Linearizing Geometry and the Einstein Tensor

A metric perturbation is a tangent vector to a family of metrics. It can contain both a physical change and a change of identification between spacetimes. Starting from a background solution $\bar g_{\mu\nu}$, this page derives the first variations of the connection, curvature, and Einstein tensor on an arbitrary spacetime. The calculation keeps background equations separate from identities, includes matter and a cosmological constant, and identifies checks that survive on curved black-hole backgrounds.

Helpful background. Black-Hole Geometry and Mechanics fixes the background and horizon data used in applications; PDEs and Green Functions for Gravitation supplies the differential-operator viewpoint; Formulations of General Relativity separates covariant equations, constraints, and evolution; Presymplectic Potentials and Currents prepares the variational language used later in this chapter.

A tangent to the space of metrics

Let $g_{\mu\nu}(\varepsilon)$ and matter fields $\psi(\varepsilon)$ be smooth one-parameter families on one differentiable manifold, with

$$g_{\mu\nu}(\varepsilon) =\bar g_{\mu\nu}+\varepsilon h_{\mu\nu}+O(\varepsilon^2), \qquad \psi(\varepsilon)=\bar\psi+\varepsilon\,\delta\psi+O(\varepsilon^2).$$

Bars denote background quantities and all indices on $h_{\mu\nu}$ are moved with $\bar g_{\mu\nu}$. Differentiating $g^{\mu\rho}g_{\rho\nu}=\delta^\mu{}_\nu$ immediately gives

$$\delta g^{\mu\nu}=-h^{\mu\nu}, \qquad \delta\sqrt{-g}=\frac12\sqrt{-\bar g}\,h, \qquad h=\bar g^{\mu\nu}h_{\mu\nu}.$$

These are variations at fixed points of the chosen background identification. A different identification changes $h_{\mu\nu}$ by a Lie derivative; Infinitesimal Diffeomorphisms and Gauge Transformations develops that gauge freedom. The distinction is essential: the calculation below linearizes geometric field equations, while a pure coordinate displacement produces no new physical spacetime.

Connection and curvature variations

Although a connection is not a tensor, the difference of two connections is. Varying the Levi-Civita formula and replacing partial derivatives by the background derivative gives the tensor

$$C^\rho{}_{\mu\nu}[h] \equiv \delta\Gamma^\rho{}_{\mu\nu} =\frac12\bar g^{\rho\sigma} \left( \bar\nabla_\mu h_{\nu\sigma} +\bar\nabla_\nu h_{\mu\sigma} -\bar\nabla_\sigma h_{\mu\nu} \right).$$

The curvature convention is the site’s $[\bar\nabla_\mu,\bar\nabla_\nu]V^\rho =\bar R^\rho{}_{\sigma\mu\nu}V^\sigma$. Varying the definition of the Riemann tensor therefore yields

$$\delta R^\rho{}_{\sigma\mu\nu} =\bar\nabla_\mu C^\rho{}_{\nu\sigma} -\bar\nabla_\nu C^\rho{}_{\mu\sigma}, \qquad \delta R_{\mu\nu} =\bar\nabla_\rho C^\rho{}_{\nu\mu} -\bar\nabla_\nu C^\rho{}_{\rho\mu}.$$

This connection-difference form is safer than memorizing a rearranged wave-operator formula: commuting its derivatives automatically produces the correct background-curvature signs. Contracting only after varying gives

$$\delta R =\bar g^{\mu\nu}\delta R_{\mu\nu} -h^{\mu\nu}\bar R_{\mu\nu},$$

and hence

$$G^{(1)}_{\mu\nu}[h] =\delta R_{\mu\nu} -\frac12\bar g_{\mu\nu}\delta R -\frac12 h_{\mu\nu}\bar R.$$

The derivation requires no vacuum or Einstein-background assumption. The connection-difference construction is also developed in Wald 1984, § 7.5, pp. 184–185, Eqs. (7.5.7)–(7.5.16); the vacuum equation imposed in that application is not needed for the geometric variation itself.

The flat-background operator

Set $\bar g_{\mu\nu}=\eta_{\mu\nu}$ in Cartesian coordinates and raise every derivative index with $\eta^{\mu\nu}$. Substitution of the connection variation gives the complete linearized Einstein, or Fierz–Pauli, operator:

$$\begin{aligned} 2G^{(1)}_{\mu\nu} ={}&\partial^\rho\partial_\mu h_{\nu\rho} +\partial^\rho\partial_\nu h_{\mu\rho} -\Box h_{\mu\nu}-\partial_\mu\partial_\nu h\\ &-\eta_{\mu\nu} \left(\partial^\rho\partial^\sigma h_{\rho\sigma}-\Box h\right). \end{aligned}$$

This is the ungauged equation, not just its wave-operator part (Wald 1984, p. 75, Eq. (4.4.5)). Its divergence vanishes identically: the first and third terms cancel after commuting derivatives, while the second and fourth give precisely the derivative of the scalar in parentheses. Substituting $h_{\mu\nu}=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu$ also gives zero, for an arbitrary smooth $\xi_\mu$. These are two different checks: the first concerns the output of the operator, and the second its gauge kernel.

Coupled metric and scalar equations

Define the complete equation tensor

$$E_{\mu\nu}[g,\psi] =G_{\mu\nu}+\Lambda g_{\mu\nu}-8\pi G T_{\mu\nu}.$$

If $(\bar g,\bar\psi)$ solves $E_{\mu\nu}=0$, the first-order equation is

$$E^{(1)}_{\mu\nu}[h,\delta\psi] =G^{(1)}_{\mu\nu}[h] +\Lambda h_{\mu\nu}-8\pi G\,\delta T_{\mu\nu}=0.$$

Here $\delta T_{\mu\nu}$ must include both the matter-field variation and the explicit metric dependence of $T_{\mu\nu}$. Holding the matter components fixed while varying the metric does not, by itself, enforce the matter equation.

For a concrete calculation take a minimally coupled real scalar with Lagrangian $L_{\mathrm m}=-(\nabla\phi)^2/2-U(\phi)$ and a smooth potential. Define $p_\mu=\bar\nabla_\mu\bar\phi$ and $\varphi=\delta\phi$. The exact stress tensor and scalar equation are

$$\begin{aligned} T_{\mu\nu}&=\nabla_\mu\phi\nabla_\nu\phi -g_{\mu\nu}\left[\frac12(\nabla\phi)^2+U(\phi)\right],\\ K&=\Box_g\phi-U'(\phi)=0. \end{aligned}$$

The massive free field is the special case $U=m^2\phi^2/2$ (Wald 1984, p. 70, Eqs. (4.3.9)–(4.3.10)). Varying the displayed expressions, including the inverse metric in $(\nabla\phi)^2$, yields

$$\begin{aligned} \delta T_{\mu\nu} ={}&2p_{(\mu}\bar\nabla_{\nu)}\varphi -h_{\mu\nu}\left(\frac12p^\rho p_\rho+U\right)\\ &-\bar g_{\mu\nu}\left( p^\rho\bar\nabla_\rho\varphi -\frac12h^{\rho\sigma}p_\rho p_\sigma+U'\varphi \right),\\ K^{(1)} ={}&\bar\Box\varphi-U''\varphi -h^{\mu\nu}\bar\nabla_\mu\bar\nabla_\nu\bar\phi\\ &-\left(\bar\nabla_\mu h^{\mu\nu} -\frac12\bar\nabla^\nu h\right)p_\nu . \end{aligned}$$

Here $U,U',U''$ are evaluated at $\bar\phi$. The last line follows from varying the connection in $\Box_g\phi$; its contracted connection term must not be dropped. Compare the explicit scalar wave-operator variation in Aksteiner and Andersson 2011, arXiv v2, p. 17, Eq. (4.5), PDF. The coupled linear problem is $E^{(1)}=0$ together with $K^{(1)}=0$.

Bianchi and matter identities

The exact scalar stress tensor obeys an off-shell identity. Apply the product rule, commute derivatives on the scalar, and differentiate $U$ to obtain

$$\nabla^\mu T_{\mu\nu}=K\nabla_\nu\phi, \qquad \nabla^\mu E_{\mu\nu}=-8\pi G K\nabla_\nu\phi.$$

Now vary on a background satisfying both $\bar E=0$ and $\bar K=0$. Variations of the connection and the inverse metric multiplying $\bar E$ vanish, as does the term $\bar K\,\bar\nabla_\nu\varphi$. What remains is

$$\bar\nabla^\mu E^{(1)}_{\mu\nu} =-8\pi G K^{(1)}p_\nu.$$

Thus the background equations alone do not make the divergence of the coupled metric equation vanish for arbitrary matter perturbations. The four differential relations involve the linearized matter equation. For a vacuum Einstein–$\Lambda$ background the corresponding identity is instead

$$\bar\nabla^\mu\left(G^{(1)}_{\mu\nu}+\Lambda h_{\mu\nu}\right)=0;$$

one may not drop the $\Lambda h_{\mu\nu}$ term when $\Lambda\ne0$. These identities underlie constraint propagation once the appropriate evolution and matter equations hold.

For a pure-gauge perturbation $h_{\mu\nu}=\mathcal L_\xi\bar g_{\mu\nu}$ and $\delta\psi=\mathcal L_\xi\bar\psi$, covariance gives $E^{(1)}=\mathcal L_\xi\bar E=0$. This is a sharper sign check than the flat limit because it retains background curvature.

Two simpler checks remain useful. First, on Minkowski spacetime with a constant perturbation, every derivative in $C^\rho{}_{\mu\nu}$ vanishes, so the linearized curvature must vanish; an algebraic term surviving that test cannot be correct. Second, the Hessian of the complete coupled action on an on-shell background is formally symmetric on two compactly supported field variations. This is a statement about the second variation, not the first-order Einstein–Hilbert density alone; boundary terms matter for noncompact variations. Quadratic Actions, Symplectic Currents, and Canonical Energy develops that variational statement. Neither check by itself tests horizon behavior: a black-hole calculation must additionally use fields regular in a horizon-penetrating chart and verify that its chosen boundary flux is finite.

The perturbation also has dimensions. In coordinates where the metric is dimensionless, $h_{\mu\nu}$ is dimensionless, $C^\rho{}_{\mu\nu}$ has inverse length, and $G^{(1)}_{\mu\nu}$ has inverse length squared. Consequently $8\pi G\,\delta T_{\mu\nu}$ has the same dimension. This catches a common mistake in which a geometrized mass length is inserted as though it were a physical mass while $G$ is still explicit.

What the derivation does and does not establish

On a Ricci-flat background, the trace-reversed field in Lorenz gauge reduces the principal part to a normally hyperbolic operator plus an algebraic Riemann-curvature coupling. That simplification is important, but it is not part of the definition of $G^{(1)}_{\mu\nu}$: matter backgrounds, $\Lambda$, and other gauges retain additional terms. Initial data must also satisfy the linearized constraints, and boundary conditions at a horizon or infinity are needed before the differential equation defines a physical problem.

Exact background construction remains in Exact Solutions of Einstein’s Equations. This page takes a smooth accepted background as input; it does not establish its maximal extension or parameter range. Likewise, full finite-amplitude evolution remains in Black-Hole Dynamics. A small linear residual is evidence about the perturbative equation, not proof that a nonlinear spacetime stays close for arbitrarily long time.

Exercise: a pure-gauge curvature check

Let $h_{\mu\nu}=\mathcal L_\xi\bar g_{\mu\nu}$ on a background that solves the vacuum Einstein equation with cosmological constant. Without expanding all second derivatives of $\xi^\mu$, show that $G^{(1)}_{\mu\nu}[h]+\Lambda h_{\mu\nu}=0$.

Solution

Let $\Phi_s$ be the diffeomorphism generated by $\xi^\mu$. Covariance of the Einstein tensor gives

$$G[\Phi_s^*\bar g]+\Lambda\Phi_s^*\bar g =\Phi_s^*(G[\bar g]+\Lambda\bar g).$$

Differentiate at $s=0$. The left side is $G^{(1)}[\mathcal L_\xi\bar g]+\Lambda\mathcal L_\xi\bar g$; the right side is $\mathcal L_\xi( G[\bar g]+\Lambda\bar g)=0$ by the background equation. The result follows without choosing coordinates. If a direct component calculation gives anything else, a curvature-commutator sign or an index variation has been mishandled.

Exercise: a constant scalar background

Suppose $\bar\phi=\phi_0$ with $U'(\phi_0)=0$ and $\bar G_{\mu\nu}+[\Lambda+8\pi G U(\phi_0)]\bar g_{\mu\nu}=0$. Find $\delta T_{\mu\nu}$ and $K^{(1)}$. Does the divergence of the metric equation imply the scalar equation in this case?

Solution

Here $p_\mu=0$, so

$$\delta T_{\mu\nu}=-U(\phi_0)h_{\mu\nu}, \qquad K^{(1)}=[\bar\Box-U''(\phi_0)]\varphi.$$

The metric divergence vanishes identically because the right side of the linearized Noether identity is proportional to $p_\nu$. It therefore contains no information about $K^{(1)}$ in this example. The scalar Klein–Gordon equation remains an independent equation, even though its fluctuations do not source the metric at first order.

References

• Aksteiner, Steffen, and Lars Andersson. “Linearized Gravity and Gauge Conditions.” Classical and Quantum Gravity 28 (2011): 065001. DOI. Open PDF, arXiv:1009.5647v2.
• Wald, Robert M. General Relativity. Chicago: University of Chicago Press, 1984. Publisher.