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Particle drift

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If there is any force acting on a particle perpendicular to a uniform magnetic field, the gyro motion will not be exactly circular. Instead (for sufficiently small forces that the gyro motion is not disrupted), the guiding center will move with time. The generalized drift for any force is given by:

$\mathbf{v}_{d}=\frac{\mathbf{F}\times\hat{b}}{\omega_c m}$

Let us suppose there is some time-independent force $\mathbf{B}$ . The equation of motion becomes:

$m\dot{\mathbf{v}}=\mathbf{F} + \frac{q}{c} \mathbf{v}\times\mathbf{B}$

We can define an arbitrary, convenient velocity $\mathbf{u}$ :

$\mathbf{u}\equiv \mathbf{v}-\frac{c}{q}(\mathbf{F}\times\mathbf{B})/B^2$

Since $\dot{\mathbf{u}}=\dot{\mathbf{v}}$ , and we can write:

$m\dot{\mathbf{u}}=\mathbf{F} + \frac{q}{c} \mathbf{u}\times\mathbf{B} + (\mathbf{F}\times\mathbf{B})\times\mathbf{B}/B^2$

We can use the vector identity :

$\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = (\mathbf{C}\times\mathbf{B})\times\mathbf{A} = (\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{A}\cdot\mathbf{B})\mathbf{C}$

To write:

$m\dot{\mathbf{u}}=\mathbf{F} + \frac{q}{c} \mathbf{u}\times\mathbf{B} + (\mathbf{E}\cdot\mathbf{B})\mathbf{B}/B^2 - \mathbf{F} = \frac{q}{c} \mathbf{u}\times\mathbf{B} + (\mathbf{E}\cdot\mathbf{B})\mathbf{B}/B^2$

This clearly has a component perpendicular to and one parallel to the magnetic field $\frac{q}{c} \mathbf{u}\times\mathbf{B}$ , is the same as was used to derive gyro motion , only with u instead of v. Therefore, shifting back to the regular frame, we gain a velocity around which the guiding center will move with drift velocity: