Question regarding Energy Interaction of two particles https://imgur.com/s6RGUKb
To give a context as to what I'm asking here ,I am talking about the energy of a two particle system (section 4.9 Taylor's Classical Mechanics) .
My question is what does $\nabla_1 U$=$\frac{\partial U}{\partial x_1}\hat{i}$+$\frac{\partial U}{\partial y_1}\hat{j}$+$\frac{\partial U}{\partial z_1}\hat{k}$ mean ?
What does the $x_1$,$y_1$ & $z_1$ mean here ?
If we are talking through the lens of basic principles ,what does it mean ?
I am not convinced that $gradient$ at $\vec{r_1}$ is $\nabla_1U$ ,I mean why can't we substitute thepoint (or) vector $r_1$ to the expression $\nabla U$?
If my mass is at $x_1$ then is $\partial{x_1}$ a small change in $x_1$ direction ? But isn't that same as a small change in x direction ?
 A: First, excuse my English, I used google translator.
Since the forces are conservative: $\nabla_1\times\vec{f_{12}}=0$ and  $\nabla_2\times\vec{f_{21}}=0$. I can set the potentials $U_1$,$U_2$, such that:
$$\vec{\nabla_1}U_{1}(\vec{\mathfrak{r}})=\vec{f_{12}}~~~and~~~\vec{\nabla_2}U_{2}(\vec{\mathfrak{r}})=\vec{f_{21}}$$
$$\vec{\mathfrak{r}}=\vec{r_1}-\vec{r_2}$$
If you have trouble accepting this, you can use the Helmholtz equation to get an idea. Since vector fields in general satisfy the equation:
$$\vec{f}=-\vec{\nabla}{U}+\vec{\nabla}\times\vec{H}$$
you can check that $\vec{\nabla}\times\vec{f}=0\leftrightarrow\vec{\nabla}\times\vec{H}=0$. Which makes it possible to define a potential function and get $\vec{f}$ as the negative of his gradient. The nabla indices are just because it occurs in the respective coordinates.
As for the meaning of $x_1,y_1,x_2,y_2$, they are the coordinates of each mass in relation to an origin. So that:
$$\vec{\mathfrak{r}}=\vec{r_1}-\vec{r_2}=(x_1,y_1)-(x_2,y_2)=(x_1-x_2,y_1-y_2)=(\mathfrak{r}_x,\mathfrak{r}_y)$$
If you go back a little in the text of the book:
$\vec{f_{12}}=-f~\mathfrak{\hat{r}}=-f~\mathfrak{\frac{\vec{r_1}-\vec{r_2}}{|\vec{r_1}-\vec{r_2}|}}$ and by the third law $\vec{f_{12}}=-\vec{f_{21}}$
To check $\vec{\nabla}_1U=\vec{\nabla}U$ just use the chain rule:
$$-\vec{\nabla}_1U(\mathfrak{r})=-\bigg(\frac{\partial{U}}{\partial{x_1}}\hat{x}+\frac{\partial{U}}{\partial{y_1}}\hat{y}\bigg)=-\bigg(\frac{\partial{U}}{\partial{\mathfrak{r}_x}}\frac{\partial{\mathfrak{r}_x}}{\partial{x_1}}\hat{x}+\frac{\partial{U}}{\partial{\mathfrak{r}_y}}\frac{\partial{\mathfrak{r}_y}}{\partial{y_1}}\hat{y}\bigg)$$
replace $\mathfrak{r}_x$ and $\mathfrak{r}_y$ to get
$$-\vec{\nabla}_1U(\mathfrak{r})=-\bigg(\frac{\partial{U(\mathfrak{r}_x,\mathfrak{r}_y)}}{\partial{\mathfrak{r}_x}}\hat{x}+\frac{\partial{U(\mathfrak{r}_x,\mathfrak{r}_y)}}{\partial{\mathfrak{r}_y}}\hat{y}\bigg)=-\vec{\nabla}U(\mathfrak{r})$$
So: $$\vec{\nabla}_1U=\vec{\nabla}U$$
if you do this for particle 2 you will see that a minus sign will appear confirming the third law. This is due to the difference in positions.
If you change the coordinates $x_1,y_1$ you are just changing the position $\vec{r_1}$ of the masses in relation to the origin.The interaction of particles depends on the separation vector $\vec{\mathfrak{r}}$ between them.
