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 | | Gauge freedom has two sides to it: we may choose a convenient gauge to simplify a proof, but if the result depends on the gauge, it may be insignificant from a physical point of view. |
 | | In the transversal gauge, the vector potential $\A$ is defined by \emph{(\ref{tg})}. |
 | | \end{cor} \subsection{The Coulomb Gauge} In the Coulomb gauge, the vector potential $\A$ is defined by \be \label{coco} \A(\x) := -\frac1{\omega_\nu}\, \int_{\R^\nu}\dfrac{\x-\y}{\x-\y^{\,\nu}} \,\times\,\B(\y)\,dy \ee with $\omega_2:=S^1=2\pi$ and $\omega_3:=S^2=4\pi$. |
| www.ma.utexas.edu /mp_arc/papers/05-242 (10211 words) |
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