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ClassicalSpinPrecessionInTheRotatingFrame

< Classical spin precession | Index | Addition of a rotating radiofrequency field >

It is instructive to redo the exercise of the previous page in a different way.

The new approach looks very intuitive, once we know already the solution: since the motion is a precession, if we look at it from a a rotating frame, that is, a turntable, rotating at an angular velocity equal to the Larmor frequency {$\omega=\gamma B$}, we will see a stationary magnetic moment.

The instructive bit is to remind ourselves that a turntable is not an inertial reference frame. In principle Newton's equations do not work on the turntable, unless we invoke fictitious forces and fictitious torques.

The fictitious torque which we must invoke here is very simple:

• it must be proportional to turntable angular velocity (let's call it {$\Omega$} to allow for any velocity);
• it must yield a stationary moment when {$\Omega=\gamma B$}, hence it must provide a total vanishing torque in addition to that provided by the uniform magnetic field.

If we write this torque, quite generally, as {$\Omega\times\mathbf{ m}/\gamma$}, with the vector {$\Omega$} parallel to the external field, we can rewrite Newton's equation in the rotating frame as

{$$ \begin{equation} \frac {d\mathbf{ L}} {dt} = \mathbf{ m}\times \left( \mathbf{ B} - \frac \Omega \gamma \right) \end{equation} $$}

which provides exactly a constant angular momentum and a stationary magnetic moment when the fictitious field {$-\Omega/\gamma$} cancels exactly the external field.

Equation (1) tells us that when we look at a precession from a rotating frame we see the precession motion that we would expect in the presence of a total effective magnetic field, which is just the superposition of the real applied field plus the fictitious field:

{$$ \begin{equation} \mathbf{ B}_{e} = \left( \mathbf{ B} - \frac \Omega \gamma \right) \end{equation} $$}

< Classical spin precession | Index | Addition of a rotating radiofrequency field >