 Solution

Let $\Psi =k_{\alpha }{x}^{\alpha }$. For a nonzero symmetric amplitude, substituting the plane wave into the wave equation gives $-(k_{\sigma }{k}^{\sigma }){ϵ}^{\mu \nu }\cos \Psi =0$. The gauge condition gives $-k_{\sigma }{ϵ}^{\nu \sigma }\sin \Psi =0$. Thus a propagating mode satisfiesThe wavevector is null, so the gravitational wave propagates at the vacuum speed of light. The amplitude is transverse as a four-dimensional tensor in Lorenz gauge in linearized gravity. This condition alone does not set all time components or the trace to zero: residual gauge symmetry of linearized gravity remains. Of the ten symmetric components, four Lorenz constraints and four independent residual coordinate choices leave two physical gravitational wave polarizations.