= Solution
Let $\Psi$ be the <gravitational potential>, with $\nabla^2\Psi=4\pi G\rho$, and let $\varpi^2=x^2+y^2$ measure distance from the rotation axis. Uniform <solid-body rotation> contributes the <centrifugal potential> $-\Omega^2\varpi^2/2$. In the corotating frame, static force balance therefore gives the <uniformly rotating stellar hydrostatic equilibrium> equations
$$
\boxed{\phi=\Psi-\frac12\Omega^2\varpi^2,\qquad
\nabla P=-\rho\nabla\phi,\qquad
\nabla^2\phi=4\pi G\rho-2\Omega^2.}
$$
The final coefficient follows because $\nabla^2(x^2+y^2)=4$.
A tangent displacement along a regular connected <equipotential surface> has zero dot product with $\nabla\phi$, and hence with $\nabla P$. Thus $P=P(\phi)$ locally. Taking the curl of the <hydrostatic equilibrium> equation gives $\nabla\rho\times\nabla\phi=0$, so $\rho=\rho(\phi)$ on such surfaces as well. The <Poisson equation> then shows that $\nabla^2\phi$ is constant on each <equipotential surface>.
\b[The magnitude $g=|\nabla\phi|$ need not be constant.] For instance $\phi=A(x^2+y^2)+Bz^2$, with positive $A\ne B$, has constant Laplacian but different pole and equator gradient magnitudes on an ellipsoidal equipotential. In a rotating star, flattening and <centrifugal acceleration> likewise make the effective gravity latitude dependent.
With uniform composition, an <equation of state> determines $T$ from $P$ and $\rho$; for the supplied <ideal gas> plus <radiation pressure> law its right side increases monotonically with $T$. Hence $T=T(\phi)$, and the <radiative conductivity> $\chi(\rho,T)$ is also constant on an <equipotential surface>. Put $f(\phi)=-\chi\,dT/d\phi$. Then $\mathbf F=f(\phi)\nabla\phi$. On the stellar surface, below mass shedding, the outward normal is $\nabla\phi/g$, so
$$
F_n=f(\phi_s)g,\qquad
\boxed{T_e=\left(\frac{f(\phi_s)}{\sigma}\right)^{1/4}g^{1/4}.}
$$
This is the <radiative gravity-darkening law>. The <effective temperature> measures emergent flux; it is not the same quantity as the interior material temperature that was constant on an equipotential.
A rapidly rotating star is oblate, with reduced effective gravity and cooler <effective temperature> around the equator. \b[The poles are hottest.] The original sketch uses a <Roche model of a uniformly rotating star> only to illustrate the shape and surface-flux variation:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-63-gravity-darkening.png]
{title=Oblate rotating star with hotter poles and cooler equator}
The obstruction to purely local <radiative equilibrium> is visible by taking the divergence:
$$
\nabla\cdot\mathbf F=f'(\phi)|\nabla\phi|^2
+f(\phi)(4\pi G\rho-2\Omega^2).
$$
On an <equipotential surface>, $\rho\epsilon$, $f$, $f'$ and the second term are constant, but $g^2$ generally varies. Unless $f'=0$ or another special circumstance removes the mismatch, $\nabla\cdot\mathbf F=\rho\epsilon$ cannot hold at every latitude. This is the <radiative-equilibrium obstruction in a rotating barotropic star>; heat advection by <stellar meridional circulation> resolves the missing energy transport in the leading hydrostatic treatment.
For the steady circulation, <mass conservation> gives $\nabla\cdot(\rho\mathbf v)=0$. The <specific enthalpy> identity and <hydrostatic equilibrium> give
$$
T\nabla s=\nabla h-\frac1\rho\nabla P=\nabla(h+\phi).
$$
Consequently
$$
\rho\mathbf v\cdot T\nabla s
=\nabla\cdot[\rho\mathbf v(h+\phi)].
$$
Both $h$ and $\phi$ are constant on the bounding <equipotential surface>. By the <divergence theorem> and steady <continuity equation>,
$$
\int_V\rho\mathbf v\cdot T\nabla s\,dV
=(h+\phi)_S\int_S\rho\mathbf v\cdot d\mathbf S=0.
$$
Integrating the given thermal energy equation therefore gives <equipotential luminosity conservation with stellar circulation>:
$$
\boxed{L=\int_S\mathbf F\cdot d\mathbf S=\int_V\rho\epsilon\,dV.}
$$
Individual circulation streams can cross the surface; only their net steady mass flux has to vanish.
Since $f$ is constant on that surface, integrating the <Poisson equation> gives
$$
L=f\int_S\nabla\phi\cdot d\mathbf S
=f(4\pi Gm-2\Omega^2V).
$$
Thus $dT/d\phi=-L/[\chi(4\pi Gm-2\Omega^2V)]$, while $dP/d\phi=-\rho$. Dividing these derivatives and inserting $\chi=4acT^3/(3\kappa\rho)$ yields the <radiative gradient in a uniformly rotating star>:
$$
\boxed{\frac{d\log T}{d\log P}
=\frac{3\kappa PL}{16\pi acGmT^4}
\left(1-\frac{\Omega^2V}{2\pi Gm}\right)^{-1}.}
$$
The bracket must be positive for the regular outward-gravity configuration being used.
\b[For an interior equipotential, both mass factors must be the enclosed mass $m$.] The printed final bracket uses $M$, but its derivation gives $m$; at the outer surface $m=M$ and the printed bracket is recovered. For a direct check, a uniform-density model has $m=\rho V$, so the true correction $\Omega^2V/(2\pi Gm)=\Omega^2/(2\pi G\rho)$ is independent of the chosen enclosed volume. Replacing $m$ by a fixed total mass would incorrectly make it volume dependent.
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