# Maxwell Equations in Friedman-Robertson-Walker metric

The Maxwell equations are relativistic. But what happens to them in an expanding space time? I assume that only the charge density $$\rho$$ is affected, i.e. only Gauss's law gets modified. Am I right with this assumption or are there further effects?

• Did you google "maxwell theory in expanding universe"? – Avantgarde Nov 7 '18 at 10:59
• Oh, that's weird, I am almost sorry for asking the question, as there is even an wikipedia-article related to my question. Of course, I searched for an answer beforehand, but DuckDuckGo didn't deliver helpful results. Very strange... Thanks a lot for your reply! – kalle Nov 7 '18 at 11:51
• Why does everyone forget about poor Lemaître? haha. Friedmann-Lemaître-Robertson-Walker. FLRW. – FGSUZ Nov 7 '18 at 11:57
• How could I forget poor Lemaître, I just skipped him for convenience and saving space in the title :P – kalle Nov 11 '18 at 11:23

The Wikipedia article "Maxwell's equations in curved spacetime" presents the equations in a particularly simple way that involves only partial derivatives rather than covariant derivatives. However, that obscures a simple procedure for making Lorentz-covariant equations generally covariant, which is to raise and lower all indices with a general metric tensor rather than the Minkowski tensor, and to replace all partial derivatives with covariant derivatives.

The Lorentz-covariant Maxwell's equations are $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$ $$\partial_\mu F^{\mu\nu}=j^\nu$$

and the generally-covariant Maxwell's equations are

$$F_{\mu\nu}=D_\mu A_\nu-D_\nu A_\mu$$ $$D_\mu F^{\mu\nu}=j^\nu$$

where $$D_\mu$$ is the covariant derivative for the metric tensor $$g_{\mu\nu}$$. (In the first equation, the terms in the covariant derivative involving Christoffel symbols simply cancel, but in the second equation they are important.)

If you prefer to start with the action, then you also have to make the 4-volume element generally invariant by the replacement $$d^4x\rightarrow\sqrt{-g}\,d^4x,$$ where $$g$$ is the determinant of the metric tensor. So

$$S=\int\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+j^\mu A_\mu\right)d^4x$$

becomes

$$S=\int\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+j^\mu A_\mu\right)\sqrt{-g}\,d^4x$$

Note: The Lagrangian density in the Wikipedia article has a $$\sqrt{-g}$$ in the first term but not in the second. This is because the article prefers to use $$J^\mu$$, a vector density of weight 1, rather than $$j^\mu$$, a vector, to represent the current density.

Thank you very much for your explanation! In the meantime I also discovered a nice solution in the 3+1 formalism from THorne and Macdonald (Mon. Not. R. astr. Soc. (1982) 198, 339-343 and Microfiche MN 198/1):

$$\nabla\cdot E=4\pi\rho_e$$ with $$E$$ the 4-vector of the EM-field. The classical result remains, that electric field lines terminate on electric charge.

$$\nabla\cdot B =0$$ with $$B$$ the 4-vector of the EM-field. The classical result remains, there are no magnetic charges, i.e. the magnetic field lines never end.

$$D_\tau E +\frac{2}{3}\theta E-\sigma\cdot E = \alpha^{-1}\nabla\times\left(\alpha B\right) - 4\pi j$$ with $$\theta$$ the expansion, $$\sigma$$ the shear, and the acceleration $$a=\nabla\log\alpha$$ implying that a fiducial observer, being in a perfectly conducting medium, would never experience an electric field (i.e. the field is frozen). Otherwise, the observed curl induced a time-changing magnetic field.

$$D_\tau B + \frac{2}{3}\theta B - \sigma\cdot B = -\alpha^{-1}\nabla\times\left(\alpha E \right)$$

Everything is in much more detail in the stated paper, worth reading to anyone who'd like to use Maxwell's equations in the 3+1 formalism. :)