Virial Theorem for Bound Systems

Derivation

Virial Theorem for Bound Systems

D-333 Home PU-308 Threads energy · force Depends on newtonian-gravitation, Equation of Stellar Hydrostatic Equilibrium

Statement

For an isolated, gravitationally bound system of point masses whose configuration remains bounded for all time, the long-time averages of the total kinetic energy \(T\) and total gravitational potential energy \(U\) satisfy the scalar virial relation \(2\langle T\rangle + \langle U\rangle = 0\). Equivalently the mean total energy obeys \(\langle E\rangle = -\langle T\rangle = \tfrac{1}{2}\langle U\rangle\), so a bound self-gravitating system is virialised with negative total energy.

Why it matters

The virial theorem converts an observable — the internal velocity spread of a star cluster, galaxy, or galaxy cluster — into a dynamical mass, without following any orbit. It is the single most-used mass estimator in astrophysics, from Zwicky's 1933 discovery of dark matter in the Coma cluster to modern velocity-dispersion masses of galaxies.

It also fixes the internal energetics of stars: because \(\langle E\rangle = -\langle T\rangle\), a gas sphere that radiates energy away must heat up. This negative specific heat drives stellar contraction, sets the Kelvin–Helmholtz timescale, and explains why gravitational collapse ignites nuclear burning.

Assumptions

The system is isolated:if drawn to it, external forces add an external virial \(\sum_i \vec r_i\cdot \vec F_i^{\,\text{ext}}\) — a surface-pressure term \(-3PV\) for a confined gas — and the relation becomes \(2\langle T\rangle+\langle U\rangle = 3PV\), not zero.

The potential is pure inverse-square gravity (\(U\propto 1/r\), homogeneous of degree \(-1\)):for a general power-law force \(F\propto r^{n}\) the coefficient changes to \(2\langle T\rangle = (n+1)\langle U\rangle\); a harmonic potential (\(n=1\)) gives \(\langle T\rangle=\langle U\rangle\) instead.

The moment of inertia \(I\) and its rate \(\dot I\) stay bounded (statistical steady state):if the system is expanding, collapsing, or evaporating, \(\langle\ddot I\rangle\neq 0\) and the vanishing of the boundary term \(\big[\dot I\big]_0^\tau/\tau\) fails, so the averaged relation carries a residual \(\tfrac12\langle\ddot I\rangle\) term.

Masses are non-relativistic point particles:if speeds approach \(c\) the kinetic virial \(2T\) is replaced by \(T\) (ultra-relativistic gas), shifting the coefficient and making the bound configuration only marginally stable.

Derivation

\[ I \equiv \sum_i m_i\, \vec r_i\cdot\vec r_i = \sum_i m_i r_i^2 \]

Define the scalar moment of inertia about a fixed origin; a smooth, positive, bounded function for a bound system. A

\[ \frac{1}{2}\frac{dI}{dt} = \sum_i m_i\, \vec r_i\cdot\dot{\vec r}_i \equiv G \]

Differentiate once in time; \(G=\sum_i \vec r_i\cdot\vec p_i\) is the virial generating function. A

\[ \frac{1}{2}\frac{d^2 I}{dt^2} = \sum_i m_i\,\dot{\vec r}_i\cdot\dot{\vec r}_i + \sum_i m_i\,\vec r_i\cdot\ddot{\vec r}_i \]

Differentiate again by the product rule; masses are constant. A

\[ \sum_i m_i\,\dot{\vec r}_i\cdot\dot{\vec r}_i = \sum_i m_i v_i^2 = 2T \]

The first sum is twice the total kinetic energy \(T=\tfrac12\sum_i m_i v_i^2\). A

\[ m_i\,\ddot{\vec r}_i = \vec F_i \quad\Rightarrow\quad \sum_i m_i\,\vec r_i\cdot\ddot{\vec r}_i = \sum_i \vec r_i\cdot\vec F_i \]

Insert Newton's second law; the right side is the Clausius virial. A

\[ \frac{1}{2}\ddot I = 2T + \sum_i \vec r_i\cdot\vec F_i \]

Combine steps 3–5. This is the exact (un-averaged) virial identity. B

\[ \vec F_i = -\nabla_i U,\qquad U = -\sum_{i<j}\frac{G m_i m_j}{\lvert \vec r_i-\vec r_j\rvert} \]

The internal forces derive from the gravitational potential energy \(U\), which is homogeneous of degree \(-1\) in the coordinates. B

\[ \sum_i \vec r_i\cdot\nabla_i U = (-1)\,U \quad\text{(Euler's theorem, degree }-1) \]

For \(U\) homogeneous of degree \(k\), \(\sum_i\vec r_i\cdot\nabla_i U = kU\); here \(k=-1\). C

\[ \sum_i \vec r_i\cdot\vec F_i = -\sum_i \vec r_i\cdot\nabla_i U = -(-1)U = U \]

The gravitational virial equals the total potential energy itself. C

\[ \frac{1}{2}\frac{d^2 I}{dt^2} = 2T + U \]

Substitute step 9 into step 6: the exact energy form of the virial identity. B

\[ \langle \ddot I\rangle \equiv \frac{1}{\tau}\int_0^{\tau}\ddot I\,dt = \frac{\dot I(\tau)-\dot I(0)}{\tau}\;\xrightarrow[\tau\to\infty]{}\;0 \]

Time-average over a long interval. Since \(\dot I\) is bounded for a bound system, the boundary term vanishes as \(\tau\to\infty\). C

\[ 2\langle T\rangle + \langle U\rangle = 0 \]

Average step 10 and drop \(\langle\ddot I\rangle\): the scalar virial theorem for a bound gravitating system. B

Result

\[ 2\langle T\rangle + \langle U\rangle = 0,\qquad \langle E\rangle = \langle T\rangle+\langle U\rangle = -\langle T\rangle = \tfrac{1}{2}\langle U\rangle \]

Reading. On average a bound self-gravitating system carries exactly twice as much kinetic energy (in magnitude) as its total energy, and half as much as its (negative) potential energy. It is virialised: the mean kinetic energy is pinned by the depth of the gravitational well. Because \(\langle E\rangle=-\langle T\rangle<0\), losing energy makes \(\langle T\rangle\) rise — the negative-specific-heat behaviour of gravity.

Units check. Every term is an energy. \(T=\tfrac12 mv^2\) has units \(\mathrm{kg\,(m/s)^2}=\mathrm{J}\); \(U=GMm/r\) has \(\mathrm{(m^3\,kg^{-1}s^{-2})(kg^2)(m^{-1})}=\mathrm{kg\,m^2\,s^{-2}}=\mathrm{J}\). The identity \(\tfrac12\ddot I=2T+U\) balances \(\mathrm{kg\,m^2\,s^{-2}}\) on both sides since \(\ddot I\) carries \(\mathrm{kg\,m^2\,s^{-2}}=\mathrm{J}\).

Limiting cases

Circular two-body orbit: \(T=\tfrac12 U\!\mid\) so \(2T+U=0\) holds instantaneously — no averaging needed for a closed orbit.
Harmonic potential (\(U\propto r^2\), degree \(+2\)): Euler's theorem gives \(2\langle T\rangle = 2\langle U\rangle\), i.e. \(\langle T\rangle=\langle U\rangle\) — energy equipartition of an oscillator.
General power law \(F\propto r^n\): \(2\langle T\rangle = (n+1)\langle U\rangle\); gravity is the special case \(n=-2\).
Confined gas (external pressure \(P\) on volume \(V\)): \(2\langle T\rangle+\langle U\rangle = 3PV\), the pressure-supported form used in the intracluster medium.
Ultra-relativistic gas (\(T\to\) rest-frame energy): kinetic virial \(2T\to T\), giving \(\langle T\rangle+\langle U\rangle=0=\langle E\rangle\) — marginal binding.

Breaks when

The system is not in a steady state. During collapse, expansion, or violent relaxation \(\langle\ddot I\rangle\neq0\); the omitted \(\tfrac12\langle\ddot I\rangle\) term dominates and a "virial mass" is meaningless until the system relaxes (a few crossing times).
The force is not inverse-square over the relevant scale. A softened potential, a dark-matter halo of different profile, or a modified-gravity law changes the degree of homogeneity, so the coefficient \(2\) in \(2\langle T\rangle+\langle U\rangle\) no longer applies.
Mass or energy is not conserved. Tidal stripping, evaporation of high-velocity stars, ongoing accretion, or radiative loss on the dynamical timescale all inject unbalanced boundary terms and bias the inferred mass.
Relativistic velocities or strong fields. Near a black hole or for a relativistic gas the Newtonian kinetic-energy form fails and the tensor virial theorem in GR must be used.

Failure modes

Using the observed line-of-sight dispersion as \(\langle v^2\rangle\). For an isotropic system \(\langle v^2\rangle = 3\sigma_{\text{los}}^2\); forgetting the factor of 3 underestimates the mass threefold.
Sign error in \(U\). Writing \(U=+GM^2/R\) gives \(2T=-U<0\), an impossible negative kinetic energy. Gravitational \(U\) is negative.
Confusing \(\langle E\rangle=-\langle T\rangle\) with \(\langle E\rangle=+\langle T\rangle\). The minus sign is the whole physics; getting it wrong predicts a system that cools as it contracts.
Applying it to an unrelaxed or interacting system (a merging pair, a spiral arm) and quoting a "virial mass" as if the theorem held.
Taking the structure coefficient \(\alpha=3/5\) as universal. \(U=-\alpha GM^2/R\) depends on the density profile; a centrally concentrated cluster has larger \(\alpha\).

Discussion

The deep content of the theorem is the homogeneity of the \(1/r\) potential. Euler's theorem turns the geometric statement "\(U\) scales as \(1/\text{length}\)" into the exact algebraic factor that links kinetic and potential energy. Nothing about the theorem cares which orbits the stars follow or whether the system is chaotic — only that it stays bounded long enough to average. This is why it is so robust as a mass estimator: it needs a snapshot of positions and speeds, not a solved dynamical model.

The negative specific heat encoded in \(\langle E\rangle=-\langle T\rangle\) is one of the strangest features of self-gravity. Remove energy from a star and it contracts, \(\langle U\rangle\) becomes more negative, and \(\langle T\rangle=-\langle E\rangle\) rises — the gas gets hotter. A self-gravitating system driven by radiative losses therefore runs away toward higher temperature and density, the "gravothermal catastrophe" that terminates in core collapse for star clusters and in nuclear ignition for protostars. It also means self-gravitating systems have no true thermodynamic equilibrium in the usual sense.

The theorem connects directly to hydrostatic equilibrium: integrating the equation of hydrostatic support \(dP/dr=-\rho GM_r/r^2\) against volume reproduces \(2T+U=0\) for a monatomic ideal-gas star, so the virial theorem is the global, energy-integrated statement of local pressure balance. It thereby links the microscopic pressure to the global energy budget and yields order-of-magnitude central temperatures without solving the stellar structure equations.

At the next level of rigour the scalar theorem is a single trace of the tensor virial theorem, \(\tfrac12\ddot I_{jk} = 2T_{jk}+U_{jk}\), whose off-diagonal and anisotropic components govern the shapes of rotating and pressure-anisotropic systems (flattened elliptical galaxies, rotating stars). Taking the trace collapses this to the scalar form derived here; the full tensor version is needed whenever the velocity dispersion is anisotropic or the figure departs from spherical.

Common misconceptions. The virial theorem is not an instantaneous energy balance — \(2T+U=0\) holds only after time-averaging (or exactly, at every instant, only for special closed orbits). And "virialised" does not mean "in equilibrium at a temperature"; it means \(\dot I\) is bounded so \(\langle\ddot I\rangle=0\), a far weaker and purely kinematic condition.

Worked examples

Virial mass of the Coma cluster from its velocity dispersion.

Model the cluster as a uniform sphere of mass \(M\), radius \(R\), isotropic velocities. B

\[ T=\tfrac12 M\langle v^2\rangle = \tfrac32 M\sigma^2,\qquad U=-\frac{3}{5}\frac{GM^2}{R} \]

Isotropy gives \(\langle v^2\rangle=3\sigma^2\) with \(\sigma\) the line-of-sight dispersion; \(\alpha=3/5\) for a uniform sphere. B

\[ 2T+U=0 \;\Rightarrow\; 3M\sigma^2 = \frac{3}{5}\frac{GM^2}{R}\;\Rightarrow\; M=\frac{5\sigma^2 R}{G} \]

Symbols rearranged before inserting numbers. B

\[ \sigma=10^{6}\,\mathrm{m\,s^{-1}},\;\; R=1.5\,\mathrm{Mpc}=4.63\times10^{22}\,\mathrm{m},\;\; G=6.67\times10^{-11}\,\mathrm{m^3kg^{-1}s^{-2}} \]

\(\sigma\approx1000\ \mathrm{km\,s^{-1}}\), \(R\approx1.5\ \mathrm{Mpc}\) for Coma. A

\[ M=\frac{5\,(10^{6})^2\,(4.63\times10^{22})}{6.67\times10^{-11}} = 3.5\times10^{45}\,\mathrm{kg} \]

Arithmetic. A

\[ M \approx 3.5\times10^{45}\,\mathrm{kg} \approx 1.7\times10^{15}\,M_\odot \]

Reading. The dynamical mass vastly exceeds the luminous stellar mass — the original signature of dark matter. Units check. \(\mathrm{(m/s)^2\cdot m / (m^3kg^{-1}s^{-2})} = \mathrm{m^2s^{-2}\cdot m\cdot m^{-3}kg\,s^{2}} = \mathrm{kg}\).

Mean internal temperature of the Sun from the virial theorem.

Treat the Sun as a self-gravitating monatomic ideal gas of mean molecular weight \(\mu\). B

\[ T_{\text{kin}}=\frac{3}{2}Nk\bar T,\qquad U=-\frac{3}{5}\frac{GM^2}{R},\qquad N=\frac{M}{\mu m_H} \]

Total kinetic energy of \(N\) particles at mean temperature \(\bar T\); \(N\) from total mass and mean particle mass. B

\[ 2T_{\text{kin}}=-U \;\Rightarrow\; 3Nk\bar T = \frac{3}{5}\frac{GM^2}{R} \;\Rightarrow\; \bar T = \frac{GM^2}{5NkR} \]

Rearrange the averaged virial \(2\langle T\rangle+\langle U\rangle=0\) for \(\bar T\). B

\[ \bar T = \frac{3}{5}\,\frac{GM\mu m_H}{3kR} = \frac{GM\mu m_H}{5kR} \]

Symbolic result before numbers. B

\[ M=1.99\times10^{30}\,\mathrm{kg},\; R=6.96\times10^{8}\,\mathrm{m},\; \mu=0.6,\; m_H=1.67\times10^{-27}\,\mathrm{kg},\; k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}} \]

Standard solar values; fully ionised H/He gas has \(\mu\approx0.6\). A

\[ \bar T = \frac{(6.67\times10^{-11})(1.99\times10^{30})(0.6)(1.67\times10^{-27})}{5\,(1.38\times10^{-23})(6.96\times10^{8})} \approx 4.6\times10^{6}\,\mathrm{K} \]

Arithmetic. A

\[ \bar T \approx 5\times10^{6}\,\mathrm{K} \]

Reading. A pure order-of-magnitude estimate of the Sun's mean interior temperature — a few million kelvin — obtained with no stellar-structure integration, correctly predicting temperatures high enough for nuclear fusion in the hotter core. Units check. \(\mathrm{\dfrac{(m^3kg^{-1}s^{-2})(kg)(kg)}{(J\,K^{-1})(m)} = \dfrac{kg\,m^2 s^{-2}}{J}\,K = K}\).

Problems

A star cluster has total kinetic energy \(T=4\times10^{41}\,\mathrm{J}\). Assuming it is virialised, find its total potential and total energy.
Solution
From \(2\langle T\rangle+\langle U\rangle=0\): \(\langle U\rangle=-2T=-8\times10^{41}\,\mathrm{J}\). Total energy \(\langle E\rangle=T+U=4\times10^{41}-8\times10^{41}=-4\times10^{41}\,\mathrm{J}=-\langle T\rangle\), confirming \(\langle E\rangle=-\langle T\rangle\). The system is bound (\(E<0\)).
A globular cluster has line-of-sight velocity dispersion \(\sigma=12\ \mathrm{km\,s^{-1}}\) and half-mass radius \(R=5\ \mathrm{pc}\). Estimate its mass using \(M=5\sigma^2R/G\).
Solution
\(\sigma=1.2\times10^4\,\mathrm{m/s}\), \(\sigma^2=1.44\times10^8\); \(R=5\times3.086\times10^{16}=1.54\times10^{17}\,\mathrm{m}\). \(M=\dfrac{5(1.44\times10^8)(1.54\times10^{17})}{6.67\times10^{-11}} = \dfrac{1.11\times10^{26}}{6.67\times10^{-11}} = 1.7\times10^{36}\,\mathrm{kg}\approx 8\times10^{5}\,M_\odot\).
Show that for a particle in a circular orbit of radius \(r\) about mass \(M\), the virial relation \(2T+U=0\) holds exactly and instantaneously.
Solution
Circular orbit: gravity supplies centripetal force, \(\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\Rightarrow v^2=\dfrac{GM}{r}\). Then \(T=\tfrac12 mv^2=\tfrac12\dfrac{GMm}{r}\) and \(U=-\dfrac{GMm}{r}\). So \(2T=\dfrac{GMm}{r}=-U\), i.e. \(2T+U=0\). Because \(r\) is constant, \(I\) is constant, \(\ddot I=0\), so the identity holds at every instant with no averaging.
For a force law \(F\propto r^n\) the potential energy is homogeneous of degree \(n+1\). Derive the generalised virial coefficient and check gravity (\(n=-2\)) and the harmonic oscillator (\(n=1\)).
Solution
Euler's theorem: \(\sum_i\vec r_i\cdot\nabla_i U=(n+1)U\), so the virial \(\sum\vec r_i\cdot\vec F_i=-(n+1)U\). The averaged identity \(2\langle T\rangle+\sum\vec r_i\cdot\vec F_i=0\)... more precisely \(0=2\langle T\rangle-(n+1)\langle U\rangle\), giving \(2\langle T\rangle=(n+1)\langle U\rangle\). Gravity \(n=-2\): \(2\langle T\rangle=-\langle U\rangle\Rightarrow 2\langle T\rangle+\langle U\rangle=0\). Harmonic \(n=1\): \(2\langle T\rangle=2\langle U\rangle\Rightarrow\langle T\rangle=\langle U\rangle\).
A gas cloud of mass \(M=10^{4}\,M_\odot\) and radius \(R=1\ \mathrm{pc}\) collapses and virialises. Estimate its equilibrium velocity dispersion \(\sigma\) (uniform sphere, \(\langle v^2\rangle=3\sigma^2\)).
Solution
Virial: \(3M\sigma^2=\tfrac35 GM^2/R\Rightarrow \sigma^2=\dfrac{GM}{5R}\). \(M=10^4\times1.99\times10^{30}=1.99\times10^{34}\,\mathrm{kg}\); \(R=3.086\times10^{16}\,\mathrm{m}\). \(\sigma^2=\dfrac{(6.67\times10^{-11})(1.99\times10^{34})}{5(3.086\times10^{16})}=\dfrac{1.33\times10^{24}}{1.54\times10^{17}}=8.6\times10^{6}\,\mathrm{m^2s^{-2}}\). \(\sigma\approx2.9\times10^{3}\,\mathrm{m/s}\approx3\ \mathrm{km\,s^{-1}}\), a typical molecular-cloud linewidth.