Gradient With Respect to a Set of Coordinates

[cwill53]

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?**

 

[anuttarasammyak]

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

A question asks me to prove that

In a simple one-dimension case of harmonic oscillator
$$U(x_1 - x_2) =\frac{k}{2} ( x_1 - x_2)^2$$
$$\frac{\partial U}{\partial x_1}=k(x_1-x_2)=-\frac{\partial U}{\partial x_2}$$
Forces from the spring are same but have opposite signs at the ends.

Do you want to extend it from 1D to 3D or more higher dimensions, or from 2 to more ends or particles, or both ?

 

[PeroK]

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

I'm not sure I follow what you are doing there. If we have two particles in 3D, then we can represent the position of each particle as:$$\mathbf{r_1} = (x_1, y_1, z_1), \ \mathbf{r_2} = (x_2, y_2, z_2)$$Technically, the potential is a function of six coordinates:$$V:\mathbb R^6 \rightarrow \mathbb R$$Where the first three coordinates represent the position of first particle and the second three coordinates represent the second particle.

Assuming, however, that the potential is a function of the difference in position between the particles, then we have a "reduced" potential function:$$U: \mathbb R^3 \rightarrow \mathbb R$$Where the argument here is the difference in position of the two particles.

I've used two different letters here to emphasise the different functions involved. Technically, we have:
$$V(\mathbf{r_1}, \mathbf{r_2}) \equiv V(x_1, y_1, z_1, x_2, y_2, z_2) = U(\mathbf{r_1} - \mathbf{r_2}) \equiv U(x_1 - x_2, y_1 - y_2, z_1 - z_2)$$Now, we have two gradient functions defined here:
$$\mathbf{\nabla_1}V = (\frac{\partial V}{\partial x_1}, \frac{\partial V}{\partial y_1}, \frac{\partial V}{\partial z_1})$$$$\mathbf{\nabla_2}V = (\frac{\partial V}{\partial x_2}, \frac{\partial V}{\partial y_2}, \frac{\partial V}{\partial z_2})$$And, now we can see what is meant by these two gradients acting on the potential function $U(\mathbf{r_1} - \mathbf{r_2})$:
$$\frac{\partial V(x_1, y_1, z_1,x_2, y_2, z_2)}{\partial x_1} = \frac{\partial U(x_1-x_2, y_1-y_2, z_1 - z_2)}{\partial x_1} = \frac{\partial U(x_1 - x_2, y_1 - y_2, z_1-z_2)}{\partial x}$$Where $\frac{\partial U}{\partial x}$ is just the usual partial derivative of $U$ with respect to its first argument (which we've called $x$). And, of course, we have the equivalent for $y_1$ and $z_1$.

And, to take the partial derivate wrt the second coordinates, we need to use the chain rule:
$$\frac{\partial V(x_1, y_1, z_1,x_2, y_2, z_2)}{\partial x_2} = \frac{\partial U(x_1-x_2, y_1-y_2, z_1 - z_2)}{\partial x_2} = -\frac{\partial U(x_1 - x_2, y_1 - y_2, z_1-z_2)}{\partial x}$$This leads us to the conclusion that:
$$\mathbf{\nabla_1}V = -\mathbf{\nabla_2}V$$In the special case where the potential can be written as a function $U$ of the difference in positions.

Finally, we can adopt the notation:$$\mathbf{\nabla_1}U(\mathbf{r_1} - \mathbf{r_2}) \equiv \mathbf{\nabla_1}V(\mathbf{r_1}, \mathbf{r_2})$$$$\mathbf{\nabla_2}U(\mathbf{r_1} - \mathbf{r_2}) \equiv \mathbf{\nabla_2}V(\mathbf{r_1}, \mathbf{r_2})$$To get the required result.

Note that a physicist would assume implicitly almost all of this supporting mathematical complexity. But, if you want to justify this fully, then this is what I've shown.

Finally, if you want to extend this to more dimensions, then the same steps apply at every stage, but the vector notation would get more complicated. Instead of $x_1, y_1, z_1$, we would need something like $\mathbf{r_1} = (x_{11}, x_{12} \dots x_{1n})$ and $\mathbf{r_2} = (x_{21}, x_{22} \dots x_{2n})$ for the $n$ coordinates of the two particles.

 

[PeroK]

PS the simpler approach is just to assume/state that:
$$\frac{\partial U}{\partial x_1} = -\frac{\partial U}{\partial x_2}$$and hence$$\mathbf{\nabla_1}U = -\mathbf{\nabla_2}U$$Where $U(\mathbf{r_1} - \mathbf{r_2})$ is the potential function. Whether that constitutes a "proof" is a moot point.