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I'm writing a piece on the nuclear force. My intention was to start with the Yukawa interaction and talk about pions, using them to lead into quarks and gluons and the residual strong force and more. However I've hit a stumbling block, and I'm struggling to find anything online.

I know that the electron magnetic dipole moment is −9284.764 × 10−27 J⋅T−1. I know that proton magnetic dipole moment is a 14.10606 × 10−27 J⋅T−1. I know that the neutron magnetic dipole moment is −9.662364 × 10−27 J⋅T−1. But I can't find anything about the pion magnetic dipole moment. So I have a question: what are the magnetic dipole moments of the charged and neutral pions π+ π- and π0?

All answers gratefully received.

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    $\begingroup$ You might benefit from reading my answer to this question: physics.stackexchange.com/questions/292913/… BTW I concur with the answer by @AccidentalFourierTransform. $\endgroup$ – Lewis Miller Jan 2 '17 at 21:58
  • $\begingroup$ @Lewis Miller : thanks very much. I looked through all your answers. There's some interesting stuff there. $\endgroup$ – John Duffield Jan 3 '17 at 14:05
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It is zero. If the magnetic dipole moment $\boldsymbol \mu$ were different from zero, the particle would have a preferred direction in space, which is not possible for a scalar.

Recall that, in general, $$ \boldsymbol \mu\propto\boldsymbol S $$ which, for a scalar, implies $\boldsymbol\mu\equiv\boldsymbol 0$.

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  • $\begingroup$ Thanks very much AFT. As for how a charged particle can be a scalar, I don't know. I imagine I'll have some further questions about pions at a later date. $\endgroup$ – John Duffield Jan 3 '17 at 14:08
  • $\begingroup$ Pions are pseudoscalar rather than scalar (still spin 0). I don't know of any charged scalar particles and I suspect there are symmetry arguments against their existence. $\endgroup$ – Lewis Miller Jan 3 '17 at 15:09
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    $\begingroup$ @Lewis Miller. The scalar $a_0(980)$ is an isovector, and it has charged avatars, S. Teige et al., PhysRev D59 (1998) 012001. They are so rare, though.... $\endgroup$ – Cosmas Zachos May 8 at 14:18

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