If I understand this right the Ricci flow on a compact manifold given by

$\partial g_{\mu \nu} = - 2R_{\mu \nu} + \frac{2}{n}\!R_{\alpha}^{\alpha} \,g_{\mu \nu}$

tends to expand negatively curved regions and to shrink positively curved regions.

Looking at the above definition I`m wondering if the parameter n can be used to achieve $\partial g_{\mu \nu} = 0 $ even if the Ricci tensor is not zero such that the validity of physics, that depends on the metric to be constant (as a precondition), could be extrapolated to curved manifolds to describe an expanding universe with a positive cosmological constant?


1 Answer 1


I get the impression that OP is referring to Normalized Ricci Flow (NRF):

$$ \frac{1}{2} \partial_t g_{\mu\nu} ~=~ -R_{\mu\nu} + \frac{\langle R \rangle}{n} g_{\mu\nu}~. $$

Here $\langle R \rangle$ is the average scalar curvature over the full space-time $M$. The average procedure is often weighted with an Einstein-Hilbert Boltzmann factor. It is just a number (as opposed to a space-time dependent scalar quantity).

Also $n$ is the space-time dimension, which is fixed, and hence cannot be easily varied as OP suggests.

  • $\begingroup$ Can someone explain to me the dimensions of the equation because the metric g is dimensionless normally and R the curvature, having units of inverse length squared. $\endgroup$ Nov 22, 2019 at 3:04
  • 1
    $\begingroup$ Apparently here $[g] = [R][t]$, $[g_{\mu\nu}] [x]^2= [g]$, $[R_{\mu\nu}] [x]^2=[R]=1$. $\endgroup$
    – Qmechanic
    Nov 22, 2019 at 22:29

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.