Estimate of the effect of quantum disturbances on a macroscopic object I am self-studying P. Davies, D. Betts, Quantum Mechanics. Exercise 4 of Chapter 1 says:
"A snooker ball of mass $0.1$kg rests on top of an identical ball and is stabilized by a dent $10^{-4}$m wide on the surface of the lower ball. Use the uncertainty principle to estimate how long the system will take to topple, neglecting all but quantum disturbances." (the answer is "About $10^{27}$s.")
I have a hard time thinking about how the uncertainty principle can be applied here. Is the ball to be considered as a "macroscopic particle", or should the principle be applied to a specific particle (or a group of particles) in the ball itself?
I would prefer to not receive the solution straightforwardly. I am looking for suggestions on the correct way to look at this problem.
 A: If one uses the position-momentum uncertainty relation $\Delta x\Delta p\sim h$, one arrives at a time of $\tau\sim 10^{25}$ s. If, on the other hand, one uses the energy-time uncertainty relation $\Delta E\Delta t\sim h$, one arrives at a time of $\tau\sim 10^{29}$ s.
Let's just take the geometric mean of these answers??
All joking aside, this is a pretty teachable moment. The uncertainty principle shouldn't really be used for anything more than incredibly rough calculations (except for rigorous circumstances where it actually provides useful information as a mathematical statement). Just as in astrophysics, this is a question where getting your answer within the right order-of-magnitude of the correct order-of-magnitude is a pretty okay result.
I hope this helps!
A: The arrangement of two hard smooth spheres balanced one on top of the other is doubly unstable. I agree with Bob Knighton : probably the lower ball should be assumed to be fixed in place while the upper ball is balanced on it.
The question asks for a time, so you should use the Uncertainty Relation between energy and time. The uncertainty in energy is the increase in potential energy required to topple out of the dent. The dent is presumably in the shape of a spherical cap, into which the upper ball fits snugly. So the CM of the upper ball has to rise by the depth of the dent in order for it to topple. The depth of the dent is much smaller than its diameter. This probably accounts for the extra 2 orders of magnitude.
