# Is there a "square root" version of the Einstein field equation?

It is well known that the Klein-Gordon equation have a kind of "square root" version : the Dirac equation.

The Maxwell equations can also be formulated in a Dirac way.

It is also well known that the metric of general relativity have a kind of "square root" version : the tetrad field (or vierbein) of components $$e_{\mu}^a(x)$$ : $$\begin{equation}\tag{1} g_{\mu \nu}(x) = \eta_{ab} \, e_{\mu}^a(x) \, e_{\nu}^b(x). \end{equation}$$ Now, a natural question to ask is if the full Einstein equations : $$\begin{equation}\tag{2} G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\, \kappa \, T_{\mu \nu}, \end{equation}$$ could be reformulated for the tetrad field only (or other variables ?), as a kind of a "Dirac version" of it ? In other words : is there a "square root" version of equation (2) ?

• I would be curious to see the "square root" version of the Friedmann-Lemaitre equations, in cosmology, and what interpretation it could have.
– Cham
Sep 22, 2017 at 13:55
• @ Cham, Given that the scaling factor $a$ is only time dependent (no space dependence), the Friedmann-Lemaitre equation is similar to the 1 dimensional Klein–Gordon (with potential terms), and the square root of which is one dimensional Dirac equation with $\psi$ dependent potential terms. Jan 12, 2021 at 22:56
• @MadMax, can you formulate this in a mathematical way, as an answer?
– Cham
Jan 13, 2021 at 0:04
• @MadMax The scale factor a is only purely time dependent in comoving coordinates, which in general won't be the same as the orthonormal frame you're choosing to calculate explicit answers in. Jan 13, 2021 at 8:16

1. Since Nature has fermionic matter we are anyway ultimately forced to rewrite the metric in GR in terms of a vielbein (and introduce a spin connection). See e.g. my Phys answer here. The fermionic matter obeys a Dirac equation in curved spacetime. This however would not amount to a square root of EFE.

2. There exist supersymmetric extensions of GR, such as, SUGRA.

3. Another idea is to consider YM-type theories as a square root of GR, or GR as a double copy of YM. See e.g. the Ashtekar formulation or the KLT relations.

By taking the "Dirac square root" of the Hamiltonian constraint for GR, you naturally end up with Supergravity...so in some appropriate sense, SUGRA "is" a "square root" of GR. For more on this, see:

• Romualdo Tabensky, Claudio Teitelboim, "The square root of general relativity". Physics Letters B 69 no.4 (1977) pp 453-456. Eprint
• Claudio Teitelboim, "Supergravity and Square Roots of Constraints". Phys. Rev. Lett. 38 (1977) 1106. Eprint
• Is there a public version of these papers, available somewhere ? The Eprint from the links above asks for a pesky paid commercial membership.
– Cham
Sep 22, 2017 at 13:49
• @Someone see here and here. Sep 22, 2017 at 13:56
• @Someone if you are unfamiliar with supergravity, a good review is Paulo Vargas Moniz's book Quantum Cosmology - The Supersymmetric Perspective, volume 1, which discusses the constraint structure of SUGRA and how the supercharge is the "squareroot" of the Hamiltonian constraint. Plus it's a decent introduction to the subject of SUGRA in general. Sep 22, 2017 at 14:03
• @AlexNelson, I'm not really familiar with supersymetry, while I've heard of it pretty often. I'm surprised that the "square root" of EFE is actually bringing supersymetry in the picture. It feels weird to me.
– Cham
Sep 22, 2017 at 14:13
• @Someone Yeah, but it's also pretty odd the "square root" of the scalar gives you fermions. The "square root" of fields isn't really as intuitive as...well, as I would like. At any rate, since you're not familiar with SUSY, I would highly recommend reading Moniz's book to learn more about SUGRA. Sep 22, 2017 at 14:24