How to show that interaction potential depends only on separation of particles in system with position translation symmetry? System
2 particles with mass moving in one spatial dimension $x$. Positions of particles are $x_1$ and $x_2$ respectively and they are only acted on by a conservative interaction force corresponding to potential energy $U(x_1,x_2)$
Question:
Show that symmetry under spatial translations $x\rightarrow x+a$ requires the potential energy to only depend on the difference in positions $x_1 - x_2$
Working/Thoughts:
Translation symmetry means that shifting both particles by the same amount along the axis doesn't affect the physics of the particles. Therefore, the particles will only have the same interaction potential energy after being shifted if it only depends on their relative separation - in one dimension this is the difference of their coordinates but if it was in $\mathbb{R}^2$ or $\mathbb{R}^3$ then it would be the norm of the difference of the position vectors. I think I understand the intuition behind this, but I don't know how to show it mathematically.
I have tried out different combinations such as: $x_1 + x_2$, $x_1-x_2$, $x_1x_2$, $(x_1)^2 - (x_2)^2$, $c_1x_1+c_2x_2$, $c(x_1-x_2)$, $(x_1-x_2)^2$ and I found only only combinations containing $(x_1-x_2)$ are invariant under $x_1\rightarrow x_1+a$ and $x_2\rightarrow x_2+a$. There is strong evidence for $U(x_1,x_2) = U(x_1-x_2)$ being the only way. However, I cannot be sure that the potential only depends on the difference in position.
My lecturer also showed me a solution that he didn't expect me to know using the method of characteristics that can be used to find this relation. I haven't learned that yet so I couldn't reproduce the same result.
Question to you: Is there a simpler way than the method of characteristics which one can use to show that $U(x_1,x_2) = U(x_1 - x_2)?$
 A: Change variables to $$y_1 := (x_1+x_2)/2 \quad y_2 = (x_1-x_2)/2\tag{0}\:.$$
Notice that the map is smooth bijective with smooth inverse $x_1 = y_1+ y_2$ and $x_2 = y_1-y_2$ so that you can indifferently use coordinates $x$ or $y$.
Next, define $$V(y_1,y_2) := U(x_1(y_1,y_2), x_2(y_1,y_2))\:,$$
so that $$ U(x_1,x_2) = V(x_1+x_2, x_1-x_2)\:.\tag{1}$$
According to (0) and the very definition of $V$, saying that $$U(x_1+a,x_2+a) = U(x_1,x_2)$$ for every $a$ is the same as saying that
$$V(y_1,y_2) = V(y_1+a, y_2)$$
for every $a$.
In other words, $V(y_1',y_2) = V(y_1,y_2)$ for every $y_1,y'_1$ and $y_2$.
In other words, $V$ is constant in the first entry:
$$V(y_1,y_2)= W(y_2)\:,$$
for some function $W$ of a single variable.
Inserting this result in (1), we have
$$U(x_1,x_2)= W(x_1-x_2)$$
for some function $W$ with a unique entry.
A: For $4$ points $x_1,x_2,y_1,y_2$, there exists $a$ such that $x_1 = a + y_1$ and $x_2 = a + y_2$ if, and only if $x_1 - x_2 = y_1 - y_2$.
This, means that if $x_1 - x_2 = y_1 - y_2$, then we must have $ U(x_1,x_2) = U(y_1, y_2)$.
Now, if we try to define a function $\tilde{U}$ of $1$ variable such that $U(x_1,x_2) = \tilde{U}(x_1 - x_2)$, the only way this can fail is if their are points $x_1,x_2,y_1,y_2$ with $x_1 - x_2 = y_1 - y_2$ and $U(x_1,x_2) \not= U(y_1,y_2)$. But, as we have seen, this cannot be true if the system is invariant by translation.
Another, maybe simpler, way to see this : let $\tilde U(x) = U(x,0)$. Then, by translation invariance :$$U(x,y) = U(x-y,y-y) = U(x-y,0) = \tilde U(x-y)$$
