An inertial reference frame $S$ and another reference frame $S^{\prime}$ have a common origin O. $S^{\prime}$ rotates with constant angular velocity $\omega$ with respect to $S$. Assuming the result that

$$\left(\frac{d \mathbf{a}}{d t}\right)_{S}=\left(\frac{d \mathbf{a}}{d t}\right)_{S^{\prime}}+\boldsymbol{\omega} \times \mathbf{a}$$

for an arbitrary vector $\mathbf{a}(t)$, show that

$$\left(\frac{d^{2} \mathbf{x}}{d t^{2}}\right)_{\mathcal{S}}=\left(\frac{d^{2} \mathbf{x}}{d t^{2}}\right)_{\mathcal{S}^{\prime}}+2 \boldsymbol{\omega} \times\left(\frac{d \mathbf{x}}{d t}\right)_{\mathcal{S}^{\prime}}+\boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{x})$$

where $\mathbf{x}$ is the position vector of a point $P$ measured from the origin.

A system of electrically charged particles, all with equal masses $m$ and charges $e$, moves under the influence of mutual central forces $\mathbf{F}_{i j}$ of the form

$$\mathbf{F}_{i j}=\left(\mathbf{x}_{i}-\mathbf{x}_{j}\right) f\left(\left|\mathbf{x}_{i}-\mathbf{x}_{j}\right|\right)$$

In addition each particle experiences a Lorentz force due to a constant weak magnetic field $\mathbf{B}$ given by

$$e \frac{d \mathbf{x}_{i}}{d t} \times \mathbf{B}$$

Transform the equations of motion to the rotating frame $\mathcal{S}^{\prime}$. Show that if the angular velocity is chosen to satisfy

$$\boldsymbol{\omega}=-\frac{e}{2 m} \mathbf{B}$$

and if terms of second order in $\mathbf{B}$ are neglected, then the equations of motion in the rotating frame are identical to those in the non-rotating frame in the absence of the magnetic field B.