# Current 3-form, current density, current vector, etc.

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1. Jul 10, 2014

### Geometry_dude

I have been thinking about the Maxwell equations lately and was wondering about their "natural" differential form formulation to get some nice geometric interpretation. This post mainly concerns the inhomogenous microscopic Maxwell equations on some spacetime $(M,g)$, as the homogenous ones $d F = 0$ just mean that the total electromagnetic flux through a $2$-surface $\Omega$ does not change under a smooth homotopy of $\Omega$, i.e.
$$\int\limits_{\Omega} F = \int\limits_{\Omega'} F \, .$$

When one checks out the literature, one often finds the following vector field:
$$j := j^\mu \, \partial _\mu = \begin{pmatrix} \frac{1}{c} \rho\\ \vec{j} \end{pmatrix} \, .$$
My first question is: What is the natural interpretation of the flow of this vector field?
There are a lot of different formulations one finds in the literature, but I have thought about it for a while and it seems like defining a $3$-form $J$ out of $j$ to get the total charge in some spacetime region $\Omega \subseteq M$
$$Q := \int\limits_{\Omega} \, J$$
is most convenient as $J$ can then naturally be interpreted as a charge density in our spacetime. My final questions are: Using this definition, how does one get $J$ from $j$ coordinate independently in SI-units s.t. the above equation holds (using the Cartan derivative, Hodge operator, metric, etc.)? How do the inhomogenous Maxwell equations look like in this form language in SI units?

I tried doing the calculation, but for some reason I always get a wrong factor in the $J$ and the continuity equation $d J = 0$ has the wrong sign.

2. Jul 10, 2014

### WannabeNewton

It is proportional to the 4-velocity field of a charged fluid, with the proportionality being the charge density, so you can just interpret the integral curves of the flow as the worldlines of a family of charged particles.

$J$ is just the Hodge dual of $j$. The inhomogenous Maxwell equations are $d \star F = 4\pi \star j = 4\pi J$. You can use dimensional analysis to convert this to SI units. It is easy to see that $Q = -\int _{\Sigma}\star j = -\int _{\Sigma} J$ where $\Sigma$ is a space-like hypersurface in space-time with boundary, which may or may not be at spatial infinity.

3. Jul 10, 2014

### pervect

Staff Emeritus
As WBN said, j is just the hodge duals of the 3-form, as long as you ignore the sign issues (density is usually taken over an unsigned volume, the signed volume of the 3-form will flip sign if you reflect your coordinates).

There's another way of looking at j though, if you have a swarm of particles. In that case it's the number-flux 4-vector, multipled by the charge.

see http://web.mit.edu/edbert/GR/gr2b.pdf

4. Jul 10, 2014

### Geometry_dude

So one could interpret the flow as a "flow of charge", that seems very reasonable, thanks!
As I haven't found this anywhere else, I'll put it together myself, in case other people are wondering the same thing:

First of all, let's recall the local coordinate expression for a k-form $\alpha$ for the hodge operator on pseudo-Riemannian manifolds with metric signature $(r,s)$ (first positive, then negative, $n=r+s$):
$$\alpha = \frac{1}{k!} \alpha_{i_1 \dots i_k} \, d x^{[i_1 \dots i_k]}$$
$$\implies$$
$$\star \alpha = \frac{(-1)^s}{k! (n-k!)} \, \alpha_{i_1 \dots i_k} \, g^{i_1 j_1} \cdots g^{i_k j_k} \, \varepsilon_{j_1 \dots j_k j_{k+1} \dots j_n} \, \sqrt{\lvert \det g \rvert} \, d x^{[j_{k+1} \dots j_n]}$$
(It's a PITA to find the correct formula for pseudo-Riemannian manifolds). Here we took the volume form to be
$$\mu := \frac{1}{n!} \, \sqrt{\lvert \det g \rvert} \, dx^{[1\dots n]} \, .$$

For our case we want $(r,s)=(1,3)$. Then, if we want the total charge on a spacelike $3$-submanifold (with boundary) $S$ to be
$$Q := c \int\limits_S \, J < \infty\, ,$$
then the current density in SI-units is
$$J = - \, \star (g \cdot j ) = \frac{1}{6} \, \varepsilon_{\nu_0 \nu_1 \nu_2 \nu_3} \, j^{\nu_0} \, \sqrt{\lvert \det g \rvert} \, d x^{[ \nu_1 \nu_2 \nu_3]}$$
with $j$ defined as above. The dot denotes tensor contraction.
For $F = \frac{1}{2} \,F_{ij} \, d x^{[ij]}$ we get the hodge-dual
$$\star F = - \frac{1}{4} \, \varepsilon^{\nu_0 \nu_1}{}_{\nu_2 \nu_3} \, F_{\nu_0 \nu_1} \, \sqrt{\lvert \det g \rvert} \, d x^{[\nu_2 \nu_3]} \, ,$$
hence
$$d \star F = - \frac{1}{4} \varepsilon_{\nu_0 \nu_1 \nu_2 \nu_3} \, \partial_{\beta} \left( F^{\nu_0 \nu_1} \sqrt{\lvert \det g \rvert} \right) \, d x^{[\beta \nu_2 \nu_3]} \, .$$
Setting
$$d \star F = b \, J$$
and using the identities for the Levi-Civita Symbol from http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf (thanks again, WannabeNewton !), we get
$$j^\beta = \frac{1}{b \, \sqrt{\lvert \det g \rvert}} \, \partial_{\gamma} \left( F^{\beta \gamma} \sqrt{\lvert \det g \rvert}\right)$$
unless I made a sign error.
For the Minkowski metric, we need
$$j^\beta = \frac{1}{\mu_0} \partial_\gamma F^{\gamma \beta}$$
and thus $b = - \mu_0$.
Hence, in this convention, the Maxwell equations are
$$d F = 0$$
$$d \star F + \mu_0 J = 0$$, which agrees with what WBN said.
$$d \star F =\mu_0 \, \star g \cdot j \, .$$

Last edited: Jul 10, 2014
5. Jul 11, 2014

### WannabeNewton

Yes that's one way to do it. You can also start straight from the covariant Maxwell equations. Then the calculation is a lot shorter:

We have $(^{\star}F)_{ab} = \frac{1}{2}\epsilon_{abcd}F^{cd}$ so $\epsilon^{abef}\nabla_{e}(^{\star}F)_{ab} = \frac{1}{2}\epsilon^{abef}\epsilon_{abcd}\nabla_{e}F^{cd} = -2\nabla_{e}F^{ef} = -8\pi j^{f}$.
Hence $\epsilon_{fjki}\epsilon^{feab}\nabla_{e}(^{\star}F)_{ab} = -6\nabla_{[j}(^{\star}F)_{ki]}= -8\pi\epsilon_{fjki} j^{f} = -8\pi(^{\star}j)_{jki}$
therefore $3\nabla_{[a}(^{\star}F)_{bc]} = d(^{\star}F)_{abc} = 4\pi(^{\star}j)_{abc}$ i.e. $d(^{\star}F) = 4\pi(^{\star}j)$.

Because $\nabla^{a}j_{a} = 0$ in any space-time, we can then apply Stokes' theorem to a space-time region $\Omega \subseteq M$ bounded by two space-like hypersurfaces $\Sigma, \Sigma'$ from a single foliation and find that $\int _{\Omega}\nabla^{a}j_{a} = 0 = \int _{\Sigma}j_{a}n^{a} -\int _{\Sigma'}j_{a}n^{a}$ i.e. the total charge $Q = -\int _{\Sigma}j_{a}n^{a}$ is conserved (here $n^{a}$ is the outward unit normal field to the space-like foliation that $\Sigma,\Sigma'$ belong to; the negative sign is to compensate for the negative sign that comes out of the inner product in the integral.