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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.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

For one target galaxy, an encounter with relative speed $u$ sweeps a cylinder of volume $\pi R^{2}udt$ in time $dt$. Multiplication by the number density gives the encounter rate $\Gamma =\pi R^{2}N⟨v_{\mathrm{rel}}⟩$. Thus the geometric galaxy-encounter rate givesThis is the expected number of mergers in the geometric model. For independent encounters, the Poisson process probability of at least one is $P_{\ge 1}=1-e^{-\mu }$, so the displayed linear probability is valid for rare encounters, $\mu \ll 1$. There is no factor of one-half for a single target; that factor would enter a count of distinct pairs across the whole population.

For an illustrative group environment, take $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N\sim 1{\mathrm{Mpc}}^{-3}$, $⟨v_{\mathrm{rel}}⟩\sim 300\mathrm{km}{\mathrm{s}}^{-1}$ and a Hubble time of order $10^{10}$ years, or $3\times 10^{17}\mathrm{s}$. The path length is about $9\times 10^{19}\mathrm{km}\simeq 3\mathrm{Mpc}$, using the stated distance conversion. HenceThe estimate scales linearly with environment density and quadratically with the adopted merger radius. It is an order-of-magnitude model estimate, not a universal observed merger fraction; reducing the illustrative density by a factor of one hundred reduces the estimate by the same factor.

For the tidal calculation, use the relative potential convention in which acceleration is $\nabla \psi$. Expand the point-mass potential about the target centre:The constant does not exert a force. The linear term accelerates the target centre and is subtracted in its freely falling frame. The remaining leading quadrupolar point-mass tidal potential isThe condition $r<r_p$ permits the convergent expansion; using only this term is the leading tidal approximation, accurate when the target radius is small compared with the closest separation.

The given impact and velocity directions imply ${\mathbf{r}}_p(t)=(p,0,vt)$. This trajectory is in the $x,z$ plane, correcting the incompatible printed $x,y$ plane. In the impulse approximation, hold the stellar position fixed while integrating the tidal tensor:For $s=vt/p$, the needed integrals arewith the mixed integral zero by oddness. Thus the integrated tidal tensor of a straight-line flyby is $2G{m}_p\mathrm{diag}(1,-1,0)/(v{p}^2)$ andThere is stretching along the impact direction, compression in the other transverse direction and no net kick along the path.

Write $\mathbf{u}$ for the star's initial velocity, distinguishing it from the perturber speed $v$. Its specific energy change is $\mathbf{u}\cdot \Delta \mathbf{v}+\frac{1}{2}|\Delta \mathbf{v}{|}^2$. The no-correlation assumption removes the first term on averaging; it is not an identity for every individual star. The mean specific heating at a given position is thereforeFor a spherical target, the mass-weighted averages satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. The tidal impulse heating of a spherical galaxy is consequentlyHere $⟨r^2⟩=m_g^{-1}\int r^2\rho d^{3}r$ is assumed finite. Notice the change from specific energy to total energy after multiplying by $m_g$.

For two equal targets, each gains the same internal energy, so $\Delta U_{\mathrm{tot}}=8G^2{m}_g^3⟨r^2⟩/(3v^2{p}^4)$. Their relative-motion reduced mass is $m_g/2$. If $v$ denotes their relative speed at infinity, their initial orbital energy is $E_{\mathrm{orb}}=\frac{1}{2}(m_g/2)v^2=m_g{v}^2/4$. Internal tidal heating comes from this orbital energy. In the tidal capture of galaxies model, capture occurs if $\Delta U_{\mathrm{tot}}>E_{\mathrm{orb}}$, which gives the equal-mass tidal-capture thresholdCapture allows further passages and eventual merger in this simplified picture.

The encounter lasts roughly $p/v$. The impulse approximation needs $\Omega p/v\ll 1$, so that a star barely moves during the tide. When $v/p<\Omega$, stellar orbits respond during the perturbation and the kicks can cancel. Adiabatic invariance of an orbital action produces adiabatic shielding of tidal encounters, rather than the impulsive heating used above. The capture inequality cannot be extrapolated into that slow-encounter regime.

A small-angle gravitational encounter between a test star and a field star of mass $m$ gives a transverse kick $\Delta v_{\perp }\simeq 2Gm/(bu)$, where $b$ is the impact parameter and $u$ the relative speed. For number density $n$, encounters in $b,b+db$ occur at rate $2\pi bdbnu$. Independent kicks add in mean square, giving the random-walk derivation of stellar relaxationThe Coulomb logarithm in stellar dynamics has $b_{\max }$ of order system size $r_s$, and $b_{\min }$ of order the strong-deflection scale $Gm/u^2$. For a virialized $N$-star system, $u^2\sim GNm/r_s$, so $\Lambda$ is of order $N$. The stellar relaxation time is the time for accumulated velocity variance to become comparable with $u^2$:Using $n\sim 3N/(4\pi r_s^3)$ gives $t_{\mathrm{rel}}\sim [N/(6\ln N)]r_s/u$. With a diameter-crossing convention $T_{\mathrm{cross}}\sim 2r_s/u$, this is $[N/(12\ln N)]T_{\mathrm{cross}}$, conventionally rounded toThe coefficient is approximate: spatial profile, velocity averages and crossing-time convention change it by factors of order unity. The robust result is $N/\ln N$ crossing times.

For $N\sim 10^{10}$ and $T_{\mathrm{cross}}\sim 10^8$ years, the estimate is about $4\times 10^{15}$ years, enormously longer than a Hubble time. Most galaxies therefore behave as collisionless stellar systems. Dense nuclei and star clusters can relax more rapidly. Negligible stellar encounters do not mean negligible collective gravitational instabilities.

Define the mass-weighted galactic distribution function by $dM=F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$, so $\rho =\int F{d}^{3}v$. Hamiltonian gravitational motion preserves phase space volume by the Liouville theorem in Hamiltonian mechanics. In the collisionless regime no encounter term redistributes stars between neighbouring phase space trajectories, so $dF/dt=0$. With acceleration $\nabla \psi$, the Collisionless Boltzmann equation is

For a steadily rotating galactic bar, write the inertial polar angle as ${\varphi }_{\mathrm{I}}=\varphi +{\Omega }_{b}t$. The inertial radial and angular equations are $\ddot{R}-R{\dot{\varphi }}_{\mathrm{I}}^2={\psi }_R$ and $R{\ddot{\varphi }}_{\mathrm{I}}+2\dot{R}{\dot{\varphi }}_{\mathrm{I}}={\psi }_{\varphi }/R$. Substitution gives the equations of motion in a rotating frame:Here the velocities are measured in the bar frame. Set $v_R=\dot{R}$, $v_{\varphi }=R\dot{\varphi }$ and define the positive-force rotating-frame relative effective potentialThe centrifugal sign is positive in this relative-potential convention. The characteristics areConsequently the rotating-frame collisionless Boltzmann equation isFor these planar equations, integrate over $d{v}_{R}d{v}_{\varphi }$ and interpret $\rho$ as the planar or vertically integrated mass density. Define $\rho ⟨A⟩=\int AFd{v}_{R}d{v}_{\varphi }$. Boundary terms in velocity vanish for a sufficiently decaying distribution.

Let the velocity accelerations above be $A_R,A_{\varphi }$. Their velocity divergence is ${\partial }_{v_R}{A}_R+{\partial }_{v_{\varphi }}{A}_{\varphi }=-v_R/R$. Thus integration by parts in the zeroth moment contributes $\rho ⟨v_R⟩/R$, producing the cylindrical continuity equation for a rotating stellar systemThis is conservation of mass: change of density balances flux through the radial and azimuthal sides of a small cylindrical element. The factor $R$ is geometric; the physical phase space measure is $RdRd\varphi d{v}_{R}d{v}_{\varphi }$.

For the radial first moment, multiply the rotating-frame collisionless Boltzmann equation by $v_R$. Its velocity integrations are $\int v_R{A}_R{F}_{v_R}{d}^{2}v=-\rho ⟨A_R⟩$ and $\int v_R{A}_{\varphi }{F}_{v_{\varphi }}{d}^{2}v=\rho ⟨v_R^2⟩/R$. Hence the radial cylindrical Jeans equations in a rotating frame areFor the azimuthal first moment, the $A_R$ velocity integral vanishes, while ${\partial }_{v_{\varphi }}(v_{\varphi }{A}_{\varphi })={\psi }_{\mathrm{eff},\varphi }/R-2{\Omega }_b{v}_R-2v_R{v}_{\varphi }/R$. This givesThe Jeans equations are local momentum-balance equations. The second moments include streaming momentum flux and random-velocity stress, while gravity, centrifugal acceleration and Coriolis acceleration supply the frame-dependent forces. They do not by themselves close the full distribution: a stress prescription or a galactic distribution function is also needed.

False. The enclosed-mass formula is a consequence of the spherical shell theorem. In a spherical galaxy, exterior shells exert no radial force and interior mass acts as though concentrated at the centre, giving $v_c^2=GM(<r)/r$.

A razor-thin axisymmetric disc instead hasBoth interior and exterior annuli contribute to that derivative. Its geometry is not determined by $M(<R)$.

A concrete counterexample compares a central point mass $M_0$ plus an exterior spherical shell with the same point mass plus a thin circular ring of mass $m$ and radius $a$. Their enclosed-mass profiles are identical: $M_0$ inside $a$, and $M_0+m$ outside. The interior gravitational force of an exterior thin ring follows by expanding its angularly averaged relative potential:For $R\ll a$, the ring's acceleration is outward, ${\partial }_R{\psi }_{\mathrm{ring}}=GmR/(2a^3)+\cdots$, whereas the spherical shell's acceleration is zero. With $M_0$ large enough to retain circular orbits, the disc-side speed isA narrow smooth annulus gives the same distinction. This proves that enclosed mass does not determine a disc rotation curve, even when the mass profiles of the spherical and disc models agree.

For finite-thickness disk relaxation, assume a well-separated range $b_{\min }\ll z_0\ll R$ and approximately virialized self-gravity. The three factors have distinct origins. Replacing the characteristic encounter speed by $u=\lambda v$ shortens the diffusion time by ${\lambda }^3$: the variance-production rate has one inverse power of speed, and the target random velocity squared has two powers. Concentrating the same number of stars into volume of order $R^2{z}_0$, rather than $R^3$, raises the number volume density by $R/z_0$ and shortens the time by $z_0/R$. Finally, only separations below the thickness sample a three-dimensional encounter geometry. The logarithmic upper cutoff becomes $z_0$, and therefore the inverse-logarithm time acquires $\log (R/b_{\min })/\log (z_0/b_{\min })$. More distant encounters have the planar geometry and provide a convergent, non-logarithmic correction. A smaller Coulomb logarithm in stellar dynamics actually partly offsets the first two shortenings.

Thus, using the same cutoff for this comparison,For actual point particles the disk cutoff is $Gm/(\lambda v{)}^2$, whereas the spherical one is $Gm/v^2$; one must use the appropriate cutoff in each logarithm rather than silently identifying them. For a softened calculation it is instead of order the larger of the weak-deflection scale and the softening length. The formula is not valid when $z_0$ approaches that cutoff.

Write $h=z_0/R$ and $q=K/\epsilon$, with simulation duration $K{t}_{\mathrm{dyn}}$ and allowed fractional diffusion $\epsilon$. Combining the paper's spherical normalization and its comparison formula gives the leading criterionThe spherical comparison logarithm cancels when its numerical convention and a common comparison cutoff are used; order-one shape factors remain approximate. In an unsoftened disk with $v^2\sim GNm/R$, $z_0/b_{\min }\sim h{\lambda }^{2}N$. Hence the implicit requirement is $N/\log (h{\lambda }^{2}N)\gtrsim 8q/({\lambda }^{3}h)$, in the regime $h{\lambda }^{2}N\gg 1$. For $\lambda =h=0.1$, the stellar relaxation time is of order $10^{-4}$ of the spherical value apart from logarithms; $N\sim 10^8$ gives a time of order 100 crossing times in this estimate. Negligible relaxation over that duration requires an additional margin. A disk generally needs far more particles than a round galaxy for the same relaxation tolerance; no universal minimum follows without duration, thickness and softening. Collective spiral or bending responses are a separate source of evolution even in a collisionless stellar system.

The common-age conditional probabilities giveHere $GW$ includes both possible component orders. Hence the binary star fractions have the concise ratioAs a check on the role of shared age, $p_{GG}=4p_G^2/3$ rather than $p_G^2$. This is covariance induced by a shared latent variable: the red giant indicator random variables have covariance equal to the variance of $g(Y)$, namely $1/97200$.

For a starburst galaxy with age $t$, put $y_b=t/(10\mathrm{Gyr})$. There is no age averaging. The component masses remain independent, so the fraction of systems with two red giants isThe last comparison is ambiguous in the printed question. Literally, the fraction of individual red giants in the first galaxy is $1/180$. Since $p_{GG}^{\mathrm{burst}}<1/8100<1/180$, no allowed starburst age satisfies that literal comparison. If the intended comparison is instead between systems containing two red giants in the two galaxies, compare with $p_{GG}=1/24300$; the answer isFor completeness, the individual red giant fraction in the burst is $g(y_b)=t/(900\mathrm{Gyr})$, which exceeds $1/180$ precisely for $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. This is a third, different comparison; it does not change the computed fraction of two-giant systems.

Ignoring gravitational focusing, a galaxy sweeps a cylinder of volume $\pi R^2{v}_{\mathrm{rel}}T$. Multiplying this by the number density $N$ of target galaxies and averaging the relative speed gives the mean encounter countFor independent encounters modeled by a Poisson process, the probability of at least one galaxy merger is $1-e^{-\mu }$. The printed expression $P=\mu$ is its rare-encounter approximation $\mu \ll 1$; it is not an exact probability for arbitrary $T$.

For an illustrative present-day field population take an effective merger impact parameter $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N=10^{-2}{\mathrm{Mpc}}^{-3}$, and $⟨v_{\mathrm{rel}}⟩=200\mathrm{km}{\mathrm{s}}^{-1}$. These are assumed order-of-magnitude inputs, not precise observational measurements. Taking a Hubble time of $T\simeq 1.4\times 10^{10}\mathrm{yr}\simeq 4.4\times 10^{17}\mathrm{s}$, the given distance conversion yieldsand henceThus the geometric field estimate is of order $10^{-5}$ to $10^{-4}$ per Hubble time. It scales as $R^{2}N⟨v_{\mathrm{rel}}⟩T$ and is very sensitive to environment and the adopted effective merger radius. Enhanced density in groups, gravitational focusing, and the evolution of the galaxy population are all omitted; this number is not a prediction of the full cosmological merger fraction.

Now use the stipulated rapid, distant encounter. The consistent rectilinear trajectory isWith the stated impact parameter along $x$ and velocity along $z$, this trajectory lies in the $xz$-plane. The original PDF's reference to the $xy$-plane is a typo; the local TeX also corrupts the impact-parameter direction. The original PDF fixes that direction as $x$.

Use the positive potential $G{m}_p/|\mathbf{R}-\mathbf{r}|$, whose gradient is the attractive acceleration under the question's sign convention. Its multipole expansion isThe first term has no force; the linear term accelerates the entire galaxy and disappears in the frame following its centre of mass. The leading internal tidal potential is thereforeThis is a quadrupole approximation valid for $r\ll p$, rather than an exact equality for every $r<p$. Its acceleration tidal tensor acts on $\mathbf{r}$ asIn the impulse approximation, each stellar position is held fixed during the flyby and the velocity kick is the integral of this acceleration. Put $s=vt$; terms odd in $s$ integrate to zero, whileThe integrated diagonal coefficients of the tidal tensor are consequently $(2,-2,0)/(v{p}^2)$, givingThe encounter stretches the galaxy along the impact parameter, compresses it along $y$, and gives no net leading kick along the flyby direction.

Denote the pre-encounter stellar velocity by $\mathbf{u}$, to distinguish it from the relative flyby speed $v$. The instantaneous change of specific kinetic energy isUncorrelated kicks with zero mean cross term imply $⟨\mathbf{u}\cdot \Delta \mathbf{v}⟩=0$, so the phase-averaged heating at a specified position isFor an individual star the cross term need not vanish: the formula is an ensemble or orbital-phase average. Integrating over a spherical galaxy of mass $m_g$, symmetry gives $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. Therefore the total tidal heating isFor two identical galaxies, each receives this heating with $m_p=m_g$. Adding both contributions gives

In the centre of mass frame, equal masses approach with speeds $v/2$ if $v$ is their relative speed at infinity. The initial orbital energy iswhere $m_g/2$ is the reduced mass. The orbital potential energy vanishes at infinite separation. Within the weak-deflection approximation this asymptotic speed is also the nearly constant speed used in the flyby calculation.

Conservation of energy transfers the positive internal heating out of the relative orbit. Tidal capture of galaxies occurs in this model if $E_{\mathrm{orb}}-\Delta E_{\mathrm{int}}<0$, which givesThe equality is marginal capture. A bound pair still needs subsequent evolution to coalesce, so the result is an approximate capture criterion used here as a merger criterion. Its right-hand side has dimensions of length times speed, as required.

The impulse approximation requires the flyby duration $\tau \sim p/v$ to be short compared with the stellar dynamical time ${\Omega }^{-1}$, namely $v/p\gg \Omega$. If $v/p<\Omega$, a star moves substantially during the encounter. Its orbital adiabatic invariance of an orbital action suppresses net heating by a slowly varying tidal field: adiabatic shielding of tidal encounters replaces the frozen-position calculation. The divergent $v^{-2}$ heating predicted by extrapolating the impulsive formula to slow encounters is therefore spurious. Close passages with $p$ comparable to the galaxy size or strong gravitational focusing also lie outside the derivation.

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.

A rotation-anisotropy diagram compares ordered rotation divided by velocity dispersion with a galaxy's flattening. An oblate isotropic rotator provides a reference curve. Falling below it can indicate anisotropic random support, but inclination, measurement apertures, and the definition of rotation change the interpretation. Intrinsic axis ratios and projected ellipticities are different observables.

The scale factor $a(t)>0$ measures how spatial lengths in an FLRW metric change with cosmic time $t$. Choose $a$ to be a dimensionless quantity with $a(t_0)=1$ at a reference epoch, usually today. For two comoving observers, their fixed radial comoving distance $\chi$ and their proper distance in cosmology $d(t)$ satisfyThe proper distance in cosmology is measured along a spatial slice at one cosmic time; it is not the distance that a travelling photon covers between emission and reception. Only ratios of the scale factor have a normalization-independent meaning: replacing $a$ by $Ca$ and $\chi$ by $\chi /C$ leaves $d$ unchanged.

For example, take $\chi =100\mathrm{Mpc}$, where $\mathrm{Mpc}$ means a million parsecs. At $a=0.5,1,1.5$, the same pair of comoving observers has proper distance in cosmology $50,100,150\mathrm{Mpc}$ respectively. Every such separation changes by the same factor, although more distant pairs gain more distance. The illustration treats the marked objects as ideal comoving observers; internal motions in galaxies need not follow this background scaling.

The beauty of the scale factor is that one function describes the changing spatial ruler for every comoving observer. Spatial homogeneity gives no preferred centre of cosmic expansion: in a flat spatial patch, relative positions obey $\mathbf{r}-{\mathbf{r}}_O=a(\mathbf{x}-{\mathbf{x}}_O)$, so recentering on any comoving observer preserves Hubble's law. The evolution of the scale factor is determined by the Friedmann equations, while its ratio between emission and reception determines cosmological redshift.

The two resolved quasar images of 0957+561 provide an observational question: are they separate sources, or different light paths from one source? In May 1979, spectrophotometry used the McDonald Observatory 2.7 m reflecting telescope, the UVITS spectrograph, and an intensified dissector scanner at the f/9 focus of the Cassegrain reflector. A 600-groove-per-millimetre diffraction grating produced 4100-7100 ångström scans sampled every 2 ångströms, with about 7 ångström resolution over most of that range. Two spectrograph slits, about 52 arcseconds apart, recorded object and sky background simultaneously for subtraction. Interleaving observations of the two images reduced instrumental and atmospheric gain changes.^[ref]

The readout was received light versus wavelength for each image. Closely matching absorption line redshifts supported a common source and absorber; differences in continua and line strengths motivated differential interstellar extinction. Spectral agreement supplied evidence for gravitational lensing, rather than proving the point-mass gravitational lens model. The authors discussed a physical double and discrepant optical and radio light ratios. Their note added in proof attributed the redder spectrum to foreground galaxy contamination and emphasized an extended mass distribution.^[ref]

A star-forming galaxy is a galaxy with ongoing star formation. Its hot young stars can supply hydrogen-ionizing radiation to the intergalactic medium, with the contribution controlled by the ionizing photon escape fraction.

A stellar halo is the extended, approximately spheroidal stellar population surrounding a galaxy. It differs from the total dark matter halo. If its outer luminosity density is $r^{-p}$, its projected light follows the projection of a spherical power-law emissivity and scales as $R^{1-p}$ away from breaks and outer truncation.

A stellar stream is a correlated collection of stars commonly originating in the disruption of a bound progenitor. Phase mixing can erase spatial contrast while leaving coherence in orbital integrals. Such fossil debris constrains a galaxy's past accretion, although not every stream is from a galaxy: star clusters also produce streams.

Structure formation is the growth of cosmological matter inhomogeneities into structures such as galaxies through gravitational instability and gas dynamics. Linear cosmological density perturbations describe the early, small-amplitude regime; nonlinear evolution is required once the density contrast becomes large.