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For repeated minor galaxy mergers with negligible internal satellite binding energy and zero orbital energy at infinity, parabolic dry-merger energy accounting leaves total energy nearly constant. The virial theorem then gives gravitational radius $R_g\propto M^2$ and mean-square speed proportional to $M^{-1}$. Small satellite mass does not by itself imply negligible internal binding energy.

In a dry galaxy merger, zero orbital energy at infinite separation and retention of all mass and energy give $E_F=E_I+E_A$. If the progenitors and remnant satisfy the virial theorem, their gravitational radii obey $R_F/R_I=(1+\eta {)}^2/(1+ϵ\eta )$, where $\eta =M_A/M_I$ and $ϵ=⟨v_A^2⟩/⟨v_I^2⟩$. The latter average is weighted by satellite mass. Nonzero orbital energy, escaping stars, or dissipative losses change the energy accounting.

Work in the centre-of-mass frame and define $I={\sum }_{i}m|{\mathbf{r}}_i{|}^2$, $T=\frac{1}{2}{\sum }_{i}m|{\mathbf{v}}_i{|}^2$. Differentiation givesFor each gravitational pair, the two contributions combine toSumming pairs therefore gives the exact scalar virial theoremIn a stationary equilibrium, or a bounded long-time average with vanishing average $\ddot{I}$, $2T=-W$. With $M=Nm$, the stated gravitational radius gives $W=-G{M}^2/R_g$, henceThe rms speed itself is $[GNm/R_g{]}^{1/2}$. This is a three-dimensional mean-square speed; for an isotropic system the one-component velocity dispersion is smaller by a factor of three in its square. No assumption about a uniform spatial density was used.

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.

Use a mass-normalized galactic distribution function $F$, so that $\rho =\int F{d}^{3}v$ and $\rho ⟨v_i{v}_j⟩=\int v_i{v}_{j}F{d}^{3}v$. Multiplying the stationary Collisionless Boltzmann equation by $v_j$ and integrating over velocity space givesAssume the galactic distribution function decays sufficiently rapidly for the velocity boundary term to vanish. Integration by parts then gives $\int v_j{\partial }_{v_i}F{d}^{3}v=-{\delta }_{ij}\rho$, and therefore the Jeans equations areThe second moment includes both ordered motion and velocity dispersion; no assumption of zero mean velocity was needed.

To obtain the tensor virial theorem, multiply the $j$th Jeans equation by $x_i$ and integrate over position. For an isolated, finite system with a vanishing spatial surface term,Since the left side is $-2K_{ij}$, $2K_{ij}+W_{ij}=0$. This convention puts all stellar second moments into $K_{ij}$; it does not separate ordered and random kinetic energies.

For a self-gravitating system, the trace $W$ is the Newtonian gravitational potential energy. Indeed, symmetrizing the pair integral givesThus the trace of the tensor virial theorem is $2K+W=0$, and the total energy obeysThese identities require self-gravity without an additional external potential or an omitted confining boundary pressure.

Let $⟨v_I^2⟩$ now denote a mass-weighted average over the entire initial system. Then $K_I=M_I⟨v_I^2⟩/2$. The gravitational radius is defined by $R_I=-G{M}_I^2/W_I$, so the virial theorem immediately yieldsThe gravitational radius measures total binding energy, rather than a particular geometric edge or half-mass radius.

For the accreted systems define the mass-weighted internal mean-square speed by $M_A⟨v_A^2⟩={\sum }_s{M}_s⟨v_s^2⟩$. If each satellite initially satisfies the virial theorem, its internal energy contributes to $E_A=-M_A⟨v_A^2⟩/2$. In parabolic dry-merger energy accounting, the orbital energy at large separation is zero. Assume a dry galaxy merger, no loss of mass or energy through escaping stars, no external work, and a final relaxed system satisfying the virial theorem. Then conservation of energy gives $E_F=E_I+E_A$ and $M_F=M_I+M_A$. Energy transferred by Chandrasekhar dynamical friction remains part of the total energy under these assumptions. Consequently,Applying the final virial theorem and dividing by the initial relation givesFor a mass doubling, $\eta =1$. A merger of two identical equilibrated galaxies has $ϵ=1$, hence $R_F/R_I=2$. For accretion through many minor galaxy mergers whose satellites have much smaller internal mean-square speeds, $ϵ\ll 1$, hence $R_F/R_I\simeq 4$, tending to $4$ in the cold-satellite limit. Equivalently, nearly fixed total energy makes the gravitational radius scale as $M^2$ during cold minor-merger size growth. Small satellite mass alone does not logically imply $ϵ\ll 1$: the factor four also requires this weak-binding assumption. Likewise, equal masses require comparable internal binding to give the factor two. These are the physical limits behind the two stated merger comparisons.

False. Stellar evaporation removes stars above the escape energy after two-body relaxation has redistributed their energies. The remaining self-gravitating cluster need not become less dense; losing mass is not the same as holding its radius fixed.

For a slowly evolving isolated cluster, apply the virial theorem at each approximate equilibrium: $E=T+W=W/2=-T$. Stars evaporating with nearly zero energy at infinity remove little total energy, so the bound cluster has nearly constant negative $E$. With gravitational radius $R_g$ defined by $W=-G{M}^2/R_g$,For homologous structure, its characteristic mean density therefore scales asIt increases as mass is lost. Escaping stars carrying positive energy make the bound energy more negative and strengthen the contraction tendency. This is virial contraction under stellar evaporation, related to the negative heat capacity of bound gravitational systems. A tidally limited globular cluster has additional radius constraints, so this scaling is not universal, but the claimed inevitable density decrease is false even in the simplest isolated evaporation model.

The Schechter function has logarithmic slope $\alpha \simeq -1$ for $M\ll M_{\ast }$ and an exponential cutoff for $M\gg M_{\ast }$. Thus a log-log plot is approximately a straight line of slope $-1$ at low mass and bends sharply downward near $M_{\ast }$.

For a collisionless self-gravitating system, define the inertia tensorTwo time derivatives, the collisionless Boltzmann equation, and integration by parts give the tensor virial theoremwhereFor an isolated steady state, or a long-time average of a bounded system, ${\ddot{I}}_{ij}=0$ and surface terms vanish, soTaking the trace gives $2K+W=0$. Therefore

Introduce the gravitational radius by $W=-G{M}^2/R$. Virial equilibrium givesAssume dissipationless accretion, negligible escaping mass, no external work, and final re-virialization. The accreted systems bringWith $\eta =M_A/M_I$ and $ϵ=⟨v_A^2⟩/⟨v_I^2⟩$,Since $M_F=M_I(1+\eta )$ and $E_F=-M_F⟨v_F^2⟩/2$,Using $E=-G{M}^2/(2R)$ also gives

Many minor galaxy mergers have dynamically cold accreted material, $ϵ\ll 1$. If they double the mass, $\eta =1$, then $R_F/R_I\simeq 4$, andThe steep low-mass side of the Schechter function supplies many small satellites. Repeated dry galaxy mergers can therefore add stars mainly at large radius, increasing size much faster than mass and lowering mean density. This explains how compact, dense redshift-two galaxies can evolve into larger present-day elliptical galaxies without requiring comparable in-situ star formation.

For an isolated quasistatic cluster with near-zero escape energies, the bound negative energy is approximately fixed. The virial theorem and gravitational radius then imply $R_g\propto M^2$. A homologous mean density scales as $M^{-5}$ and increases as mass evaporates. Tidal limitation or finite positive escape energies modifies this simple model.

Motion can weigh a system whose mass is otherwise inaccessible. For a self-gravitating isolated system in virial equilibrium, let $K$ and $U$ denote equilibrium averages and use its gravitational radius $R_g=-G_N{M}^2/U$ and a mass-weighted mean squared speed relative to its centre of mass. Then $K=M⟨v^2⟩/2$ gives $M=R_g⟨v^2⟩/G_N$. If random velocities are isotropic and mean streaming has been removed, $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$, where ${\sigma }_{\mathrm{los}}$ is the one-dimensional line-of-sight velocity dispersion. Observations sample many objects at one epoch; replacing a time average by that sample requires a representative, approximately stationary population, beyond the trajectory identity itself.^[ref]

A practical measurement route uses an optical telescope to locate members and a spectrograph to measure Doppler shifts of identified spectral lines. Their shifts relative to the system's mean give radial velocities; angular separations plus a distance estimate constrain its size. Subtracting instrumental broadening and measurement variance, testing membership, and looking for coherent rotation or subgroups help control the inference. The projected luminous size need not equal $R_g$, and luminous tracers need not have the same distribution as all gravitating mass. Unknown velocity-anisotropy parameters, interlopers, mergers, external tidal forces and appreciable boundary stresses can bias a naive estimate. A large velocity dispersion alone does not establish virial equilibrium.