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Removing helium from a hydrogen-rich envelope lowers He/H and raises its reciprocal H/He. A helium-rain explanation of atmospheric depletion must keep those two ratio conventions distinct. Settling helium-rich material can release gravitational energy while changing atmospheric composition; excess luminosity alone does not uniquely determine the interior process.

Two point masses $m_1,m_2$ separated by distance $r$ have potential energyThe corresponding attractive force is $-\nabla V$.

This is the Newtonian point-mass specialization of gravitational energy, with the reference value chosen at infinite separation.

For a regular, spherical star, hydrostatic equilibrium gives $dP/dr=-Gm(r)\rho (r)/r^2$. Multiply by $4\pi r^3$ and integrate to the surface. Integration by parts givesThe central boundary term vanishes for finite central pressure. Since $dm=4\pi r^2\rho dr$, the right-hand side of the integrated hydrostatic equation isThe last expression is the gravitational energy: adding a thin mass shell $dm$ to the interior mass $m$ contributes $-Gmdm/r$, counting each interacting pair once. HenceWith the specified zero surface pressure this proves the stellar virial theorem in the required form,

Let $\Omega =-{\int }_0^{M}Gmdm/r$ be the gravitational energy. Multiply the stellar hydrostatic equation by $4\pi r^3$ and integrate through the star. Integration by parts and zero surface pressure givewhile its hydrostatic right-hand side is $-{\int }_0^{R}4\pi Gmr\rho dr=\Omega$. Hence the stellar virial theorem is

There is a genuine reference-sheet error to resolve. A monatomic ideal gas with adiabatic exponent $5/3$ has specific internal energy of an ideal gas $u=P/[\rho (5/3-1)]=3P/(2\rho )$, not the printed $3P/\rho$ on PDF page 2. The latter would give total internal energy $-\Omega$ and zero total energy, inconsistent with radiating contraction. The following derivation uses the consistent monatomic value.

A fully convective star is approximately isentropic and has an adiabatic stellar polytrope with index $3/2$. The given polytropic gravitational-energy formula yields $\Omega =-6G{M}^2/(7R)$. The virial relation gives total internal energy $U=-\Omega /2$, so total energy is $U+\Omega =-3G{M}^2/(7R)$. At constant mass with no nuclear heating, Kelvin-Helmholtz contraction obeys $d(U+\Omega )/ds=-L$, whenceHere $s$ denotes physical time.

In the plane-parallel atmosphere, the inward optical depth satisfies $d\tau =-\kappa \rho dr$, so hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. Neglecting radiation pressure, $\rho =\mu P/(\mathcal{R}T)$ and the specified opacity becomes $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{16}$. ThereforeThe grey atmosphere relation and the boundary $P(0)=0$ integrate toTo apply the Schwarzschild criterion, put $w=2+3\tau$. Logarithmic differentiation givesThis increases inward, starting from zero. Marginal convection has ${\nabla }_{\mathrm{rad}}=2/5$, so $w_b^3=136/5$. Since $T_b^{12}/T_e^{12}=w_b^3/64$, the grey-atmosphere convection onset with seventeenth-power opacity isThe purely radiative atmosphere is used only up to this boundary.

The convective relation $P=K{T}^{5/2}$ is spatially uniform in its entropy coefficient $K$ at each instant; $K$ can change as the star radiates and contracts. At the convection boundary $w_b$ is a fixed number, so $P_b\propto g^{1/2}{T}_e^{-8}$ at fixed composition and opacity normalization. ThusOn the other hand, the central scalings for the $n=3/2$ stellar polytrope give $P_c\propto G{M}^2/R^4$ and $T_c=\mu P_c/(\mathcal{R}{\rho }_c)\propto \mu GM/(\mathcal{R}R)$. Hence, at fixed composition,Equating the interior and atmosphere expressions yields $T_e^{21/2}\propto M{R}^{1/2}$. Applying the Stefan–Boltzmann law then proves the Hayashi contraction with seventeenth-power opacity relation

At fixed mass write $L=A{M}^{8/21}{R}^{\alpha }$, with constant $A$ and $\alpha =46/21$. The contraction equation is $\dot{R}=-C{R}^{\alpha +2}$, where $C=7A{M}^{8/21}/(3G{M}^2)$. Direct integration givesMultiplying through by $R^{\alpha +1}$ and substituting the current luminosity givesFor the stated very large initial radius, or after that initial-radius term becomes negligible,This is the luminosity-age relation within the contraction model, rather than an assertion that a physical star begins at literally infinite radius.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

A cold, self-gravitating region in a molecular cloud can collapse when gravity overcomes thermal, turbulent and magnetic support. Increasing central density produces a protostar, while radiation removes part of the released gravitational energy. The cloud initially carries angular momentum; material that cannot fall radially to the centre settles into a rotating protoplanetary disk. Collisions and cooling reduce random vertical motion, while angular-momentum transport allows some material to accrete and some to move outwards. Bipolar outflows and magnetic stresses can remove additional angular momentum.

The schematic sequence is $\text{rotating cloud}\to \text{protostar plus disk}\to \text{accreting solids and gas}\to \text{planetary system plus cleared disk}$. In the cool disk, grains condense and collide; settling and concentration can produce planetesimals, whose gravitational growth leads to planetary embryos. Massive cores may capture gas before the disk disperses; sufficiently massive unstable disks can also form bound condensations. Migration and interactions modify the final orbits. Radiation, accretion and outflows eventually disperse the remaining gas. Thus planet formation uses the same rotating reservoir produced by ordinary star formation, although the efficiency and final architecture are not universal and individual growth stages remain active research topics.

Helium becoming partially immiscible in dense hydrogen can form helium-rich droplets that settle inward. Gravitational energy release and associated thermal evolution can supply extra luminosity, while the upper hydrogen envelope becomes helium depleted. Different interior temperature-pressure histories can make separation stronger in Saturn than in Jupiter.

The source's ratio is reversed for this intended explanation. Depleting helium lowers He/H and raises H/He. For a concrete example, keeping ten hydrogen units while reducing one helium unit to one half changes H/He from ten to twenty, not to a lower value. Thus the literal statement that Saturn has lower H/He cannot simultaneously be used as evidence for the stated upper-envelope helium-depletion mechanism. If the intended comparison is lower He/H, the observations and extra heat are mutually consistent with helium rain. Composition measurements and interior modeling are still needed; excess heat alone does not uniquely prove that process.^[ref]