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For a Newtonian gravitational potential obeying the Poisson equation for Newtonian gravity, define $T_{ij}=[{\partial }_i\Phi {\partial }_j\Phi -{\delta }_{ij}|\nabla \Phi {|}^2/2]/(4\pi G)$. Commutation of mixed partial derivatives gives ${\partial }_j{T}_{ij}=\rho {\partial }_i\Phi$. Thus the gravitational force density can be put in conservative form as $-{\partial }_j{T}_{ij}$. The sign is chosen so that $T_{ij}$ is added to the momentum flux; reversing the stress convention reverses the divergence sign.

For a weak, nearly static conformal factor in Nordstrom theory of gravitation, the trace of the dust stress-energy tensor is negative in signature $(-+++)$, so its scalar field equation reduces to the Poisson equation for the Newtonian gravitational potential. The geodesic equation for a conformally flat metric gives $\ddot{\mathbf{x}}=-\nabla \Phi$ at small speeds. The constant one in $\varphi$ has no effect on the force.

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

For a self-gravitating isothermal slab with $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$ and $H^2={\sigma }^2/(2\pi G{\rho }_0)$, hydrostatic equilibrium gives the displayed Newtonian gravitational potential. A test particle has constant in-plane velocity and conserved vertical energy $E_z={\dot{z}}^2/2+\Phi (z)-\Phi (0)$. Its vertical motion is periodic for every finite $E_z>0$, with turning height $H\mathrm{arcosh}\exp [E_z/(2{\sigma }^2)]$. Near the plane it is a harmonic oscillator of squared frequency $4\pi G{\rho }_0$; at large height the force tends to the constant $2\pi G\Sigma$, where $\Sigma =2{\rho }_{0}H$. The ideal infinite slab has no finite vertical escape energy.

The pressure tensor has Cartesian components $p_{ab}=\int f(v_a-u_a)(v_b-u_b)d^{3}v$. With zero mean velocity and an isotropic galactic distribution function, off-diagonal components vanish by reflection symmetry and diagonal components are equal by rotational symmetry. ThereforeHere $p$ is a tracer number-weighted random-motion stress, rather than thermodynamic pressure per volume. Its tensor divergence is $(\mathrm{div}\mathbf{p}{)}_a={\partial }_b(p{\delta }_{ab})={\partial }_{a}p$. The time-independent, nonstreaming Jeans equation consequently reduces toThe gravitational acceleration convention is $\nabla \psi$: this $\psi$ differs by a sign from the usual Newtonian gravitational potential $\Phi$.

Let $z$ measure distance along the line of sight. Spherical density projection givesBecause isotropy makes the local second stellar velocity moment along every line of sight equal to $p/n$, and because there is no streaming motion, the projected moment isThis is the forward relation for isotropic stellar pressure deprojection.

Both inversions follow from one Abel transform calculation. Write $F(u)=N(\sqrt{u})$ and $a(v)=n(\sqrt{v})$, so $F(u)={\int }_u^{\infty }a(v)(v-u{)}^{-1/2}dv$. Compose the transforms, using Tonelli's theorem for nonnegative densities, or absolute convergence for signed profiles:Assume sufficient decay to make these integrals finite and sufficient regularity for the last derivative. Then $J^{\prime }(x)=-\pi a(x)$. Setting $x=r^2$ gives the spherical Abel deprojectionReplacing $F(u)$ by $N(\sqrt{u}){\sigma }^2(\sqrt{u})$ and $a(v)$ by $p(\sqrt{v})$ provesThese are the requested expressions, with $u=R^2$. The composed-transform argument avoids an invalid separate differentiation of a divergent lower-end kernel.

In the dominant central Newtonian gravitational potential, conservation of energy for gas falling from rest at infinity gives $u^2/2=G{M}_c/R$, hence $u=a{R}^{-1/2}$ with $a=(2G{M}_c{)}^{1/2}$. The characteristic stellar speed has the same radial scaling. For random stellar directions, the cross term in $|\mathbf{u}-\mathbf{v}{|}^2$ averages to zero, so the typical relative speed is also $V=b{R}^{-1/2}$. With the stated kinetic-energy estimate, an rms convention gives $V^2\simeq 4G{M}_c/R$; another convention changes only the constant $b$.

In the steady spherical inflow approximation, mass conservation gives $A=4\pi R^2\rho u$. ThusThis leading calculation treats the inward gas flux as essentially undepleted over the region used. Substantial distributed removal by stars would make the local flux a function of $R$, an important limitation discussed below.

Insert these scalings in the Bondi--Hoyle--Lyttleton accretion rate. The factors $R^{3/2}$ from $V^{-3}$ and $R^{-3/2}$ from $\rho$ cancel:This is quadratic accretion in a central Keplerian potential. Write its fixed coefficient as $\lambda =\Lambda$. Integrating $d(1/M)/dt=-\lambda$ givesThe cancellation of radius means that orbital variation of $R$ alone does not change this coefficient, provided the assumed steady gas profile, central mass and relative-velocity scaling remain fixed.

Conserve the number of stars along these deterministic mass trajectories. Inverting the map and taking its Jacobian determinant givesThe mass-spectrum transport under quadratic accretion is thereforeFor a narrow but nonzero initial mass interval, the mapping stretches the high end especially strongly. If its endpoints are $m_1<m_2$, their evolved ratio is $(m_2/m_1)(1-\lambda m_{1}t)/(1-\lambda m_{2}t)$, which can become large as $t$ approaches $1/(\lambda m_2)$ from below. In an intermediate high-mass range with $\lambda Mt\gg 1$, mapped back into a portion of the initial interval where $F_0$ changes little,Here $M_{\ast }$ is near $1/(\lambda t)$ on the populated side of the initial interval. Finite initial support gives a finite upper mass cutoff at every time before the earliest blow-up; this is an approximate broad-range power law, not a literal infinite tail of a finite pre-blow-up population. A strictly monochromatic initial population remains monochromatic under identical accretion coefficients and does not generate a spectrum.

The limitations of this accretion model are substantial. The star cluster must remain centrally dominated, and both the central mass and gas supply must remain effectively fixed. Stellar capture depletes the gas flux and eventually invalidates the $R^{-3/2}$ profile. Gas pressure, angular momentum, shocks, radiative feedback and winds can reduce capture relative to the cold Bondi–Hoyle accretion estimate; a broad relative-speed distribution is not faithfully represented by one typical speed, especially because capture scales as $V^{-3}$. A growing star can perturb the surrounding potential, overlap capture regions, change its orbit through momentum loading, or compete with its neighbors for the same gas. Finally, number conservation excludes mergers, new star formation and stellar losses. The formal finite-time mass divergence cannot persist under a finite available gas mass, so the inferred mass-spectrum slope applies only while these approximations remain adequate.

Write ${\sigma }_0^2=⟨v_z^2⟩=v_{\mathrm{rms}}^2/3$ for the one-dimensional thermal velocity dispersion of an isotropic monatomic ideal gas. An isothermal process keeps ${\sigma }_0$ fixed, and the vertical pressure is $P=\rho {\sigma }_0^2$. Hydrostatic equilibrium in the Newtonian gravitational potential $\Phi (z)$ givesPlanar symmetry reduces the Poisson equation for Newtonian gravity to ${\Phi }^{\prime \prime }=4\pi G\rho$. Combining the two equations yieldsThe velocity factor here is a velocity second moment, not the square of a particular particle's instantaneous vertical velocity.

Set $H^2={\sigma }_0^2/(2\pi G{\rho }_0)$ and try the symmetric self-gravitating isothermal slab profile $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$. Its logarithmic derivatives areThe chosen $H$ makes the second expression satisfy the required equation. The central conditions $\rho (0)={\rho }_0$ and ${\rho }^{\prime }(0)=0$ fix the smooth even solution uniquely. In the scale-height convention used here,The undefined constant printed as $a$ in the scale-height denominator must be $G$; this follows both from substitution and dimensions.

Integrating hydrostatic equilibrium, with the additive potential constant chosen so that $\Phi (0)=0$, givesThe orbits in a self-gravitating isothermal slab separate into horizontal and vertical motions. There is no horizontal force, so $x=x_i+v_{x}t$ and $y=y_i+v_{y}t$, with constant horizontal velocities. For collisionless test particles, or between gas collisions, conservation of energy fixes the vertical energyEvery finite $E_z>0$ has two turning points $\pm A$, whereThus the general orbit drifts uniformly in the plane while oscillating periodically across it. A particle with $E_z=0$ stays in the midplane; an orbit with nonzero horizontal velocity is generally not closed.

Near the plane, $\Phi (z)={\sigma }_0^2{z}^2/H^2+O(z^4)$, so the vertical motion is a harmonic oscillator with ${\nu }^2=2{\sigma }_0^2/H^2=4\pi G{\rho }_0$. Far from the plane, $\Phi \sim 2{\sigma }_0^2|z|/H$ and the force tends to a constant magnitude. Indeed the surface density is $\Sigma =\int \rho dz=2{\rho }_{0}H$, and that magnitude is $2\pi G\Sigma$. Large-amplitude vertical motion is therefore approximately uniformly accelerated on each side, rather than sinusoidal. Because this ideal slab extends infinitely in the horizontal directions and its potential grows without bound as $|z|\to \infty$, no finite-energy particle escapes vertically.

For an isolated smooth mass distribution with spherical symmetry, assemble each shell $dM$ at radius $r$ against the Newtonian gravitational potential of its interior mass. The spherical shell theorem givesAssume finite total mass, finite binding energy and no central point-mass self-energy, so $M(r{)}^2/r$ vanishes at both endpoints. Integration by parts then givesFor example, $M(r)\propto r^3$ near a regular center guarantees the central boundary condition, while finite total mass guarantees the one at infinity. A divergent central binding energy would fall outside this formula's finite-energy hypotheses.

For the discrete system use the centre of mass frame, let $N$ be the particle number, and write $M_{\mathrm{tot}}=Nm$. The printed $M$ in the particle-count factor and the subsequent $n$ must be understood consistently as $N$. DefineBy Newton's second law, differentiation givesEach pair contributesTherefore the exact dynamical virial theorem is $\ddot{I}/2=2K+W$. In a stationary statistical state, or on a bounded-motion time average for which the mean $\ddot{I}$ vanishes, it becomesThere is no external pressure, tidal field or boundary work in this isolated point-particle derivation. For a time-averaged system, the following $W$ and gravitational radius refer to that same averaged binding energy.

Using $W=-G(Nm{)}^2/R_g$ and $2K=Nm⟨v_{\mathrm{eq}}^2⟩$, virial equilibrium givesThis is the three-dimensional mean squared speed about the centre of mass. For an isotropic system each one-dimensional velocity dispersion squared is one third of it.

Set the Newtonian gravitational potential to zero at infinity and exclude each particle's own field. The local escape velocity in the instantaneous frozen potential obeys $v_{\mathrm{esc},i}^2=-2{\Phi }_i$, where ${\Phi }_i=-Gm{\sum }_{j\ne i}{r}_{ij}^{-1}$. Since every pair occurs twice, ${\sum }_{i}m{\Phi }_i=2W$. Averaging over the identical particles proves the gravitational rms escape and collapse speeds relationAn actual encounter in a changing potential can exchange energy, so this is the usual local escape-speed definition, not a claim about every many-body trajectory.

Finally, dissipation-free collapse from rest at infinite separation has total energy zero. Conservation of energy at the specified binding energy therefore gives $K_{\mathrm{coll}}=-W$, twice the equilibrium kinetic energy at the same $W$. HenceThis mean squared speed includes coherent infall; a local random velocity dispersion after subtracting the infall velocity need not obey that ratio.

Write $H=\dot{R}/R$. A fluid element in the homologous spherical flow obeys $\dot{\mathbf{r}}=H\mathbf{r}$, so $\mathbf{r}/R$ remains constant. Also $\nabla \cdot {\mathbf{u}}_0=3H$. Along every element, the continuity equation therefore gives $D{\rho }_0/Dt=-3H{\rho }_0$. Every element starts with the same mass density, and hence has the same mass density at subsequent times:This proves preservation of uniformity, rather than assuming it when conserving the total mass.

By the shell theorem, only the mass interior to radius $r$ contributes to its gravitational acceleration. That mass is $4\pi {\rho }_0{r}^3/3$. The same result follows directly from the Poisson equation for Newtonian gravity: spherical symmetry gives $(r^2{\Phi }_r{)}_r=4\pi G{\rho }_0{r}^2$, and regularity at the centre sets the integration constant to zero. Hence ${\Phi }_r=4\pi G{\rho }_{0}r/3$ andThe material derivative of the background velocity is $(\dot{H}+H^2)\mathbf{r}=(\ddot{R}/R)\mathbf{r}$. There is no pressure force, and the momentum equation consequently reduces toMultiplication by $\dot{R}$ and integration, using collapse from rest, givesChoose the negative square root for contraction. Put $C^2=2K/R_0$ and $R=R_0{\cos }^2(\varphi /2)$. For $0<\varphi <\pi$, the energy integral then gives $\dot{R}=-C\tan (\varphi /2)$, while $dR/d\varphi =-R_0\sin \varphi /2$. Their ratio isIntegrating from the initial turning point proves the cycloid parametrization of pressureless homologous spherical collapse:It covers the entire contracting branch, with complete collapse at $\varphi =\pi$ and $t_c=\sqrt{3\pi /(32G{\rho }_{00})}$. At $R=0$ the uniform dust configuration is singular, so the smooth calculation is restricted to $t<t_c$.

Each Fourier mode among the comoving perturbations of a homologously collapsing cloud has its phase fixed on each fluid element. In fact, for $\mathbf{k}=\mathbf{q}/R$,Its physical wavelength contracts in proportion to $R$, while its wavelength in the comoving coordinate $\mathbf{r}/R$ stays fixed. This is the physical reason to use a time-dependent wavenumber; holding the physical wavenumber fixed would not follow the contracting pattern. Boundary effects are being excluded in this bulk Fourier mode calculation.

Let ${\rho }^{\prime }={\rho }_1(t)e^{i\mathbf{k}\cdot \mathbf{r}}$ and let the perturbed velocity and Newtonian gravitational potential have amplitudes ${\mathbf{u}}_1(t)$ and ${\Phi }_1(t)$. Linearizing the continuity equation, the uniform background eliminates ${\mathbf{u}}_1\cdot \nabla {\rho }_0$. The advected phase eliminates its own time derivative, leavingFor the density contrast $\delta ={\rho }_1/{\rho }_0$, use ${\dot{\rho }}_0=-3H{\rho }_0$ to obtain $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$. The linear momentum equation and Poisson equation for Newtonian gravity areDifferentiating the density contrast equation must include $\dot{\mathbf{k}}=-H\mathbf{k}$:Thus the compressive Fourier mode with $k\ne 0$ satisfiesThe pressureless approximation removes the usual acoustic restoring term of Jeans instability. A transverse velocity perturbation instead obeys ${\dot{\mathbf{u}}}_{1\perp }+H{\mathbf{u}}_{1\perp }=0$ and does not drive this density contrast.

Since $d/dt=(C/R)d/d\varphi$, the density contrast equation becomesFor a direct verification, put $s=\tan (\varphi /2)$. As $d/d\varphi =(1+s^2)d/(2ds)$, the first-derivative terms cancel, reducing the equation toThe function ${\delta }_g=s(1+s^2)/2$ has second derivative $3s$ and satisfies this equation. Transforming back gives the required growing branch:To establish what the other solution does, use reduction of order, not just the growth of this example. The coefficient of ${\delta }_{\varphi }$ is $d\log R/d\varphi$, so an independent solution is proportional to ${\delta }_g\int d\varphi /(R{\delta }_g^2)$. Choose the integration endpoint at complete collapse to select the decaying branch:For example, evaluating the integral with $a=\tan (s/2)$ gives integrand $4/[a^2(1+a^2{)}^2]$ and antiderivative $-4/a-6\arctan a-2a/(1+a^2)$. Its endpoint value gives the displayed expression. Its Wronskian with ${\delta }_g$ is $-(1+\cos \varphi {)}^{-1}$, so the two solutions are independent on $0<\varphi <\pi$.

For $ϵ=\pi -\varphi \to 0$, $R\sim R_0{ϵ}^2/4$, ${\delta }_g\sim 4{ϵ}^{-3}$ and the integral in ${\delta }_d$ is ${ϵ}^5/40+O({ϵ}^7)$. Consequently the two density modes of a collapsing pressureless sphere haveEquivalently, with $\tau =t_c-t$, the local powers are ${\tau }^{-1}$ and ${\tau }^{2/3}$. There is one independent growing branch: every solution is $A{\delta }_g+B{\delta }_d$, and only the branch with $A=0$ decays at collapse. The uniqueness is up to normalization and addition of the decaying solution; it does not mean that only one literal function can grow. A fixed nonzero initial perturbation will eventually violate $|\delta |\ll 1$, so the divergent linear result is not a nonlinear solution through collapse.

Write $s=(r^2+z^2{)}^{1/2}$. The Paczyński-Wiita potential is a spherical Newtonian gravitational potential with ${\Phi }^{\prime }(s)=GM/(s-a{)}^2$ and ${\Phi }^{\prime \prime }(s)=-2GM/(s-a{)}^3$. At $z=0$, $s=r$, soSubstituting into the orbital formulas givesChoosing positive orbital and vertical frequencies, $\Omega ={\Omega }_z=\sqrt{GM}/[\sqrt{r}(r-a)]$. For $r>3a$, $\kappa =\Omega \sqrt{(r-3a)/(r-a)}$. For $a<r<3a$, ${\kappa }^2<0$ and radial disturbances grow at rate $\sqrt{-{\kappa }^2}$, even though the vertical motion remains stable. Thus circular orbits are radially unstable for $a<r<3a$, and $r=3a$ is the marginal orbit. This recovers the innermost stable circular orbit of the Paczyński-Wiita circular orbit model.

For a circular mid-plane orbit, radial force balance in the Newtonian gravitational potential is $r{\Omega }^2={\Phi }_{,r}$. HenceAxisymmetry conserves the specific angular momentum $j=r^2\dot{\varphi }$. At fixed $j$, radial motion obeys $\ddot{r}=-{\partial }_r{\Phi }_{\mathrm{eff}}$, where the effective potential is ${\Phi }_{\mathrm{eff}}=\Phi +j^2/(2r^2)$. Linearize around a circular radius $r_0$, writing $r=r_0+\xi$. Since $j^2=r_0^3{\Phi }_{,r}(r_0,0)$ there,This is the radial epicyclic frequency; equivalently ${\kappa }^2=r^{-3}d(r^4{\Omega }^2)/dr$.

Reflection symmetry gives ${\Phi }_{,z}(r,0)=0$ and ${\Phi }_{,rz}(r,0)=0$, so vertical and radial displacements decouple at linear order. The vertical equation is $\ddot{z}=-{\Phi }_{,zz}(r_0,0)z$, giving the vertical epicyclic frequencyPositive squared frequencies give oscillatory stability; a negative squared frequency gives exponential instability rather than a real oscillation frequency.

The separated disk modes. Away from the plane of a razor-thin disk, there is no volume source, so the Newtonian gravitational potential obeys the Laplace equation in cylindrical coordinates. Separation of variables givesHere $k>0$ is a radial wavenumber, with units of inverse length. Regularity on the axis selects the Bessel function of the first kind $J_0(kR)$ rather than the singular second solution. Decay away from the disk, continuity across it and reflection symmetry select $Z=e^{-k|z|}$. These conditions apply to each nonzero-wavenumber mode; a uniform sheet is a separate zero-mode problem.

Integrate the Poisson equation for Newtonian gravity across the disk to obtain the jump conditionFor the unit-amplitude separated mode, the jump is $-2k{J}_0(kR)$. ThusAn individual basis density is signed; positive physical surface densities are built by superposition.

An arbitrary axisymmetric surface density. Define its order-zero Hankel transform byMatching the coefficients of the mode densities gives the Hankel representation of thin-disk gravity,The boundary conditions exclude added external fields. For sufficiently decaying surface density, this is the isolated-disk potential; for infinite scale-free disks, only potential differences or regularized forces need be finite.

The Mestel disk. Put $C={\Sigma }_0{R}_0$. The Bessel function integral ${\int }_0^{\infty }{J}_0(k{R}^{\prime })d{R}^{\prime }=1/k$, understood with a decaying regulator, gives $\hat{\Sigma }=C/k$. Differentiate the potential before removing the regulator and use $J_0^{\prime }=-J_1$:The last integral is $1/R$, since ${\int }_0^{\infty }{J}_1(u)du=[-J_0(u){]}_0^{\infty }=1$. Therefore the circular speed and enclosed mass areConsequently $v_c^2=GM(R)/R$. This equality follows from the specific Mestel disk calculation; the spherical shell theorem cannot be applied to general disks. Its total mass and absolute potential referenced to infinity diverge. A convenient finite midplane reference is $\Phi (R,0)=v_0^2\log (R/R_0)$.

The rotating distribution function. Use specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. For the printed positive exponential, take $\epsilon =-E$ and $s:=\sigma >0$; $s$ has units of squared velocity. If instead $\epsilon$ means ordinary total energy and $\sigma >0$, the velocity integral diverges, so that literal convention cannot define the requested galactic distribution function. Equivalently one can retain ordinary energy and choose the printed $\sigma$ negative.

With the binding-energy convention, write the amplitude as $A$ to distinguish it from the function:Integration over both radial velocities and positive azimuthal velocities givesThe half-line integral follows by setting $u=v_{\varphi }^2/(2s)$ in the defining Gamma function integral and requires $q>-1$. Matching its radial power and normalization to the Mestel disk yields the rotating Mestel disk distribution function parametersSince $v_0^2,s>0$, the required integrability condition is automatic. The amplitude uses the stated potential reference; changing the additive potential constant changes $A$ without changing the physical distribution.

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.

For a photon, $p^i{p}_i=0$ and the null displacement along the ray is parallel to $p^i$. Consequently $p_{j}d{x}^j=0$, and the first term of the momentum law vanishes. Write $d{x}^i=p^{i}d\lambda$ with an appropriate path parameter. ThenFor a static potential, a photon emitted and received by laboratories at rest in the background frame has energy proportional to $p^0$. Using $E=h_{\mathrm{P}}\nu$, the scalar-force gravitational redshift isRestoring $c$, the leading fractional shift is $\Delta \nu /\nu =-\Delta {\Phi }_N/c^2$, where ${\Phi }_N$ is the dimensional Newtonian gravitational potential. Near Earth's surface, an upward displacement $H$ has $\Delta {\Phi }_N=gH>0$, hence $\Delta \nu /\nu =-gH/c^2$: the receiver sees a redshift.