Past exam of the general relativity II course of the University of Oxford / 2021 / 2

For the plane gravitational wave in linearized gravity, the chain rule gives ${\partial }_{\kappa }{\tilde{h}}_{\mu \nu }=n_{\kappa }{\tilde{h}}_{\mu \nu }^{\prime }$. The Lorenz gauge in linearized gravity, called wave gauge in the paper, imposes zero divergence of the trace-reversed metric perturbation. HenceThe wave equation itself holds for any smooth profile of $u$, because $\square {\tilde{h}}_{\mu \nu }=(n^{\kappa }{n}_{\kappa }){\tilde{h}}_{\mu \nu }^{\prime \prime }=0$ for the constant null vector $n$.

Use the active sign convention for the linearized coordinate gauge transformation, $h_{\mu \nu }\mapsto h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$. The paper's prescribed generator has precisely this sign; using the passive convention would require its negative. Direct partial differentiation givesThe middle terms cancel when the indices are symmetrized. ThereforeTo test both equations (3), put $H(u)=\tilde{h}(u)={\eta }^{\kappa \sigma }{\tilde{h}}_{\kappa \sigma }(u)$, the trace of the original metric perturbation. Since $n$ is a null vector, $\hat{h}=0$ and its trace-reversed metric perturbation equals $\hat{h}$. Differentiating the result of part (a) givesIn the second identity, the term with $n^2$ is zero. ThusIn general $\hat{h}$ does not satisfy (3). On an open interval of $u$, both equations hold exactly when $H^{\prime \prime }=0$ there: the first equation forces this for $u\ne 0$, and smoothness gives it at $u=0$ as well. A linearized coordinate gauge transformation preserves the linearized vacuum Einstein equations, but only a residual gauge symmetry of linearized gravity preserves the imposed Lorenz gauge in linearized gravity.

For an explicit failure, use the null coordinates of part (c) and take only ${\tilde{h}}_{uv}={\tilde{h}}_{vu}=u^2/4$ nonzero. Then $H=u^2$, the original field obeys (3), but $\varphi =uv/2$ gives ${\partial }^{\mu }{\overline{\hat{h}}}_{\mu u}=u$ and $\square {\hat{h}}_{uu}=2$.

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.