I am having trouble understanding when/why we can sometimes use partial derivatives in place of covariant derivatives for electrodynamics in a curved spacetime. And how to interpret / intuitively understand what is going on.

Below are lengthy details, but the summary is:

  1. Why can we swap covariant for partial derivatives in the source equations and the charge conservation equation?

  2. Also I'm having trouble wrapping my head around what exactly it means for Maxwell's equations to seem independent of the Christoffel symbols, yet light curves. And even more confusingly in terms of the potential suddenly we have to use the covariant derivative and the Ricci curvature even comes in explicitly! Are Maxwell's equations kind of "misleading" / "hiding" subtleties from us? What is going on?

In the wikipedia article Maxwell's equations in curved spacetime, it states without calculation that despite the use of partial derivatives, the equations are invariant under arbitrary curvilinear coordinate transformations. Because of symmetry, it is not too hard to see that:

$$ F_{ab} \, = \, \nabla_a A_b \, - \, \nabla_b A_a = (\partial_a A_b -{\Gamma^c}_{ab} A_c) - (\partial_b A_a - {\Gamma^c}_{ba} A_c) = \partial_a A_b - \partial_b A_a - {\Gamma^c}_{ab} A_c + {\Gamma^c}_{ba} A_c $$

And the last two terms cancel because the Christoffel symbols are symmetric in the lower indicies. Similarly, it is not too hard to see it also works for the Faraday–Gauss equation $$ \nabla_\lambda F_{\mu \nu} + \nabla_\mu F_{\nu \lambda} + \nabla_\nu F_{\lambda \mu} = \partial_\lambda F_{\mu \nu} + \partial _\mu F_{\nu \lambda} + \partial_\nu F_{\lambda \mu} = 0$$ if I expand out in Christoffel symbols using:

$$\nabla_\gamma F_{\alpha \beta} \, = \, \partial_\gamma F_{\alpha \beta} - {\Gamma^{\mu}}_{\alpha \gamma} F_{\mu \beta} - {\Gamma^{\mu}}_{\beta \gamma} F_{\alpha \mu}. $$

But I can't figure out how it works for the source equation:

$$\mathcal{D}^{\mu\nu} \, = \, \frac{1}{\mu_{0}} \, g^{\mu\alpha} \, F_{\alpha\beta} \, g^{\beta\nu} \, \sqrt{-g} \\ J^{\mu} \, = \, \partial_\nu \mathcal{D}^{\mu \nu}$$

And later they also use the partial derivative, instead of covariant derivative, of the source equation as well:

$$\partial_\mu J^{\mu} \, = \, \partial_\mu \partial_\nu \mathcal{D}^{\mu \nu} = 0$$

Can someone show me why we can interchange covariant derivatives for partial derivative here?

Secondly, I have trouble understanding how this is even possible. If Maxwell's equations don't depend on the Christoffel symbols, then how can light curve? And why when we write it with the potentials does it seem to now depend on the Christoffel symbols and even the Ricci curvature?

For instance the flat space-time, inertial frame equations: $$\partial^\mu \partial_\mu A^\nu = - \mu_0 J^\nu $$ apparently turn out to generalize to: $$\nabla^\mu \nabla_\mu A^\nu = - \mu_0 J^\nu + {R^{\nu}}_{\mu} A^{\mu} $$ So now we need the covariant derivatives and even an explicit curvature term. I'm having trouble understanding / reconciling this with Maxwell's equations seeming to be independent of that. Are Maxwell's equations kind of "misleading" / "hiding" subtleties from us? What is going on?

  • 1
    $\begingroup$ Regarding the source terms, and to answer with the notation you seem comfortable with, I think you may have just forgotten that the tensor is weight +1. So there is an extra term when expanding with the Christoffel symbols: $\nabla_a D^{ab} = \partial_a D^{ab} + {\Gamma^a}_{ca} D^{cb} + {\Gamma^b}_{ca} D^{ac} - {\Gamma^c}_{ca} D^{ab}$. The second Christoffel term goes away because of symmetry in the Christoffel vs antisymmetry in $D$. The other two cancel each other after considering arbitrariness of repeated index labels and the symmetry of the Christoffel symbol. $\endgroup$ – PPenguin Sep 12 '17 at 14:46

Replacing covariants with partials:

The source equation you cite involves a skew-symmetric tensor density.

You may know that if $\nabla$ is the Levi-Civita connection, you can calculate divergences of vector fields without using the connection coefficients/Christoffel symbols: $$ \nabla_\mu X^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(X^\mu\sqrt{-g}). $$

Now, let $\mathcal{J}^\mu=X^\mu\sqrt{-g}$ be a vector density. Then $$ (\nabla_\mu X^\mu)\sqrt{-g}=\nabla_\mu(X^\mu\sqrt{-g})=\nabla_\mu\mathcal{J}^\mu=\partial_\mu(X^\mu\sqrt{-g})=\partial_\mu\mathcal{J}^\mu, $$ so vector density fields can be differentiated partially.

However, as it turns out the situation is the same for arbitrary (k,0)-type antisymmetric tensor fields: Let $F^{\mu\nu}$ be such a field. Then $$ \nabla_\nu F^{\mu\nu}=\partial_\nu F^{\mu\nu}+\Gamma^\mu_{\ \nu\sigma}F^{\sigma\nu}+\Gamma^{\nu}_{\ \nu\sigma}F^{\mu\sigma}, $$ but the second term here vanishes because $\Gamma$ is symmetric in the lower indices, but $F$ is skew-symmetric in the upper indices, so we're left with $$ \nabla_\nu F^{\mu\nu}=\partial_\nu F^{\mu\nu}+\Gamma^\nu_{\ \nu\sigma}F^{\mu\sigma}=\partial_\nu F^{\mu\nu}+\partial_\sigma\ln\sqrt{-g}F^{\mu\sigma} \\ =\frac{1}{\sqrt{-g}}\partial_\nu(F^{\mu\nu}\sqrt{-g}). $$

Defining $\mathcal{F}^{\mu\nu}=F^{\mu\nu}\sqrt{-g}$ gives $$ \nabla_\nu \mathcal{F}^{\mu\nu}=\partial_\nu \mathcal{F}^{\mu\nu}. $$

Dependence of Maxwell's equations on the (pseudo-)Riemannian structure:

Remember that the fundamental field here is $A_\mu$, which is a covector field. The only Maxwell equation that doesn't depend on the Riemannian structure is $\partial_{[\mu} F_{\nu\sigma]}=0$, because you can replace the covariants here with partials do to skew-symmetry.

Also do remember that $F_{\mu\nu}=\partial_\mu A_\nu -\partial_\nu A_\mu$ is also well-defined without covariants.

When we get to the source equation things change, however, because

  • to define divergence we need upper indices, however $F$ is lower-indiced by nature, so we need the metric to raise indices;
  • you can replace covariants with partials there only if you multiply with $\sqrt{-g}$ to create densities. Which, of course, depends on the metric.

So you can weasel out of using the connection (which is great for computations), but you cannot weasel out of using the metric, therefore Maxwell's equations are absolutely not topological.

An aside on differential forms:

You should read about differential forms. I am trying to think of reference that should be quick-readable by physicist. Probably Flanders' book is a good one. Otherwise Anthony Zee's General Relativity and QFT books also contain differential forms but only in a heuristic manner. Sean Carroll's GR book also has an OK recount of them.

Basically, differential forms are totally antisymmetric covariant tensor fields. Instead of using index notation as in, say $\omega_{\mu_1,...,\mu_k}$ to denote them, usually 'abstract' notation is used as $\omega=\sum_{\mu_1<...<\mu_k}\omega_{\mu_1...\mu_k}\mathrm{d}x^{\mu_1}\wedge...\wedge\mathrm{d}x^{\mu_k}$, where the basis is written out explicitly. The wedge symbols are skew-symmetric tensor products.

Differential forms are good because they generalize vector calculus to higher dimensions, arbitrary manifolds and also to cases when you don't have a metric. Differential forms can be differentiated ($\omega\mapsto\mathrm{d}\omega$), where the "$\mathrm{d}$" operator, called the exterior derivative, turns a $k$-form into a $k+1$ form without the need for a metric or a connection, and generalizes grad, div and curl, all in one.

The integral theorems of Green, Gauss and Stokes are also generalized.

The point is, if you also have a metric, the theory of differential forms is enriched. You get an option to turn $k$-forms into $n-k$ forms ($n$ is the dimension of your manifold), and also to define a "dual" operation to the exterior derivative, called the codifferential. The codifferential essentially brings the concept of divergence to differential forms.

Written with differential forms, Maxwell's equations are given by $$ \mathrm{d}F=0 \\ \mathrm{d}^\dagger F=kJ, $$ where $\mathrm{d}^\dagger$ is the codifferential, and $k$ is some constant I care not about right now.

I am noting two things:

  • The $F$ field strength 2-form is given by $F=\mathrm{d}A$, where $A$ is ofc the 4-potential. The exterior derivative satisfies $\mathrm{d}^2=0$ (think of $\text{div}\ \text{curl}=0$ and $\text{curl}\ \text{grad}=0$), so with potentials, the first equation is $\mathrm{d}F=\mathrm{d}^2A=0$, which is trivially true.
  • The first equation contains only $\mathrm{d}$, which is well-defined without a metric. The second equation depends on the codifferential $\mathrm{d}^\dagger$, which does depend on the metric. There is your metric dependance!
  • $\begingroup$ Explain the downvote please. The answer addresses the question and is factually correct. $\endgroup$ – Bence Racskó Sep 24 '17 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.