# What are the analogues of $F_{\mu\nu}$ in General Relativity?

In electromagnetism, the measurable gauge-invariant quantities are the electric and magnetic fields or the six independent components of the field strength tensor $$F_{\mu\nu}$$. What are the analogues of $$F_{\mu\nu}$$ in General Relativity? I have a feeling that $$g_{\mu\nu}$$ is not the analogue of $$F_{\mu\nu}$$ but I am not sure. Any help?

Warning: this answer takes the perspective of the second table, rather than the first, in a question @knzhou linked.

We need to first explain what's analogous to $$A_\mu$$. The Christoffel symbols are. We can then ask what's analogous to $$F_{\mu\nu}$$; the Riemann tensor is. I suspect someone else's answer will provide a rationale for why that differs from what I'll say here, because I will look at it, perhaps unfortunately, from a perspective outside of GR.

Classical electromagnetism and general relativity are both gauge theories; the latter's equivalent of local $$U(1)$$ transformations is general coordinate transformations. In both cases, a very limited symmetry that preserves the partial derivatives of those tensors it preserves is expanded to something for which the preserved notion of a derivative is partial-plus-extra. One need only compare the gauge covariant derivative $$\partial_\mu\phi+qA_\mu\phi$$ (or in a non-Abelian Yang-Mills theory, $$\partial_\mu\phi_a+qf_a^{\:bc}A_{\mu b}\phi_c$$) with the Riemannian connection (aka "covariant derivative") $$\partial_\mu V^\nu+\Gamma_{\mu\rho}^\nu V^\rho$$ to appreciate the analogy.

We'll respectively denote the linear operators used above here as $$D_\mu,\,\nabla_\mu$$. This gives us commutators: $$F_{\mu\nu}\propto[D_\mu,\,D_\nu]$$, while $$[\nabla_\mu,\,\nabla_\nu]V_\rho=R_{\mu\nu\rho\sigma}V^\sigma$$. (The latter lacks a derivative of the vector field, because GR has zero torsion.)

In the above analogy, what's actually analogous to $$g_{\mu\nu}$$ is the above scalar field $$\phi$$.

• Isn't $A$ analogous to the Levi-Civita connection? Sep 11, 2021 at 19:36
• @Filippo if you write $\nabla = \partial + A$, the symbol $\nabla$ is the "Levi-Civita connection" and $A$ is the "Christoffel". Am I missing something? (or maybe it's a matter of semantics?). See this: physics.stackexchange.com/a/1950/226902 Feb 17 at 12:44
• @Quillo If $A$ denotes a covariant derivative, then it is in some sense analogous to the Levi-Civita connection, since it is a covariant derivative, too, right? Feb 17 at 13:28
• @Filippo the covariant derivative is typically written as $\nabla$ or $D$, see physics.stackexchange.com/q/618415/226902 physics.stackexchange.com/q/466911/226902 . Probably your doubt is addressed here: physics.stackexchange.com/a/8152/226902 Feb 17 at 13:33
• @Quillo Thank you for the links! Regarding the last link: I'd say that a connection defines a covariant derivative, not the other way around. Feb 17 at 13:59

I think it's worth saying that there is, in a way, not a great 1-1 analogue here:

the Maxwell action is proportional to $$F^{ab}F_{ab}$$, while, while JG is right that in a lot of ways, the "analogue" to choose for GR is $$R_{abcd}$$, the Hilbert action is proportional to $$g^{ab}R_{ab}$$ not $$R^{ab}R_{ab}$$. The coupling to matter happens via a $$g^{ab}$$ term, too, and not a $$\Gamma$$ term. General covariance just makes nailing down the "real" degrees of freedom of GR much more complex than it is for Electromagnetism, and there are just a lot of terms to worry about.

• It gets even murkier when you seek an analogoue for the source term $A_\mu j^\mu$; the answer then ends up contingent on what other physics we embed in GR.
– J.G.
Sep 10, 2021 at 20:24
• Also, of note here, in the weak-field limit (and the "right coordinates"), the "classical potential" is $|g_{tt}| - 1$ Sep 11, 2021 at 22:21

This answer says roughly the same things as @JG's answer, but phrased slightly differently.

The analogy between the Riemann tensor $$R$$ and the electromagentic field strength tensor $$F$$ is not so apparent when we dress them up with indices $$F_{\mu\nu}$$ and $$R^\rho_{\,\,\mu\nu\sigma}$$. However, things become more apparent if we look slightly more abstractly.

• $$F$$ can be thought of as a $$2$$-form on the spacetime manifold, and it satisfies $$dF=0$$ (the exterior derivative). Or in components, for all $$a,b,c$$, $$\frac{\partial F_{bc}}{\partial x^a}+\frac{\partial F_{ca}}{\partial x^{b}}+\frac{\partial F_{ab}}{\partial x^{c}}=0$$. This equation encodes $$\nabla \cdot B=0$$ and $$\nabla \times E=-\frac{\partial B}{\partial t}$$

• Similarly, on any vector bundle with a linear connection $$(E,\pi,M, \nabla)$$, the curvature $$R$$ of the connection is an $$\text{End}(E)$$-valued $$2$$-form on $$M$$, i.e it is a smooth vector bundle morphism $$R:\bigwedge^2(TM)\to \text{End}(E)$$. Roughly speaking, this says to each point $$x\in M$$, and each pair of vectors $$h_x,k_x\in T_xM$$, we consider the plane/bivector $$h_x\wedge k_x$$, and to such a bivector, we have an endomorphism $$R(h_x\wedge k_x)\in \text{End}(E_x)$$. I explain the intuition more in this MSE answer. Now, one can also prove that $$d_{\nabla}R=0$$; i.e the exterior covariant derivative of the curvature vanishes. In components, this says $$\nabla_a(R_{bc})+\nabla_b(R_{ca})+\nabla_c(R_{ab})=0$$, which is none other than the differential Bianchi identity. In the case where $$E=TM$$ is the tangent bundle (a very common special case), then the description of $$R$$ as an $$\text{End}(TM)$$-valued $$2$$-form on $$M$$ is equivalent to saying the curvature is a $$(1,3)$$-tensor field on $$M$$.

So, both $$F,R$$ are morally speaking the same type of object (a $$2$$-form, the only difference is one is scalar-valued, the other is endomorphism-valued), and both satisfy a form of Bianchi's identity ($$dF=0$$ vs $$d_{\nabla}R=0$$). This is also why you may hear $$F$$ being referred to as a curvature.

From the perspective of Principal Lie-group (G) bundle, the curvature 2-form $$\textbf{F}=\textbf{dA}+\frac{1}{2}[\textbf{A,A}]$$corresponding to Lie group $$G=SU(N)$$ is essentially the Yang-Mills field $$F_{\mu\nu}$$. For $$G=GL(n;C)$$ or $$GL(n;R)$$, the components of curvature 2-form are the Riemann tensor components $$R_{\alpha\beta\gamma\delta}$$.

The propagating d.o.f. for gravitational field is contained in Weyl curvature components $$C_{\alpha\beta\gamma\delta}$$ (traceless part of Riemann tensor) and hence from physics point of view, the analogue for $$F_{ab}$$ in GR can be taken as the Weyl curvature $$C_{abcd}$$. There are certain similarities b/w Maxwell tensor $$F_{ab}$$ and Weyl tensor $$C_{abcd}$$:

Both $$F_{ab}$$ and $$C_{abcd}$$ are traceless satisfying the source free equations $$\nabla ^aF_{ab}=0$$ (Maxwell) and $$\nabla ^aC_{abcd}=0$$ (Einstein). In language of representation theory, one can decompose $$F_{ab}$$ as the $$(1,0)\oplus (0,1)$$ irreducible representation of $$su(2)_L\times su(2)_R$$. This can be expressed using the symmetric spinor $$\phi_{AB}$$: $$F_{ab} =F^{+}_{ab}+F^{-}_{ab} \leftrightarrow \phi_{AB}\epsilon_{A'B'}+\bar{\phi}_{A'B'}\epsilon_{AB}$$Similarly one can decompose Weyl tensor as the $$D(2,0)$$ and $$D(0,2)$$ irreducible representation of $$SL(2,C)$$: $$C_{abcd}=C^{+}_{abcd}+C^{-}_{abcd}\leftrightarrow \Psi_{ABCD}\epsilon_{A'B'}\epsilon_{C'D'}+c.c.$$where $$\Psi_{ABCD}$$ (symmetric) is the gravitational spinor. This kind of decomposition is not possible for Riemann tensor. One can go on further to define analogue of other physical quantities such as:

1. Bel-Robinson tensor $$T_{abcd}=\Psi_{ABCD}\bar{\Psi}_{A'B'C'D'}$$ as the gravitational analogue for free electromagnetic stress energy tensor: $$T_{ab}=\frac{1}{2\pi}\phi_{AB}\bar{\phi}_{A'B'}$$ , both of them satisfying the Rainich's condition and "conservation law".

2. Electric and Magnetic parts of $$C_{abcd}$$ can be defined wrt some unit timelike vector $$u^a$$: $$E_{ab}=C_{abcd}u^cu^d$$ (electric part) and $$H_{ab}=\frac{1}{2}\eta_{ade}{C^{de}}_{bc}u^c$$ (magnetic part). Note, the similarity with Electric and magnetic vectors in Maxwell theory: $$E_a=F_{ab}u^b$$ , $$H_a=\frac{1}{2}\eta_{abc}F^{bc}$$.

I will add a different point of view by using tetrad formalism. As @J.G. said, the Christoffel symbols are roughly the analogs of the gauge connection. In fact, the tetrad formalism has a more accurate analog: the spin connection $$\omega$$. This spin connection is $$\mathfrak{spin}(1,3)$$ valued, just like a gauge connection is valued in some gauge group (for example, in QCD one has a $$\mathfrak{su}(3)$$-valued gauge connection). So one has the following connection: $$\begin{equation} \omega \in \Omega^1_{\mathfrak{spin}(1,3)}(\mathcal{M}) \equiv\Gamma(\mathfrak{spin}(1,3))\otimes_{\Omega^0(\mathcal{M})}\Omega^1(\mathcal{M})\simeq \Gamma(\mathfrak{spin}(1,3)\otimes \wedge^1 T^\ast \mathcal{M}) \end{equation}$$ Where $$\Omega^i(\mathcal{M})\equiv \Gamma(\wedge^i T^\ast \mathcal{M})$$. The curvature of the spin connection is then obviously defined as: $$\begin{equation} \Omega^2_{\mathfrak{spin}(1,3)}(\mathcal{M}) \ni \Omega\equiv d_\omega\omega \stackrel{!}{=}d\omega+[\omega \stackrel{\wedge}{,} \omega]_{\mathfrak{spin}(1,3)} \end{equation}$$ Adding a torsion-free condition $$d_\omega e=0$$, where $$e$$ is the tetrad, one can uniquely define $$\omega$$ in terms of $$e$$. Concretely, $$\Omega$$ defined by $$e$$ is nothing but the analog of the Riemann tensor. But I think that it is more clear why $$\Omega$$ is analogous to $$F$$ in the usual gauge theory framework, using the tetrad formalism. One has the following analogies: $$\begin{equation} \begin{array}{ccc} \omega=\omega_\mu^{IJ}\sigma_{IJ}dx^\mu &\longleftrightarrow &A=A_\mu^a \tau^a dx^\mu \\ \Omega= \Omega_{\mu \nu}^{IJ}\sigma_{IJ}dx^\mu \wedge dx^\nu &\longleftrightarrow &F=F_{\mu \nu}^a \tau^a dx^\mu \wedge dx^\nu \end{array} \end{equation}$$ To go back to the usual formalism in GR, usually, the magnetic part and the electric part of the curvature of space-time are defined by the use of the Weyl tensor, just like @KP99 said. One can ask for a theory where the analogy is stronger: can one has a Lagrangian density defined as the contraction of the Weyl tensor with itself? The answer is Yes, it is the conformal gravity, but it is not strictly equivalent to the Einstein-Hilbert action.

In U(1) gauge theory magnetic and the electric field are gauge invariant quantities. In non-abelian gauge theory however, their analouges are not gauge invariant and therefore not measureable (except if the coupling goes to zero). The only things that are, are the tuneling rate, topological charge and the Lagrangian itself.