 Solution

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.