The Newtonian Limit of the Einstein Equation

Newtonian gravity emerges from Einstein’s equation only after several approximations are imposed together: the metric is weak, sources and test bodies move slowly, stresses are small compared with energy density, and the field changes slowly on a light-crossing time. Under those assumptions, the time-time metric component becomes $g_{00}=-(1+2\Phi)$ and the field equation becomes $\nabla^2\Phi=4\pi G\rho$. Both the minus sign and the factor $4\pi$ are consequences to be checked, not conventions to be inserted afterward.

Required background. We use The Einstein Equation with its $8\pi G$ normalization, Events, Intervals, and Minkowski Geometry for the almost-inertial expansion, and The Newtonian Gravitational Field and Potential for the acceleration and Poisson equations being recovered.

Helpful background. Tensor Fields and Natural Operations clarifies which statements depend on the background split, Affine and Variational Geodesics supplies the test-particle equation, and The Wave Equation and Characteristics explains why slow time variation turns the wave operator into a spatial Laplacian.

The Newtonian hierarchy

Introduce a bookkeeping parameter $\varepsilon\ll1$ and a source scale $L$. The Newtonian regime requires

$$\frac{v_{\rm test}}{1}=O(\varepsilon), \qquad \frac{v_{\rm source}}{1}=O(\varepsilon), \qquad \lvert\Phi\rvert=O(\varepsilon^2), \qquad \partial_t=O(\varepsilon\,\partial_i).$$

For ordinary slow matter,

$$T_{00}=\rho\,[1+O(\varepsilon^2)], \qquad T_{0i}=O(\rho\varepsilon), \qquad T_{ij}=O(\rho\varepsilon^2).$$

Thus pressure, momentum density, retardation, and nonlinear gravitational self-coupling are subleading. The metric perturbations have the consistent orders $h_{00},h_{ij}=O(\varepsilon^2)$ and $h_{0i}=O(\varepsilon^3)$. These assumptions and their order-by-order consequences are given in Reall 2013, § 8.2, printed pp. 80–83, Eqs. (8.19)–(8.35), PDF.

This hierarchy excludes relativistic pressure, rapidly varying sources, strong fields, and precision post-Newtonian binary dynamics. It also assumes a region in which a Minkowski background and suitable decay or matching conditions are available.

The assumptions are logically distinct. A field can be weak but rapidly varying, as for a small-amplitude wave, so stationarity does not follow from $\lvert h\rvert\ll1$. A test particle can move relativistically through a weak field, in which case spatial metric and velocity terms enter at leading observable order. Matter can also move slowly while carrying pressure comparable to its energy density, invalidating the dust-like source hierarchy. The Newtonian limit requires all the stated suppressions together.

Slow geodesics fix the potential sign

Let $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$. For a slowly moving test body in a stationary field, the spatial geodesic equation at leading order is

$$\frac{d^2x^i}{dt^2}=-\Gamma^i{}_{00}+O(\varepsilon^4/L), \qquad \Gamma^i{}_{00}=-\frac12\partial_i h_{00}.$$

Matching this equation to the Newtonian law $d^2x^i/dt^2=-\partial_i\Phi$ requires

$$h_{00}=-2\Phi, \qquad g_{00}=-(1+2\Phi).$$

For an isolated positive mass, $\Phi=-GM/r<0$. Hence $g_{00}=-1+2GM/r$ and

$$-\boldsymbol\nabla\Phi=-\frac{GM}{r^2}\,\mathbf n,$$

which points inward. This is an independent sign check: choosing $g_{00}=-(1-2\Phi)$ with the same Newtonian definition of $\Phi$ would predict repulsion.

The spatial perturbation is not needed for the leading acceleration of a slow particle, but Einstein’s equation determines it at the same amplitude order. In an isotropic weak-field gauge,

$$g_{ij}=(1-2\Phi)\delta_{ij}+O(\varepsilon^4).$$

That spatial curvature becomes essential for null propagation, which is why a time-time-only argument misses half of the general-relativistic light deflection.

Einstein’s equation becomes Poisson’s equation

In harmonic gauge, the trace-reversed perturbation satisfies

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

Stationarity makes $\Box_\eta\simeq\nabla^2$. Slow matter and asymptotically decaying boundary data then give

$$\nabla^2\bar h_{00}=-16\pi G\rho, \qquad \bar h_{0i}=O(\varepsilon\bar h_{00}), \qquad \bar h_{ij}=O(\varepsilon^2\bar h_{00}).$$

Since $\bar h=-\bar h_{00}+O(\varepsilon^4)$, trace reversal gives

$$h_{00}=\frac12\bar h_{00}+O(\varepsilon^4).$$

Using $h_{00}=-2\Phi$ therefore yields

$$\boxed{\nabla^2\Phi=4\pi G\rho}$$

at leading Newtonian order. Equivalently, the trace-reversed form of the full field equation gives $R_{00}\simeq4\pi G\rho$, while direct stationary linearization gives $R_{00}\simeq\nabla^2\Phi$. Agreement of these two routes checks both the factor of two from trace reversal and the sign inherited from the curvature convention.

Dimensions provide a separate check. With $c=1$, $\Phi$ is dimensionless, $G\rho$ has dimension $L^{-2}$, and so does $\nabla^2\Phi$. For a compact isolated source, the Green-function solution is

$$\Phi(\mathbf x) =-G\int\frac{\rho(\mathbf x')}{\lvert\mathbf x-\mathbf x'\rvert}\,d^3x',$$

which reduces to $-GM/r$ outside a spherical source. A non-negligible cosmological constant, relativistic stress, source multipoles, or time dependence changes this leading problem and must be treated explicitly rather than hidden inside $\rho$.

Boundary conditions fix the additive and homogeneous freedom in $\Phi$. For an isolated source, $\Phi\to0$ at spatial infinity selects the displayed Green-function solution. Adding a constant changes the normalization of the stationary time coordinate but not the acceleration, while adding a nontrivial harmonic function represents external tidal data rather than matter inside the chosen region. Poisson’s equation alone therefore does not specify a unique gravitational field without its physical boundary or matching problem.

Check your understanding

A uniform-density sphere has radius $R$, density $\rho_0$, and total mass $M=4\pi\rho_0R^3/3$. Solve Poisson’s equation inside the sphere, requiring regularity at the center and matching to $\Phi=-GM/r$ at $r=R$.

Solution

Spherical symmetry gives $r^{-2}(r^2\Phi')'=4\pi G\rho_0$. Regularity removes the $1/r$ integration term, so $\Phi'=4\pi G\rho_0r/3=GMr/R^3$. Integrating and matching at $R$ gives

$$\Phi(r)=-\frac{GM}{2R^3}(3R^2-r^2), \qquad 0\le r\le R.$$

The acceleration is $-GM\mathbf r/R^3$, directed toward the center. At $r=R$ both the potential and its first derivative match the exterior solution, confirming the sign and normalization.

References

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