# Why is the Ricci scalar non-zero in this case?

The Einstein equations can be written as (1): $$R_{ab}-\frac{1}{2}Rg_{ab} = -8\pi GT_{ab}$$

or by contracting the above equation with the metric tensor and resubstituting: (2) $$R_{ab}=8\pi G(\frac{1}{2}Tg_{ab}-T_{ab}).$$

In a vacuum, equation (1) reduces to $R_{ab}-\frac{1}{2}Rg_{ab}=0$ and equation (2) reduces to $R_{ab}=0$, which implies that in a vacuum, $R=0$.

However, if I explicitly calculate $R$ for a plane wave of the form $$h_{ab} = A_{ab}\exp(ikx)$$ (the Minkowski metric $\eta_{ab}$ perturbed by $h_{ab}$), I obtain: $$R=k^ak^bh_{ab}-k^\lambda k_\lambda h\ \ ,$$ where $h=\eta_{ab}h^{ab}$, which looks like some sort of wave equation, but is nonzero. It's supposed to be $0$ but is not. Why?

• The Ricci scalar is zero for a gravitational wave, as is any curvature scalar. See Hans-Juergen Schmidt, "Why do all the curvature invariants of a gravitational wave vanish?" arxiv.org/abs/gr-qc/9404037 . The fact that you get a nonvanishing Ricci scalar tells us that your metric is not a gravitational wave in vacuum, or that you made a mistake in your calculation. – Ben Crowell Mar 6 '18 at 1:28

Gravitational waves are transverse and travel along null rays. Thus, you have to have $k^{a} h_{ab} = 0$ and $k^a k_a = 0$. (More accurately, the transversality condition can be viewed as a gauge condition: we can always apply a local diffeomorphism such that this first condition is true.) Under these conditions, the Ricci scalar does indeed vanish—as should the Ricci tensor, if you've done everything correctly. But a completely arbitrary wave-like metric with an arbitrary polarization $h_{ab}$ and propagation vector $k_a$ will not, as you've found, satisfy Einstein's equation in vacuum.
• Hmm but $R$ is gauge invariant, so you shouldn't need to impose any transversality condition to ensure it vanishes, right? What am I missing here? – AccidentalFourierTransform Mar 5 '18 at 22:31
• @AccidentalFourierTransform The plane wave ansatz given in the question is only a solution to the (linearised) vacuum Einstein equations in transverse gauge. If we move out of this gauge, the form of the metric changes, and so $R$ picks up extra terms that cancel the now non-vanishing $k^a h_{ab}$. – gj255 Mar 6 '18 at 0:47