I have some confusion on potential energy of a two particle system. I'm using Section 4.9 from 'Classical Mechanics' by John R. Taylor as reference.
Assume two particles are at location $\vec{r_1}$ and $\vec{r_2}$. My first question is that the section claims that the force (and potential energy) only depends on $\vec{r_1} - \vec{r_2}$ because the two particle interaction should be translationally invariant. Why not depending on $\lvert \vec{r_1} - \vec{r_2}\rvert$? Surely if you rotate your view point, the force and potential energy should be the same?
The second question is I don't know how to interpret the potential energy $U(\vec{r_2} - \vec{r_1})$ as function of $\vec{r_2} - \vec{r_1}$. For single particle's potential energy, $U(\vec{r})$ is defined to be the negation of the work from a reference point $r_0$, that is $U(\vec{r}) = -\oint_{r_0}^r \vec{F} \cdot d\vec{r}$. How to interpret the two particle potential energy in the same way? Where is the reference point? How is the integral defined? Or do we just say that work integral definition is not available in a two particle system, so that the potential energy is defined to be the function such that $\nabla_{r_1}U(\vec{r_2} - \vec{r_1}) = F_{12}$?
Finally I would like some clarity on the $\nabla$ operator with subscript. In the book it is defined to be
$$ \nabla_{r_1} = \frac{\partial}{\partial x_1} \hat{\vec{x}} + \frac{\partial}{\partial y_1} \hat{\vec{y}} + \frac{\partial}{\partial z_1} \hat{\vec{z}} $$
That operator seems should apply to a scalar function $U(\vec{r_1})$ instead of $U(\vec{r_1} - \vec{r_2})$. Maybe $U(\vec{r_1} - \vec{r_2})$ is a composite function of $U(\vec{r_1}, \vec{r_2})$ and $d(\vec{r_1}, \vec{r_2}) = \vec{r_1} - \vec{r_2}$? Some more mathematical precision would be appreciated here.
Thanks in advance!