{d}{dr}*\frac{dr}{dt}U_{ab}=\frac{dr}{dt}\frac{d}{dr}U_{ab}=\frac{dr}{dt}[-F_{ab}]=\frac{d}{dt}[r_{a}-r_{b}][-F_{ab}]=[\frac{d}{dt}r_{a}\nabla_{a}+\frac{d}{dt}r_{b}\nabla_{b}]U_{ab}$$

My simple question is just whether my derviation is correct, for I assume $r=r_{a}-r_{b}$ at here.

Answer 1

Yes, the $r$-argument really is $r_{ik}:=|r_i-r_k|$, as he writes two pages earlier at the beginning of "Systeme von endlich vielen Teilchen". But then you don't need the force to show the relation, it's just the chain rule, which makes derivatives of $U$ into a two term expression and notice that $U(r_{ik})=U(|r_i-r_k|)=U(|r_k-r_i|)=U(r_{ki})$.

Also, it's not so good, that you write "$\frac{d}{dt}r_{a}\nabla_{a}U_{ab}$" for "$\frac{dr_{a}}{dt}\nabla_{a}U_{ab}$", because it suggests that you mean "$\frac{d}{dt}(r_{a}\nabla_{a}U_{ab})$". Moreover, the books name is not the autors name. And I would change the title to something readable, and by that I don't mean the problem with the total derivative, but a title which is a sentence, not a formula. E.g. "A problem deriving the energy conservation for radial two particle potentials".