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For a quasistatic spherical star with fixed total mass, the stellar structure equations in radius areHere $s$ is the specific entropy of a moving mass shell, and the entropy term accounts for release or storage of thermal and gravitational energy. In thermal equilibrium it vanishes. An equation of state, opacity, energy-generation law and composition evolution close the equations. The absence of mass loss fixes the outer mass coordinate; it does not require the star to be in thermal equilibrium.

The radiative temperature gradient needed to carry the full luminosity isHere the actual temperature gradient is $\nabla =d\log T/d\log P$. For uniform composition the Schwarzschild criterion says that a radiative layer is stable if ${\nabla }_{\mathrm{rad}}<{\nabla }_{\mathrm{ad}}$, with marginal stability at equality, where the adiabatic temperature gradient is $(\partial \log T/\partial \log P{)}_s$. Use $\nabla ={\nabla }_{\mathrm{rad}}$ there. If ${\nabla }_{\mathrm{rad}}>{\nabla }_{\mathrm{ad}}$, convection carries some of the flux. Efficient convection gives $\nabla \simeq {\nabla }_{\mathrm{ad}}$; inefficient surface convection needs a transport prescription and can be superadiabatic. For composition gradients, use the Ledoux criterion, with threshold ${\nabla }_{\mathrm{ad}}+\varphi {\nabla }_{{\mu }_{\mathrm{mol}}}/{\delta }_{\mathrm{th}}$ instead. Here ${\nabla }_{{\mu }_{\mathrm{mol}}}=d\log {\mu }_{\mathrm{mol}}/d\log P$, $\varphi =(\partial \log \rho /\partial \log {\mu }_{\mathrm{mol}}{)}_{P,T}$ and ${\delta }_{\mathrm{th}}=-(\partial \log \rho /\partial \log T{)}_{P,{\mu }_{\mathrm{mol}}}$, using the mean molecular weight ${\mu }_{\mathrm{mol}}$.

Above the thin burning shell, take constant luminosity $L$, constant mean molecular weight ${\mu }_{\mathrm{mol}}$, ideal-gas pressure $P=\mathcal{R}T\rho$ with $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$, and the Kramers' opacity law $\kappa ={\kappa }_0\rho T^{-7/2}$. Then the temperature equation isTo implement the printed assumption that $P,\rho$ and $M_r$ are all power laws while retaining mass conservation, write $M_r\propto r^m$, $\rho \propto r^b$, $P\propto r^p$ and $T\propto r^d$. The mass equation gives $m=b+3$; the ideal gas law gives $d=p-b$; hydrostatic equilibrium gives $d=m-1$. The diffusion equation gives $d-1=2b-2-13d/2$. Solving this linear system gives the self-gravitating Kramers power-law envelope:The coefficients can also be matched consistently. If $r_c$ is the core boundary and $M_c$ its enclosed mass,Radiative transfer fixes the constant luminosity throughwhose right-hand side is independent of radius for these exponents. Thus this is a solution of all four envelope equations, not just a dimensional estimate.

Matching the boundary temperature to the isothermal core and extrapolating the idealized power law to the specified photosphere givesFor reference $M_R/M_c=(R/r_c{)}^{1/11}=2500^{1/10}\simeq 2.19$, so the envelope mass is not negligible in this particular solution. The given $M_c\simeq M_{\odot }$ fixes the core-radius normalization $r_c=(11/42)G{M}_c/(\mathcal{R}10^7\mathrm{K})$; a numerical absolute radius needs the unspecified molecular weight. The exponent ratio also gives $\nabla =d/p=5/21<2/5$, so the fully ionized monatomic version of this idealization is convectively stable.

A different common approximation neglects the envelope's self-gravity and sets $M_r\simeq M_c$ in the force equation. It cannot obey the exact mass equation with a nonzero density and exactly constant $M_r$. Under that additional approximation, the same calculation instead givesThis is the constant-core-mass power-law opacity radiative envelope limit, not the full power-law solution with changing enclosed mass. Stating which approximation is used resolves the otherwise different numerical radius ratios.

A compact nearly isothermal core, a thin luminosity-producing shell and a much larger cool envelope describe a shell-burning red giant. The full power-law profile is a deliberately simple model. The Kramers' opacity law and an ionized ideal gas are not reliable at a $4000\mathrm{K}$ photosphere; partial ionization, other opacity sources and a convective envelope usually modify real cool giants. The quoted radius ratios are extrapolations within the stipulated model.

Write $Q={\dot{N}}_{\mathrm{ion}}$ for the constant hydrogen-ionizing photon production rate. Assume an initially neutral gas of hydrogen atoms, a sharp spherical ionization front, complete ionization inside it, and a fixed gas density. The Electron and proton number densities inside are then both $n$. Take the recombination coefficient $\alpha$ to be constant, with the usual Case B recombination interpretation: photons from recombination directly to the ground state are absorbed locally. Neglect absorption by dust, pressure-driven gas motion, and light-travel delays.

The total rate of Case B recombination within radius $R$ isAt photoionization equilibrium, $Q=\mathcal{R}(R_S)$, giving the Strömgren radiusThe square of the density appears because recombination requires encounters between an Electron and a proton.

During Ionization-front growth of a Strömgren sphere, each photon either replaces an ion lost through Case B recombination or ionizes an additional atom. There are $N_i=(4\pi /3)n{R}^3$ ions inside the ionization front, so photon counting givesThe recombination time is $t_{\mathrm{rec}}=(\alpha n{)}^{-1}$. Solving this first-order differential equation with $R(0)=0$ givesFor an initial radius $R_0$, the corresponding result is $R^3(t)=R_S^3+(R_0^3-R_S^3)e^{-t/t_{\mathrm{rec}}}$. Initially, the idealized photon-counting solution has $R\simeq [3Qt/(4\pi n){]}^{1/3}$; at late times, $R/R_S\simeq 1-\frac{1}{3}{e}^{-t/t_{\mathrm{rec}}}$. The initially divergent front speed in this approximation is a consequence of neglecting light-travel delays; the formula describes the growth once that approximation is applicable.

For a power law density, specify a reference radius $r_0$ so that $n(r)=n_0(r/r_0{)}^{-\beta }$. With an inner cutoff $r_{\mathrm{in}}>0$, the recombination integral for a power-law nebula isEquivalently, a logarithmic interval in radius contributes $d\mathcal{R}/d\ln r=4\pi \alpha n_0^2{r}_0^{2\beta }{r}^{3-2\beta }$. For $\beta <3/2$, the central integral converges as $r_{\mathrm{in}}\to 0$, and the recombination rate grows as $R^{3-2\beta }$. At $\beta =3/2$, each logarithmic interval contributes equally and the centre gives a logarithmic divergence. For $\beta >3/2$, the unmodified profile gives a power divergence at the centre, while its outer recombination integral converges. The change at $\beta =3/2$ is set by the competition between the spherical volume element and the squared density.

To regularize the centre, choose the following explicit cored power-law density profile, taking $r_0=r_c$ and $n_0$ as the core density:For $R\le r_c$, $\mathcal{R}(R)=(4\pi /3)\alpha n_0^2{R}^3$. For $R>r_c$, put $u=R/r_c$ and integrate the core and envelope separately:This rate is continuous and strictly increasing with $R$. If $\beta >3/2$, it has a finite supremum, the critical ionizing photon rate of a cored power-law nebula:For $0<Q<Q_{\mathrm{crit}}$ there is a unique finite radius at photoionization equilibrium. At $Q=Q_{\mathrm{crit}}$ the equality is approached only as $R\to \infty$; for $Q>Q_{\mathrm{crit}}$ even ionizing the whole profile cannot consume all photons through Case B recombination, so no equilibrium radius exists. For $\beta \le 3/2$, the envelope integral grows without bound and there is no finite critical rate in this infinite-extent model.

For a concrete choice, take any $n_0>0$, $r_c>0$, and $\beta =2$. Then $Q_{\mathrm{crit}}=16\pi \alpha n_0^2{r}_c^3/3$. In the envelope, defining $q=Q/(4\pi \alpha n_0^2{r}_c^3)$ gives $R=r_c/(4/3-q)$ for $1/3<q<4/3$, explicitly showing the divergence as the critical rate is approached. A different core shape changes the numerical coefficient of $Q_{\mathrm{crit}}$, while the outer-slope condition $\beta >3/2$ remains the same.

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

For a statistically homogeneous and isotropic density contrast,Using the Fourier transform conventionand translational invariance giveswithAt one point,

For a spherical top-hat volume $V=4\pi R^3/3$,The overlap of two radius-$R$ balls separated by $r\le 2R$ isTherefore

For $\xi (r)=(r/r_0{)}^{-1.8}$,ThusThis calculation assumes that the observed galaxies trace the matter field with unit, scale-independent galaxy bias, and that the fitted power law remains valid throughout the top-hat separations.

Write the given pressure-dependent ratio of permeability of a porous medium to sponge thickness aswhere $K$ is constant. The fluid pressure drop across the sponge is the hydrostatic pressure of the current,because the air pressures above and below are equal. Therefore Darcy's law gives the downward volume flux per unit horizontal area asThus the sponge drainage obeys a power law in the current thickness, as described by pressure-dependent Darcy drainage.

For a power law $f(u)\sim u^n$,Independence of $\rho$ requires $n=1$, givingThis is viscous hydrodynamic coarsening. The viscous and capillary terms both scale as $u^{-2}$, whereas the convective inertial term scales as $u^{-1}$. It is therefore subleading for $u\ll 1$.

Independence of $\eta$ requires $n=2/3$, givingThis is inertial hydrodynamic coarsening. Both inertial terms and the capillary term scale as $u^{-4/3}$, while the viscous term scales as $u^{-5/3}$ and is subleading for $u\gg 1$.

The crossover occurs when $u$ is order one. Hencewith a dimensionless order-one constant $C_X$ determined by the coefficients and morphology.

A galactic distribution function $f(\mathbf{x},\mathbf{v},t)$ is the stellar mass or number per six-dimensional phase space volume,Its velocity moments give the spatial density, mean velocity, and velocity dispersion; integrating those quantities along the line of sight and weighting by luminosity produces surface-brightness and line-of-sight-velocity observables. A model is compared with data only after the same projection, selection function, and instrumental convolution have been applied.

Jeans theorem states that every steady solution of the Collisionless Boltzmann equation depends on phase-space coordinates only through integrals of motion. Conversely, every nonnegative function of isolating integrals is a steady collisionless distribution function on the region where those integrals are defined.

For the stated power law in relative energy, isotropy givesWith the requested change of variables $v=\sqrt{2\Psi }\cos \theta$, the density becomeswhere the last equality uses the Beta function and Gamma function. Thusfor $n>1/2$.

The normalized second velocity moment isThe same substitution and the Beta-function recurrence giveConsequently the one-dimensional isotropic velocity dispersion is ${\sigma }^2=\overline{v^2}/3=\Psi /(n+1)$, proving the required linear dependence on the relative potential.

Let $w(x)$ be the downward velocity between the vertical plates. A steady vertical momentum balance givesSymmetry requires ${\tau }_{xz}(0)=0$, so the stress magnitude isThe wall stress first reaches the yield stress when $\rho g{l}_c={\tau }_y$. The onset measurement therefore gives

For $l>l_c$, define $\Delta =l-l_c$. The central region $|x|\le l_c$ is a plug flow of a yield-stress fluid, while the layers $l_c<|x|\le l$ are yielded. The Herschel–Bulkley fluid law gives, for $x\ge l_c$,Integrating from the no-slip wall $w(l)=0$ givesin the yielded layer, and the plug moves atThe volumetric flow rate per unit span is consequently

Near onset, the plug term dominates andThe observed exponent two therefore corresponds to $n=1$, a linear post-yield constitutive law at small shear rate. Far above onset, the fully yielded contribution scales asThe observed exponent four corresponds to $n=1/2$, a shear-thinning square-root law at high shear rate.

A plausible stress curve is therefore odd in $\dot{\gamma }$, starts at the two yield-stress values $\tau =\pm {\tau }_y$, and has finite linear slope just after yield:It then crosses smoothly to the concave high-rate behaviorwith the negative-rate branch fixed by odd symmetry. This combines the two power laws inferred independently from the measured flux limits.

Cold dark matter has $c_s=0$. During matter domination,The density equation becomesTrying the power law $\delta =t^p$ giveswhose roots are $p=2/3$ and $p=-1$. Thus the cosmic-time matter density modes areand the leading growing mode is $\delta \propto t^{2/3}\propto a(t)$.

For either component $j\in \{h,s\}$ and $0<\gamma <3$, direct integration of the power law density givesLet $B_j=4\pi G{A}_j/(3-\gamma )$. Since $d{\Phi }_j/dr=G{M}_j/r^2=B_j{r}^{1-\gamma }$,The additive constants depend on the chosen reference and cannot generally be set by requiring $\Phi \to 0$ at infinity for an untruncated power law.

For a satellite on a circular orbit, the linearized effective radial force gives the tidal radiusHere ${\Omega }^2=G{M}_h(<r_o)/r_o^3=B_h{r}_o^{-\gamma }$ and ${\Phi }_h^{\prime \prime }=(1-\gamma ){\Omega }^2$, soSubstitution of the two mass profiles yieldsand equivalently

Now set $\gamma =1$. ThenThe host circular speed is $v=B\sqrt{r_o}$ with $B=\sqrt{2\pi G{A}_h}$. The magnitude of Chandrasekhar dynamical friction becomesa constant. The specific angular momentum is $j=r_{o}v=B{r}_o^{3/2}$. Since the drag torque gives $dj/dt=-r_o{a}_{\mathrm{df}}$,and thereforeIntegration giveswhich is finite.

If stripping is switched off, hold the satellite mass at its initial value $M_0=\mu r_{o0}^2$. The frictional acceleration is then $a_{\mathrm{df}}=a_0{r}_{o0}^2/r_o^2$, where $a_0$ is the constant acceleration in the stripped calculation. The same torque equation givesThusWithout tidal stripping, the satellite retains its mass while the background density increases inward, so dynamical friction strengthens rapidly. Stripping instead gives $M_s\propto r_o^2$ and removes the very mass that creates the gravitational wake.

For a singular isothermal host, ${\rho }_h=A_h{r}^{-2}$, $M_h\propto r$, and $v=B$ is constant. The tidal formula with a satellite ${\rho }_s=A_s{r}^{-1}$ givesConsequently $a_{\mathrm{df}}\propto M_s{\rho }_h/v^2\propto r_o^2$. Since now $j=B{r}_o$, the torque equation yieldsIts solution satisfies $r_o^{-2}=r_{o0}^{-2}+2C_{2}t$, so $r_o$ approaches zero only as $t^{-1/2}$ and never arrives in finite time. This is dynamical-friction stalling by tidal stripping: the inward tidal field strips the satellite so aggressively that its wake and drag disappear. Observationally, disrupted satellites should deposit stars in streams and the stellar halo, surviving low-mass remnants can remain at finite radii for very long times, and merger times inferred from a constant satellite mass can be severe underestimates.

A power law is a relation $f(x)=C{x}^p$, with fixed constants $C$ and $p$. For positive $x$, any real number $p$ is allowed. Its useful feature is a simple scaling rule:The same proportional change of input produces the same proportional change of output, independently of the starting value. The exponent controls that response; the coefficient fixes the magnitude.

For example, the area of a circle is $A(r)=\pi r^2$. Doubling the radius multiplies its area by four, and tripling it multiplies the area by nine. The volume of a sphere is $V(r)=4\pi r^3/3$: the same changes multiply its volume by eight and twenty-seven. The distinction explains why similarly shaped objects do not preserve their ratio of surface area to volume when their size changes.

When the output is positive, a logarithm turns the power law into a straight-line relation. Using positive reference values $x_0$ and $f_0=f(x_0)$,These ratios are dimensionless quantities, so the formula also applies to physical measurements with units. Plotting both ratios on a logarithmic scale gives a line of slope $p$. Changing a unit changes the numerical coefficient, while the exponent stays the same. Positive $p$ describes increasing values; negative $p$ describes decreasing values; $p=0$ gives a constant. The zero coefficient is a degenerate case with no useful logarithmic slope.

Exact scaling is stronger than regular variation. The function $x^2\log x$ has a doubling ratio tending to four, but its logarithmic correction keeps growing. The diagram contrasts it with the exact power law $x^2$.

A power law for a probability density and a power law for a tail probability have different exponents. For a Pareto distribution with $\alpha >0$ and lower endpoint one,The tail probability has exponent $-\alpha$; differentiating it and changing sign gives the probability density exponent $-\alpha -1$. Confusing them changes the interpretation of a fitted exponent. The normalizing coefficient and lower endpoint also matter: a decreasing power law is not automatically a valid probability density over the entire positive real line.

The appearance of a line over a finite logarithmic range is evidence to investigate, rather than a proof of an exact power law or its mechanism. In their study of power-law distributions, Clauset, Shalizi and Newman combine maximum likelihood estimation, goodness-of-fit tests and comparisons with alternative models; their abstract explains why ordinary least-squares fitting can give misleading results for tail data. See Aaron Clauset, Cosma Rohilla Shalizi and M. E. J. Newman, Power-law distributions in empirical data, SIAM Review 51 (2009), 661–703. The scaling rule is mathematically exact when it defines a model; whether a dataset supports that model is a question of statistical inference.

A power-law distribution uses a power law for a probability mass function or a probability density on a specified range. For a continuous model with $p_X(x)=C{x}^{-\beta }$ on $x\ge x_{\min }>0$ and zero below it, the total integral is finite exactly when $\beta >1$. Normalization then requires $C=(\beta -1)x_{\min }^{\beta -1}$, giving a Pareto distribution with shape parameter $\alpha =\beta -1$. Discrete models instead normalize a sum. The support and normalizing coefficient are part of the model, and a power law for its tail probability uses a different exponent from its probability density.