4 ms·
This reminds me of a similar problem I came across whilst studying quantum mechanics in my undergrad. The problem is stated as follows 'imagine a perfectly upri
by ganonm 8y ago
This reminds me of a similar problem I came across whilst studying quantum mechanics in my undergrad. The problem is stated as follows 'imagine a perfectly upright pencil balancing on its perfectly sharpened tip. How long does it take to fall over?'. At first this seems surprising. Classically speaking, the pencil would never fall over, but knowing that we live in a quantum world, we can conclude that it would in fact fall over. The reason for this is the famous Heisenberg Uncertainty Principle which says that the product of the uncertainty in a particle's momentum and its position is always greater than or equal to hbar/2. You can derive a similar expression for uncertainty in angular momentum and angular position.
Now, these expressions don't imply it's impossible for the angle (theta) and angular momentum (L) both to be simultaneously zero, just that it's infinitely improbable. If you plug in reasonable values for both parameters, say by assuming that Δtheta = ΔL = sqrt(hbar) and that the actual values took a figure corresponding to an expected value assuming these uncertainties, you can work out that the pencil will actually fall over in a few seconds.
- bsder 8y agoI don't find this quantum argument convincing. We know that the better you isolate an atom, the longer it takes for an excited electron to come back to ground state. The issue is that the electron must dump a full quantum of energy, and it tends to do so as part of an exchange with the outside world. If you isolate an atom really well, that electron can take a really long time to decay. See: "Collective Electrodynamics" https://mitpress.mit.edu/books/collective-electrodynamics https://mitpress.mit.edu/books/collective-electrodynamics Brownian motion and standard thermodynamics almost certainly makes a stronger case.