# Verifying the relation $\nabla_1U(\mathbf{r}_1- \mathbf{r}_2 )=-\nabla_2U(\mathbf{r}_1- \mathbf{r}_2 )$ for a scalar function $U$

In physics there is a notation $$\nabla_i U$$ to refer to the gradient of the scalar function $$U$$ with respect to the coordinates of the $$i$$-th particle, or whatever the case may be.

A question asks me to prove that

$$\nabla_1U(\mathbf{r}_1- \mathbf{r}_2 )=-\nabla_2U(\mathbf{r}_1- \mathbf{r}_2 )$$

How do you actually extend this idea to $$\mathbb{R}^n$$ though? I'm not sure if everything I have written below makes sense or is correct, I'm hoping to get some clarification. The weakness is in my mathematics.

Suppose I have $$X$$ be an open set in $$(\mathbb{R}^n )^m$$ and $$f:X\rightarrow \mathbb{R}$$ a scalar valued function; and let $$A=\left \{ \mathbf{r}_1, \mathbf{r}_2, ..., \mathbf{r}_m \right \}\subseteq X$$.

Then

$$\nabla f= \begin{bmatrix} \frac{\partial f}{\partial x _{1_{1}}}& \frac{\partial f}{\partial x _{2_{1}}} &\cdots & \frac{\partial f}{\partial x _{n_{m}}} \end{bmatrix}^T \in (\mathbb{R}^n )^m$$

$$\nabla_k f= \begin{bmatrix} \frac{\partial f}{\partial x _{1_{k}}}& \frac{\partial f}{\partial x _{2_{k}}} &\cdots & \frac{\partial f}{\partial x _{n_{k}}} \end{bmatrix}^T \in \mathbb{R}^n$$

where the subscript $$k$$ denotes that the derivatives are taken with respect to the $$n$$ coordinates of the $$k$$-th element of the indexed set $$A=\left \{ \mathbf{r}_1, \mathbf{r}_2, ..., \mathbf{r}_m \right \}\subseteq X$$. Let $$\mathbf{r}_i, \mathbf{r}_j \in A$$, and let

$$\mathbf{r}_i= \begin{bmatrix} x_{1_{i}} & x_{2_{i}} & \cdots & x_{n_{i}} \end{bmatrix}$$ $$\mathbf{r}_j= \begin{bmatrix} x_{1_{j}} & x_{2_{j}} & \cdots & x_{n_{j}} \end{bmatrix}$$

and let $$\mathbf{x}= \mathbf{r}_i- \mathbf{r}_j$$.

My questions are:

1. Is it true that $$\nabla_k f \in \mathbb{R}^n$$ ?
2. Given the above, how should $$\nabla _i f(\mathbf{x})$$ be written? Can someone give explicit steps for the chain rule manipulations that need to occur?
• Rotate your variables' axes by π/4 to ${\mathbf r}_1 -{\mathbf r}_2= {\mathbf r}$ and ${\mathbf r}_1 +{\mathbf r}_2=2 {\mathbf R}$. Observe $\nabla_1+\nabla_2=\nabla_R$. Commented Feb 13, 2023 at 18:05
• @CosmasZachos which first definition? Commented Feb 14, 2023 at 4:22
• The one following “Then”. Commented Feb 14, 2023 at 9:51
• To give another physical interpretation: If 1 and 2 denote the positions of two particles of the same mass, then Cosmas’s R-vector is the COM. The identity then indicates that the potential energy is indifferent to the COM. Commented Feb 14, 2023 at 23:27

The credit isn't mine in this answer, but I want to post it for completeness:

You can be extra explicit. Let $$g:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n$$ be the function $$g(\mathbf{x}_1, \mathbf{x_2}) = \mathbf{x}_1-\mathbf{x}_2$$. Then you are computing the derivatives of $$U\circ g$$.

By the chain rule, $$d(U\circ g)_{(\mathbf{x_1}, \mathbf{x_2})} = dU_{g(\mathbf{x_1}, \mathbf{x_2})} dg_{(\mathbf{x_1}, \mathbf{x_2})}$$

In index notation,

$$\partial_{x_1^i}(U\circ g)(\mathbf{x}_1, \mathbf{x}_2) = \frac{\partial U}{\partial x^k}\frac{\partial g}{\partial x_1^i}(\mathbf{x}_1, \mathbf{x}_2)$$

Finally, we have the fact that

$$\frac{\partial g}{\partial x^i_1}(\mathbf{x_1}, \mathbf{x}_2) = 1,$$ and $$\frac{\partial g}{\partial x_2^i}(\mathbf{x}_1, \mathbf{x}_2) = -1$$

I should also clarify, by $$dF$$ i mean the matrix of derivatives

$$dF^i_j = \frac{\partial F^i}{\partial x^j}$$