Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 56 / 3

Keep only first order in the metric perturbation and use the Minkowski metric to raise and lower indices; set $G=c=1$. The Levi-Civita connection hasThe quadratic products of Christoffel symbols in the Riemann curvature tensor can be dropped at this order. Contracting the remaining derivatives giveswhere $\square ={\eta }^{\rho \sigma }{\partial }_{\rho }{\partial }_{\sigma }=-{\partial }_t^2+{\nabla }^2$ is the d'Alembert operator. In four dimensions the trace-reversed metric perturbation satisfies $\bar{h}=-h$ and $h_{\mu \nu }={\bar{h}}_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }\bar{h}$. Writing the Linearized Einstein equations in terms of it givesIn Lorenz gauge in linearized gravity, the first, second and fourth terms vanish. Therefore $G_{\mu \nu }^{(1)}=8\pi T_{\mu \nu }$ becomesThe divergence of this equation also gives ${\partial }^{\mu }{T}_{\mu \nu }=0$, the leading-order stress-energy conservation required by the contracted Bianchi identity.

For a nonzero, nonconstant vacuum plane gravitational wave in linearized gravity with real wave covector $k$, the Linearized Einstein equations and Lorenz gauge in linearized gravity requireThe zero-amplitude solution places no restriction on $k$; a constant mode $k=0$ is not a propagating wave. Under the sign convention for the infinitesimal change specified here, the trace-reversed metric perturbation changes byThus the residual gauge symmetry of linearized gravity is characterized by $\square {\xi }_{\nu }=0$. The proposed plane-wave form obeys this for every constant complex $X_{\nu }$, because $k$ is null. Changing the sign used to name $\xi$ would reverse both gauge formulas, with no physical consequence.

Rotate the spatial axes so the wave propagates along $+z$, choosing $k_{\mu }=(-\omega ,0,0,\omega )$ with $\omega >0$. The Lorenz gauge in linearized gravity says $H_{3\nu }=-H_{0\nu }$; in particular $H_{03}=-H_{00}$, $H_{33}=H_{00}$ and $H_{3i}=-H_{0i}$ for $i=1,2$. The amplitude change isChooseIndeed, $\delta H_{00}=i\omega (X_3-X_0)$ and $\delta H_{0i}=-i\omega X_i$ cancel the time components, while $\delta H_{11}=\delta H_{22}=-i\omega (X_0+X_3)$ cancels the transverse trace. The preserved Lorenz gauge in linearized gravity then cancels all longitudinal components. This is an explicit plane-wave reduction to transverse-traceless gauge, yieldingThe trace-reversed metric perturbation has zero trace in this transverse-traceless gauge, so $h^{\prime }={\bar{h}}^{\prime }$. The null dispersion relation is ${\omega }^2=|\mathbf{k}{|}^2$, giving the speed of light. Only components perpendicular to the direction of propagation remain, and the two independent amplitudes give the plus and cross gravitational wave polarizations.

These are physical transverse tidal distortions, rather than just a convenient display of the metric perturbation: in transverse-traceless gauge, $R_{0i0j}^{(1)}=-\frac{1}{2}{\partial }_t^2{h}_{ij}^{\prime }$, so a freely falling detector has ${\ddot{S}}^i=\frac{1}{2}{\ddot{h}}_{ij}^{\prime }{S}^j$ at first order. There is no longitudinal tidal acceleration. The plus polarization stretches one transverse axis while compressing the other; the cross polarization does the same along axes rotated by $\pi /4$.

The equivalence principle permits Riemann normal coordinates at any event, in which the first derivatives of the metric tensor, and hence the Christoffel symbols, vanish. A putative universal local gravitational stress-energy tensor constructed from the metric and its first derivatives would vanish in that freely falling description. If it were a genuine tensor, it would then vanish in every description, even where other coordinates assign nonzero gravitational energy density. The gravitational field therefore does not have a local stress-energy tensor analogous to that of matter, with coordinate-independent energy density and the usual energy-balance interpretation.

Gravitational energy is not represented by a universal local covariant stress-energy tensor. This does not assert that all local curvature tensors vanish or are forbidden: the Riemann curvature tensor measures tidal gravity, and higher-curvature tensors can be constructed. They do not provide the sought ordinary local gravitational energy density. Boundary charges, quantities associated with a background, and suitable averaged wave energies have different definitions and evade that local requirement. This is the absence of a local gravitational stress-energy tensor.

Expand the Einstein tensor about the Minkowski metric, keeping a second-order metric perturbation as well:Here $G^{(2)}[h,h]$ includes all quadratic terms from the Einstein tensor, including the inverse-metric corrections in its contractions. In vacuum, the first-order Linearized Einstein equations give $G^{(1)}[h]=0$, while the next order gives $G^{(1)}[j]=-G^{(2)}[h,h]$. Define the quadratic gravitational effective stress-energy tensor byIt is symmetric because the Einstein tensor is symmetric, and is quadratic in $h$ and its derivatives. The linearized contracted Bianchi identity is the off-shell identity ${\partial }^{\mu }{G}_{\mu \nu }^{(1)}[j]=0$. The second-order vacuum equation therefore impliesOne can obtain the same conclusion directly by expanding the full contracted Bianchi identity to second order: every correction to the flat divergence multiplies $G^{(1)}[h]$ and vanishes on the first-order vacuum solution. Thus conservation holds for vacuum solutions of linearized gravity, rather than for an arbitrary off-shell $h$. Relative to the fixed Minkowski background, $t$ is a tensor under background changes of coordinates, but this does not make it invariant under the perturbative gauge freedom.

The quadratic gravitational effective stress-energy tensor depends on the split between the fixed background and the metric perturbation. A residual gauge symmetry of linearized gravity changes $h$ without changing the first-order tidal field, but generally changes $G^{(2)}[h,h]$ and hence the local value of $t$. At second order the compensating change of $j$ restores the same physical geometry; assigning only the quadratic term to an energy density loses this compensation.

Moreover, a stress-energy superpotential can change the local energy expression while preserving its divergence and, with suitable boundary behavior, its integrated charges. Symmetry and flat conservation do not make $t$ a unique, gauge-independent local gravitational energy density. Averaging in an appropriate short-wavelength regime can lead to a useful physical gravitational-wave energy flux, but that is a qualified approximation, not a cure for the local definition requested here.