According to Mach's Principle, angular momentum (as in the Newton bucket experiment) is relative to the overall mass distribution of the universe, or perhaps some other characteristic of the universe's overall space-time metric.

Also, according to the Einstein-de Haas experiment, electron spin angular momentum is made of the same "stuff" as classical angular momentum, in that the "spin" of the electrons can affect the rotation of a macroscopic body.

Does this mean that electron spin is also somehow connected to the mass distribution of the universe? Are there any effects related to electron spin that are analogous to the Newton bucket experiment?


1 Answer 1


Your question is written as if Mach's principle were true, but in fact it's false, in the sense that general relativity doesn't obey Mach's principle, and GR is verified by experiment, while more machian theories like Brans-Dicke gravity are falsified by experiment.

So if we were to answer your question, we would have to (1) come up with some new theory of gravity that works at least as well as GR and is more machian than GR; (2) quantize that theory.

If you want to know how intrinsic spin transforms when you go into a rotating frame, then it's not a big mystery. This is a common thing to do in nuclear physics. We add a term $-\omega J$ to the Hamiltonian, where the angular frequency acts like a Lagrange multiplier that can be used to control $\langle J \rangle$. This new Hamiltonian has different eigenstates with different structure, but spin $-1/2$ particles are still spin $1/2$.

  • $\begingroup$ AFAIK, Brans-Dicke theory has not been falsified. Rather, certain values of its coupling constant have been ruled out. $\endgroup$
    – user76284
    Aug 9, 2020 at 20:49
  • $\begingroup$ According to GR, the distribution of matter and field energy-momentum in the universe determines the inertial frame at each point in the universe. Is this not a different way to state Mach's principle -- that the inertial frames are determined by the structure of the universe? $\endgroup$
    – David H
    Aug 9, 2020 at 22:54

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