Im familiar with both formulations of GR - standard with metric and connection coefficients and that based on orthonormal frames and differential forms (Cartan's structure eqns) in solving Einstein's Field Equations. But i wonder how to relate curvature between them and how to switch between them without starting from "scratch" (vierbeins and structural equations). In one of articles i found relation for energy-momentum tensor :

$ T^{(a)(b)} = e^{(a)}_{\mu}e^{(b)}_{\nu}(G^{\mu\nu}+\kappa\lambda Rg^{\mu\nu}) $

in Rastall gravity, so from "educated guessing" in Einstein case it'll be

$T^{(a)(b)} = e^{(a)}_{\mu}e^{(b)}_{\nu}G^{\mu\nu} $

and therefore:

$G^{(a)(b)} = e^{(a)}_{\mu}e^{(b)}_{\nu}G^{\mu\nu} $

with $(a)$ indices in orthonormal frame and latin indices for general spacetime metric. Is this right way to do it? If so is it standard tensor contraction (i tried it in xact Mathematica notebook and i get wrong results)?

Edit (from comment): Here's part of my computation in xAct package for Mathematica:

DefManifold[M, 4, IndexRange[a, n]];

DefChart[ch, M, {0, 1, 2, 3}, {t[], r[], [Theta][], [Phi][]}, ChartColor -> Orange];




Metric (Morris Thorne wormhole [spherically symmetric and time independent] $g_{\mu\nu}=diag[-Exp[-2a(r)],1/(1-b(r)/r),r^{2},(rsin\theta)^{2}]$):

met3 = CTensor[ DiagonalMatrix[{-Exp[2*aa[r[]]], 1/(1 - bb[r[]]/r[]), r[]^2, (r[]^2) (Sin[[Theta][]])^2}], {-ch, -ch}]

SetCMetric[met3, ch, SignatureOfMetric -> {3, 1, 0}];


cd = CovDOfMetric[met3]

G = Einstein[cd];(einstein tensor for given metric)

That's my transformation matrix/vierbein $e^{\mu}_{(a)} = diag[Exp[-a(r)],(1-b(r)/r)^{1/2},1/r,1/(r sin\theta)]$:

vierb = CTensor[ DiagonalMatrix[{Exp[-aa[r[]]], (1 - bb[r[]]/r[])^(1/2), 1/r[], 1/(r[]*Sin[[Theta][]])}], {ch, -ch}]

ortG = vierb[[Alpha], -c] vierb[[Beta], -d] G[-[Alpha], -[Beta]] (my orthonormal einstein tensor with greek letters for orginal metric $G_{(a)(b)}$)

  • $\begingroup$ hi Adam, welcome to Phys.SE . I am one of the developers of xact. what did you exactly try? Anyway yes, all the repeated indices are contracted as usual. The 'e' s are simply the tranformation matrices. $\endgroup$
    – magma
    Oct 31, 2018 at 23:44
  • $\begingroup$ I've edited answer and added my code. Everything works fine - i just mistype one character in metric definition and that's why i get confused $\endgroup$ Nov 1, 2018 at 18:58
  • $\begingroup$ Great! I guess you read this: arxiv.org/abs/1711.11500 $\endgroup$
    – magma
    Nov 1, 2018 at 22:00
  • $\begingroup$ I suggest you try my xPrint package to avoid typos :-) sites.google.com/site/xprintforxact , but is not yet completely compatible with CTensor framework $\endgroup$
    – magma
    Nov 1, 2018 at 22:10
  • $\begingroup$ I also suggest to look at the xact group for support groups.google.com/forum/m/?hl=en#!forum/xAct $\endgroup$
    – magma
    Nov 5, 2018 at 14:23

1 Answer 1


The only nontrivial relationship is for the connection, as it is not a tensor.

Remember that sometimes the Cartan-Chern formalism (the orthonormal frame approach) is mystified a bit compared to the usual coordinate basis method. The reason for that mystification has to do with the similarities to the principal bundle approach to Riemannian geometry, but let us agree that the difference between the coordinate basis approach and the orthonormal frame approach is just a case of bases.

Given any local frame $e_a$ and dual coframe $\theta^a$ (we do not need these to be anholonomic or orthonormal), a type $(k,l)$ tensor field $T$ has components $$ T^{a_1...a_k}{}_{b_1...b_l}=T(\theta^{a_1},...,\theta^{a_k},e_{b_1},...,e_{b_l}). $$

To avoid cluttering, let us consider a vector field $X$ first. Assume that we have a coordinate basis $\partial_\mu$ with dual frame $dx^\mu$, and an orthonormal frame $e_a$ with dual frame $\theta^a$. Assume the relationship is $$ e_a=e^\mu_a\partial_\mu \ \text{and}\ \theta^a=\theta^a_\mu dx^\mu,$$ with the matrices $e^\mu_a$ and $\theta^a_\mu$ being inverses of one another. The the components of $X$ are related as $$ X^a=X(\theta^a)=X(\theta^a_\mu dx^\mu)=\theta^a_\mu X(dx^\mu)=\theta^a_\mu X^\mu, $$ so you can see that the "coordinate basis" and "orthonormal basis" components are related in a simple way via the matrices $e^\mu_a$ and $\theta^a_\mu$.

So yes, you are right. For example, the curvature tensor in the Cartan-Chern formalism is given by $$ \Omega^a_{\ b}=d\omega^a_{\ b}+\omega^a_{\ c}\wedge\omega^c_{\ b}=\frac{1}{2}R^a_{\ b\mu\nu}dx^\mu\wedge dx^\nu, $$ and if we focus at the coefficients $R^a_{\ b\mu\nu}$, if $R^\rho_{\ \sigma\mu\nu}$ are the curvature components in the coordinate frame, then we have $$ R^\rho_{\ \sigma\mu\nu}=R^a_{\ b\mu\nu}e^\rho_a\theta^b_\sigma. $$

For the connection, the situation is more complicated because it is nontensorial. We can easily see that if we define $$ \omega^a_{\ b}(X)e_a=\nabla_Xe_b=\nabla_X(e^\mu_b\partial_\mu)=X(e^\mu_b)\partial_\mu+e^\mu_b\nabla_X\partial_\mu \\ =de^\mu_b(X)\partial_\mu+e^\mu_b\Gamma^\nu_{\ \mu}(X)\partial_\nu=de^\mu_b\partial_\mu+e^\nu_b\Gamma^\mu_{\ \nu}(X)\partial_\mu=\omega^a_{\ b}(X)e^\mu_a\partial_\mu. $$

Comparing the sides of the last equality gives $$ \omega^a_{\ b}=\theta^a_\mu\Gamma^\mu_{\ \nu}e^\nu_b+\theta^a_\mu de^\mu_b. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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