Weak Metric Perturbations and Gauge Freedom

A weak gravitational field is described by a metric close to a fixed background, but the perturbation components are not themselves observables. Around Minkowski spacetime, an infinitesimal coordinate change can alter $h_{\mu\nu}$ while leaving the linearized curvature unchanged. The calculation below identifies that gauge freedom, derives the first-order connection and curvature, and states the weak-field regime in which the split $g=\eta+h$ is meaningful.

Required background. We use The Einstein Equation for the geometric field law, Events, Intervals, and Minkowski Geometry for the background metric, and The Newtonian Gravitational Field and Potential for the later static limit.

Helpful background. Tensor Fields and Natural Operations clarifies the background-dependent tensor split, and The Wave Equation and Characteristics supplies the hyperbolic operator used in harmonic gauge.

Linear geometry about Minkowski spacetime

Choose an almost-inertial chart on a region where

$$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}, \qquad \eta_{\mu\nu}=\operatorname{diag}(-1,1,1,1), \qquad \lvert h_{\mu\nu}\rvert=O(\varepsilon),$$

with $0<\varepsilon\ll1$. Derivatives need not be small merely because amplitudes are small, so linearization also requires products such as $h\,\partial^2h$ and $(\partial h)^2$ to be negligible compared with the retained second derivatives. Indices on first-order quantities are raised with $\eta^{\mu\nu}$. Multiplying the metric and its inverse fixes the first normalization check,

$$g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}+O(\varepsilon^2).$$

The Levi-Civita connection is therefore

$$\Gamma^{(1)\rho}{}_{\mu\nu} =\frac12\eta^{\rho\sigma} \left( \partial_\mu h_{\sigma\nu} +\partial_\nu h_{\sigma\mu} -\partial_\sigma h_{\mu\nu} \right).$$

Keeping only derivatives of this connection in the site’s curvature convention gives

$$R^{(1)}_{\rho\sigma\mu\nu} =\frac12\left( \partial_\mu\partial_\sigma h_{\rho\nu} +\partial_\nu\partial_\rho h_{\sigma\mu} -\partial_\mu\partial_\rho h_{\sigma\nu} -\partial_\nu\partial_\sigma h_{\rho\mu} \right).$$

Contracting yields

$$R^{(1)}_{\mu\nu} =\frac12\left( \partial_\rho\partial_\mu h^\rho{}_\nu +\partial_\rho\partial_\nu h^\rho{}_\mu -\Box_\eta h_{\mu\nu} -\partial_\mu\partial_\nu h \right),$$

where $h=\eta^{\mu\nu}h_{\mu\nu}$ and $\Box_\eta=\eta^{\rho\sigma}\partial_\rho\partial_\sigma$. These formulas, including the trace reversal below, are developed in Reall 2013, § 8.1, printed pp. 77–80, Eqs. (8.1)–(8.18), PDF.

Gauge transformations and invariant curvature

Consider the passive infinitesimal coordinate change $x'^\mu=x^\mu+\xi^\mu(x)$, with $\xi=O(\varepsilon)$. Re-expressing the same physical metric at the same background coordinate value gives

$$h'_{\mu\nu} =h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu =h_{\mu\nu}-2\partial_{(\mu}\xi_{\nu)}.$$

By contrast, the active pullback along the flow generated by the same $\xi^\mu$ gives $\delta_\xi h_{\mu\nu}=+2\partial_{(\mu}\xi_{\nu)}$. The two conventions describe the same gauge equivalence with opposite signs for their respective maps. Substitution into $R^{(1)}_{\rho\sigma\mu\nu}$ produces terms containing three derivatives of $\xi$. They cancel pairwise because partial derivatives commute. Thus

$$\delta_\xi R^{(1)}_{\rho\sigma\mu\nu}=0.$$

This is the independent physical check: a pure-gauge perturbation can make the connection nonzero in a chart, but it cannot create first-order tidal curvature. The statement relies on the flat background. On a curved background, the perturbation of curvature changes by the Lie derivative of the background curvature rather than generally vanishing.

The ten functions $h_{\mu\nu}$ therefore contain four coordinate-gauge functions. This does not mean that every remaining component is independently observable: field equations and constraints impose further relations, and an observable must be tied to curvature, boundary data, or a specified measurement.

There is also a useful converse check. If $R^{(1)}_{\rho\sigma\mu\nu}$ is nonzero in a simply connected local region, no infinitesimal coordinate transformation can set every component of $h_{\mu\nu}$ to zero throughout that region. One can still choose coordinates that simplify $h$ or make the connection vanish at one event, but curvature obstructs eliminating the perturbation over a finite neighborhood. Conversely, zero linearized curvature makes a perturbation locally pure gauge under the usual regularity and topology assumptions; global identifications or boundary conditions can retain information not visible in a local curvature test.

Trace reversal and harmonic gauge

Define the trace-reversed perturbation

$$\bar h_{\mu\nu}=h_{\mu\nu}-\frac12\eta_{\mu\nu}h, \qquad h_{\mu\nu}=\bar h_{\mu\nu}-\frac12\eta_{\mu\nu}\bar h.$$

Under the passive transformation,

$$\partial^\nu\bar h'_{\mu\nu} =\partial^\nu\bar h_{\mu\nu}-\Box_\eta\xi_\mu.$$

Locally, with appropriate initial or boundary conditions, one may solve $\Box_\eta\xi_\mu=\partial^\nu\bar h_{\mu\nu}$ and impose harmonic gauge,

$$\partial^\nu\bar h_{\mu\nu}=0.$$

For vanishing cosmological constant, the linearized Einstein equation then becomes

$$\Box_\eta\bar h_{\mu\nu}=-16\pi G\,T_{\mu\nu}.$$

Gauge fixing is not gauge elimination. Transformations with $\Box_\eta\xi_\mu=0$ preserve harmonic gauge, so boundary conditions and residual gauge freedom still matter. Nor is the background split exact or unique: when $h$ is not uniformly small, when nonlinear terms compete with $\partial^2h$, or when no suitable almost-inertial region exists, this Minkowski linearization is not controlled.

Stress–energy conservation provides one more normalization check. Taking a divergence of the harmonic-gauge equation gives $\Box_\eta(\partial^\nu\bar h_{\mu\nu})=-16\pi G\,\partial^\nu T_{\mu\nu}$. The gauge condition and appropriate boundary data are consistent only with the leading conservation law $\partial^\nu T_{\mu\nu}=0$, the flat-background limit of covariant conservation.

Check your understanding

Let $\xi_0=0$ and $\xi_i=\partial_i\chi(\mathbf x)$ for a smooth, time-independent scalar $\chi$. Starting from $h_{\mu\nu}=0$, find the transformed spatial perturbation and decide whether it describes a Newtonian tidal field.

Solution

The passive transformation gives $h'_{ij}=-2\partial_i\partial_j\chi$ and $h'_{00}=h'_{0i}=0$. Direct substitution in the linearized Riemann tensor gives zero because all fourth derivatives of $\chi$ cancel under index exchange. The new components are therefore pure gauge, not a Newtonian tidal field. A genuine static Newtonian field has $R^{(1)}_{0i0j}=\partial_i\partial_j\Phi$ and cannot be removed when that Hessian is nonzero.

References

• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes (2013), §§ 8.1–8.2, printed pp. 77–83. PDF.