When we try to construct a covariant derivative, we impose several conditions on it so that the resulting derivative is unique. However, I can't make sense of the condition of metric compatibility. I realize how we use it to get the connection from the metric or write it in terms of the metric. But I don't see any physical significance for it on its own. It seems like an unduly assumption that just makes the math easier, but doesn't have any manifestation in physical reality. At least not one that's obvious to me.


2 Answers 2


There is a formulation of GR where metric compatibility follows from the equations of motion. If you think of the Einstein-Hilbert action as being a functional of both the connection and the metric (this is called a first-order formulation), and the connection will be determined using the equations of motion. In the absence of matter, or for "standard" matter fields like a scalar field or electromagnetic field with minimal coupling, and if you assume that the connection is torsion-free (symmetric on the lower indices), then the equations of motion for the connection end up being the metric compatibility condition, and the theory is equivalent to general relativity. (See https://en.wikipedia.org/wiki/Palatini_variation). The assumption that the connection is torsion-free is needed for this to work; it is consistent (but more complicated and not necessary to explain observations) to assume there is some non-dynamical torsion.

I think I can detect in your question an undercurrent of dissatisfaction with making a "purely mathematical" assumption without physical reasoning in the usual way GR is presented. I share that feeling, and to that end I'd also like to note that there are other ways to arrive at GR, starting from a more physical set of principles. In particular, you can think of GR as the unique low energy theory of a massless spin-2 particle, with local and Lorentz invariant interactions. See, eg, [1, 2, 3]. Relatedly, Weinberg's textbook on GR takes the point of view that GR is a consequence of the equivalence principle, and does not start from differential geometry.

[1] https://www.amazon.com/Feynman-Lectures-Gravitation-Frontiers-Physics/dp/0813340381

[2] https://inspirehep.net/literature/1461

[3] https://arxiv.org/abs/gr-qc/0411023

  • $\begingroup$ I thought that torsion arises if we give up the torsion-free condition where the connection coefficients are symmetric with respect to the lower two indices. Are those two conditions (metric compatibility and torsion-free connection) related? If so, why do we include them as two independent conditions, if one can be begotten from the other? $\endgroup$
    – user626542
    May 3 at 2:30
  • $\begingroup$ @user626542 You are 100% right. Sorry about that. I went back to the books and edited my answer. I shouldn't have brought up torsion at all. What I should have said (and now say) is that if you assume the connection is torsion-free, then metric compatibility follows from the Einstein-Hilbert action in the Palatini formalism. You don't have to assume the metric is torsion-free, but it is simpler to assume there is no torsion and adding torsion doesn't help to explain any observations. I left in my sales-pitch at the end for a more particle physics way of understanding GR. $\endgroup$
    – Andrew
    May 3 at 2:50
  • $\begingroup$ Admittedly I am not giving a physical intuition for what metric compatibility means, just arguing that the condition can be understood as following from an action principle so isn't completely arbitrary. $\endgroup$
    – Andrew
    May 3 at 2:52
  • $\begingroup$ Do you have any idea why most textbooks mention it as a condition on the derivative then? If it follows from another one, then it's not really a condition, right? That's what I understood from your answer. It's also the same problem I have with demanding that the derivative commutes with contractions. It can be obtained from the condition on the derivative to reduce to normal partial derivatives when acting on scalars, as far as I can tell. $\endgroup$
    – user626542
    May 3 at 3:00
  • $\begingroup$ @user626542 Part of it is that there's a mix of math and physics. From the differential geometry point of view, metric compatibility is a condition, since you can freely choose to assume it or not. From the point of view of physics, it depends on whether you consider the Einstein-Hilbert action in a first-order or second order form. In first order form, it follows from the EOMs. Ultimately I think the justification for using it in GR is that it is the simplest assumption you can make, and it works. I'm sure someone has written a paper exploring what happens if you don't assume it. $\endgroup$
    – Andrew
    May 3 at 3:06

Metric compatibility follows from the Einstein equivalence principle, which asserts that the laws of special relativity hold for local non-gravitational phenomena in a freely falling laboratory. Here, "local" means up to tidal effects that are second-order in the size of the laboratory. A freely falling laboratory is a reference frame in which $\partial_\lambda g_{\mu\nu} = 0$ at a point, so that the metric locally takes the flat-spacetime Minkowski form as in special relativity. Likewise, we must have $\Gamma^\lambda_{\mu\nu} = 0$ in such a frame so that equations involving the covariant derivative reduce to those of special relativity. The requirement that $\Gamma^\lambda_{\mu\nu} = 0$ when $\partial_\lambda g_{\mu\nu} = 0$ is sufficient for metric compatibility.

Note, one can mathematically define many different connections on a given spacetime. For the question about "the" connection to be meaningful, I take it as referring to the connection that matter couples to, i.e., the one that governs physical parallel transport and that appears in covariant equations such as $\nabla_\mu T^{\mu\nu} = 0$.

Define $\Gamma' = \Gamma - \Gamma_{\mathrm{LC}}$ as the difference between an arbitrary connection $\Gamma$ and the Levi-Civita connection $\Gamma_{\mathrm{LC}}$. Since $\Gamma'$ is a tensor, it cannot be made to vanish by a choice of reference frame, and it could be directly measured if matter coupled to it. This would violate the Einstein equivalence principle because it would affect local non-gravitational phenomena in a freely falling laboratory.

Even with metric compatibility, $\Gamma'$ could contain torsion, which would likewise violate the Einstein equivalence principle if coupled to matter, as in Einstein-Cartan theory. On the other hand, teleparallel gravity defines a connection with torsion but, in a special case of the theory that is equivalent to general relativity, $\Gamma'$ drops out of physical predictions and is unobservable.

  • 2
    $\begingroup$ I'm not sure I follow this argument. As far as I'm aware, it's perfectly fine to have non-metricity whilst also respecting the EEP. You cannot uniquely determine $\Gamma^{\mu}_{\nu \lambda}$ just from locally requiring Minkowski. I'm quite sure that the statement 'Metric compatibility follows from the Einstein equivalence principle' is incorrect. $\endgroup$
    – Eletie
    May 3 at 13:10
  • $\begingroup$ @Eletie: An argument that the EEP implies the existence of a metric formulation of gravity is given in Will's "Theory and Experiment in Gravitational Physics", Section 2.3. $\endgroup$ May 3 at 19:53
  • $\begingroup$ @MichaelSeifert In Will's work, how 'metric' theories are defined does not actually rule out theories with non-metricity. You can read about this in the general metric-affine gravity setting here sciencedirect.com/science/article/abs/pii/037015739400111F $\endgroup$
    – Eletie
    May 3 at 20:05
  • $\begingroup$ @MichaelSeifert whether the EEP is violated comes down to things like matter couplings for your specific theory. Just the existence of non-metricity doesn't imply violations of EEP, and hence the argument that EEP $\rightarrow$ metric-compatibility doesn't convince me yet. $\endgroup$
    – Eletie
    May 3 at 20:12
  • 1
    $\begingroup$ @Eletie: I think you're right; I got myself mixed up between a "metric theory" as Will defines it and a theory in which the only derivative operators present are metric-compatible covariant derivatives. $\endgroup$ May 3 at 20:12

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