Two masses $m_1$ and $m_2$ interact via a potential energy $U = U(|x_1-x_2|)$. Suppose that the Equations of motion are given by: $$m_1x_1'' = - \frac{\partial U}{\partial x_1},\quad m_2x_2'' = - \frac{\partial U}{\partial x_2}$$ Let $x=x_1-x_2$ and $m=\frac{m_1m_2}{m_1+m_2}$. Show that: $$mx'' = - \frac{\partial U}{\partial x}(|x|)$$
My attempt:
We have: $$m_2m_1x_1'' = - m_2\frac{\partial U}{\partial x_1},\quad m_1m_2x_2'' = - m_1\frac{\partial U}{\partial x_2}$$
$$\Rightarrow m_1m_2(x_1''-x_2'') = - m_2\frac{\partial U}{\partial x_1} + m_1\frac{\partial U}{\partial x_2} $$
From the third Newton law: $$m_1m_2(x_1''-x_2'') = - (m_1+m_2)\frac{\partial U}{\partial x_1}$$
I couldn't prove that $\frac{\partial U}{\partial x}(|x|) = \frac{\partial U}{\partial x_1}$. How could I do it?
Thanks in advance
Temporarily, assume $x_1 > x_2$.
Since $U(x_1,x_2) = U(x_1-x_2)$, $$ \frac{\partial U}{\partial x_2} = - \frac{\partial U}{\partial x_1} $$ This is the concept you missed above.
Then when you do your two steps of algebra, you get $$ m_1m_2(x_1''-x_2'') = -m_2\frac{\partial U}{\partial x_1} -m_1\frac{\partial U}{\partial x_1} $$ Next, notice that keeping $x_2$ fixed, and $x=x_1-x_2$, $$ \frac{\partial U}{\partial x_1}= \frac{\partial U}{\partial x} $$ Plugging that in, and noting that $x_1''-x_2'' = x''$ you get $$ x'' = -\frac{m_1m_2}{m_1+m_2} \frac{\partial U}{\partial x} $$
If you then look at the case of $x_2 > x_1$ the same form of reasoning gives the same equation. Combining those two facts gives the desired result (although when $x_1=x_2$ you have to deal with that case carefully).