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The Interior Schwarzschild metric has central pressureIt diverges when $M/R=4/9$, the limiting compactness in Buchdahl's theorem.

Curvature and conservation. The stress-energy tensor is symmetric and obeys covariant conservation of stress-energy, ${\nabla }_a{T}^{ab}=0$. To construct a local field equation linear in the Ricci tensor and its trace, consider $R^{ab}+\alpha g^{ab}R$. The contracted Bianchi identity and metric compatibility giveThe divergence must vanish identically, rather than constrain every matter source to have constant scalar curvature. Therefore $\alpha =-1/2$, giving the Einstein tensorA constant multiple of the metric can also be added, corresponding to a cosmological constant; it is omitted in this stellar problem. Proportionality to matter by itself does not exclude arbitrary higher-curvature theories. The stated linear combination assumes the usual second-order metric field equation, linear in curvature, with its coupling fixed by the Newtonian limit.

The source's displayed curvature components use the convention $G_{ab}=-8\pi G{T}_{ab}/c^4$, opposite to another common curvature convention. Retain both that sign and those components together: replacing only the sign would give an incorrect negative stellar mass.

Fluid frame and mass function. Because the metric and the Einstein tensor are diagonal, the mixed time-space components of the stress-energy tensor vanish. For the perfect fluid in general relativity, $T^{0i}=(\rho +p/c^2)u^0{u}^i$. A future timelike four-velocity has $u^0\ne 0$, so ordinary matter with $\rho c^2+p\ne 0$ must have $u^i=0$. Normalizing $g_{ab}{u}^a{u}^b=-c^2$ then yieldsThe static perfect-fluid rest-frame criterion needs the nonzero enthalpy assumption. A vacuum-like component with $p=-\rho c^2$ has $T^{ab}=p{g}^{ab}$ for every normalized timelike $u^a$, so diagonality cannot pick its rest frame. That is a concrete exception to the unqualified inference about velocity; it does not describe the ordinary stellar fluid considered next.

Lowering the indices of the static-fluid tensor gives $T_{00}=\rho c^2{e}^{2\Phi }$ and $T_{11}=p{e}^{2\Lambda }$. The time component of the field equation therefore impliesRegularity at the centre fixes the integration constant to zero, givingA central point mass would introduce an additional constant, but is excluded by the regular-star boundary condition. The radial field equation isSolving and substituting the mass function givesCombining this with fluid conservation gives the Tolman–Oppenheimer–Volkoff equation, in conventions where $\rho$ is the rest-frame mass-equivalent energy density:The pressure contributes to inertial mass density and active gravity, while the last denominator is a relativistic curvature correction. The weak-field, low-pressure limit recovers $p^{\prime }\simeq -G\rho M/r^2$.

Constant-density integration. Set $w=p/(\rho c^2)$ and $\alpha =8\pi G\rho /(3c^2)$. Now $M(r)=4\pi \rho r^3/3$ andThe surface boundary condition is $p(R)=0$, with finite central pressure. Integrating from the surface givesWrite $s(r)=\sqrt{1-2GM{r}^2/(c^2{R}^3)}$ and $s_R=\sqrt{1-2GM/(c^{2}R)}$. Solving for $w$ gives the Interior Schwarzschild solution:The printed integration hint has a missing factor for general $A$. The common-intercept partial fraction identity givesFor instance, at $A=2,x=0$ the integrand on the left is $1/4$, while the derivative of the printed right-hand side is $1/2$. The normalized $w$ calculation above uses $A=1$, where that missing factor happens not to matter.

Positive finite central pressure requires $3s_R>1$. ConsequentlyEquality is a singular limiting solution with divergent pressure. The constant-density model is incompressible and is not a realistic causal neutron-star equation of state. Nevertheless, Buchdahl's theorem extends this compactness bound to regular static spherical isotropic fluids with nonincreasing density, under its usual assumptions.

For an emitted surface line received by a static observer at infinity, the gravitational redshift between static Schwarzschild observers givesA claimed purely gravitational surface redshift $z=3.1$ would require compactness $1-1/(4.1{)}^2\simeq 0.9405$, exceeding $8/9$. It is inconsistent with the regular isotropic static neutron-star interpretation, even though the surface would formally remain outside its Schwarzschild radius. Misidentification, additional kinematic shifts or departures from the model would need examination before treating that line as a neutron-star surface measurement.

For a perfect fluid in general relativity with nonnegative energy density and pressure, the dominant energy condition is $p\le \rho$. Pressure decreases outward, so it is enough to test the central pressure. Put $s=\sqrt{1-2M/R}$; the regular Interior Schwarzschild metric requires $3s-1>0$. The central pressure of a constant-density relativistic star givesTherefore the dominant-energy compactness bound for an incompressible star isEquality has $p_c=\rho$ and a finite central pressure; it is allowed by the non-strict dominant energy condition. This is stronger than the regular-star Buchdahl's theorem limit $M/R<4/9$.

For the constant-density model, $m(r)=4\pi \rho r^3/3=M{r}^3/R^3$. This is the incompressible Interior Schwarzschild metric, replacing the earlier finite-slope barotropic stellar equation of state. The Tolman–Oppenheimer–Volkoff equation separates asOn the left, partial fractions giveOn the right the antiderivative is $\frac{3}{4\rho }\log (1-8\pi \rho r^2/3)$. Applying the surface condition $p(R)=0$, and writing $s(r)=\sqrt{1-2M{r}^2/R^3}$, $s_R=\sqrt{1-2M/R}$, givesSolving for the pressure recovers the Interior Schwarzschild solution:In particular the central pressure of a constant-density relativistic star isThe denominator must be positive for a regular positive-pressure centre, requiring $M/R<4/9$. As that limiting compactness is approached, the central pressure diverges, in agreement with Buchdahl's theorem.

For the constant density ${\rho }_0$,The definition of $m$ then givesSubstitution in the Tolman–Oppenheimer–Volkoff equation yieldsThe minus sign follows both from the stated TOV equation and from the outward decrease of pressure; a version of the requested intermediate formula without that sign is a typographical error.

SetSeparating variables givesSinceintegration and the surface condition $P(R)=0$ giveWritingone obtains the pressure of the Interior Schwarzschild metric:In particular, the central pressure of a constant-density relativistic star isIt diverges whenor equivalentlyThis is the limiting equality case of Buchdahl's theorem.

Set $C=M/R$ and $s=\sqrt{1-2C}$. At the center,The barotropic bound $p_c\le \omega {\rho }_0$ impliesThereforeAs $\omega \to \infty$, this approaches $4/9$, the limiting value in Buchdahl's theorem. Constant-density stars with increasingly large allowed central pressure can therefore approach the Buchdahl bound arbitrarily closely.