Does a Buckyball spin like an electron or like a baseball?

We are often told that an electron does not really spin like a baseball. Only one (or two, if you count up and down) spin states, for example.

How about a Buckyball? Does it spin more like an electron, or more like a baseball?

Where is the dividing line? How can you measure the difference?

  • $\begingroup$ A buckyball is a fairly large molecule. Why couldn't it spin like a baseball? $\endgroup$ – Asher Aug 27 '15 at 21:49

The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic angular momentum $\vec{S}$ is fixed, which is the sense in which an electron has "only one" spin state. (The direction is not fixed, so, as you say, the spin can be up or down.)

A buckyball, like a baseball, has internal structure; the carbon atoms can be set in motion around the center of mass, giving it angular momentum. (The fermions inside the carbon atoms have intrinsic angular momentum, but in the ground state of the molecule these cancel out.) The magnitude of the buckyball's angular momentum $\vec{J}$ is not fixed, so in this sense it is more like a baseball.

But angular momentum is quantized, and while this is utterly irrelevant for a baseball, it has measurable consequences for even large molecules like fullerenes. The total angular momentum obeys $\lvert\vec{J}\rvert^2 = \hbar^2 J(J+1)$, where $J$ is an integer (assuming your buckyball has an even number of fermions, e.g., 60 ${}^{12}$C atoms), while the projection on any axis is restricted to integers from $-J$ to $+J$.

The kinetic energy associated with rotation is $\frac{\lvert\vec{J}\rvert^2}{2I}$, where $I$ is the moment of inertia, so this implies (unequally spaced) steps in the allowed energy. For C$_{60}$, $I$ is small enough that these steps have been measured using Raman spectroscopy.

  • $\begingroup$ I have accepted this answer, but I am still thinking about the boundary. A proton and a neutron are not point particles, so I would say they also spin like baseballs. Besides the electron, the photon which is not a point particle, but is a zero (rest) mass “particle” would not spin like a baseball. Similarly the gluon, the quark and the graviton would be non-baseball spinners. This leaves the neutrino, sdomewhat more uncertain, but still probably on the not baseball-like side. But everything else would fall on the spins like a baseball side. $\endgroup$ – Jim Graber Aug 28 '15 at 14:47
  • $\begingroup$ All the standard-model particles are "point particles"—in the sense of having no constituents—including photons, neutrinos, and the rest. In a baryon (proton, neutron, etc.), there can be contributions from the quarks' intrinsic angular momentum and also from their orbital angular momentum (their relative motion). I believe it's the intrinsic contribution that dominates, but this is well outside my area of expertise. (In the equivalent case of an atom, the different contributions from intrinsic and orbital angular momenta depend on the type of atom and its excitation state.) $\endgroup$ – Stephen Powell Sep 3 '15 at 13:43
  • $\begingroup$ Yes, one way to answer the question is to say that only point particles spin like an electron. So quarks, yes, but protons and neutrons, no. Then you have the photon. It has intrinsic spin, I think, but I hesitate to call it a point particle. Also, there is the little matter of spin one versus spin one-half. $\endgroup$ – Jim Graber Sep 4 '15 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.