I'm porting Maxwell's equations to curved spacetime and am having trouble reconciling the tensor and forms treatments. I think the problem boils down to a misunderstanding on my part concerning the exterior covariant derivative rather than just the exterior derivative, but that's only a guess. So in Minkowski spacetime we have

$$ F_{\mu \nu} = A_{\nu,\mu}-A_{\mu,\nu}\text{, }F_{[\mu \nu , \kappa]}=0 $$


$$ F=\text{d}A\text{, }\text{d}F=0, $$

the equivalence of which is easy to show. However, suppose we're working in curved spacetime. We get

$$ F_{\mu \nu} = A_{\nu;\mu}-A_{\mu;\nu}\text{, }F_{[\mu \nu ; \kappa]}=0. $$

However, the connection coefficients in the covariant derivatives neatly cancel out, so both tensor expressions are equivalent to each other. However, suppose we attempt to bring the flat space forms expression into curved spacetime, by turning the exterior derivative into an exterior covariant derivative. We start with

$$F = \text{d}_DA$$

and then (naively?) find

$$ \text{d}_DF =\text{d}_D \text{d}_D A = \Omega \wedge A \neq 0 $$

where $\Omega$ is the curvature 2-form. So suddenly the tensor and forms methods seem to give differing versions of Maxwell's laws. Is one not obligated to employ the covariant exterior derivative in this situation? Maybe I am seriously missing something but it seems like the exterior covariant derivative gets much less mention than it should in relativity, if it is required to deploy forms at all on curved manifolds! The Hodge star of course contains information about the metric, but it seems here like we arrive at an inconsistency before needing to invoke it to find the other two Maxwell's equations.

Thanks for any insight.


In the language of differential forms, the Maxwell-Lorentz equations are simply $$\begin{eqnarray*}\mathrm{d}\!\star\!F = J/\lambda_0 &\text{,}\quad&\mathrm{d}F = 0\text{,}\end{eqnarray*}$$ where $1/\lambda_0$ is the characteristic impedance of free space, and can be fixed to $1$ in appropriate units.

From that point of view, all you need to "move" to curved spacetime is realize that the only place where the metric comes up is the in the Hodge dual operator $\star$.

This was what I expected initially, but I got scared by the existence of the exterior covariant derivative -- I can't figure out why it is needed if the exterior derivative already behaves geometrically.

As already said, the differential structure does not care about either the metric or the manifold-connection, and hence not the curvature. That's simply a matter of how they are defined, and therefore curved spacetime wouldn't care about the exterior derivative, but only have a different Hodge dual.

However, there is an informal way to motivate this by analogy, which is what I gather you're asking for (this is not clear, so I'm guessing). Recall that the notion of covariant derivative for scalar fields is particularly trivial and independent of curvature: $$\nabla_u \phi = u^\mu\phi_{,\mu}\text{.}$$ The same kind of thing happens with the covariant exterior derivative. Recall the $k$-forms are certain types of maps in the form $$\omega_p: (T_pM)^k\to\mathbb{R}\text{,}$$ and are therefore scalar-valued.

On the other hand, one can consider a kind of generalized form that takes values in some vector bundle over over the manifold, and try to have an "exterior derivative" with $\mathrm{d}$ acting on each component in an arbitrary basis. This generally does not work without a connection on that vector bundle, with a notable exception where the values taken are are actually scalars, since then all possible bases for this $1$-dimensional vector space are just proportional to one another.

So the short of it is that the exterior derivative makes sense regardless of curvature and the covariant exterior derivative doesn't do anything interesting for ordinary scalar-valued forms.


The calculus of forms is already well defined on curved manifolds, so you can use $d$ right off the bat.

  • $\begingroup$ This was what I expected initially, but I got scared by the existence of the exterior covariant derivative -- I can't figure out why it is needed if the exterior derivative already behaves geometrically. $\endgroup$ – user2275987 Nov 30 '13 at 20:40
  • $\begingroup$ @user2275987: it doesn't behave geometrically. The differentiable structure is (algebraically) topological. It's definable regardless of whether curvature is. $\endgroup$ – Stan Liou Nov 30 '13 at 20:47

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