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I want to know when I am asked to find the surface current density and the space current density what should I do? Also what is the difference between the surface and space current? Can the space current density help me to find the latter if I know it?

Adam
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1 Answers1

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The following discussion refers to electrodynamics in three dimensions.

Definitions.

Let a two-dimensional surface $\Sigma\subset\mathbb R^3$ be given, then a surface current density on $\Sigma$ is a function $\mathbf K:\Sigma\to \mathbb R^3$. In other words, it is a vector field on the surface. For each point $p$ on the surface, it physically represents the charge per unit time passing through a unit cross-sectional length on the surface.

You can think of surface current density like this. Consider some point $p$ on the surface $\Sigma$. Let $\gamma$ be a short curve segment on $\Sigma$ passing through $p$. Let $\mathbf n$ denote the unit vector tangent to $\Sigma$ at $p$ but normal to $\gamma$, then $$ (\mathbf K(p)\cdot\mathbf n)\mathrm{len}(\gamma) $$ approximates the charge per unit time flowing on the surface and passing through $\gamma$.

On the other hand, a current density (or space current density as you call it) is a function $\mathbf J:\mathbb R^3\to\mathbb R^3$. In other words, it is a vector field. For each point $\mathbf x$, the current density at that point represents the charge per unit time passing through a unit cross-sectional area.

You can think of the current density like this. Consider some point $\mathbf x\in\mathbb R^3$, and let $\alpha$ denote a small surface segment passing through $\mathbf x$. Let $\mathbf n$ denote the unit vector at $\mathbf x$ normal to $\alpha$, then $$ (\mathbf J(\mathbf x)\cdot \mathbf n)\mathrm{area}(\alpha) $$ approximates the charge per unit time passing through $\alpha$.

Computing $\mathbf J$ from $\mathbf K$.

A surface current density $\mathbf K$ can be written as a current density $\mathbf J$ that includes Dirac deltas. For example, the surface current density of a spinning sphere of radius $R$ with angular velocity $\omega$ and with uniform surface charge density $\sigma$ fixed to it is given in spherical coordinates by \begin{align} \mathbf K = \sigma\omega R\sin\theta\,\hat{\boldsymbol\phi} \end{align} which can equivalently be written as a current density \begin{align} \mathbf J = \sigma\omega R\sin\theta\,\delta(r-R)\,\hat{\boldsymbol\phi} \end{align}

joshphysics
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    @Adam Sure thing. – joshphysics Mar 17 '14 at 00:33
  • @joshphysics: can you gimme any reference where electrodynamics is formulated in the way you answered the question here? – rainman Sep 15 '14 at 15:52
  • @ome Griffiths EM uses this notation, and I'm fairly certain Jackson does as well. – joshphysics Sep 15 '14 at 16:15
  • @joshphysics Actually Griffiths uses these notations but his treatment is not so mathematically rigorous like yours. Thats why I am asking any reference that presents materials rigorously like yours. – rainman Sep 16 '14 at 08:38
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    @Ome Ah I see. Then in that case I can't think of a reference. My answer was mostly just their treatments in more mathematically dressed-up language. Perhaps there are mathematical physicists who have written EM books...I'm not sure. – joshphysics Sep 16 '14 at 12:50
  • Perfect! Sadly that my EM II professor don't have a clue. He insist that the reason of $\mathbf K =0 $ in pg. 397 (Griffiths's textbook) is because some effect of the response of the boundary surface with incident EM wave. – Nogueira Apr 16 '15 at 01:24