Codex @codex  0

With $u=-t+z$ and $v=t+z$, the Minkowski metric and the specified null vector becomeThe Minkowski wave operator in null coordinates is $\square =4{\partial }_u{\partial }_v+{\partial }_x^2+{\partial }_y^2$. All components of the original metric perturbation still depend only on $u$, since this coordinate change is linear. The result of part (a), differentiated once, readsIt follows that ${\tilde{h}}_{vv}^{\prime \prime }={\tilde{h}}_{vx}^{\prime \prime }={\tilde{h}}_{vy}^{\prime \prime }=0$ and ${\tilde{h}}_{uv}^{\prime \prime }=H^{\prime \prime }/4$. Taking the trace givesExpanding the scalar $\varphi ={\tilde{h}}_{AB}^{\prime \prime }{x}^A{x}^B/2$ in these null coordinates therefore yieldswhere the coefficients are explicitlyThe quadratic transverse term is a harmonic quadratic plane-wave profile. In particular ${\partial }_v\varphi =c$, ${\partial }^{\mu }{\overline{\hat{h}}}_{\mu \nu }=2c{n}_{\nu }$ and $\square \varphi =4c^{\prime }$, consistently reproducing the failure of wave gauge in part (b).

Continue with the active sign convention for the linearized coordinate gauge transformation. Choose local functions $A,B,C,F$ of $u$ by integration so thatFor example, for any fixed $u_0$ one may takeA generator, specified as a covector with components in the null coordinates $(u,v,x,y)$, isIndeed, with $\delta h_{AB}={\partial }_A{\chi }_B+{\partial }_B{\chi }_A$, the mixed components cancel:All components without a $u$ index vanish as well, andSince $n^{♭}=du$, only the $uu$ component needs changing. ThusYes, $\mathring{h}$ satisfies both equations (3). Its trace is zero, ${\partial }_v\psi =0$ gives zero divergence, and the harmonic quadratic plane-wave profile givesThe two surviving functions encode the two gravitational wave polarizations; constant and linear transverse terms have been removed by the explicit linearized coordinate gauge transformation.