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Be careful: The spin contribution to the number current is proportional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition""Gordon Decomposition" for more details.

Be careful: The spin contribution to the number current is proportional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition" for more details.

Be careful: The spin contribution to the number current is proportional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition" for more details.

Added link. Dear mike stone: If this is the wrong link please roll back.
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Qmechanic ♦
Qmechanic ♦
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Be careful: The spin constributioncontribution to the number current is proprotionalproportional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition""Gordon Decomposition" for more details.

Be careful: The spin constribution to the number current is proprotional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition" for more details.

Be careful: The spin contribution to the number current is proportional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition" for more details.

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mike stone
mike stone
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Be careful: The spin constribution to the number current is proprotional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition" for more details.