Consider a particle $\sf{P}$ of mass $m$ that is at rest or in slow motion so that its mechanical energy is just $E= m c^{2}$. Let $\sf{P}$ be characterized by its quark coefficients $n$. These coefficients determine the total angular momentum quantum number $π$, orbital radius $R$ and also $\sf{P}$'s cross-sectional area $\, A$. If $\sf{P}$ is rotating then the **current** $\, I \,$ due to the motion of any induced charge $\, \mathcal{Q}$ is

$I = \mathcal{Q} \, / \, \hat{\tau}$

where $\hat{\tau}$ is period of $\sf{P}$. The current is measured in Coulombs per second, or *Amperes*, and abbreviated by (A). Definition: the **magnetic moment** due to the rotation of the $\zeta$-type quarks in $\sf{P}$ is

$\begin{align} \overline{\mu} ^{\, \zeta} \equiv A I^{\zeta} \, \hat{z} \end{align}$

where $\hat{z} \equiv (0, 0, 1)$ notes the polar axis of $\sf{P}$. The norm of a moment is written without an overline as $\mu \equiv \left\| \, \overline{\mu} \, \right\|$. By this definition the magnetic moment is given by the product of a current and an area, so the measurement units used for $\mu$ are abbreviated as (Aβm^{2}). The magnetic moment of the whole particle $\sf{P}$ is given by a sum over quark moments

$\begin{align} \overline{\mu} ^{\, \sf{P}} \equiv \sum_{\zeta =1}^{10} \overline{\mu} ^{\, \zeta} \end{align}$

$\begin{align} \mu ^{\sf{P}} = \frac{ e h }{4 \pi } \frac{ π }{ m} \sum_{\zeta =1}^{10} \chi_{m} ^{\zeta} \Delta n^{\zeta} \end{align}$

The forgoing expression summarizes all thirteen known nuclear magnetic moments to within experimental error.

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^{3}The representation uses ten adjustable parameters, i.e. the magnetic susceptibilities of the ten different types of thermodynamic quark. For more detail about comparisons with measurement click here.

As discussed earlier, the angular momentum, mass and magnetic susceptibilities of ordinary-quarks and anti-quarks are the same as each other. Also, the net number of quarks $\Delta n$ in particle $\sf{P}$ and its anti-particle $\overline{\sf{P}}$ are related as $\rm{\Delta} \it{n} ^{\zeta} \left( \sf{P} \right) = - \rm{\Delta} \it{n}^{\zeta} \left( \sf{\overline{P}} \right)$. So the theorem above implies that the magnetic moments of particles and anti-particles are related as

$\overline{\mu} \left( \sf{P} \right) = - \overline{\mu} \left( \sf{\overline{P}} \right)$