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ClassicalPrecessionInTheRotatingFrame

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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 turntable rotating at the very same angular velocity {$\omega=\gamma B$} we will see a stationary magnetic moment.

The instructive bit is to remind 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 value;
• it must yield a stationary moment, hence it must provide a total vanishing torque in addition to that provided by the uniform magnetic field.

If we write it quite generally as {$\Omega\time\mbox{\it\bf m}/\gamma$}, with the vector {$\Omega$} parallel to the external field, we can rewrite Newton's equation in the turntable frame as

{$ (1) \qquad\qquad \frac {d\mbox{\it\bf L}} {dt} = \mbox{\it\bf m}\times \left( \mbox{\it\bf B} - \frac \Omega \gamma \right)$}

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