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ClassicalSpinPrecession

< The top and the spin | Index | Classical spin precession in the rotating frame >

The spin precession in a uniform magnetic field is the direct result of Newton's law for a rigid body. This can be seen directly by calculating the motion around the centre of mass, which is determined by the torque of Eq. (2), previous page, through:

{$$ \begin{equation}\frac {d\mathbf{ L}} {dt} = \mathbf{m}\times \mathbf{B} \end{equation} $$},

Since {$\mathbf{ L} = \mathbf{ m}/\gamma$} (Eq. (3), previous page) we can rewrite this equation as

{$$ \begin{equation} \frac {d\mathbf{ m}} {dt} = \mathbf{ m}\times \gamma \mathbf{ B}\end{equation} $$}

Spelling out Eq. (2) into three Cartesian components, with {$\mathbf{ B} = B\hat z$} and the Larmor frequency {$ \omega=-\gamma B $}, we get: {$$ \begin{align} \frac {dm_x} {dt} &= \omega m_y \\ \frac {dm_y} {dt} &= -\omega m_x \\ \frac {dm_z} {dt} &= 0 \end{align} $$} It is easy to check that the solution of these equations reads: {$$ \begin{align} m_x(t) &= m_{0\perp} \sin(\omega t + \phi) \\ m_y(t) &= m_{0\perp} \cos(\omega t + \phi) \\ m_z(t) &= m_{0\parallel} \end{align}$$}, which describes precisely the precessional motion of a vector of length {$\sqrt{m_{0\perp}^2 + m_{0\parallel}^2}$} along the surface of a cone.

Notice the minus sign in the Larmor equation: a field along the positive {$ \hat z$} direction produces a precession along the opposite direction for a positively charged particle. Since {$\gamma=eg/2m$} has the same sign as the charge, negatively charged electrons precess with a Larmor frequency vector parallel to the field.

This is consistent with the rotation of the linear momentum due to the Lorentz force in a uniform magnetic field: a positively charged particle will turn according to the force {$ e \mathbf{v}\times\mathbf{B}/m$}. For e.g. {$\mathbf v$} along {$\hat x$} and, again, {$\mathbf B$} along {$\hat z$} this results in a centripetal force along {$-\hat y$}, i.e. in a clockwise, negative rotation around the field axis. The reverse applies to negatively charged particles. Consequently, recalling that {$g_S=2$}, the spin (i.e. the magnetic moment) of a free charged particle rotates together with the linear momentum, at the same Larmor frequency and in the same direction around the applied magnetic field.

< The top and the spin | Index | Classical spin precession in the rotating frame >