Discussion of Note 386(1)

Many thanks. The fact that you have solved Eqs. (5) to (7) for the spin connection is of key importance, so I will try a few ideas based on that. Eq. (8.352) of PECE2 is a useful expression for the vector potential of a current loop. If it is assumed in the first approximation that this can be used in the antisymmetry equations ((5) to (7), then the spin connection for a current loop can be found and graphed. The magnetic field is then

B = curl A – omega x A


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