Here's my solution to the problem of finding the distance of closest approach for two atoms approaching each other in a Lennard-Jones potential, starting with velocities (+/-) v_0: They start with energy T0 = (1/2)m(2v02) which is conserved. Thus, E = T + V = (1/2)m(2v2) + A/r12 - B/r6 = T0. At the distance of closest approach, v=0 and we have: T0 = A/rt12 - B/rt6. If we let x = rt6 and rearrange: T0x2 + Bx - A = 0 which has the solutions x = (1/2T0)(-B +/- sqrt(B2 + 4AT_0) My question is - what, physically, do each of these roots correspond to? They're both real, but of opposite sign - is the negative one a kind of 'virtual' turning point for the atoms passing through each other (tunnelling?) and coming out the other side?