With the de Sitter–Schwarzschild metric: $$ds^2=-\left(1-\frac{2M}{r}-\frac{\Lambda r^2}{3}\right)dt^2+\left(1-\frac{2M}{r}-\frac{\Lambda r^2}{3}\right)^{-1}dr^2+r^2d\theta+r^2\sin^2\theta d\phi$$

I can calculate the Riemann curvature tensor, Ricci tensor, Ricci Scalar, Weyl tensor explicitly and find all these quantities depend on $\Lambda$. For example, $$R=4\Lambda,\quad C_{0202}=-\frac{M(6M-3r+r^3\Lambda)}{3r^2}.$$

While the Weyl curvature conjecture (cf. Gron & Hervik 2002), $$C_{abcd}C^{abcd}=\frac{48M^2}{r^6},$$ doesn't depend on $\Lambda$, therefore this quantity is same as the counterpart of Schwarzschild spacetime.

Are there some intuitive or physical reasons engendering this result?

  • 1
    $\begingroup$ Well, did you just calculate that contraction at random, or did you have a physical purpose in mind for it? $\endgroup$
    – ACuriousMind
    Aug 10, 2015 at 15:35
  • $\begingroup$ It would be interesting take $-f(r)\,\mathrm dt^2+\frac{1}{f(r)}\mathrm d r^2+r^2\mathrm d\theta+r^2\sin^2\theta\,\mathrm d\theta \mathrm \,d\phi$, compute the scalar, set it to $c\,r^n$ and solve the ODE. Challenge accepted? $\endgroup$
    – Nikolaj-K
    Aug 10, 2015 at 16:28
  • $\begingroup$ @ACuriousMind arxiv.org/abs/gr-qc/0205026v1 Weyl Curvature Conjecture $\endgroup$
    – 346699
    Aug 10, 2015 at 16:37
  • $\begingroup$ @NikolajK This quantity maybe have some physical interest, see Penrose’s Weyl curvature hypothesis $\endgroup$
    – 346699
    Aug 10, 2015 at 16:41
  • $\begingroup$ @user34669: I'm not sure if this is intended as a response to what I propose, is it? $\endgroup$
    – Nikolaj-K
    Aug 10, 2015 at 17:26

1 Answer 1


Yes. The reason is the following.

For any diagonal, 4-D spacetime, such as what you are considering, the de Sitter-Schwarzschild metric, the Riemann tensor has 20 independent components.

To do General Relativity, one must obtain the Ricci tensor which is obtained from contracting the Riemann tensor to obtain 10 independent components:

$R_{ab} - \frac{1}{2}g_{ab}R + \Lambda g_{ab} = k T_{ab}$.

As you can see, the Ricci tensor is "related" to the cosmological constant, in the sense that it appears alongside it in the Einstein equations.

But, the other 10 components from the Riemann tensor are missing, which do not take part in Einstein's equations, but form the basis for the Weyl tensor. In other words, the Weyl tensor represents the free gravitational field, and is not coupled to any matter distribution/cosmological constant.

The Weyl tensor is itself traceless, so, it will not contribute to $R_{ab}$. It is the part of $R_{abcd}$ which is not determined by Einstein's field equations. For example, in a Schwarzschild spacetime, $R_{ab} = 0$, but the non-zero components of $R_{abcd}$ must correspond to non-zero components of $C_{abcd}$.


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