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Reading T violation from Perkins High Energy Physics, 4th ed, pp 82. enter image description here

Here I don't understand the definition of last 4 quantities, magnetic and electric dipole moments, longitudinal and transverse polarisations. Why they are defined in that way?

I know the definition of dipole moments when a charge/current placed in an external electric/magnetic field. Are these definitions for intrinsic dipole moments of elementary particles?

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    $\begingroup$ Yes, they are all quantities for elementary particles with a given spin and momentum. $\endgroup$
    – Cosmas Zachos
    Commented Jan 6 at 15:48
  • $\begingroup$ Is there any particular reason behind the definition dipole moments like $\sigma.B$ and $\sigma.E$ ? $\endgroup$
    – Sagar K. Biswal
    Commented Jan 6 at 18:44

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A matter of language: it is not the (vector) moments defined as the scalar operators $\vec \sigma \cdot \vec B$ and $\vec \sigma \cdot \vec E$; these are the magnitude-stripped (scalar) energy shifts induced by such moments in magnetic and electric fields, respectively.

Wikipedia reminds you that these moments are proportional to the spin for elementary and spherically symmetric (composite hadron) particles, where gyromagnetic constant factors and the signs of the shift are irrelevant to the argument and dropped in your text. You see that magnetic-moment energy shifts are comfortably stable under P and T, but Electric dipole ones flip under T!

So, a non-vanishing moment for the elementary particle assayed is a hallmark of T violation (P is broken in the weak interactions and is no big deal!), and your text outlines their non-observation limits constraining T-breaking theories.

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  • $\begingroup$ What about those last two quantities. Longitudinal and transverse polarisation? Are those only for photons? And what it means to have two momentum vectors $p_1$ and $p_2$ ? $\endgroup$
    – Sagar K. Biswal
    Commented Jan 8 at 4:27
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    $\begingroup$ Maybe they can extend beyond photons. P1 and p2 are two momentum components transverse to the direction of motion. $\endgroup$
    – Cosmas Zachos
    Commented Jan 8 at 8:20

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