1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.