When discussing the average kinetic energy, we noted that kinetic energy is a relative quantity, dependent on the relative velocities of the system and the observer. Resolve the following paradox. We consider two molecules, each with mass $m$, with a mutual attractive interaction.
a. In the center of mass frame, the molecules move toward each other with equal speed. At a distance $R_1$ between the molecules, each has speed $v_1$, and at a closer distance $R_0$, the speed of each molecule has doubled to $v_0=2 v_1$.
b. Now we follow exactly the same process, but making our measurements from the standpoint of one of the molecules, which we call the reference. The reference molecule appears to be still, with no kinetic energy, and the other molecule moves toward it at speed $2 v_0$ when at a distance $R_0$, and at a distance $2 v_1=4 v_0$ when at a distance $R_0$.
If energy is conserved in frame (a), the center of mass frame, the potential energy must have decreased by an amount equal to the rise in the kinetic energy:
$$
\begin{aligned}
E_1= & U_1+\frac{m v_1^2}{2}+\frac{m v_1^2}{2}=U_1+m v_1^2 \\
= & E_0=U_0+\frac{m v_0^2}{2}+\frac{m v_0^2}{2}=U_0+4 m v_1^2 ; \\
& U_1-U_0=3 m v_1^2 .
\end{aligned}
$$
Applying the same logic to our measurements in frame (b), however, we find
$$
\begin{gathered}
E_1=U_1+\frac{m\left(2 v_1\right)^2}{2}=U_1+2 m v_1^2 \\
=E_0=U_0+\frac{m\left(2 v_0\right)^2}{2}=U_0+4 m v_1^2 ; \\
U_1-U_0=2 m v_1^2 .
\end{gathered}
$$
Does the shape of the potential energy curve also depend on the choice of reference frame?