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 $V(x) = V(\Delta^i_{ab}(x))$.

Now, using the chain rule, we write:

\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 $\frac{d}{dt} p = -\frac{d}{dt} p = 0$

  • 1
    $\begingroup$ 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. $\endgroup$ – King Nerd the Third Mar 30 at 0:07

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