Maxwell's Equations in Curved Spacetime - Electric Current

Electric Current

The electric current is the divergence of the electromagnetic displacement. In a vacuum,

If magnetization-polarization is used, then this just gives the free portion of the current

This incorporates Ampere's Law and Gauss's Law.

In either case, the fact that the electromagnetic displacement is antisymmetric implies that the electric current is automatically conserved

because the partial derivatives commute.

The Ampere-Gauss definition of the electric current is not sufficient to determine its value because the electromagnetic potential (from which is was ultimately derived) has not been given a value. Instead, the usual procedure is to equate the electric current to some expression in terms of other fields, mainly the electron and proton, and then solve for the electromagnetic displacement, electromagnetic field, and electromagnetic potential.

The electric current is a contravariant vector density, and as such it transforms as follows

Verification of this transformation law

\begin{align} \bar{J}^{\mu} \, & = \, \frac{\partial}{\partial \bar{x}^{\nu}} \left( \bar{\mathcal{D}}^{\mu \nu} \right) \, = \, \frac{\partial}{\partial \bar{x}^{\nu}} \left( \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial \bar{x}^{\nu}}{\partial x^{\beta}} \, \mathcal{D}^{\alpha \beta} \, \det \left \right) \\ & = \, \frac{\partial^2 \bar{x}^{\mu}}{\partial \bar{x}^{\nu} \partial x^{\alpha}} \, \frac{\partial \bar{x}^{\nu}}{\partial x^{\beta}} \, \mathcal{D}^{\alpha \beta} \, \det \left \, + \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial^2 \bar{x}^{\nu}}{\partial \bar{x}^{\nu} \partial x^{\beta}} \, \mathcal{D}^{\alpha \beta} \, \det \left \\ & + \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial \bar{x}^{\nu}}{\partial x^{\beta}} \, \frac{\partial \mathcal{D}^{\alpha \beta}}{\partial \bar{x}^{\nu}} \, \det \left \, + \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial \bar{x}^{\nu}}{\partial x^{\beta}} \, \mathcal{D}^{\alpha \beta} \, \frac{\partial}{\partial \bar{x}^{\nu}} \det \left \\ & = \, \frac{\partial^2 \bar{x}^{\mu}}{\partial x^{\beta} \partial x^{\alpha}} \, \mathcal{D}^{\alpha \beta} \, \det \left \, + \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial^2 \bar{x}^{\nu}}{\partial \bar{x}^{\nu} \partial x^{\beta}} \, \mathcal{D}^{\alpha \beta} \, \det \left \\ & + \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial \mathcal{D}^{\alpha \beta}}{\partial x^{\beta}} \, \det \left \, + \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial \bar{x}^{\nu}}{\partial x^{\beta}} \, \mathcal{D}^{\alpha \beta} \, \det \left \frac{\partial \bar{x}^{\rho}}{\partial x^{\sigma}} \frac{\partial^2 x^{\sigma}}{\partial \bar{x}^{\nu} \partial \bar{x}^{\rho}}\\ & = \, 0 \, + \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \frac{\partial^2 \bar{x}^{\nu}}{\partial \bar{x}^{\nu} \partial x^{\beta}} \, \mathcal{D}^{\alpha \beta} \, \det \left \\ & + \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, J^{\alpha} \, \det \left \, + \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \mathcal{D}^{\alpha \beta} \, \det \left \frac{\partial \bar{x}^{\rho}}{\partial x^{\sigma}} \frac{\partial^2 x^{\sigma}}{\partial x^{\beta} \partial \bar{x}^{\rho}} \\ & = \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, J^{\alpha} \, \det \left \, + \, \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \, \mathcal{D}^{\alpha \beta} \, \det \left \left( \frac{\partial^2 \bar{x}^{\nu}} {\partial \bar{x}^{\nu} \partial x^{\beta}} \, + \, \frac{\partial \bar{x}^{\rho}}{\partial x^{\sigma}} \frac{\partial^2 x^{\sigma}}{\partial x^{\beta} \partial \bar{x}^{\rho}} \right) \end{align}

So all that remains is to show that

which is a version of a known theorem (see Inverse functions and differentiation#Higher derivatives).

\begin{align} \frac{\partial^2 \bar{x}^{\nu}}{\partial \bar{x}^{\nu} \partial x^{\beta}} \, + \, \frac{\partial \bar{x}^{\rho}}{\partial x^{\sigma}} \frac{\partial^2 x^{\sigma}}{\partial x^{\beta} \partial \bar{x}^{\rho}} & = \, \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} \frac{\partial^2 \bar{x}^{\nu}}{\partial x^{\sigma} \partial x^{\beta}} \, + \, \frac{\partial \bar{x}^{\nu}}{\partial x^{\sigma}} \frac{\partial^2 x^{\sigma}}{\partial x^{\beta} \partial \bar{x}^{\nu}} \\ & = \, \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} \frac{\partial^2 \bar{x}^{\nu}}{\partial x^{\beta} \partial x^{\sigma}} \, + \, \frac{\partial^2 x^{\sigma}}{\partial x^{\beta} \partial \bar{x}^{\nu}} \frac{\partial \bar{x}^{\nu}}{\partial x^{\sigma}} \, = \, \frac{\partial}{\partial x^{\beta}} \left( \frac{\partial x^{\sigma}}{\partial \bar{x}^{\nu}} \, \frac{\partial \bar{x}^{\nu}}{\partial x^{\sigma}} \right) \\ & = \, \frac{\partial}{\partial x^{\beta}} \left( \, \frac{\partial \bar{x}^{\nu}}{\partial \bar{x}^{\nu}} \right) \, = \, \frac{\partial}{\partial x^{\beta}} \left( \mathbf{4} \right) \, = \, 0 \,. \end{align}

Read more about this topic:  Maxwell's Equations In Curved Spacetime

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