The Constant-Density Interior Solution

The constant-density interior is the simplest exact relativistic fluid sphere. Its geometry can be integrated in closed form and matched smoothly to Schwarzschild vacuum, yet its central pressure diverges already at $2M/R=8/9$, before the surface could reach $R=2M$. The example cleanly separates a coordinate horizon from a matter-supported curvature singularity and demonstrates the sharpness of Buchdahl’s compactness bound: equality is approached only through this singular infinite-central-pressure limit, not attained by a regular fluid star.

Required background. Hilbert Stress–Energy from Metric Variation supplies the perfect-fluid stress tensor.

Helpful background. The Einstein Equation fixes the field equations; The Riemann Curvature Tensor supplies curvature tests; Exact Solutions of Einstein’s Equations locates this family; GR Formulations treats constraints and evolution; and Continuum Kinematics and Balance Laws supplies fluid variables.

This page uses geometrized units $G=c=1$. The symbol $M$ is the length $M_{\mathrm{geom}}=G M_{\mathrm{phys}}$, so $2M/R=8/9$ means $2G M_{\mathrm{phys}}/R=8/9$ with dimensions restored.

Integrating the uniform-density sphere

Let $\rho(r)=\rho_0>0$ for $0\le r\le R$, with $p(R)=0$. The mass equation gives

$$m(r)=\frac{4\pi}{3}\rho_0r^3 =M\frac{r^3}{R^3}, \qquad M=\frac{4\pi}{3}\rho_0R^3.$$

Define

$$A=\sqrt{1-\frac{2M}{R}}, \qquad B(r)=\sqrt{1-\frac{2Mr^2}{R^3}}.$$

Integrating hydrostatic balance and fixing the time normalization by matching at $r=R$ yields

$$ds^2=-\frac14[3A-B(r)]^2dt^2 +\frac{dr^2}{1-2Mr^2/R^3}+r^2d\Omega^2,$$ $$p(r)=\rho_0\frac{B(r)-A}{3A-B(r)}.$$

At the surface $B(R)=A$, so $p(R)=0$ and $g_{tt}(R)=-(1-2M/R)$. The radial metric coefficient likewise equals the exterior Schwarzschild value. These are the elementary matching checks; continuity of the extrinsic curvature is equivalent here to vanishing surface pressure.

The density discontinuity at $R$ is not a thin shell. It makes $m'$ jump, but $m$ and the first and second fundamental forms remain continuous. The idealization is mathematically consistent even though an exactly incompressible relativistic material would have an infinite sound speed and is not a realistic equation of state.

Why the pressure diverges first

The pressure is largest at the center because $B$ decreases with $r$. Since $B(0)=1$,

$$p_c=\rho_0\frac{1-A}{3A-1}.$$

For positive finite pressure one needs $3A-1>0$. The limiting value is therefore

$$A=\frac13 \quad\Longleftrightarrow\quad 1-\frac{2M}{R}=\frac19 \quad\Longleftrightarrow\quad \frac{2M}{R}=\frac89.$$

Equivalently, $R=9M/4$, which is larger than the Schwarzschild radius $2M$. As this limit is approached from below in compactness, $p_c\to+\infty$. Because the trace equation gives

$$R=8\pi(\rho_0-3p),$$

the divergence is invariant. It is not removed by changing the static time coordinate. For $2M/R>8/9$, the denominator $3A-B(r)$ vanishes at a nonzero interior radius, producing a pressure and curvature pole before the center is reached.

This behavior illustrates, but does not by itself prove, Buchdahl’s theorem. For a static, isotropic, spherically symmetric perfect fluid with nonnegative density that does not increase outward, regularity and matching imply $2M/R\le8/9$; equality is attained only as the uniform-density central pressure becomes infinite Buchdahl 1959, pp. 1027–1034. Relaxing isotropy, density monotonicity, or the matter assumptions can change the bound, so $8/9$ is not a universal limit for every exact interior.

Independent limits and geometric checks

For weak compactness $C=M/R\ll1$, expand $A=1-C+O(C^2)$ and $B=1-Cr^2/R^2+O(C^2)$. Then

$$p(r)=\frac{3M^2}{8\pi R^4} \left(1-\frac{r^2}{R^2}\right)+O(C^3/R^2),$$

which is exactly the Newtonian pressure profile of a uniform-density sphere. This checks the normalization of both $M$ and $p$.

At $r=0$, $m=O(r^3)$ and every metric coefficient is an even analytic function of $r$ when $2M/R<8/9$. The apparent polar-coordinate degeneracy of $r^2d\Omega^2$ is therefore a regular center. At $r=R$, substituting $m(R)=M$ recovers the exterior metric coefficients. Finally, setting $M\to0$ at fixed $R$ sends $\rho_0,p\to0$ and the metric to Minkowski space.

The solution’s domain should be stated as $0\le r\le R$ with $0\le2M/R<8/9$ for a regular configuration with finite nonnegative pressure. The formula may be manipulated outside this interval, but it no longer represents the same regular perfect-fluid body.

Common pitfalls

Calling the density jump a surface mass. A step in $\rho$ is not a Dirac delta. A surface stress arises from a jump in extrinsic curvature, which is absent when the pressure vanishes and the matching is performed as above.

Interpreting the solution as realistic incompressible matter. Constant density is an exact geometric benchmark. It does not supply a causal microscopic equation of state or a stability theorem.

Exercises

Verify that the central pressure reduces to the Newtonian value $p_c=3M^2/(8\pi R^4)$ at leading order in $M/R$.

Solution

With $C=M/R$, $A=\sqrt{1-2C}=1-C+O(C^2)$. Hence

$$\frac{p_c}{\rho_0} =\frac{1-A}{3A-1} =\frac{C+O(C^2)}{2-3C+O(C^2)} =\frac{C}{2}+O(C^2).$$

Using $\rho_0=3M/(4\pi R^3)$ gives

$$p_c=\frac{3M}{4\pi R^3}\frac{M}{2R} +O(C^3/R^2) =\frac{3M^2}{8\pi R^4}+O(C^3/R^2).$$

References

• Hans A. Buchdahl, “General Relativistic Fluid Spheres,” Physical Review 116, 1027–1034 (1959). DOI.

Further reading

• Richard C. Tolman, “Static Solutions of Einstein’s Field Equations for Spheres of Fluid,” Physical Review 55, 364–373 (1939). DOI.