[Wald General Relativity] 4. Einstein's Equation

This article is one of commentary project.

4.1 The Geometry of Space in Prerelativity Physics; General and Special Covariance

In prerelativity physics it is assumed that space has the manifold structure of $\mathbb{R}^3$. The coordinates of space are Catesian coordinates $(x^1,x^2,x^3)$, and it has rigid grids. Then distance $D$ between two points $x$ and $\bar{x}$  defined $$\begin{align}D^2=(x^1-\bar{x})^2+(x^2-\bar{x}^2)^2+(x^3-\bar{x}^3)^2\end{align}$$ is independent of the choice of Cartesian coordinate system and thus can be viewed as describing an intrinsic property of space.

Distance between two "nearby" points is $$\begin{align}(\delta D)^2=(\delta x^1)^2+(\delta x^2)^2+(\delta x^3)^2.\end{align}$$ This suggests that the metric of space is given by $$\begin{align}ds^2=(dx^1)^2+(dx^2)^2+(dx^3)^2,\end{align}$$ or in index notation, in the Cartesian coordinate basis, we have $$\begin{align}h_{ab}=\sum_{\mu,\nu}h_{\mu\nu}(dx^\mu)_a(dx^\nu)_b\end{align}$$ with $h_{\mu\nu}=\mbox{diag}(1,1,1)$. This definition of $h_{ab}$ is independent of choice of Cartesian coordinate system.

Since the components of the metric in the Cartesian coordinate basis are constants, the ordinary derivative operator of this coordinate system satisfies $$\begin{align}\partial_ab_{bc}=0.\end{align}$$ Then $\Gamma^a_{bc}$ vanishes, so ordinary derivatives commute on all tensors, it means curvature vanishes. Also geodesics is "straight lines".

Special Covariance: The metric of space, equation (4), has a nontrivial group of isometries. (Only metric)

General covariance: Special covariance of the physic law under isometries follows. (Metric & Tensors combination)

4.2 Special Relativity

Spacetime has the manifold structure of $\mathbb{R}^4$. Inertial observers (normal coordiante) can set up a rigid grid coordinates $x^1,x^2,x^3,t$ for each event. This map of spacetime into $\mathbb{R}^4$ is called a global inertial coordinate system. Many different global inertial coordinate systems are possible. The different systems can be put into one-to-one correspondence with elements of the $10$-parameter Poincare group. Thus, the labels at $t=x^0$, $x^1$, $x^2$, $x^3$ of a given event do not have intrinsic meaning. Spacetime interval $I$ between two events $x$ and $\bar{x}$ defined (in units where $c=1$) by $$\begin{align}I=-(x^0-\bar{x}^0)^2+(x^1-\bar{x}^1)^2+(x^2-\bar{x}^2)^2+(x^3-\bar{x}^3)^2\end{align}$$ has the same value of all global inertial coordinate system and thus can be viewed as representing an intrinsic property of spacetime.

(1) suggests that we define the metric of spacetime $\eta_{ab}$ by $$\begin{align}\eta_{ab}=\sum^3_{\mu,\nu=0}\eta_{\mu\nu}(dx^\mu)_a(dx^\nu)_b\end{align}$$ with $\eta_{\mu\nu}=\mbox{diag}(-1,1,1,1)$, where $\{x^\mu\}$ is any global inertial coordinate system. Again, the tensor field $\eta_{ab}$ thereby obtained is independent of the choice of global inertial coordinates. Again, the ordinary derivative operator, $\partial_a$, of the global inertial coordinates satisfies $$\begin{align}\partial_a\eta_{bc}=0.\end{align}$$ Then the curvature of $\eta_{ab}$ vanishes, the geodesics of $\eta_{ab}$ are straight lines. 

Curves are classified as timelike, null, or spacelike according to whether the norm $\eta_{ab}T^aT^b$ of their tangent, is, respectively, negative, zero, or positive. Special relativity asserts that that the paths in spacetime of material particles are always timelike curves. We may parameterize timeliker curves by the proper time $\tau$ defined by $$\begin{align}\tau=\int (-\eta_{ab}T^aT^b)^{1/2}dt,\end{align}$$ where $t$ is an arbitrary parametrization of the curve (arc length parametrization), and $\tau$ is the tangent to the curve in this parametrization. The maximum elapsed time between two events is given by geodesic (i.e. inertial) motion.

The tangent vector $u^a$ to a timelike curve parameterized by $\tau$ is called the 4-velocity of the curve. It follows directly from the definition of $\tau$ that the 4-velocity has unit length, $$\begin{align}u^au_a=-1.\end{align}$$ As already mentioned above, a particle subject to no external forces will travel on a geodesic; i.e., its 4-velocity will satisfy the equation of motion, $$\begin{align}u^a\partial_a u^b=0,\end{align}$$ where $\partial_a$ is the derivative operator associated with $\eta_{ab}$, i.e., the ordinary derivative operator of a global inertial coordinate system. If forces are present, the right-hand side of equation (6) will be nonzero.

All material particles have an attribute known as "rest mass" $m$, which appears as a parameter in the equations of motion when forces are present. The energy-momentum 4-vector, $p^a$, of a particle of mass $m$ is defined by $$\begin{align}p^a=mu^a.\end{align}$$ The energy of a particle as measured by an observer -present at the site of the particle- whose 4-velocity is $v^a$ is defined by $$\begin{align}E=-p_av^a.\end{align}$$ Thus, in special relativity, energy is recognized to be the "time component" of the 4-vector $p^a$.

Continuous matter distributions in special relativity are described by a symmetric tensor $T_{ab}$ called the stress-energy-momentum tensor. For an observer with 4-velocity $v^a$, the component $T_{ab}v^av^b$ is interpreted as the energy density, i.e., the mass-energy per unit volume, as measured by this observer. For normal matter, this quantity will be nonnegative, $$\begin{align}T_{ab}v^av^b\ge 0.\end{align}$$ If $x^a$ is orthogonal to $v^a$, the component $-T_{ab}v^ax^b$ is interpreted as the momentum density of the matter in the $x^a$-direction. If $y^a$ also is orthogonal to $v^a$, then $T_{ab}x^ay^b$ represents the $x^a$-$y^b$ component of the stress tensor of the material. 

A perfect fluid is defined to be a continuous distribution of matter with stress-energy tensor $T_{ab}$ of the form $$\begin{align}T_{ab}=\rho u_au_b+P(\eta_{ab}+u_au_b),\end{align}$$ where $u^a$ is a unit timelike vector field representing the 4-velocity of the fluid.

The equation of motion of a perfect fluid subject to no external forces is simply $$\begin{align}\partial^aT_{ab}=0.\end{align}$$ Writing out equation (11) in terms of $\rho$, $P$, and $u^a$, and projecting the resulting equation parallel and perpendicular to $u^b$, we find: $$\begin{align}u^a\partial_a\rho+(\rho+P)\partial^a u_a=0,\\ (P+\rho)u^a\partial_au_b+(\eta_{ab}+u_au_b)\partial^aP=0.\end{align}$$ In the nonrelativistic limit, $P\ll \rho$, $u^\mu=(1,\vec{v})$, and $vdP/dt\ll |\vec{\nabla}P|$, these equations become $$\begin{align}\frac{\partial\rho}{\partial t}+\vec{\nabla}\cdot(\rho\vec{v})=0,\\ \rho\left\{ \frac{\partial\vec{v}}{\partial t}+(\vec{v}\cdot\vec{\nabla})\vec{v}\right\}=-\vec{\nabla}P,\end{align}$$ so we see that equation (11) reduces to the conservation of mass, equation (14), and Euler's equation (15).

Equation (11) has an important physical interpretation. Consider a family of inertial observers with parallel 4-velocities $v^a$, so that $\partial_bv^a=0$. According the the above interpretation of $T_{ab}$, the quantity $$\begin{align}J_a=-T_{ab}v^b\end{align}$$ represent the mass-energy current density 4-vector of the fluid as measured by these observers. Equation (11) implies $$\begin{align} \partial^aJ_a=0.\end{align}$$ Using Gauss's law equation (17) implies that over the three dimensional boundary, $S$, of any four-dimensional spacetime volume $V$, we have $$\begin{align}\int_SJ_an^adS=0,\end{align}$$ where $n^a$ is the unit normal whose direction. Applying this to the volume shown in Figure 4.1, we see that the energy change in the fluid in this volume equals the time integrated energy flux into the volume. Thus, equation (18) implies conservation of energy. Conversely, conservation of energy as measured by all inertial observers requires equation (11). Thus, more generally, conservation of energy implies that equation (11) must hold for all continuous matter distributions, not just for perfect fluids.

To illustrate the description of fields in special relativity, we will give two examples: the scalar field and the electromagnetic field. Although no classical scalar field exists in nature, it is instructive for many purpose to consider a field $\phi$ satisfying the Klein-Gordon equation, $$\begin{align}\partial^a\partial_a\phi-m^2\phi=0.\end{align}$$ The stress-energy tensor of this scalar field is $$\begin{align}T_{ab}=\partial_a\phi\partial_b\phi-\frac{1}{2}\eta_{ab}(\partial^c\phi\partial_c\phi+m^2\phi^2).\end{align}$$ Again, $T_{ab}$ satisfies the energy condition, equation (9), and is conserved, equation (11). by virtue of the field equation (19).

In prerelativity physics, the electric field $\vec{E}$ and magnetic field $\vec{B}$ each are spatial vectors. In special relativity these fields are combined into a single spacetime tensor field $F_{ab}$ which is antisymmteric in its indices, $F_{ab}=-F_{ba}$. Thus $F_{ab}$ has six independent components. For an observer moving with 4-velocity $v^a$, the quantity $$\begin{align}E_a=F_{ab}v^b\end{align}$$ is interpreted as the electric field measured by that observer, while $$\begin{align}B_a=-\frac{1}{2}\epsilon_{ab}^{cb}F_{cd}v^b\end{align}$$ is interpreted as the magnetic field, where $\epsilon_{abcd}$ is the totally antisymmetric tensor of positive orientation with norm $\epsilon_{abcd}\epsilon^{abcd}=-24$ so that in a right-handed orthonormal basis we have $\epsilon_{0123}=1$.

In terms of $F_{ab}$, Maxwell's equations take the simple and elegant form, $$\begin{align}\partial^aF_{ab}=-4\pi j_b,\\ \partial_{[a}F_{bc]}=0,\end{align}$$ where $j^a$ is the current density 4-vector of electric charge. Note that the antisymmetry of $F_{ab}$ implies that $$\begin{align}0=\partial^b\partial^aF_{ab}=-4\pi\partial^bj_b.\end{align}$$ Thus, Maxwell's equations imply $\partial^bj_b=0$, which, by the same argument as given above for $J_a$, states that electric charge is conserved. The equation of motion of a particle of charge $q$ moving in the electromagnetic field $F_{ab}$ is  $$\begin{align}u^a\partial_au^b=\frac{q}{m}F^b_cu^c,\end{align}$$ which reformulates the usual Lorentz force law in terms of $F_{ab}$.

The stress-energy tensor of the electromagnetic field is $$\begin{align}T_{ab}=\frac{1}{4\pi}\left\{F_{ac}F_b^c-\frac{1}{4}\eta_{ab}F_{de}F^{de}\right\}.\end{align}$$ Again, $T_{ab}$ satisfies the energy condition, equation (9), and if $j^a=0$, we have $\partial^aT_{ab}=0$ by virtue of Maxwell's equations. If $j^a\ne 0$, then the stress-energy $T_{ab}$ of the electromagnetic field alone is not conserved, but the total stress-energy of the field and the charged matter is still conserved.

By the converse of the Poincare lemma, equation (24) implies that there exists a vector field $A^a$ (called the vector potential) such that $$\begin{align}F_{ab}=\partial_aA_b-\partial_bA_a.\end{align}$$ In terms of $A^a$, Maxwell's equations become $$\begin{align}\partial^a(\partial_aA_b-\partial_bA_a)=-4\pi j_b.\end{align}$$ We have the gauge freedom of adding the gradient $\partial_a\chi$ of a function $\chi$ to $A_a$ since, by equation (28), this leaves $F_{ab}$ unchanged. By solving the equation $$\begin{align}\partial^a\partial_a\chi=-\partial^bA_b\end{align}$$ for $\chi$, we may make a gauge transformation to impose the Lorentz gauge condition, $$\begin{align}\partial^aA_a=0,\end{align}$$ in which case, using the commutativity of derivatives in flat spacetime, equation (29) becomes $$\begin{align}\partial^a\partial_aA_b=-4\pi j_b.\end{align}$$

We may seek solutions of Maxwell's equations of the form of a wave oscillating with constant amplitude, $$\begin{align}A_a=C_a\exp (iS),\end{align}$$ where $C_a$ is a constant vector field and the function $S$ is called the phase of the wave. To yield a solution with $j^a=0$, the phase must satisfy $$\begin{align}\partial^a\partial_aS=0,\\ \partial S\partial^aS=0,\end{align}$$ and $$\begin{align}C_a\partial^aS=0.\end{align}$$ Now, for any function $f$ on any manifold with a metric, the vector $\nabla^af$ is normal (i.e., orthogonal) to the surfaces of constant $f$, since for any vector $t^a$ tangent to the surface we have $t^a\nabla_af=0$. Equation (35) states that the normal $k^a=\partial^aS$ to the surfaces of constant $S$ is a null vector, $k_ak^a=0$. We call such a surface a null hypersurface. Note that null hypersurfaces satisfy the property that their normal vector is tangent to the hypersurface. Differentiation of equation (35) yields $$\begin{align}0=\partial_b(\partial_aS\partial^aS)=2(\partial^aS)(\partial_b\partial_aS)=2(\partial^aS)(\partial_a\partial_bS)=2k^a\partial_ak_b;\end{align}$$ i.e., the integral curves of $k^a$ are null geodesics(???). Indeed, since the derivation of equation (37) from equation (35) is valid in curved spacetime as well (with $\nabla_a$ replacing $\partial_a$ everywhere), it follows that all null hypersurface in a Lorentz spacetime are generated by null geodesics. The frequency of the wave (i.e., minus the rate of change of the phase of the wave) as measured by an observer with 4-velocity $v^a$ is given by $$\begin{align}w=-v^a\partial_aS=-v^ak_a.\end{align}$$

The most important solutions of the form, equation ((33), are the plane waves, $$\begin{align}S=\sum^3_{\mu=0}k_\mu x^\mu,\end{align}$$ where $\{x^\mu\}$ are global inertial coordinates and $k_\mu$ are constants (and thus $k^a$ is a constant vector field). 

4.3 General Relativity

Just $\eta_{ab}$ and $\partial_a$ become $g_{ab}$ and $\nabla_a$. Let's define the 4-velocity, $u^a$, of a particle to be the unit tangent to its world line. A free particle satisfies the geodesic equation of motion, $$\begin{align}u^a\nabla_au^b=0,\end{align}$$ where $\nabla_a$ is the derivative operator associated with $g_{ab}$. If the acceleration $a^b=u^a\nabla_au^b$ of the particle is nonvanishing, we say that a force $f^b=ma^b$ acts on the particle, where $m$ is its (rest) mass. For example, if the particle has (rest) mass $m$ and charge $q$, and is placed in an electromagnetic field $F_{ab}$, it satisfies the Lorentz force equation, $$\begin{align}u^a\nabla_au^b=\frac{q}{m}F^b_cu^c.\end{align}$$ The 4-momentum of the particle is defined by $$\begin{align}p^a=mu^a.\end{align}$$ The energy of the particle as determined by an observer who is present at the event on the particle's world line at which the energy is measured is again $$\begin{align}E=-p_av^a\end{align}$$ where $v^a$ is the 4-velocity of the observer. However, there is one important difference here: Because spacetime is curved, there is no well defined notion of vectors at different points being parallel; parallel transport is curve dependent. Thus, there is no natural 'global family of inertial observers, and a given observer cannot, in general, defined the energy of a distant particle.

In general relativity, continuous matter distributions and fields again are described by a stress-energy tensor $T_{ab}$. The stress tensor of a perfect fluid is given by $$\begin{align}T_{ab}=\rho u_au_b+P(g_{ab}+u_au_b),\end{align}$$ and it satisfies the equations of motion $$\begin{align}\nabla^aT_{ab}=0,\end{align}$$ which yield $$\begin{align}u^a\nabla_a\rho+(\rho+P)\nabla^au_a=0,\\ (P+\rho)u^a\nabla_au_b+(g_{ab}+u_au_b)\nabla^aP=0.\end{align}$$ However, the interpretation of equation is altered now. A family of observers is represented by a unit timeliker vector field $v^a$. If one could find such a vector field which is covariantly constant, i.e., $\nabla_av_b=0$ -or for which merely $\nabla_{(a}v_{b)}=0$- then we would have $\nabla^a(T_{ab}v^b)=0$. Applying the curved spacetime version of Gauss's law, we again would obtain strict conservation of energy in the form (18) for the energy-momentum four-vector $J_a=-T_{ab}v^b$ measured by the observers represented by $v^b$, but there is no way to find $v^b$ in global, it fails.

The most natural generalization of the equation satisfies by a Klein-Gordon scalar field to curved spacetime is given by $$\begin{align}\nabla^a\nabla_a\phi-m^2\phi=0.\end{align}$$ The stress tensor of the field is $$\begin{align}T_{ab}=\nabla_a\phi\nabla_b\phi-\frac{1}{2}g_{ab}(\nabla_c\phi\nabla^c\phi+m^2\phi^2)\end{align}$$ and satisfies $\nabla^aT_{ab}=0$.

Maxwell's equations in curved spacetime become $$\begin{align}\nabla^aF_{ab}=-4\pi j_b,\\ \nabla_{[a}F_{bc]}=0.\end{align}$$ The electromagnetic stress tensor again is $$\begin{align}T_{ab}=\frac{1}{4\pi}\left\{F_{ac}F^c_b-\frac{1}{4}g_{ab}F_{de}F^{de}\right\}.\end{align}$$ Again, equation (13) allows us to introduce a vector potential $A_a$ (at least locally). However, Maxwell's equations for $A_a$ in the Lorentz gauge, contains an explicit curvature term resulting from the commutation of derivatives in the derivation of equation (32); we find $$\begin{align}\nabla^a\nabla_aA_b-R^d_bA_d=-4\pi j_b.\end{align}$$ We can choose other gauge to throw $R$.

In situations where the spacetime scale of variation of the electromagnetic field much smaller than that of the curvature, one would expect to have solutions of Maxwell's equations of the form of a wave oscillating with nearly constant amplitude, i.e., solutions of the form $$\begin{align}A_a=C_ae^{iS},\end{align}$$ where derivatives of $C_a$ are "small". Substituting equation (16) into equation (15) with $j_b=0$ and neglecting the "small" term $\nabla_b\nabla^bC_a$ as well as the Ricci tensor term yields the condition $$\begin{align}\nabla_a S\nabla^aS=0;\end{align}$$ i.e., we find again that the surfaces of constant phase are null, and thus $k_a=\nabla_aS$ is tangent to null geodesics. This suggests that, in this approximation, light travels on null geodesics, a suggestion which can be confirmed by studies of the Green's function.

The equation describes the relation between spacetime geometry and the matter distribution is from Newtonian gravity. In the Newtonian theory, the gravitational field may be represented by a potential, $\phi$, and the tidal acceleration of two nearby particles is given by $-(\vec{x}\cdot\vec{\nabla})\vec{\nabla}\phi$, where $\vec{x}$ is the separation vector of the particles. On the other hand, in general relativity, from equation (18) the tidal acceleration of two nearby particle is given by $-R_{cbd}^av^cx^bv^d$, where $v^a$ is the 4-velocirt of the particles and $x^a$ is the deviation vector. This suggests that we make the correspondence, $$\begin{align}R_{cbd}^av^cv^d\leftrightarrow \partial_b\partial^a\phi.\end{align}$$ However, Poisson's equation tells us that $$\begin{align}\nabla^2\phi=4\pi\rho,\end{align}$$ where $\rho$ is the mass (i.e., energy) density of matter and we remind the reader that we use units where $G=c=1$ here and throughout the text. Furthermore, in special and general relativity the energy properties of matter are described by a stress-energy tensor $T_{ab}$, and we have the correspondence $$\begin{align}T_{ab}v^av^b\leftrightarrow \rho,\end{align}$$ where $v^a$ is the 4-velocity of the observer.

The correspondences (18) and (20) together with equation (10) suggest that we have $R_{cad}^av^cv^d=4\pi T_{cd}v^cv^d$, which suggests the field equation $R_{cd}=4\pi T_{cd}$. Indeed, this equation was originally postulated by Einstein. However, it has a serious defect. As discussed above, the stress tensor satisfies $\nabla^cT_{cd}=0$. On the other hand, the contracted Bianchi identity, equation (31), tells us that $\nabla^c(R_{cd}-\frac{1}{2}g_{cd}R)=0$. Hence equality of $R_{cd}$ and $4\pi T_{cd}$ would imply $\nabla_dR=0$, i.e., that $R$, and hence, $T=T^a_a$, is constant throughout the universe. This is a highly unphysical restriction on the matter distribution, and it forces us to reject this equation, as Einstein quicklt realized.

However, this difficulty also suggests its resolution. If instead we consider the equation $$\begin{align}G_{ab}\equiv R_{ab}-\frac{1}{2}Rg_{ab}=8\pi T_{ab},\end{align}$$ then it's good. Taking the trace of equation (21), we find $$\begin{align}R=-8\pi T,\end{align}$$ and thus, $$\begin{align}R_{ab}=8\pi (T_{ab}-\frac{1}{2}g_{ab}T).\end{align}$$ In situations where Newtonian theory should be applicable, the energy of matter as measured by an observer who is roughly "at rest" with respect to the masses will be much greater than the material stresses, so we have $T\sim -\rho=-T_{ab}v^av^b$. Thus, in this case, equation (23) still leads to $R_{ab}v^av^b\sim 4\pi T_{ab}v^av^b$.

Equation (21) is the desired field equation of general relativity. It was written down by Einstein in 1915 and is known as Einstein's equation. The entire content of general relativity may by summarized as follows" Spacetime is a manifold $M$ on which there is defined a Lorentz metric $g_{ab}$. The curvature of $g_{ab}$ is related to the matter distribution in spacetime by Einstein's equation (21).

4.4 Linearized Gravity: The Newtonian Limit and Gravitational 

We simply shall assume that the deviation, $\gamma_{ab}$, of the actual spacetime metric $$\begin{align}g_{ab}=\eta_{ab}+\gamma_{ab}\end{align}$$ from a flat metric $\eta_{ab}$ is "small".We mean by "linearized gravity" that approximation to general relativity which is obtained by substituting equation (1) for $g_{ab}$ in Einstein's equation and retaining only the terms linear in $\gamma_{ab}$.

We denote by $\partial_a$ the derivative operator associated with the flat metric $\eta_{ab}$. Inverse metric is $$\begin{align}g^{ab}=\eta^{ab}-\gamma^{ab}.\end{align}$$

The linearized Einstein equation can be obtained in a straightforward manner as follows. In a global inertial coordinate system, to linear order in $\gamma_{ab}$ the Christoffel symbol is $$\begin{align}\Gamma^c_{ab}=\frac{1}{2}\eta^{cd}(\partial_a\gamma_{bd}+\partial_b\gamma_{ad}-\partial_d\gamma_{ab}).\end{align}$$ To linear order in $\gamma_{ab}$, the Ricci tensor (5) is $$\begin{align}R^{(1)}_{ab}=\partial_c\Gamma^c_{ab}-\partial_a\Gamma^c_{cd}=\partial^c\partial_{(b}\gamma_{a)c}-\frac{1}{2}\partial^c\partial_c\gamma_{ab}-\frac{1}{2}\partial_a\partial_b\gamma,\end{align}$$ where $\gamma=\gamma^c_c$. Hence, the Einstein tensor to linear order is $$\begin{align}G^{(1)}_{ab}=R^{(1)}_{ab}-\frac{1}{2}\eta_{ab}R^{(1)}=\partial^c\partial_{(b}\gamma_{a)c}-\frac{1}{2}\partial^c\partial_c\gamma_{ab}-\frac{1}{2}\partial_a\partial_b\gamma-\frac{1}{2}\eta_{ab}(\partial^c\partial^d\gamma_{cd}-\partial^c\partial_c\gamma).\end{align}$$ This expression can be simplified by defining $$\begin{align}\bar{\gamma}_{ab}=\gamma_{ab}-\frac{1}{2}\eta_{ab}\gamma.\end{align}$$ In terms of $\bar{\gamma}_{ab}$, the linearized Einstein equation is found to be $$\begin{align}G^{(1)}_{ab}=-\frac{1}{2}\partial^c\partial_c\bar{\gamma}_{ab}+\partial^c\partial_{(b}\bar{\gamma}_{a)c}-\frac{1}{2}\eta_{ab}\partial^c\partial^d\bar{\gamma}_{cd}=8\pi T_{ab}.\end{align}$$

There is gauge freedom in general relativity corresponding to the group of diffeomorphims: If $\phi:M\rightarrow M$ is a diffeomorphism of spacetime, the metric $g_{ab}$ and $\phi^*g_{ab}$ represent the same spacetime geometry. In linear approximation, this implies that two perturbations $\gamma_{ab}$ and $\gamma'_{ab}$ represent the same physical perturbation if they differ by the action of an "infinitesimal diffeomorphism" on the flat metric $\eta_{ab}$. An "infinitesimal diffeomorphism" is generated by a vector field ,$\xi^a$, the change in a tensor field induced by such an infinitesimal diffeomorphism defines the Lie derivative. This means that $\gamma_{ab}$ and $\gamma_{ab}+\mathcal{L}_\xi\eta_{ab}$ describe the same physical perturbation. We can express $\mathcal{L}_\xi\eta_{ab}$ in terms of the flat derivative operator $\partial_a$ as $$\begin{align}\mathcal{L}_\xi\eta_{ab}=\partial_a\xi_b+\partial_b\xi_a.\end{align}$$ This means that linearized gravity has a gauge freedom given by $$\begin{align}\gamma_{ab}\rightarrow \gamma_{ab}+\partial_a\xi_b+\partial_b\xi_a\end{align}$$ which is closely analogous to the electromagnetic gauge freedom  $A_a\rightarrow A_a+\partial_a\chi$. 

We may use this gauge freedom to simplify the linearized Einstein equation. By solving the equation $$\begin{align}\partial^b\partial_b\xi_a=-\partial^b\bar{\gamma}_{ab}\end{align}$$ for $\xi^a$, we can make a gauge transformation, equation (9), to obtain $$\begin{align}\partial^b\bar{\gamma}_{ab}=0,\end{align}$$ which is the analog of the Lorentz gauge condition. In this gauge, the linearized Einstein equation simplifies to become $$\begin{align}\partial^c\partial_c\bar{\gamma}_{ab}=-16\pi T_{ab}\end{align}$$ and is closely analogous to Maxwell's equation (32).

4.4a The Newtonian Limit

When gravity is weak, the linear approximation to general relativity should be valid. There exists a global inertial coordinate system of $\eta_{ab}$ such that $$\begin{align}T_{ab}\approx \rho t_at_b,\end{align}$$ where $t^a=(\partial/\partial x^0)^a$ is the "time direction" of this coordinate system. Since the sources are "slowly varying", we expect the spacetime geometry to change slowly as well, and thus we seek solutions of equation (12) where the time derivatives of $\bar{\gamma}_{ab}$ are negligible.

With these assumptions, the components of equation (12) in our global inertial coordinate system become $$\begin{align}\nabla^a \bar{\gamma}_{\mu\nu}=0\end{align}$$ for all $\mu,\nu$ except $\mu=\nu=0$, while $$\begin{align}\nabla^a\bar{\gamma}_{00}=-16\pi\rho,\end{align}$$ where $\nabla^2$ denotes the usual Laplace operator of space. The unique solution of equation (14) which is well behaved at infinity is $\bar{\gamma}=0$.(By gauge) Thus, in the Newtonian limit out solution for the perturbed metric $\gamma_{ab}$ is $$\begin{align}\gamma_{ab}=\bar{\gamma}_{ab}-\frac{1}{2}\eta_{ab}\bar{\gamma}=-(4t_at_b+2\eta_{ab})\phi,\end{align}$$ where $\phi\equiv -\frac{1}{4}\bar{\gamma_{00}}$ satisfies Poisson's equation, $$\begin{align}\nabla^2\phi =4\pi \rho.\end{align}$$

The motion of test bodies in this curved spacetime geometry is governed by the geodesic equation, $$\begin{align}\frac{d^2x^\mu}{d\tau^2}+\sum_{\rho,\sigma}\Gamma^\mu_{\rho\sigma}\left(\frac{dx^\rho}{d\tau}\right)\left(\frac{dx^\sigma}{d\tau}\right)=0,\end{align}$$ where $x^\mu(\tau)$ is the world line of the particle in globale inertial coordinates. For motion much slower that speed of light, we may approximate $dx^a/d\tau$ as $(1,0,0,0)$ in the second term, and the proper time $\tau$ may be approximated by the coordinate time $t$. Thus, we find $$\begin{align}\frac{d^2x^\mu}{dt^2}=-\Gamma^\mu_{00}.\end{align}$$ From out solution, equation (16), we have, for $\mu=1,2,3$: $$\begin{align}\Gamma^\mu_{00}=-\frac{1}{2}\frac{\partial \gamma_{00}}{\partial x^\mu}=\frac{\partial \phi}{\partial x^\mu},\end{align}$$ where, again, time derivatives of $\phi$ have been neglected. Thus, the motion of test bodies is governed by the equation, $$\begin{align}\vec{a}=-\vec{\nabla}\phi,\end{align}$$ where $\vec{a}=d^2\vec{x}/dt^2$ is the acceleration of the body relative to global inertial coordinates of $\eta_{ab}$.

The stress tensor is approximated to linear order in velocity by $$\begin{align}T_{ab}=2t_{(a}J_{b)}-\rho t_at_b,\end{align}$$ where $J_b=-T_{ab}t^a$ is the mass-energy current density 4-vector. The linearized Einstein equation again predicts that the space-space components of $\bar{\gamma}_{ab}$ satisfy the source free wave equation, but the space-time and time-time components now satisfy $$\begin{align}\partial^a\partial_a\bar{\gamma}_{0\mu}=16\pi J_\mu.\end{align}$$ Thus, $A_a\equiv -\frac{1}{4}\bar{\gamma}_{ab}t^b$ satisfies precisely Maxwell's equations in the Lorentz gauge with source $J_a$. If again we assume that the time derivatives of $\bar{\gamma}_{ab}$ are negligible, then the space-space components of $\bar{\gamma}_{ab}$ vanish, and we find that to linear order in the velocity of the test body, the geodesic equation now yields $$\begin{align}\vec{a}=-\vec{E}-4\vec{v}\times \vec{B},\end{align}$$ where $\vec{E}$ and $\vec{B}$ are defined in terms of $A_a$ by the same formulas as in electromagnetism.

4.4b Gravitational Radiation

In the linear approximation the propagation of gravitational radiation is governed by the source-free, linearized Einstein equation. $$\begin{align}\partial^a\bar{\gamma}_{ab}=0,\\ \partial^c\partial_c\bar{\gamma}_{ab}=0.\end{align}$$

In obtaining these equations at the beginning of this section, the gauge choice (25) was made. However, there remains the freedom to make further gauge transformations $\gamma_{ab}\rightarrow \gamma_{ab}+\partial_a\xi_b+\partial_b\xi_a$ provided that $$\begin{align}\partial^b\partial_b\xi^a=0,\end{align}$$ as such transformations leave equation (25) unchanged. This is closely analogous to the fact that in electromagnetism the Lorentz gauge condition does not uniquely fix the vector potential $A_a$; we have the restricted gauge freedom $A_a\rightarrow A_a+\partial_a\chi$ with $$\begin{align}\partial^a\partial_a\chi=0.\end{align}$$ When treating electromagnetic radiation it is convenient to use the remaining gauge freedom to set the component $A_0$ in some global inertial coordinate system equal to zero in a source free region ($j_a=0$). This gauge condition, called the Coulomb or radiation gauge, can be achieved as follows. On a constant time surface $t=t_0$ of our global inertial coordinate system we solve $$\begin{align}\nabla^2\chi=-\vec{\nabla}\cdot\vec{A}.\end{align}$$ We define $\chi$ throughout spacetime to be the solution of equation (28) whose initial value on the $t=t_0$ surface is given by equation (29), and whose initial time derivative is $\partial \chi/\partial t=-A_0$. Then the function $f$ defined by $$\begin{align}f=A_0+\partial \chi/\partial t\end{align}$$ will satisfy $$\begin{align}\partial^a\partial_af=-4\pi j_0\end{align}$$ by equations (32) and (28). Furthermore, on the initial surface $t=t_0$ we have $$\begin{align}f=0,\\ \frac{\partial f}{\partial t}=\frac{\partial A_0}{\partial t}+\frac{\partial^2 \chi}{\partial t^2}=\vec{\nabla}\cdots\vec{A}+\nabla^2\chi=0.\end{align}$$ Hence, if no sources are present in the region under consideration. the unique solution of equation (31) with initial data (32) and (33) is $f=0$, and the gauge transformation $A_a\rightarrow A_a+\partial\chi$ achieves the desired condition $A_0=0$ while maintaining the Lorentz gauge condition.

In a vert similar manner, in the case of linearized gravity we can use the restricted gauge freedom, equation (27), to achieve the radiation gauge $\gamma=0$, $\gamma_{0\mu}=0$ for $\mu=1,2,3$ in a source-free region ($T_{ab}=0$). As an extra bonus, we also obtain $\gamma_{00}=0$ if no sources are present throughout spacetime and good behavior at infinity is assumed. To achieve the radiation gauge, we solve on the initial surface $t=t_0$ the equations $$\begin{align}2\left(-\frac{\partial\xi_0}{\partial t}+\vec{\nabla}\cdot\vec{\xi}\right)=-\gamma,\\ 2[-\nabla^2\xi_0+\vec{\nabla}\cdot(\partial\vec{\xi}/\partial t)]=-\partial \gamma/\partial t,\\ \frac{\partial \xi_\mu}{\partial t}+\frac{\partial \xi_0}{\partial x^\mu}=-\gamma_{0\mu}\ (\mu=1,2,3),\\ \nabla^2\xi_\mu+\frac{\partial}{\partial x^\mu}\left( \frac{\partial \xi_0}{\partial t}\right)=-\frac{\partial \gamma_{0\mu}}{\partial t}\ (\mu=1,2,3),\end{align}$$ to obtain the initial values of $\xi_0,\xi_1,\xi_2,\xi_3$, and their first time derivatives. Then we define $\xi^a$ to be the solution of equation (27) with these initial date. By the same argument as used in the electromagnetic case, the gauge transformation generated by $\xi^a$ will achieve $\gamma=0$ and $\gamma_{0\mu}=0$ ($\mu=1,2,3$) in a source-free region while preserving the gauge condition, equation (25).

Our bonus, $\gamma_{00}=0$, comes about as follows. Since $\gamma=0$, we have $\gamma_{ab}=\bar{\gamma}_{ab}$. Since $\gamma_{0\mu}=0$ for $\mu=1,2,3$, the gauge condition, equation (25), yields $$\begin{align}\frac{\partial \gamma_{00}}{\partial t}=0.\end{align}$$ The linearized Einstein equation (12) then yields $$\begin{align}\nabla^2\gamma_{00}=-16\pi T_{00}.\end{align}$$ But if $T_{00}=0$ throughout the spacetime, the only solution of equation (36) which is well behaved at infinity is $\gamma_{00}=$constant. A further gauge transformation then achieves $\gamma_{00}=0$ without violating any of the previous conditions. 

We employ this radiation gauge to seek solutions of the source-free linearized Einstein equation. Plane waves, $$\begin{align}\gamma_{ab}=H_{ab}\exp \left( i\sum^3_{\mu=0}k_\mu x^\mu\right),\end{align}$$ where $H_{ab}$ is a constant tensor field, will satisfy equation (26) if and only if $$\begin{align}\sum_\mu k^\mu k_\mu=\sum_{\mu,\nu}\eta^{\mu\nu}k_\mu k_\nu=0.\end{align}$$ The radiation gauge condition require (for $\nu=0,1,2,3$) $$\begin{align}\sum^3_{\mu=0}k^\mu H_{\mu\nu}=0,\\ H_{0\nu}=0,\\ \sum^3_{\mu=0}H^\mu_\mu=0.\end{align}$$ Since equations (39a) and (39b) both imply $\sum_\nu H_{0\nu}k^\nu=0$, only eight of these nine equations are independent. Since there are 10 independent components, $H_{\mu\nu}$, this leaves two linearly independent solutions for $H_{ab}$. These two solutions describe the two independent polarization states of plane gravitational waves. An arbitrary well behaved solution of the vacuum linearized Einstein equation, i.e., an arbitrary packet of  gravitational radiation, can be expressed as a superposition of these plane wave solutions.

Green's function as used for a scalar field and in electromagnetism, $$\begin{align}\bar{\gamma}_{\mu\nu}(x)=4\int_\Lambda \frac{T_{\mu\nu}(x')}{\left| \vec{x}-\vec{x}'\right|}dS(x'),\end{align}$$ where $\Lambda$ denotes the past light cone of the point $x$ and the volume element on the light cone is $dS=r^2drd\Omega$.

To analyze this limit in the gravitational case, we Fourier transform all quantities in the time variable, $r$, of our global inertial coordinate system of $\eta_{ab}$, leaving the space variables untransformed. We define $$\begin{align}\hat{\bar{\gamma}}_{\mu\nu}(\omega,\vec{x})=\frac{1}{\sqrt{2\pi}}\int^\infty_{-\infty}\bar{\gamma}_{\mu\nu}(t,\vec{x})e^{i\omega t}dt,\end{align}$$ and we similarly Fourier transform $T_{\mu\nu}$. From equation (42) it follows that $$\begin{align}\hat{\bar{\gamma}}_{\mu\nu}(\omega,\vec{x})=4\int \frac{\hat{T}_{\mu\nu}(\omega,\vec{x}')}{\left| \vec{x}-\vec{x}'\right|}\exp (i\omega \left| \vec{x}-\vec{x}'\right| )d^3x',\end{align}$$ where the extra factor of $\exp (i\omega \left| \vec{x}-\vec{x}'\right|)$ arises from the fact that the original integral (42) was over the past light cone. We need only solve for the space-space components of $hat{\bar{\gamma}}_{\mu\nu}$ since the components $\hat{\bar{\gamma}}_{0\mu}$ are readily obtained in terms of them from the gauge condition (11) which yields $$\begin{align} -i\omega \hat{\bar{\gamma}}_{0\mu}=\sum^3_{\nu=1}\frac{\partial\hat{\bar{\gamma}}_{\nu\mu}}{\partial x^\nu}.\end{align}$$ Since we are interested in calculating the radiation, it suffices to obtain our solution in the "far zome", $R\gg 1/\omega$, where $R$ denotes the distance from the source. For the limit in which we are interseted, the frequencies of interset are sufficiently small that the factor $\exp (i\omega \left| \vec{x}-\vec{x}'\right|)$ varies negligibly over the source, so we may replace $\exp ((i\omega \left| \vec{x}-\vec{x}'\right|)/\left|\vec{x}-\vec{x}'\right|)$ by $\exp (i\omega R)/R$ and pull it out of the integral. We evaluate the remaining integral of the space-space components of $\hat{T}_{ab}$ as follows: $$\begin{align}\int \hat{T}^{\mu\nu}d^3x=\sum^3_{\alpha=1}\left\{ \int \frac{\partial}{\partial x^\alpha}(\hat{T}^{\alpha\nu}x^\mu)-\int \frac{\partial\hat{T}^{\alpha\nu}{\partial x^\alpha}x^\mu\right}=-i\omega \int \hat{T}^{0\nu}x^\mu=-\frac{i\omega}{2}\int (\hat{T}^{0\nu}x^\mu +\hat{T}^{0\mu}x^\nu)=-\frac{i\omega}{2}\sum^3_{\beta=1}\left\{ \int  \frac{\partial}{\partial x^\beta}(\hat{T}^{0\beta}x^\mu x^\nu)-\int \frac{\partial \hat{T}^{0\beta}}{\partial x^\beta}x^\mu x^\nu\right\}=-\frac{\omega^2}{2}\int \hat{T}^{00}x^\mu x^\nu d^3x,\end{align}$$ where in the second and fifth lines we used Gauss's law to get rid of the total divergence and also used the conservation of $T_{ab}$ to express the divergence of spatial components in terms of the time derivative of time components. Thus, we obtain our far-zone solution, $$\begin{align} \hat{\bar{\gamma}}_{\mu\nu}(\omega,\vec{x})=-\frac{2\omega^2}{3}\frac{e^{i\omega R}}{R}\hat{q}_{\mu\nu}(\omega)\quad (\mu,\nu=1,2,3),\end{align}$$ where $\hat{q}_{\mu\nu}$ is the Fourier transform of the quadrupole moment tensor, $$\begin{align}q_{\mu\nu}=3\int T^{00}x^\mu x^\nu d^3x.\end{align}$$ The inverse Fourier transform of equation (47) yields $$\begin{align}\bar{\gamma}_{\mu\nu}(t,\vec{x})=\frac{2}{3R}\left.\frac{d^2q_{\mu\nu}}{dt^2}\right|_{ret}\quad (\mu,\nu=1,2,3),\end{align}$$ where the derivative is evaluated at the retarded time $t'=t-R$. 

The total energy and energy flux of the gravitational field will be quadratic in the field $\gamma_{ab}$. A formula for this energy and energy flux is suggested by the following considerations. The linearized vacuum Einstein equation $$\begin{align}G^{(1)}_{ab}[\gamma_{cd}]=0\end{align}$$ states that the Einstein tensor for the metric $\eta_{ab}+\gamma_{ab}$ vanishes to first order in $\gamma_{ab}$. However, to second order in $\gamma_{ab}$, the vacuum Einstein equation will, in general, fail to be satisfied. Indeed, the terms in the Ricci tensor quadratic in $\gamma_{ab}$ are $$\begin{align}R^{(2)}_{ab}=\frac{1}{2}\gamma^{cd}\partial_a\partial_b\gamma_{cd}-\gamma^{cd}\partial_c\partial_{(a}\gamma_{b)d}+\frac{1}{4}(\partial_a\gamma_{cd})\partial_b\gamma^{cd}+(\partial^d\gamma^c_b)\partial_{[d}\gamma_{c]a}+\frac{1}{2}\partial_d(\gamma^{dc}\partial_c\gamma_{ab})-\frac{1}{4}(\partial^c\gamma)\partial_c\gamma_{ab}-(\partial_d\gamma^{cd}-\frac{1}{2}\partial^c\gamma)\partial_{(a}\gamma_{b)c}.\end{align}$$ Thus, in order to maintain a solution of the vacuum Einstein equation to second order, we must correct $\gamma_{ab}$ by adding to it the term $\gamma^{(2)}_{ab}$, where $\gamma^{(2)}_{ab}$ satisfies $$\begin{align}G^{(1)}_{ab}[\gamma^{(2)}_{cd}]+G^{(2)}_{ab}[\gamma_{cd}]=0,\end{align}$$ where (in the case $R^{(1)}_{ab}=0$) we have $G^{(2)}_{ab}=R^{(2)}_{ab}-\frac{1}{2}\eta_{ab}R^{(2)}$. We may write equation (52) in the form $$\begin{align}G^{(1)}_{ab}[\gamma^{(2)}_{cd}]=8\pi t_{ab},\end{align}$$ where $$\begin{align}t_{ab}=-\frac{1}{8\pi}G^{(2)}_{ab}[\gamma_{cd}].\end{align}$$

$$\begin{align}E=\int_\Sigma t_{00}d^3x\end{align}$$