Conservative Force: Translational Invariance I have a question about the following.
Why if there are two masses, $m_1$ and $m_2$ respectively, and the only force acting on them is from their mutual interaction which is conservative and central, the following is true?
$U(\vec{r_1},\vec{r_2})=U(\vec{r_1}-\vec{r_2})$
(It says this is from Translational invariance.)
Also, it says since it is central, meaning it is along the line joining them, why is it
$U(\vec{r_1}-\vec{r_2})=U(\lvert\vec{r_1}-\vec{r_2}\rvert)$?
 A: Imagine the situation you have described. Let the potential be denoted as $U(\vec{r_1},\vec{r_2})$. Note that this picture shouldn't change if I decide to observe what they do in one place or the other. Therefore, the potential should be invariant if I decide to displace the particles in the same direction by the same amount. This means the potential should satisfy the following behaviour:
$U(\vec{r_1} +\vec{a},\vec{r_2} +\vec{a}) = U(\vec{r_1},\vec{r_2})$
for any arbitrary vector $\vec{a}$. This means that the potential should be a function of $\vec{r_1} - \vec{r_{2}}$ as this is invariant under displacements. Physically, this corresponds to the fact that the relevant information for the equations of motion is encoded in the distance between the two particles. If that wasn't true, the motion would depend on the position of my experiment.
When a force is central, this means that there is no "angular dependence". The systems should look the same if I rotate them by equal amount. Therefore, in the potential you shouldn't have any terms depending on the angle between the two particles. This along with our previous result imply that the potential $U$ should be a function of $|\vec{r_1} - \vec{r_{2}}|$ as this has no relative angular dependence between the particles.
Hope this helps.
A: Here is a slightly artificial but mathematical proof. I will look at scalars, not vectors, to make the derivation slightly easier. But it should be easily extendable to vectors. Define
\begin{align}
r&=r_2-r_1&\iff &&r_1&=\tfrac 12(R-r)\\
R&=r_2+r_1&&&r_2&=\tfrac 12(R+r)\tag{1}
\end{align}
These quantities have the same information as $(r_1,r_2)$. If you know $(r_1,\vec r_2)$ you can uniquely determine $(r,R)$ and vice versa. So without loss of generality we can define $U$ in terms of $r,R$
$$U=U(r_1(r,R),r_2(r,R))=U(\tfrac 12(R-r),\tfrac 1 2(R+r))\tag 2$$
Translation invariance is defined by
$$U(r_1+a,r_2+a)=U(r_1,r_2)\quad \text{for all }a.\tag 3$$
Let us now show that $U$ does not depend on $R$ by calculating $\frac{\partial U}{\partial R}$ and seeing that it is zero.
\begin{align}
\frac{\partial U}{\partial R}&=\lim_{h\rightarrow 0}\frac{U\rvert_{R+h}-U\rvert_R}{h}\\
&=\lim_{\color{red}h\rightarrow 0}\frac{U(\tfrac 12(R+\color{red}h-r),\tfrac 1 2(R+\color{red}h+r))-U(\tfrac 12(R-r),\tfrac 1 2(R+r))}{h}\\
&=\lim_{\color{red}h\rightarrow 0}\frac{U(\tfrac 12(R-r),\tfrac 1 2(R+r))-U(\tfrac 12(R-r),\tfrac 1 2(R+r))}{h}&\text{(trans. inv.)}\\
&=0
\end{align}
A more direct proof would be to write $U$ in these new coordinates: $U(r,R)$. If we translate by $a$ we get $r\rightarrow r$ and $R\rightarrow R+2a$. So we should have
$$U(r,R+2a)=U(r,R)\quad\forall a$$
Where $\forall$ means 'for all'. If we focus on the $R$ dependence we can also read this as $f(x+a)=f(x)\ \forall a$. From this we can conclude that $f$ is a constant and likewise $U$ is constant if we vary only $R$.
This last proof we can easily reuse for rotational invariance by writing the potential as $U(r,\theta,\phi)$ and noting that rotational invariance means that
\begin{align}U(r,\theta+\Delta \theta,\phi)&=U(r,\theta,\phi)\ \forall \Delta\theta\\
U(r,\theta,\phi+\Delta \phi)&=U(r,\theta,\phi)\ \forall \Delta\phi
\end{align}
