The importance of metric signature in Ricci scalar

Question

I have read this question Different signatures of the metric in Einstein field equations (and related posts) on the invariance of Einstein field equations under metric signature change. However, there is still something not clear to me.

In Carroll's book (which uses (- + + + )), the first-order Ricci scalar is (eq. (7.7)): \begin{equation}\tag{1} R^{(1)} = \partial_{\mu}\partial_{\nu}h^{\mu\nu}- \Box h . \end{equation}

The point is that D'Inverno's book (which uses (+ - - -)) gives exactly the same expression for $R^{(1)}$ (eq. 20.12), while I would have expected the opposite, since $R$ should change sign. The further thing is that $\Box$ actually changes, because it is $-\partial_t+ \nabla^2$ in the signature (- + + +) and the opposite in the other signature. So we have that the second term on the rhs of (1) has an opposite sign, while the first one does not change. In other words, $R^{(1)}$ is neither equal (as it seems) nor opposite (as I would expect) !

What's wrong?

Answer 1

Under a change of signature, the Ricci scalar should change sign, since it's defined as

$$R = g^{\mu\nu} R^\rho{}_{\mu\rho\nu}$$

and the Riemann tensor doesn't change sign. Both terms of the RHS should then change sign under a change of signature.

Fortunately, this does happen. The first term changes sign because if we take the opposite sign for $g_{\mu\nu}$, we also do so for $\eta_{\mu\nu}$ and $h_{\mu\nu}$. In the second term, the D'Alambertian is $\Box = \eta^{\mu\nu} \partial_\mu \partial_\nu$, so indeed it changes sign under a change of signature. The trace $h = \eta^{\mu\nu} h_{\mu\nu}$ has two metrics, so it doesn't change, and everything is fine.