- #1

cwill53

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- TL;DR Summary
- Prove that ##\nabla_1U(\mathbf{r}_1- \mathbf{r}_2 )=-\nabla_2U(\mathbf{r}_1- \mathbf{r}_2 )##

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?**

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?**