Suppose we have a two-particle system with particle 1 and particle 2 that obeys Newton's laws. Further assume that the force on particle 1 due to particle 2, $F_{1, 2}$, is conservative. I know then that for some potential energy function $V \left(\bf{r}_1 - \bf{r}_2\right)$, \begin{equation} F_{1, 2} = -\nabla_1 V \end{equation} where \begin{equation} \nabla_1 \equiv \hat{\bf{x}}\frac{\partial}{\partial x_1} + \hat{\bf{y}}\frac{\partial}{\partial y_1} + \hat{\bf{z}}\frac{\partial}{\partial z_1}. \end{equation} How then does one show that \begin{equation} F_{2, 1} = -\nabla_2 V \; ? \end{equation}
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2 Answers
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@Zarathustra gives a physical answer as to why this holds. However, there's also a mathematical answer. The potential is given as a function, not of $\mathbf{r}_1$ and $\mathbf{r}_2$ separately, but of their difference $\mathbf{r} := \mathbf{r}_1 - \mathbf{r}_2$:
$$V(\mathbf{r}_1 - \mathbf{r}_2) = V(\mathbf{r})$$
Equivalently, given the function $V$ in terms of the difference vector going between the two, you can consider the above expression as representing a function in terms of each vector individually:
$$V_\mathrm{two\ point}(\mathbf{r}_1, \mathbf{r}_2) := V(\mathbf{r}_1 - \mathbf{r}_2)$$
The rules of calculus, in particular, the chain rule, will tell you that the derivative (gradient or partials) of $V_\mathrm{two\ point}$ with respect to $\mathbf{r}_2$ has to involve multiplication of the derivative of the function-of-single-vector $V$ by the derivative of the internal differencing function $(\mathbf{r}_1, \mathbf{r}_2) \mapsto \mathbf{r}_1 - \mathbf{r}_2$ with respect to $\mathbf{r}_2$. And as you probably know, the derivative of an expression like $a - x$ with respect to $x$ is $-1$. Thus there will be a multiplication by -1 on taking the gradient with respect to $\mathbf{r}_2$.
Or equivalently, you could say, your expression for $V$ effectively presumes Newton's third law.
answered Nov 28, 2018 at 6:58
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The potential Energy is defined as a the function whose gradient is equal and opposite to the force on an object, i.e. for two objects
$$F_{1} = - \nabla_{1} V(r_1, r_2) \qquad F_{2} = - \nabla_{2} V(r_1, r_2)\ .$$
If the System obeys newton's third law we know that $F_{2} = -F_{1}$. We can introduce this constraint into our potential simply by making it a function of the difference $r_1 - r_2$. This is because
$$ \nabla_{1} V(r_1 - r_2) = - \nabla_{2} V(r_1 - r_2) \ .$$
So the fact that the Potential can not only we written as a function of $r_1$ and $r_2$ separately but as a function of $r_1 - r_2$ is just a consequence of newtons third law.
