Einstein’s field equation

The stress-energy tensor describes matter. The Einstein tensor describes a particular combination of curvature. Einstein’s field equation relates them:

This is a local differential equation for the spacetime metric, coupled to the matter equations. To make a prediction, solve for a metric and matter configuration together, with suitable initial or boundary conditions. Then use that metric to calculate clock readings, light signals, and free-fall trajectories.

12.1 Reading the field equation#

In local axes with all coordinates measured in length, curvature components have units $\mathrm{m^{-2}}$ and $T_{\mu\nu}$ has units $\mathrm{J/m^3}$. Since

the right-hand side has curvature units. In angular or differently normalized coordinates, individual component units follow their coordinate bases; the tensor equation remains dimensionally consistent.

The equation is nonlinear. The inverse metric appears in contractions; the connection contains $g^{-1}\partial g$; curvature contains $\partial\Gamma+\Gamma\Gamma$. The object being solved for helps define the differential operator acting on itself.

For example, the field equation relates a fluid’s density and pressure to the metric. The fluid equation in Chapter 11 also contains that metric through its connection. Changing the geometry changes how the fluid moves, while changing the fluid changes the source of the geometry. These equations must be solved consistently.

12.2 Why subtract half the trace?#

The contracted Bianchi identity derived in Chapter 9 is

Suppose we seek a particularly simple symmetric curvature tensor of the form

with constant coefficients. Metric compatibility gives $\nabla g=0$, so its divergence is

To make this vanish for arbitrary metrics, choose $B=-A/2$. Rescale the overall equation to set $A=1$. The surviving combination is

The factor $1/2$ is required for the divergence to vanish within this chosen form of the equation. The coefficient multiplying the matter tensor still has to be fixed by measurement.

This does not establish that “the equivalence principle uniquely proves Einstein’s equation.” We selected a metric theory with a particular low-derivative curvature structure. More general curvature actions, extra fields, independent connections, or other assumptions can change the dynamics while retaining coordinate covariance. The equivalence principle guides the local relation between matter and geometry; it does not provide every dynamical postulate by itself.

The action principle in Chapter 14 will supply a second route to precisely the same trace subtraction. That derivation will identify which part of the metric variation produces the trace term.

12.3 Trace reversal: the most useful algebraic rearrangement#

Define

Contract Einstein’s equation with $g^{\mu\nu}$. In four dimensions $g^{\mu\nu}g_{\mu\nu}=4$, so

Thus

Substitute this back into the original equation:

This is the trace-reversed form. It is mathematically equivalent in four dimensions, but often much easier to use. For a perfect fluid, $T=-\epsilon+3p$. Therefore

which explains the pressure result from Chapter 11.

If $T_{\mu\nu}=0$ and $\Lambda=0$, then $R_{\mu\nu}=0$. This does not force the entire Riemann tensor to vanish: Weyl curvature can remain. Black-hole exteriors and gravitational waves provide examples with nonzero vacuum curvature.

If $T=0$ but $T_{\mu\nu}\ne0$, then $R=4\Lambda$ while Ricci curvature still responds to matter. An electromagnetic field is the standard counterexample to the false statement “zero scalar curvature means empty, flat spacetime.”

The coefficient changes with dimension. In $d>2$ dimensions,

The four-dimensional $1/2$ in trace reversal is dimension dependent. The $1/2$ in the definition of the Einstein tensor is not.

12.4 Matching Newtonian gravity#

PREPARATION FOR THIS SECTION

Find the source by enclosing it

How can a field have zero divergence outside a star and still reveal the star’s mass?

Builds onFlux and divergence Mechanics

See the idea

The gravitational field outside a spherical mass weakens as area grows. The outward flux of its inward-pointing acceleration stays negative and constant on every enclosing sphere. A local equation must reproduce that integral fact, including at a source that is too small to resolve.

Work it out

1. 01
   Fix the normalization on a sphere

   For point mass $M$ at the origin, Newtonian acceleration is $\mathbf g=-G_NM\mathbf e_r/r^2$. A sphere has area $4\pi r^2$ and outward normal $\mathbf e_r$. Its flux is independent of radius.

   $$\oint\mathbf g\cdot d\mathbf A=-4\pi G_NM.$$

   Why this step works The inverse-square falloff cancels the growth of sphere area.

2. 02
   Pass from total mass to density

   For smooth mass density $\rho$, enclose a volume $V$ and write $M_V=\int_V\rho\,d^3x$. The divergence theorem replaces the boundary flux by $\int_V\nabla\cdot\mathbf g\,d^3x$. Since arbitrary volumes must agree, $\nabla\cdot\mathbf g=-4\pi G_N\rho$. With $\mathbf g=-\nabla\Phi$, obtain Poisson’s equation.

   $$\nabla^2\Phi=4\pi G_N\rho.$$

   Why this step works A relation for every small volume determines the local density relation.

3. 03
   Keep the point source

   For $\Phi=-G_NM/r$, direct differentiation gives $\nabla^2\Phi=0$ at $r>0$. That calculation excludes the origin. Define the three-dimensional delta distribution by $\int\delta^{(3)}(\mathbf x)f(\mathbf x)d^3x=f(0)$ for smooth compactly supported test functions. The source has mass density $M\delta^{(3)}(\mathbf x)$.

   $$\nabla^2\frac1r=-4\pi\delta^{(3)}(\mathbf x).$$

   Why this step works The distribution records the missing flux at the puncture; it is not an ordinary function of infinite height.

Go deeper

For a smooth localized density and the isolated boundary condition $\Phi\to0$ at infinity, superpose point-source potentials: $\Phi(\mathbf x)=-G_N\int\rho(\mathbf x')/|\mathbf x-\mathbf x'|\,d^3x'$. Applying the distributional Laplacian collapses the integral to $4\pi G_N\rho(\mathbf x)$. The inverse-distance kernel is a Green function for this boundary problem. Different boundaries can require an additional harmonic function with $\nabla^2\Phi_h=0$; the differential equation alone does not choose it.

Further readingDavid Tong · General relativity David Tong · Electromagnetism

Test the idea

FIRST, PREDICT

You find $\nabla^2(-G_NM/r)=0$ for $r>0$. What can you conclude?

The region away from the origin is vacuum.There is no mass anywhere.The inverse-square field violates Poisson’s equation.

Compare the reasoning

The region away from the origin is vacuum.

The computation excludes the singular source, whose mass is recovered by enclosing flux.

There is no mass anywhere.

The omitted origin carries a distributional source.

The inverse-square field violates Poisson’s equation.

Its distributional Laplacian has exactly the point-source normalization.

A hint

Ask which points were in the domain of the differentiation.

NOW CHANGE THE EXAMPLE

Inside a uniform-density ball, write $\Phi=ar^2+b$. Since $\nabla^2r^2=6$, what is $a/(\pi G_N\rho)$?

A hint

Set $6a=4\pi G_N\rho$.

Work through the solution

$a=2\pi G_N\rho/3$, so the requested ratio is $2/3$.

Vacuum at a point and the total source enclosed by a surface are different statements.

We still owe an explanation of $8\pi G_N/c^4$. Temporarily write an unknown coupling $\kappa$ on the right-hand side.

Assume a weak, nearly static field, slow matter and test particles, negligible pressure compared with rest energy, and a region where coordinates are approximately Minkowskian. Let $\Phi$ denote the ordinary Newtonian gravitational potential, with dimensions of velocity squared.

First establish what $\Phi$ has to do with the metric. For a slow particle, the spatial geodesic equation is dominated by its two temporal velocities:

This follows from the proper-time equation using $dt/d\tau\simeq1$ at leading order. Terms involving spatial velocities and the associated nonaffine corrections in $t$ are higher order in this limit.

With $g_{0i}$ negligible and time derivatives negligible,

To reproduce Newton’s $d^2x^i/dt^2=-\partial_i\Phi$, we require

This step is kinematical: it identifies the Newtonian potential through the behavior of slow free fall. It has not yet used Einstein’s field equation.

Next calculate $R_{00}$. With our curvature convention,

Staticity removes the time-derivative term; weakness lets us discard products of first-order connections. What remains is

For slow, pressureless matter,

The trace-reversed source is consequently

Setting $\Lambda=0$ for the local matching calculation,

We also need the Newtonian equation relating potential to matter density. Its normalization follows from the inverse-square force law. For a spherical mass, $\partial_r\Phi=G_NM/r^2$, so the outward flux of $\boldsymbol\nabla\Phi$ through a sphere is $(G_NM/r^2)(4\pi r^2)=4\pi G_NM$.

In Newtonian gravity the contributions from separate masses add. Summing them and applying the divergence theorem from §11.5 gives, for an enclosing volume,

Away from a point source, the flux through a small box is zero because the Hessian trace from §10.5 vanishes. Each enclosed point source contributes its spherical flux. Passing to a smooth density and requiring the relation for every small volume gives Poisson’s equation:

Comparing coefficients yields

The $4\pi$ came from spherical flux. Trace reversal supplied the additional factor of two. The powers of $c$ came from relating temporal curvature to acceleration and energy density to mass density.

Keeping the cosmological constant gives, in this same static weak-field approximation,

For example, a local vacuum solution includes $\Phi_\Lambda=-\Lambda c^2r^2/6$. Its acceleration is $-\boldsymbol\nabla\Phi_\Lambda=+\Lambda c^2\mathbf r/3$: positive $\Lambda$ produces an outward contribution in this approximation. This is not a Newtonian description valid across an arbitrary cosmological spacetime.

12.5 Why the spatial metric matters#

The preceding calculation needed $R_{00}$ at leading order. Computing $G_{00}$ also needs the scalar curvature $R$, which includes spatial metric derivatives. We can see their effect explicitly.

Use two independent small, time-independent functions $\varphi$ and $\psi$:

Slow-particle motion identifies $\varphi=\Phi/c^2$, as just derived. We have not yet assumed a relation between the temporal change $\varphi$ and spatial change $\psi$.

Keeping only first-order terms in these functions, the nonzero connection types are

The indices on spatial derivatives are raised with $\delta^{ij}$ at this order. Products of connection coefficients are second order, so the Ricci formula uses only their derivatives. Substitution gives

For example, the trace of the spatial connection is $\Gamma^k{}_{ki}=-3\partial_i\psi$. Together with $\Gamma^0{}_{0i}=\partial_i\varphi$, it supplies the second derivative of $\psi-\varphi$ in $R_{ij}$. Taking the Einstein combination yields

In the leading static, pressureless Newtonian approximation, $T_{ij}$ is negligible. With $\Lambda=0$, set the displayed $G_{ij}$ to zero. Taking its spatial trace gives $\boldsymbol\nabla^2(\psi-\varphi)=0$, and substitution gives $\partial_i\partial_j(\psi-\varphi)=0$. Thus the difference is at most a constant plus a linear function. Boundary conditions that make both perturbations decay away from an isolated source set this difference to zero: $\psi=\varphi$.

The consistent metric therefore has both $g_{00}\simeq-(1+2\Phi/c^2)$ and $g_{ij}\simeq(1-2\Phi/c^2)\delta_{ij}$. It gives $G_{00}\simeq2\boldsymbol\nabla^2\Phi/c^2$.

If instead we set $\psi=0$ while retaining a nonzero $\varphi$, then $G_{00}$ vanishes to first order. That metric still predicts the chosen slow-particle acceleration, but it fails the density-sourcing part of Einstein’s equation. An approximation sufficient for one measurement can omit terms essential for another calculation.

12.6 Coordinate choices and dependent equations#

A symmetric four-by-four tensor has ten independent components. Einstein’s equation supplies ten component equations, but the Bianchi identity imposes four differential relations among the geometric expressions. We may also choose the four coordinate functions used to label events. This freedom is called coordinate gauge freedom: different labels can describe the same physical geometry. The resulting initial-value system contains constraint equations as well as evolution equations; Chapter 20 will unpack it.

The component equations must satisfy these relations together. In particular, specifying an arbitrary $T_{\mu\nu}$ that fails $\nabla_\mu T^{\mu\nu}=0$ is incompatible with the geometric identity on the other side. The matter equations are part of the problem.

Finally, $T_{\mu\nu}$ here does not contain a universal local gravitational stress tensor added by hand. Gravitational self-interaction is already present in the nonlinear left-hand side. We will return to the important distinction between that fact and the existence of physically meaningful gravitational-wave energy or total mass.

12.7 Checking the size and units of the coupling#

Curvature in an orthonormal frame has units $\mathrm{m^{-2}}$. Energy density has units $\mathrm{J/m^3}=\mathrm{N/m^2}$. To turn the latter into the former, the coupling must have units $\mathrm{N^{-1}}$:

Using $G_N\simeq6.67430\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ gives $c^4/(8\pi G_N)\simeq4.82\times10^{42}\,\mathrm N$. Multiplying curvature by this factor produces the energy-density scale on the other side of Einstein’s equation.

For a concrete scale, matter with negligible pressure and mass density $3000\,\mathrm{kg/m^3}$ has rest energy density about $2.70\times10^{20}\,\mathrm{J/m^3}$. Multiplying by $8\pi G_N/c^4$ gives about $5.60\times10^{-23}\,\mathrm{m^{-2}}$. This is the source of the time-time field-equation component in the matter’s rest frame. Determining the full curvature still requires solving the field equation, but the units and the scale of this contribution are now explicit.