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The orbital magnetic dipole moment of a particle with mass $m$ and charge $q$ can be shown to be related to the orbital angular momentum through the equation

$$\displaystyle \boldsymbol\mu_L=\frac{q}{2m}\bf{L}.$$

One of the quantum mechanics textbooks has instead the same equation but with the speed of light in the denominator,

$$\displaystyle \boldsymbol\mu_L=\frac{q}{2mc}\bf{L}.$$

The same book also defined the spin magnetic dipole moment for the electron with a $c$,

$$\displaystyle \boldsymbol\mu_S=-g\frac{e}{2m_ec}\bf{S},$$

where $g$ is the usual Lande factor (~2).

Is the speed of light appearing in those equations just a typo? (but it is not in the book errata)

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There is a typo in first sentence. – user10001 Jul 28 '12 at 21:54
@dushya OOPS! Corrected, thanks a lot. – Revo Jul 28 '12 at 22:21
up vote 3 down vote accepted

This is difference in the unit system. The former uses SI units, where the latter uses cgs/Gauss system.

In Gaussian units, unlike SI units, the electric field E and the magnetic field B have the same dimension. This amounts to a factor of c difference between how B is defined in the two unit systems, on top of the other differences.


This extra factor of $c$ extends to many magnetic quantities in Gaussian system.

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Bohr magneton has different expressions in different unit systems, one with c and one without, but not the magnetic dipole moment. The magnetic dipole moment by definition equals the current times the area enclosed by the current loop. If one starts from this definition and try to write it in terms of angular momentum, one will never get a factor of c! – Revo Jul 29 '12 at 23:05
@Revo: All formulas in this article are correct in SI units, but in other unit systems, the formulas may need to be changed. For example, in SI units, a loop of current with current I and area A has magnetic moment I×A (see below), but in Gaussian units the magnetic moment is I×A/c.( – Siyuan Ren Jul 30 '12 at 0:47

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