# Extending Maxwell's Equations from Flat Spacetime To Curved Spacetime

Assume we are working on a Minkowski (i.e. flat) spacetime.

Let $$A^{\mu} = ( \phi/c, \textbf{A})$$ be the contravariant potential four-vector. Then, assuming a covariant Minkowski metric of $$\eta_{\mu \nu} = \textrm{diag}[+, -, -, -]$$, we have that $$A_{\mu} = ( \phi/c, -\textbf{A})$$ is the covariant potential-four vector.

We also have that $$\alpha = A_{\mu} dx^{\mu}$$ is the potential one-form.

We then define $$F = d\alpha = \frac{1}{2} (\partial_\mu A_\nu - \partial_\nu A_\mu) dx^\mu \wedge dx^\nu$$ to be the electromagnetic two-form.

Now, let $$J^{\mu} = (c\rho, \textbf{J})$$ be the contravariant current four-vector.

Then, $$J = \frac{1}{6} J^\mu \epsilon_{\mu \alpha \beta \gamma}dx^\alpha \wedge dx^\beta \wedge dx^\gamma$$ is the current three-form.

With these definitions, Maxwell's equations become

$$$$dF = 0 \; \; \; (\textrm{i})$$$$ $$$$d(*F) = J \; \; \; (\textrm{ii})$$$$

(Recall that $$*$$ is the Hodge Star operator).

Now, to extend these Maxwell equations to a curved spacetime, it appears that we must alter the current three-form:

$$$$J = \frac{1}{6} \sqrt{|g|} J^\mu \epsilon_{\mu \alpha \beta \gamma}dx^\alpha \wedge dx^\beta \wedge dx^\gamma \; \; \; (\textrm{iii})$$$$

Here, $$\sqrt{|g|}$$ is the square root of the absolute value of the determinant of the covariant metric on the Riemannian manifold were are working with.

With this new definition of $$J$$, Maxwell's equations are just equations (i) and (ii).

My question is the following. Why does simply modifying the current three-form to include the "natural" pseudo-Riemannian volume form $$\sqrt{|g|} dx^\alpha \wedge dx^\beta \wedge dx^\gamma$$ allow us to use the flat spacetime formulation of Maxwell's equations in curved spacetime?

• Your curved-spacetime expression for $J$ doesn’t reduce to your flat-spacetime expression. Nor does it have even the same dimensions. So something is wrong. Commented Feb 18, 2020 at 16:50
• The question here is really about the generalization of the Levi-Civita tensor to curved spacetime, not about current density. See Wikipedia. Commented Feb 18, 2020 at 17:40
• When generalizing to curved spacetime you take the simplest generally-covariant equation that reduces in flat-spacetime to the known flat-spacetime equation. Commented Feb 18, 2020 at 18:05
• There could conceivably be extra terms involving curvature tensors, but the simplest prescription is to just make things generally covariant on a manifold without adding curvature terms. For example, in index notation you just replace derivatives with covariant derivatives. In form notation you use the natural volume form and use the exterior derivative or whatever that $d$ is called. (Sorry, I learned this stuff a long time ago using index notation rather than differential forms.) Commented Feb 18, 2020 at 18:21
• I’m not sure how much, if any, experimental evidence we currently have that this prescription is correct. Additional terms with curvature would only be significant close to black holes, neutron stars, etc. As far as I know, field theories in curved spacetime are more of theoretical interest than an experimental reality at this point. Occam’s Razor is the best justification for the minimalist prescription. Commented Feb 19, 2020 at 0:30

The idea of generalizing laws to curved space-time is to notice that we actually live in a curved space-time ourselves. What we know as "flat space-time equations" are, in fact, equations in curved space-time derived/discovered in our local (almost-)inertial frame. We can then derive their curvilinear form by simply transforming to a general frame. This is done mostly by replacing any use of the Minkowski metric structure by a general pseudo-Riemannian one.

Specifically in the case of Maxwell equations the differential-geometry form is almost covariant already. But notice that you are using a metric structure at two points, and both can be characterized as using the Hodge dual. I use a definition of a Hodge dual that takes me from $$\Lambda^{k} T^*\! \mathcal{M}$$ to $$\Lambda^{(n-k)}T^{*}\!\mathcal{M}$$, where $$n$$ is the dimension of the manifold $$\mathcal{M}$$ (this is unlike the definition used in the wikipedia page). The most practical way to define this Hodge dual for any form $$\alpha \in \Lambda^{k} T^*\! \mathcal{M}$$ is to require that $$(*\alpha) \wedge \beta = \beta(\alpha^{\#k}) \omega, \forall \beta \in \Lambda^{k} T^*\! \mathcal{M},$$ where $$\alpha^{\#k}\in T^k\! \mathcal{M}$$ is obtained by raising all the indices of $$\alpha$$, and $$\omega$$ is the pseudo-Riemannian volume form $$\omega = \sqrt{|g|} \mathrm{d}x^1\wedge...\wedge \mathrm{d}x^n$$ (note that $$\sqrt{|\eta|} = 1$$ in Cartesian/Minkowski coordinates and we specialize to $$n=4$$). Now you can see that the Hodge dual can be obtained by contracting the volume form $$\omega$$ with $$\alpha^{\#k}$$.

Back to the Maxwell equations. What you call the current 3-form is in fact the Hodge dual of the current 1-form $$j = j_\mu \mathrm{d}x^\mu, J = *j$$. In your statement you use the generation of the dual by contracting with the volume form, which would usually be stated as $$J \equiv *j = \omega (j^{\#},\cdot,\cdot,\cdot) = \iota_{j^{\#}}\omega = \frac{1}{3!} j_\mu g^{\mu\nu} \sqrt{|g|}\epsilon_{\nu\lambda\kappa\gamma} \mathrm{d}x^\lambda\wedge\mathrm{d}x^\kappa\wedge\mathrm{d}x^\gamma$$ Here you can identify $$j^\# \equiv j_\mu g^{\mu\nu} \partial_\nu = J^\nu \partial_\nu$$ as your current vector (but beware that on metric manifolds objects with raised and lowered indices are considered as identical objects expressed in a different way).

In summary, the covariant statement of the Maxwell equations is $$\mathrm{d}F = 0\,,$$ $$\mathrm{d}(*F) = *j\,,$$ where you have to remember that the Hodge dual is now generated by the general metric $$g$$. The last line is actually very often written as $$*[\mathrm{d}(*F)] = j$$ (which is equivalent to the one above since the Hodge star is a dual).

• Note that this treats the Hodge star as a map from $\bigwedge^k TM \to \bigwedge^{n−k} T^*M$ instead of $\bigwedge^k T^*M \to \bigwedge^{n−k} T^*M$. I also believe the factor $\frac1{3!}$ is unnecessary. With the latter convention and assuming I'm right about the factorial, we'd have $J = \star(j^\flat)=\star(\iota_j\,g) = \iota_j\,\omega$. Commented Feb 19, 2020 at 12:16
• Right, thanks. It would not make sense to write $d(*F)$ like this. And the various factors $1/n!,...$ appear only for coordinate components, true (abstract-indices are my usual choice of poison).
– Void
Commented Feb 19, 2020 at 14:20
• @ Void Perhaps you could clarify on the operations occurring in the expression: "$\beta (\alpha^{k}) \omega$" Commented Feb 19, 2020 at 21:46
• @JG123 It is a contraction ($\beta$ is a k-form and $\alpha^{\#k}$ is a k-vector).
– Void
Commented Feb 19, 2020 at 22:15
• @ Void Ah I see. Commented Feb 19, 2020 at 22:18

Per se, this doesn't have anything to do with relativity or curvature: The factor of $$\sqrt{|g|}$$ comes in for a similar reason that has the determinant of the Jacobian pops up in the substitution formula for integration of multiple variables: When integrating, you need to account for the volume of the unit cell spanned by your coordinate frame. So if you're using generic curvilinear coordinates instead of pseudo-Euclidean ones, you need to add it to the expression for Minkowski spacetime as well.

• @ Christoph So the $\sqrt{|g|}$ term is just accounting for a change of coordinates? Commented Feb 18, 2020 at 23:20
• Yes. Essentially, when integrating, you need to account for the volume of the unit cell spanned by your coordinate frame. It gets a bit confusing because there are different ways to slice this particular pie of defining measurable quantities - eg consider the scalar $\rho$ vs the scalar density $\mathfrak{r}=\rho\sqrt{|g|}$ vs the volume form $\omega = \rho\sqrt{|g|} dx^1\wedge\dots\wedge dx^n$ vs the volume element/pseudo-form $|\omega| = \rho\sqrt{|g|}d^nx$ Commented Feb 19, 2020 at 9:17