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A finite spherical subvolume has radial-pressure virial surface term $4\pi R^3\rho (R){\sigma }_r^2$. The infinite singular isothermal sphere has divergent total mass and energies, so a finite isolated virial theorem cannot omit this term. Different orbital distributions with the same density can have different radial boundary pressures.

For the $r^{-2}$ singular isothermal sphere, the isotropic Spherical Jeans equation gives ${\sigma }_r^2={\sigma }_{\theta }^2={\sigma }_{\varphi }^2=v_0^2/2$, so $⟨v^2⟩=3v_0^2/2$. Randomly oriented circular orbits instead have ${\sigma }_r^2=0$, each transverse second moment $v_0^2/2$, and total mean square speed $v_0^2$.

For a singular isothermal sphere of dispersion $\sigma$ and gas fraction $f_g$, placing all gas inside radius $R$ onto a hole in time $R/\sigma$ gives ${\dot{M}}_{\mathrm{dyn}}\sim 2f_g{\sigma }^3/G$. This is an optimistic supply bound that omits angular momentum transport, star formation, and active-galactic-nucleus feedback. A numerical comparison with the Eddington accretion rate requires a gas fraction, reference radiative efficiency, opacity, and crossing-time convention.

For a low-mass singular isothermal sphere satellite on a circular orbit of radius $D$ in a singular isothermal sphere host, ${\Omega }^2=2{\sigma }_h^2/D^2$ and ${\Phi }_h^{\prime \prime }=-2{\sigma }_h^2/D^2$. The radial differential effective acceleration is $({\Omega }^2-{\Phi }_h^{\prime \prime })r$. Balancing it with satellite gravity $2{\sigma }_d^2/r$ gives the displayed Jacobi tidal radius and bound mass $m_t=\sqrt{2}D{\sigma }_d^3/(G{\sigma }_h)$. Equal mean density gives a different order-one approximation, $r_t=D{\sigma }_d/{\sigma }_h$; eccentric orbits require an orbital-history prescription.

For the singular isothermal sphere, integrate its projected surface mass density inside a circle of proper radius $\xi$:With $\xi =D_d\theta$ this gives the reduced deflection magnitudeindependent of $\theta$. Hence the singular isothermal sphere lens hasThe spherical three-dimensional mass $2{\sigma }^{2}r/G$ would give the wrong coefficient here; the thin gravitational lens equation needs the projected cylindrical mass.

Using the same ${\theta }_E=3\times 10^{-4}$ and $D_{ds}/D_s=0.5$, the one-dimensional velocity dispersion isThus $\sigma \simeq 2100\mathrm{km}{\mathrm{s}}^{-1}$ within the idealized SIS model.

A complete Einstein ring requires an axisymmetric gravitational lens and exact alignment of the source with the lens centre and observer: $\beta =0$. A finite aligned circular source gives a ring of finite width. The background source must lie behind the lens, so the lensing distance factor is positive.

With $\beta =0$ the thin gravitational lens equation givesThe Einstein radius here is angular, and the mass is projected inside the corresponding aperture. For an extended mass distribution the equation is implicit because the enclosed mass depends on ${\theta }_E$.

Using the supplied approximations, $D_{ds}/(D_s{D}_d)=1/(3\times 10^{25}\mathrm{m})$ and ${\theta }_E=3\times 10^{-4}$, the inferred mass isThusThis measures the projected aperture mass without requiring the singular isothermal sphere model used subsequently.

An integral of motion is a function of phase space that is constant along a trajectory. Put $\varphi =-\psi +\mathrm{constant}$, so the stellar acceleration is $\dot{\mathbf{v}}=\nabla \psi$. In a time-independent spherical Newtonian gravitational potential,The second equality uses the radial direction of $\nabla \psi$. Thus both the binding energy and the full specific angular momentum vector are integrals of motion.

The Collisionless Boltzmann equation isFor a stationary galactic distribution function, this says $df/dt=0$ on every characteristic, so $f$ is constant on each stellar orbit. Locally choose a complete set of orbit labels $I_j$, themselves integrals of motion; constancy along the characteristic means $f=F(I_1,\dots ,I_s)$. Conversely, for any such differentiable $F$, the chain rule gives $df/dt={\sum }_j{F}_{,j}{\dot{I}}_j=0$. This proves Jeans theorem in its orbit-invariance form. It does not claim that just two particular integrals classify every orbit in an arbitrary potential. For the specified spherical model, $F(E,L)$ is a rotationally invariant stationary choice.

For the velocity integration, write $v_r=v\cos \eta$, $v_{\perp }=v\sin \eta$, $L=rv\sin \eta$, and set $k=r/r_a$. At fixed position the velocity-dependent factor isUse ${\int }_0^{\infty }{v}^2{e}^{-A{v}^2/(2{\sigma }^2)}dv=\sqrt{\pi /2}{\sigma }^3{A}^{-3/2}$ and integrate over $\chi$. The mass density isSince ${\rho }_c=\rho (0)={\rho }_1{e}^{\psi (0)/{\sigma }^2}$,These integrals use the stated untruncated galactic distribution function over all velocities; an additional escape-energy cutoff would change the result.

In local orthogonal velocity components the same exponent is $-[v_r^2+(1+k^2)(v_{\theta }^2+v_{\varphi }^2)]/(2{\sigma }^2)$. Each component is a centred Gaussian distribution. The one-dimensional identity $\int v^2{e}^{-A{v}^2/(2{\sigma }^2)}dv/\int e^{-A{v}^2/(2{\sigma }^2)}dv={\sigma }^2/A$ yieldsIn particular the velocity-anisotropy parameter is ${\beta }_{\mathrm{an}}=1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)=r^2/(r_a^2+r^2)$. This is Gaussian angular-momentum suppression in a stellar distribution.

The radial laws also require self-consistency through the Poisson equation for Newtonian gravity:For $r\ll r_a$, outside a sufficiently small core, the model is almost isotropic and isothermal. Seek its scale-free envelope with $\rho =A/r^2$. Its distribution-density relation gives ${\psi }^{\prime }=-2{\sigma }^2/r$, and the Poisson equation for Newtonian gravity gives $A={\sigma }^2/(2\pi G)$. Thus the intended intermediate singular isothermal sphere behaviour iswhere a core scale is $r_c=\sigma /\sqrt{4\pi G{\rho }_c}$.

The stated $r^{-2}$ law cannot be the actual central limit of a regular finite-density model. For finite ${\rho }_c$ and $\psi (0)$, regularity and the Poisson equation for Newtonian gravity instead giveand $v_c^2=(4\pi G{\rho }_c/3)r^2+O(r^4)$. A broad intermediate $r^{-2}$ region exists only if $r_c\ll r_a$; the radius $r_a$ alone does not ensure it.

For the outer radially anisotropic isothermal halo asymptotics, put $s=\log (r/r_a)$, $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. At $r\gg r_a$ the density relation and Poisson equation becomeThe slow outer balance is $y_s\sim -\lambda e^y$, so $d(e^{-y})/ds\sim \lambda$, giving $e^y\sim 1/(\lambda s)$. Its consistency check is $y\sim -\log (\lambda s)$, for which $y_{ss}/y_s\sim -1/s\to 0$. ThereforeThe reference radius $r_{\ast }$ changes only subleading logarithmic terms. The circular speed falls as $(\log r{)}^{-1/2}$, rather than remaining exactly flat.

Physically, the model has an almost isotropic central region and, if the scales separate, a flat-rotation isothermal envelope. Beyond $r_a$, suppression of large specific angular momentum leaves predominantly radial orbits: radial velocity dispersion remains $\sigma$, while transverse dispersion falls as $r^{-1}$. The outer density acquires a logarithmic suppression and the galaxy rotation curve declines slowly. The enclosed mass still grows approximately as $r/\log r$, so this untruncated model has infinite total mass.

Integrate the dwarf singular isothermal sphere only to its truncation radius:ConsequentlyThe dimensions are $(\text{length})(\text{speed}{)}^2/G$, as required for mass. At fixed host dispersion and separation the mass scales as ${\sigma }_d^3$, rather than the luminosity's fourth-power scaling used later. Under the rough equal-density tidal radius convention, the corresponding mass is $2D{\sigma }_d^3/(G{\sigma }_h)$, larger by $\sqrt{2}$.

An orbital prescription is needed to fix the coefficient in a tidal radius. Adopt a low-mass satellite on a circular orbit of separation $D$, with $r_t\ll D$. The host singular isothermal sphere hasIn the corotating frame, subtract the host force on the satellite centre and add the differential centrifugal force. Along the line joining the centres, the outward effective acceleration to first order in displacement $r$ is $({\Omega }^2-{\Phi }_h^{\prime \prime })r=4{\sigma }_h^{2}r/D^2$. Equate this to the dwarf's inward acceleration at its edge:This gives the Jacobi truncation of two singular isothermal spheresIt is the distance to the radial escape saddle, not an exact spherical surface: the actual tidal lobe is distorted. For an eccentric orbit, separation alone does not specify an instantaneous effective angular speed or the stripping history; a pericentric estimate is commonly used with further assumptions.

A simpler order-of-magnitude convention equates mean internal and host densities, $m_d/r_t^3=M_h(D)/D^3$, giving $r_t\sim D{\sigma }_d/{\sigma }_h$. Equivalently it retains the host gradient without the circular-orbit centrifugal contribution. Both reproduce the scaling $r_t\propto D{\sigma }_d/{\sigma }_h$, but their coefficients differ by $\sqrt{2}$. The familiar point-mass factor three is not appropriate for this extended host, whose enclosed mass grows linearly with radius. The circular-orbit Jacobi tidal radius will be used consistently below, with the rough convention's numerical values also given.

For a singular isothermal sphere, spherical integration of the mass density givesThe spherical shell theorem makes the radial attraction $GM(<R)/R^2$. Balancing it with circular centripetal acceleration yieldsThus the model has a flat galaxy rotation curve, resembling the outer galaxy rotation curves of many spiral galaxies. Central rotation curves usually rise, and not all real curves are flat throughout. A pressure-supported elliptical galaxy does not itself rotate at this test-particle circular speed.

The model has divergent central mass density, although finite enclosed mass at each finite radius, and an infinite total mass because $M(<R)\propto R$. Its potential is logarithmic, requiring an outer cutoff for a finite escape energy. A globally constant isotropic velocity dispersion, spherical symmetry, and one common profile for all gravitating components are further idealizations. Stellar light commonly has a quite different radial profile from the dark matter halo; real disks are flattened, and halos are neither exact isothermal spheres nor infinite. Tidal truncation also changes the distribution near the satellite boundary.

The host mass inside $D$ follows from the same singular isothermal sphere:Using the supplied host luminosity givesThis treats essentially all the stated optical light as contained in that aperture; a measured luminosity profile would refine that convention. Keeping consistent unrounded quantities,Thus the dwarf ratio is about twenty times the parent ratio; the rough equal-density tidal convention would give a factor thirty.

There is no logical contradiction in two objects sharing $L\propto {\sigma }^4$ but having different mass-to-light ratios. A virial theorem scaling gives $M\propto {\sigma }^{2}R$, so that $M/L\propto R/{\sigma }^2$. Equal ratios additionally require homologous relevant radii scaling as $R\propto {\sigma }^2$. Here the satellite's tidal radius scales as ${\sigma }_d$ in the fixed host field, not as ${\sigma }_d^2$. Equating the two ratios would require $r_t=D({\sigma }_d/{\sigma }_h{)}^2\simeq 56\mathrm{pc}$, far smaller than the derived tidal radius.

The heuristic derivation of the Faber-Jackson relation with common mass-to-light ratio and common surface brightness made auxiliary structural assumptions; the empirical relation alone cannot enforce them. Furthermore, the aperture dependence of mass-to-light ratio matters: a satellite's whole tidal mass and a host's $50\mathrm{kpc}$ halo mass are not homologous half-light-aperture masses. Dark matter need not follow stellar light, and a dwarf spheroidal galaxy is not automatically a scaled giant elliptical galaxy. The model calculates total gravitating mass: reading its density literally as stars alone would imply an implausibly large purely stellar ratio, exposing the limits of that interpretation.

A two-body annihilation requires two nearby particles, so the dark-matter annihilation radiation has emissivity proportional to the square of the mass density. With a spatially constant rate coefficient, writefor identical self-conjugate particles, where $E_a$ is the radiated energy per annihilation; the factor one-half counts unordered pairs. Particle and antiparticle populations instead require their respective number densities, without that identical-pair factor. For the singular isothermal sphere, the surface brightness becomesThe substitution $z=R\tan \theta$ gives $R^{-3}{\int }_{-\pi /2}^{\pi /2}{\cos }^2\theta d\theta =\pi /(2R^3)$. Hence annihilation gives the steeper projected profileFor the illustrative cored profile used above, $I_a=C_a{A}^2/[8(R^2+r_c^2{)}^{3/2}]$. These expressions assume a smooth dark matter halo, negligible absorption and constant $C_a$; unresolved density clumps or a velocity-dependent annihilation coefficient change the projected shape. The projection of a spherical power-law emissivity summarizes why three-dimensional powers $r^{-2}$ and $r^{-4}$ produce $R^{-1}$ and $R^{-3}$.

Use a steady, isotropic, self-gravitating dark matter halo, with $V$ its constant one-dimensional velocity dispersion. The Spherical Jeans equation and spherical mass conservation readFor the scale-free outer region, put $\rho =A{r}^{-p}$. Its enclosed mass scales as $r^{3-p}$, so $GM(r)/r$ is constant only if $p=2$; the Spherical Jeans equation then gives $2V^2=GM(r)/r=4\pi GA$. Thus the singular isothermal sphere hasIts circular speed is $v_c=\sqrt{2}V$. If a speed denoted $V$ were instead the circular speed, the coefficient would be $V^2/(4\pi G)$, but the printed variable is a velocity dispersion.

The assumptions matter. A constant velocity dispersion and spherical shape alone do not specify a unique central mass density or orbital distribution. Even with isotropy, the Spherical Jeans equation integrates to $\rho ={\rho }_0\exp [-(\Phi -{\Phi }_0)/V^2]$, and the self-gravitating Poisson equation for Newtonian gravity givesThere are finite-central-density isothermal solutions as well as the singular one; the $r^{-2}$ expression is the intended outer scale-free halo approximation. A finite halo must also end or steepen at large radius, since the untruncated singular isothermal sphere has infinite total mass.

Using the given constants, the mass is $2\times 10^{41}\mathrm{kg}$ and the velocity dispersion is $2\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$. The nominal galactic dynamical time isThus the collapse clock is a few tens of millions of years, of order $10^8$ years, with $R/\sigma \simeq 3\times 10^7$ years in the interior singular isothermal sphere convention. The nominal radius $GM/{\sigma }^2\simeq 3.5\times 10^{20}\mathrm{m}\simeq 12\mathrm{kpc}$ is a plausible galactic scale. A full stellar orbit takes several crossing times, giving order $10^8$ years near a galactic disc radius, whereas the age of the Milky Way is of order $10^{10}$ years. Many orbital passages therefore fit into the lifetime of the galaxy, as expected; a galactic dynamical time need not equal either its full orbital period or its age.

Let $R$ be a characteristic radius enclosing mass $M$, and let $\sigma$ denote a one-dimensional velocity dispersion. Gravity and random-motion support give a relation $GM/R=\eta {\sigma }^2$, where the dimensionless structure coefficient $\eta$ depends on the chosen radius and density profile. A crossing-based galactic dynamical time is thereforeThe scaling follows by equating acceleration scales ${\sigma }^2/R$ and $GM/R^2$, not by identifying a complete orbital period with a crossing time. An explicit scale-free example checks the coefficient: constant circular speed and spherical symmetry give $M(<r)=v_c^{2}r/G$ and $\rho =v_c^2/(4\pi G{r}^2)$. With isotropic constant velocity dispersion, the Spherical Jeans equation becomesso $v_c^2=2{\sigma }^2$, $M(<R)=2{\sigma }^{2}R/G$, and $R/\sigma =GM/(2{\sigma }^3)$. This is the singular isothermal sphere scaling, valid over an interior radial range; a literal infinite model has no finite total mass, and a hard finite boundary needs a pressure term or an outer transition. A free-fall time of a uniform sphere or orbital period changes the numerical factor. The subsequent estimates use the nominal scale $GM/{\sigma }^3$, keeping these order-unity structural uncertainties explicit.

For an isolated final system with isotropic one-dimensional velocity dispersion $\sigma$, the random kinetic energy is $K=\frac{1}{2}M⟨v^2⟩=\frac{3}{2}M{\sigma }^2$. The virial theorem gives $W=-2K$ and total energy $E_f=K+W=-K$. Starting from a diffuse, almost unbound gas reservoir with $E_i\simeq 0$, conservation of energy therefore requires the radiative energy of virialized gravitational collapseIt is the decrease in total energy, not the whole final magnitude $|W|$, that is lost; half of the binding-energy magnitude remains as kinetic energy. Assuming radiation carries most of this loss, the average luminosity during the nominal galactic dynamical time isusing $L_{\odot }\simeq 4\times 10^{26}\mathrm{W}$. This is comparable in order of magnitude to the stellar luminosity of a characteristic $L_{\ast }$ galaxy in the Schechter function, often of order $10^{10}{L}_{\odot }$ with a band-dependent definition. The equality is an energy-scale comparison: ordinary present starlight is powered mainly by stellar nucleosynthesis, whereas this estimate concerns gravitational energy lost during formation. Appreciable initial binding energy, mass loss or external stresses alter the coefficient, and the finite-boundary singular isothermal sphere cannot be used as an isolated exact virial equilibrium without its boundary term.

False. The singular isothermal sphere is infinite in extent and mass. The isolated finite-system virial theorem cannot be applied to its global divergent energies without boundary terms.

A direct Spherical Jeans equation counterexample is enough. For isotropic velocities with constant one-dimensional dispersion ${\sigma }^2$, its density slope givesThus ${\sigma }^2=v_0^2/2$ and the circular and isotropic speed comparison in a singular isothermal sphere isThis is realized by a positive isotropic distribution $F\propto \exp (E/{\sigma }^2)$, which gives a Maxwell-Boltzmann distribution of velocities and $\rho \propto \exp (\psi /{\sigma }^2)\propto r^{-2}$. Circular orbits have instead $⟨v_r^2⟩=0$ and $⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=v_0^2/2$, so their total mean square speed is $v_0^2$. The isotropic model contains eccentric orbits except for a measure-zero subset.

The boundary virial term of a singular isothermal sphere explains the different energy budgets of a finite spherical region. At radius $R$, its radial-pressure surface term is $4\pi R^3\rho (R){\sigma }_r^2$. It is nonzero in the isotropic model and zero in the circular-orbit model. Equality of the density therefore does not enforce equality of the velocity energy through an incorrectly omitted surface term.

The singular isothermal sphere has one-dimensional velocity dispersion $\sigma$, density ${\rho }_{\mathrm{tot}}(r)={\sigma }^2/(2\pi G{r}^2)$, and enclosed mass $M(<R)=2{\sigma }^{2}R/G$. The gas mass inside $R$ is thereforeTaking the dynamical crossing time to be $t_{\mathrm{dyn}}\simeq R/\sigma$, the dynamical upper bound on black-hole fuelling isThe cancellation of $R$ is special to the singular isothermal sphere. At $\sigma =200\mathrm{km}{\mathrm{s}}^{-1}$, this is approximately

Use the stated M-sigma relation to set $M_{\mathrm{BH}}\simeq 3\times 10^8{M}_{\odot }$. Defining the Eddington accretion rate with a reference radiative efficiency of black-hole accretion ${\eta }_0$,Combining this with $M_{\mathrm{BH}}\propto {\sigma }^4$ givesThus a gas fraction of order $0.1$--$0.16$ and the usual ${\eta }_0=0.1$ give roughly $60$--$90$ times the Eddington accretion rate, below $100$ at the requested dispersion. Using a crossing time $R/(\sqrt{2}\sigma )$ instead changes the coefficient by $\sqrt{2}$, so this is an order-of-magnitude estimate rather than a sharp universal bound. The problem does not specify $f_{\mathrm{gas}}$ or the rate convention: without these choices, the numerical claim cannot be proved for every gas fraction. Defining the rate as $L_{\mathrm{Edd}}/c^2$ would make this ratio ten times larger.

This is extremely optimistic because it gives every gas element a direct path to the hole on a crossing time. In a real galaxy, gas must shed angular momentum, avoid conversion into stars through star formation, cool sufficiently to flow inward, and survive active-galactic-nucleus feedback. A rotating accretion disk generally transports material on a viscous timescale, much longer than the orbital crossing time. Most available gas need not reach the central hole. If the hole is supplied at approximately the Eddington accretion rate, only about one percent of this optimistic dynamical supply reaches it; the actual percentage depends on its luminosity and duty cycle, so the bound alone is not a measurement of the fuelling efficiency.

For the temperature jump, use Bondi accretion as the local hot-gas estimate:Assuming the initial gas had $c_s\sim \sigma$, its density has not yet changed, and its polytropic coefficient remains comparable, heating to $c_{s,\mathrm{feed}}\sim 10\sigma$ givesThe capture radius $G{M}_{\mathrm{BH}}/c_s^2$ also shrinks by about $100$. Subsequent expansion can lower the density and suppress the mass accretion rate further. The result illustrates thermal suppression of black-hole accretion and negative active-galactic-nucleus feedback: vigorous activity can shut off its own gas supply. The factor $10^{-3}$ is conditional on the fixed-density comparison; the previous dynamical upper bound is not itself an actual pre-heating Bondi rate.