2005 II 2

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Pressure balance gives:

\frac{1}{c}\mathbf{J}\times\mathbf{B}=\nabla p

Where:

\mathbf{J}=nq\mathbf{v}

So:

\frac{nq}{c}\mathbf{v}_{\perp}\times\mathbf{B}=\nabla p

Crossing with \mathbf{B}:

\mathbf{v}_{\perp}=\frac{c}{nqB_{0}^{2}}\mathbf{B}\times\nabla p

This is called the diamagnetic velocity because it only seems to exist, since with a pressure gradient there are more or faster ions spinning on one side than the other.

The acceleration of the particle is, to zero order:

\frac{d\mathbf{v}}{dt}=\frac{q}{m}\frac{1}{c}\mathbf{v}\times\mathbf{B}

We can write the velocity as:

\mathbf{v}=v_{\|}\hat{n}+\mathbf{v}_{\perp}

Over a gyro orbit, there is no net perpendicular velocity, so that:

\left\langle \mathbf{v}_{\perp}\right\rangle =0

Going to next order, we write the velocity:

\mathbf{v}=v_{\|}\hat{n}+\mathbf{v}_{\perp}+\mathbf{v}_{E}

Then the cyclotron averaged acceleration is:

0=\left\langle \frac{q}{m}\left(\mathbf{E}+\frac{1}{c}\left(\mathbf{v}_{\perp}+\mathbf{v}_{E}\right)\times\mathbf{B}\right)\right\rangle

So:

\mathbf{E}=-\frac{1}{c}\mathbf{v}_{E}\times\mathbf{B}

And:

\mathbf{v}_{E}=-\frac{c}{B}\hat{n}\times\mathbf{E}_{\perp}

Going to next order:

\mathbf{v}=v_{\|}\hat{n}+\mathbf{v}_{\perp}+\mathbf{v}_{E}+\mathbf{v}_{pol}

The averaged acceleration becomes:

\left\langle \frac{d\mathbf{v}_{E}}{dt}\right\rangle =\left\langle \frac{q}{m}\left(\mathbf{E}+\frac{1}{c}\left(\mathbf{v}_{\perp}+\mathbf{v}_{E}+\mathbf{v}_{pol}\right)\times\mathbf{B}\right)\right\rangle

Solving for \mathbf{v}_{pol}:

\mathbf{v}_{pol}=\frac{mc}{qB}\hat{n}\times\frac{d\mathbf{v}_{E}}{dt}=\frac{1}{\Omega}\hat{n}\times\frac{d}{dt}\left(-\frac{c}{B}\hat{n}\times\mathbf{E}_{\perp}\right)=-\frac{c}{\Omega B}\hat{n}\times\left(\hat{n}\times\frac{d\mathbf{E}_{\perp}}{dt}\right)

This page was recovered in October 2009 from the Plasmagicians page on Generals_2005_II_2 dated 20:26, 15 May 2007.

Retrieved from "https://qed.princeton.edu/main/PlasmaWiki/Generals/2005_II_2"
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