= Solution
\b[Curvature and conservation.] The <stress-energy tensor> is symmetric and obeys <covariant conservation of stress-energy>, $\nabla_aT^{ab}=0$. 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> give
$$
\nabla_aR^{ab}=\frac12\nabla^bR,\qquad
\nabla_a(R^{ab}+\alpha g^{ab}R)=\left(\frac12+\alpha\right)\nabla^bR.
$$
The divergence must vanish identically, rather than constrain every matter source to have constant scalar curvature. Therefore $\alpha=-1/2$, giving the <Einstein tensor>
$$
\boxed{G^{ab}=R^{ab}-\frac12g^{ab}R,\qquad \nabla_aG^{ab}=0}.
$$
A 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 $G_{ab}=-8\pi GT_{ab}/c^4$, 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.
\b[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>, $T^{0i}=(\rho+p/c^2)u^0u^i$. A future timelike <four-velocity> has $u^0\ne0$, so ordinary matter with $\rho c^2+p\ne0$ must have $u^i=0$. Normalizing $g_{ab}u^au^b=-c^2$ then yields
$$
\boxed{u^0=ce^{-\Phi},\qquad u_0=-ce^{\Phi}}.
$$
The <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 $T_{00}=\rho c^2e^{2\Phi}$ and $T_{11}=pe^{2\Lambda}$. The time component of the field equation therefore implies
$$
\frac{d}{dr}\left[r(1-e^{-2\Lambda})\right]=\frac{8\pi G}{c^2}\rho r^2=\frac{2G}{c^2}\frac{dM}{dr}.
$$
Regularity at the centre fixes the integration constant to zero, giving
$$
\boxed{e^{-2\Lambda(r)}=1-\frac{2GM(r)}{c^2r},\qquad
\Lambda(r)=-\frac12\ln\left(1-\frac{2GM(r)}{c^2r}\right)}.
$$
A central point mass would introduce an additional constant, but is excluded by the regular-star boundary condition. The radial field equation is
$$
\frac{e^{2\Lambda}-1}{r^2}-\frac{2\Phi'}r=-\frac{8\pi G}{c^4}pe^{2\Lambda}.
$$
Solving and substituting the mass function gives
$$
\boxed{\Phi'=\frac{G[M(r)+4\pi p r^3/c^2]}{c^2r^2[1-2GM(r)/(c^2r)]}}.
$$
Combining this with fluid conservation gives the <Tolman–Oppenheimer–Volkoff equation>, in conventions where $\rho$ is the rest-frame mass-equivalent energy density:
$$
\boxed{\frac{dp}{dr}=-\frac{G(\rho+p/c^2)[M(r)+4\pi p r^3/c^2]}{r^2[1-2GM(r)/(c^2r)]},\qquad M'=4\pi\rho r^2}.
$$
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 $p'\simeq-G\rho M/r^2$.
\b[Constant-density integration.] Set $w=p/(\rho c^2)$ and $\alpha=8\pi G\rho/(3c^2)$. Now $M(r)=4\pi\rho r^3/3$ and
$$
\frac{dw}{(1+w)(1+3w)}=-\frac\alpha2\frac{r\,dr}{1-\alpha r^2}.
$$
The surface boundary condition is $p(R)=0$, with finite central <pressure>. Integrating from the surface gives
$$
\frac12\ln\frac{1+3w}{1+w}=\frac14\ln\frac{1-\alpha r^2}{1-\alpha R^2}.
$$
Write $s(r)=\sqrt{1-2GMr^2/(c^2R^3)}$ and $s_R=\sqrt{1-2GM/(c^2R)}$. Solving for $w$ gives the <Interior Schwarzschild solution>:
$$
\boxed{p(r)=\rho c^2\frac{s(r)-s_R}{3s_R-s(r)},\qquad
p_0=\rho c^2\frac{1-s_R}{3s_R-1}}.
$$
The printed integration hint has a missing factor for general $A$. The <common-intercept partial fraction identity> gives
$$
\frac1{(A+3x)(A+x)}=\frac1{2A}\left[\frac3{A+3x}-\frac1{A+x}\right]\qquad(A\ne0).
$$
For 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 $3s_R>1$. Consequently
$$
\boxed{\frac{2GM}{c^2R}<\frac89,\qquad M<\frac{4c^2R}{9G}}.
$$
Equality 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> gives
$$
1+z=\left(1-\frac{2GM}{c^2R}\right)^{-1/2}<3,\qquad \boxed{z<2}.
$$
A claimed purely gravitational surface redshift $z=3.1$ would require compactness $1-1/(4.1)^2\simeq0.9405$, 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.
\Image[/past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2016-incompressible-star-pressure.png]
{title=Constant-density relativistic pressure profiles and the divergence at the compactness bound}
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