 Solution

Use the plus-sign transformation rule for $h_{\mu \nu }$ stated in this part. A generator ${\xi }^{\mu }=b^{\mu }\sin \Psi$ with constant $b^{\mu }$ produces the pure-gauge plane wave in linearized gravityBecause $k^2=0$, it obeys both $\square {\xi }^{\mu }=0$ and $k_{\mu }\delta {ϵ}^{\mu \nu }=0$, preserving the imposed Lorenz gauge in linearized gravity.

For an explicit example choose $b^{\mu }=B{k}^{\mu }/(2{\omega }^2)$, with constant small $B$. Since $k\cdot b=0$, this givesIndeed, direct differentiation gives $\delta h_{\mu \nu }=B{k}_{\mu }{k}_{\nu }\cos \Psi /{\omega }^2$, whose trace is zero, so $\delta \varphi =\delta h$. This wave-shaped metric disturbance is only a coordinate description of flat spacetime: its linearized Riemann curvature operator vanishes. All its second-derivative terms are proportional to four factors of $k$, and the antisymmetric curvature combinations cancel.