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Mechanical balance and equipotentials. Let $\Phi$ be the Newtonian gravitational potential and $\varpi$ the cylindrical distance from the rotation axis. For rigid angular velocity $\Omega$, a fluid element at rest in the rotating frame has no Coriolis acceleration. The centrifugal acceleration is ${\Omega }^2\varpi \hat{\varpi }$, so hydrostatic equilibrium becomesUsing the Poisson equation for gravity and ${\nabla }^2(x^2+y^2)=4$ gives

On any regular connected equipotential surface, every tangent displacement satisfies $d\varphi =0$ and therefore $dP=0$. Locally $P=P(\varphi )$, with $dP/d\varphi =-\rho$. This makes $\rho$ a function of $\varphi$ as well. Equivalently, taking the curl of the hydrostatic equation gives $\nabla \rho \times \nabla \varphi =0$. Thus $P,\rho$ and ${\nabla }^2\varphi$ are constant on each surface. The last conclusion follows from the modified Poisson equation; it does not constrain the magnitude of the first derivative $|\nabla \varphi |$ to be constant there.

For uniform composition, an equation of state $P=P(\rho ,T)$ determines $T$ uniquely at fixed $P,\rho$ in the usual stellar regime. For the gas-plus-radiation equation, the derivative with respect to $T$ is positive. Hence $T=T(\varphi )$, and the opacity and radiative conductivity are also functions of $\varphi$.

Surface flux and the requested sketch. Define $f(\varphi )=-\chi dT/d\varphi$, positive for the outward radiative flux. ThenOn an outer equipotential surface, $f$ is constant and the outward normal is parallel to $\nabla \varphi$. Its effective gravity is $g=|\nabla \varphi |$, so $F_n=fg$. The Stefan–Boltzmann law $F_n={\sigma }_{\mathrm{SB}}{T}_e^4$ gives the radiative gravity-darkening law,A rapidly rotating star is oblate. Centrifugal support reduces the effective gravity near the equator, which is therefore cooler and dimmer; the poles are hotter. The original sketch below uses a subcritical Roche model of a uniformly rotating star to show this geometry and the corresponding temperature variation. The material temperature on an interior equipotential is not the same quantity as the emergent effective temperature defined by surface flux.

Why purely local radiative balance fails. Differentiating the flux givesOn one equipotential, $f,f^{\prime },{\nabla }^2\varphi$ and the proposed source $\rho ϵ(\rho ,T)$ are all constant, but $|\nabla \varphi {|}^2$ generally varies with latitude. Unless $f^{\prime }=0$ or there is another special cancellation, no single value of $\rho ϵ$ can match this flux divergence at all points on the surface. This is the radiative-equilibrium obstruction in a rotating barotropic star. Slow stellar meridional circulation can transport heat to compensate; it is distinct from abandoning hydrostatic balance at leading order.

Integrated energy balance with circulation. The steady continuity equation is $\nabla \cdot (\rho \mathbf{v})=0$. Combining the entropy–enthalpy relation with hydrostatic balance givesIntegrate the energy equation over $V$ and apply the divergence theorem:Uniform composition makes $h=h(\rho ,T)$ constant on the enclosing equipotential, and $\varphi$ is constant there by definition. Pull $(h+\varphi )$ out of the last integral; the remaining net mass flux is zero by steady continuity. Thus the equipotential luminosity conservation with stellar circulation isThis does not require $\mathbf{v}\cdot \hat{n}=0$ pointwise on $S$: inward and outward circulation may cancel in the net mass flux.

Finally, $f$ is constant on $S$ and the divergence theorem applied to the effective potential givesTherefore $-\chi dT/d\varphi =L/(4\pi Gm-2{\Omega }^{2}V)$. Since $dP/d\varphi =-\rho$,The mass $m$ is the mass enclosed by this surface, not necessarily the total stellar mass. The denominator is positive for the regular hydrostatic surfaces used here; near loss of centrifugal confinement the assumed stellar model must be reconsidered. The derivation uses slow circulation, retaining hydrostatic/equipotential thermodynamics to leading order while allowing a nonzero advective heat flux.

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.

The divergence theorem and Poisson equation give ${\int }_S\nabla \varphi \cdot d\mathbf{S}=4\pi Gm-2{\Omega }^{2}V$ on a bounding equipotential surface. Equipotential luminosity conservation with stellar circulation then fixes $f=L/(4\pi Gm-2{\Omega }^{2}V)$. Combining $dT/d\varphi =-f/\chi$, $dP/d\varphi =-\rho$ and radiative conductivity gives the displayed logarithmic temperature-pressure gradient. Both appearances of mass mean the enclosed mass $m$; only on the outer boundary can both be replaced by the total mass.

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.