# Show that momentum is conserved for a potential that depends on the difference in positions

I have the following problem:
Let $$N$$ particles interact according to
$$m_{a} \frac{d^{2}x^{i}_{a}}{dt^{2}} = - \frac{\partial V(x)}{\partial x^{i} _{a}}$$ With $$a = 1, \dots, N$$. Suppose $$V(x_{1}, \dots, x_{n})$$ depends only on the differences $$x^{i}_{a} - x^{i}_{b}$$, with $$a, b = 1, \dots, N$$. Show that the total momentum $$\sum_{a} m_{a} \frac{dx^{i}_{a}}{dt}$$ is conserved.

I know that I need to take the time derivative of $$p$$ and show that it equals zero: \begin{align*} \frac{d}{dt} p &= \frac{d}{dt} \sum_{a} m_{a} \frac{dx^{i}_{a}}{dt}\\ &= \sum_{a} \frac{d}{dt} (m_{a} \frac{dx^{i}_{a}}{dt})\\ &= \sum_{a} m_{a} \frac{d^{2}x^{i}_{a}}{dt^{2}} \\ &= \sum_{a} (-\frac{\partial V(x)}{\partial x^{i}_{a}}) \end{align*}

What I'm stuck on is how to use the information about the dependency of $$V(x)$$ on the difference $$x^{i}_{a} - x^{i}_{b}$$ to get this sum to be zero. I think I'm messing up a chain rule somewhere. I also think that I might be confused a little bit about the notation.

Let's define $$\Delta^i_{ab} = x^i_a - x^i_b$$, then $$V(x) = V(\Delta^i_{ab}(x))$$.
\begin{align*} \frac{d}{dt} p &=-\sum_{a}\frac{\partial V(x)}{\partial x^{i}_{a}}\\ &=-\sum_{a,b; \,a \neq b}\frac{\partial V(x)}{\partial \Delta^{i}_{ab}} \frac{\partial \Delta^{i}_{ab}}{\partial x^i_a} \\ &=-\sum_{a,b; \,a \neq b}\frac{\partial V(x)}{\partial \Delta^{i}_{ab}} \end{align*}
Now $$\Delta^{i}_{ab}$$ is antisymetric in $$a,b$$, hence $$\frac{d}{dt} p = -\frac{d}{dt} p = 0$$
• Thank you! I had $V(x) = V(f(x^{i}_{a} - x^{i}_{b}))$ but I didn't have $f$ written explicitly which I think is where I was stuck. – King Nerd the Third Mar 30 at 0:07