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For steady stellar meridional circulation, $T\nabla s=\nabla (h+\varphi )$ and $\nabla \cdot (\rho \mathbf{v})=0$. Thus $\rho \mathbf{v}\cdot T\nabla s=\nabla \cdot [\rho \mathbf{v}(h+\varphi )]$. In the leading uniformly rotating stellar hydrostatic equilibrium, $h+\varphi$ is constant on an equipotential surface; its total advected flux vanishes because the net mass flux vanishes. Integrating the energy equation then identifies the outward luminosity with enclosed nuclear energy generation, even though local heat advection is nonzero.

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 equationsThe 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 $\nabla \varphi$, and hence with $\nabla P$. Thus $P=P(\varphi )$ locally. Taking the curl of the hydrostatic equilibrium equation gives $\nabla \rho \times \nabla \varphi =0$, so $\rho =\rho (\varphi )$ on such surfaces as well. The Poisson equation then shows that ${\nabla }^2\varphi$ is constant on each equipotential surface.

The magnitude $g=|\nabla \varphi |$ need not be constant. For instance $\varphi =A(x^2+y^2)+B{z}^2$, 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 $T=T(\varphi )$, and the radiative conductivity $\chi (\rho ,T)$ is also constant on an equipotential surface. Put $f(\varphi )=-\chi dT/d\varphi$. Then $\mathbf{F}=f(\varphi )\nabla \varphi$. On the stellar surface, below mass shedding, the outward normal is $\nabla \varphi /g$, soThis 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. 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:

The obstruction to purely local radiative equilibrium is visible by taking the divergence:On an equipotential surface, $\rho ϵ$, $f$, $f^{\prime }$ and the second term are constant, but $g^2$ generally varies. Unless $f^{\prime }=0$ or another special circumstance removes the mismatch, $\nabla \cdot \mathbf{F}=\rho ϵ$ 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 giveConsequentlyBoth $h$ and $\varphi$ are constant on the bounding equipotential surface. By the divergence theorem and steady continuity equation,Integrating the given thermal energy equation therefore gives equipotential luminosity conservation with stellar circulation: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 givesThus $dT/d\varphi =-L/[\chi (4\pi Gm-2{\Omega }^{2}V)]$, while $dP/d\varphi =-\rho$. Dividing these derivatives and inserting $\chi =4ac{T}^3/(3\kappa \rho )$ yields the radiative gradient in a uniformly rotating star:The bracket must be positive for the regular outward-gravity configuration being used.

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 ${\Omega }^{2}V/(2\pi Gm)={\Omega }^2/(2\pi G\rho )$ is independent of the chosen enclosed volume. Replacing $m$ by a fixed total mass would incorrectly make it volume dependent.

In uniformly rotating stellar hydrostatic equilibrium, composition-uniform thermodynamic quantities are functions of the total potential. Radiative conductivity gives $\mathbf{F}=f(\varphi )\nabla \varphi$, with $f=-\chi dT/d\varphi$. On one surface, $f$ is constant, so the outward flux is $F_n=fg$. The Stefan–Boltzmann law therefore gives $T_e\propto g^{1/4}$. This is a surface-flux law: the material temperature constant on an interior equipotential is not the latitude-dependent emergent effective temperature.