= Solution
\b[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 $\Omega^2\varpi\,\widehat\varpi$, so <hydrostatic equilibrium> becomes
$$
\boldsymbol\nabla P=-\rho\boldsymbol\nabla\phi,\qquad
\phi=\Phi-\frac12\Omega^2\varpi^2.
$$
Using the <Poisson equation> for gravity and $\boldsymbol\nabla^2(x^2+y^2)=4$ gives
$$
\boxed{\boldsymbol\nabla^2\phi=4\pi G\rho-2\Omega^2.}
$$
On any regular connected <equipotential surface>, every tangent displacement satisfies $d\phi=0$ and therefore $dP=0$. Locally $P=P(\phi)$, with $dP/d\phi=-\rho$. This makes $\rho$ a function of $\phi$ as well. Equivalently, taking the curl of the hydrostatic equation gives $\boldsymbol\nabla\rho\times\boldsymbol\nabla\phi=0$. Thus $P,\rho$ and $\boldsymbol\nabla^2\phi$ are constant on each surface. The last conclusion follows from the modified <Poisson equation>; it does not constrain the magnitude of the first derivative $|\boldsymbol\nabla\phi|$ 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 $T=T(\phi)$, and the <opacity> and <radiative conductivity> are also functions of $\phi$.
\b[Surface flux and the requested sketch.] Define $f(\phi)=-\chi\,dT/d\phi$, positive for the outward radiative flux. Then
$$
\mathbf F=f(\phi)\boldsymbol\nabla\phi,\qquad\chi=\frac{4acT^3}{3\kappa\rho}.
$$
On an outer <equipotential surface>, $f$ is constant and the outward normal is parallel to $\boldsymbol\nabla\phi$. Its effective gravity is $g=|\boldsymbol\nabla\phi|$, so $F_n=fg$. The <Stefan–Boltzmann law> $F_n=\sigma_{\rm SB}T_e^4$ gives the <radiative gravity-darkening law>,
$$
\boxed{T_e\propto g^{1/4}.}
$$
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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-70-gravity-darkening.png]
{title=Oblate rotating star with hot poles, cool equator and radiative gravity darkening}
\b[Why purely local radiative balance fails.] Differentiating the flux gives
$$
\boldsymbol\nabla\cdot\mathbf F=f'(\phi)|\boldsymbol\nabla\phi|^2+f(\phi)\boldsymbol\nabla^2\phi.
$$
On one equipotential, $f,f',\boldsymbol\nabla^2\phi$ and the proposed source $\rho\epsilon(\rho,T)$ are all constant, but $|\boldsymbol\nabla\phi|^2$ generally varies with latitude. Unless $f'=0$ or there is another special cancellation, no single value of $\rho\epsilon$ 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.
\b[Integrated energy balance with circulation.] The steady <continuity equation> is $\boldsymbol\nabla\cdot(\rho\mathbf v)=0$. Combining the <entropy>–<enthalpy> relation with hydrostatic balance gives
$$
\rho T\mathbf v\cdot\boldsymbol\nabla s
=\rho\mathbf v\cdot\boldsymbol\nabla h-\mathbf v\cdot\boldsymbol\nabla P
=\rho\mathbf v\cdot\boldsymbol\nabla(h+\phi)
=\boldsymbol\nabla\cdot[\rho\mathbf v(h+\phi)].
$$
Integrate the energy equation over $V$ and apply the <divergence theorem>:
$$
\int_S\mathbf F\cdot d\mathbf S
=\int_V\rho\epsilon\,dV-\int_S\rho(h+\phi)\mathbf v\cdot d\mathbf S.
$$
Uniform composition makes $h=h(\rho,T)$ constant on the enclosing equipotential, and $\phi$ is constant there by definition. Pull $(h+\phi)$ out of the last integral; the remaining net mass flux is zero by steady continuity. Thus the <equipotential luminosity conservation with stellar circulation> is
$$
\boxed{L=\int_S\mathbf F\cdot d\mathbf S=\int_V\rho\epsilon\,dV.}
$$
This does not require $\mathbf v\cdot\widehat n=0$ 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 gives
$$
L=f\int_S\boldsymbol\nabla\phi\cdot d\mathbf S
=f\int_V(4\pi G\rho-2\Omega^2)\,dV
=f(4\pi Gm-2\Omega^2V).
$$
Therefore $-\chi\,dT/d\phi=L/(4\pi Gm-2\Omega^2V)$. Since $dP/d\phi=-\rho$,
$$
\begin{aligned}
\frac{d\ln T}{d\ln P}&=-\frac{P}{\rho T}\frac{dT}{d\phi}
=\frac{PL}{\rho T\chi(4\pi Gm-2\Omega^2V)}\\
&=\boxed{\frac{3\kappa PL}{16\pi acGmT^4}
\left(1-\frac{\Omega^2V}{2\pi Gm}\right)^{-1}}.
\end{aligned}
$$
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.
Back to article page
