In mechanics, the virial theorem provides a general equation that relates the average over time of the total kinetic energy of a stable system of discrete particles, bound by a conservative force, with that of the total potential energy of the system. Mathematically, the theorem states that
⟨ T ⟩ = − 1 2 ∑ k = 1 N ⟨ F k ⋅ r k ⟩ , {\displaystyle \langle T\rangle =-{\frac {1}{2}}\,\sum _{k=1}^{N}\langle \mathbf {F} _{k}\cdot \mathbf {r} _{k}\rangle ,}
where T {\displaystyle T} is the total kinetic energy of the N {\displaystyle N} particles, F k {\displaystyle F_{k}} represents the force on the k {\displaystyle k} th particle, which is located at position rk, and angle brackets represent the average over time of the enclosed quantity. The word virial for the right-hand side of the equation derives from vis, the Latin word for "force" or "energy", and was given its technical definition by Rudolf Clausius in 1870. The significance of the virial theorem is that it allows the average total kinetic energy to be calculated even for very complicated systems that defy an exact solution, such as those considered in statistical mechanics; this average total kinetic energy is related to the temperature of the system by the equipartition theorem. However, the virial theorem does not depend on the notion of temperature and holds even for systems that are not in thermal equilibrium. The virial theorem has been generalized in various ways, most notably to a tensor form. If the force between any two particles of the system results from a potential energy V ( r ) = α r n {\displaystyle V(r)=\alpha r^{n}} that is proportional to some power n {\displaystyle n} of the interparticle distance r {\displaystyle r} , the virial theorem takes the simple form
2 ⟨ T ⟩ = n ⟨ V TOT ⟩ . {\displaystyle 2\langle T\rangle =n\langle V_{\text{TOT}}\rangle .}
Thus, twice the average total kinetic energy ⟨ T ⟩ {\displaystyle \langle T\rangle } equals n {\displaystyle n} times the average total potential energy
⟨ V TOT ⟩ {\displaystyle \langle V_{\text{TOT}}\rangle } . Whereas V ( r ) {\displaystyle V(r)} represents the potential energy between two particles separated by distance r {\displaystyle r} , V TOT {\displaystyle V_{\text{TOT}}} represents the total potential energy of the system, i.e., the sum of the potential energy V ( r ) {\displaystyle V(r)} over all pairs of particles in the system. A common example of such a system is a star held together by its own gravity, where n = − 1 {\displaystyle n=-1} .
History In 1870, Rudolf Clausius delivered the lecture "On a Mechanical Theorem Applicable to Heat" to the Association for Natural and Medical Sciences of the Lower Rhine, following a 20-year study of thermodynamics. The lecture stated that the mean vis viva of the system is equal to its virial, or that the average kinetic energy is one half of the average potential energy. The virial theorem can be obtained directly from Lagrange's equation as applied in classical gravitational dynamics, the original form of which was included in Lagrange's "Essay on the Problem of Three Bodies" published in 1772. Carl Jacobi's generalization of the identity to N {\displaystyle N} bodies and to the present form of Laplace's identity closely resembles the classical virial theorem. However, the interpretations leading to the development of the equations were very different, since at the time of development, statistical dynamics had not yet unified the separate studies of thermodynamics and classical dynamics. The theorem was later utilized, popularized, generalized and further developed by James Clerk Maxwell, Lord Rayleigh, Henri Poincaré, Subrahmanyan Chandrasekhar, Enrico Fermi, Paul Ledoux, Richard Bader and Eugene Parker. Fritz Zwicky was the first to use the virial theorem to deduce the existence of unseen matter, which is now called dark matter. Richard Bader showed that the charge distribution of a total system can be partitioned into its kinetic and potential energies that obey the virial theorem. As another example of its many applications, the virial theorem has been used to derive the Chandrasekhar limit for the stability of white dwarf stars.
Illustrative special case Consider N = 2 {\displaystyle N=2} particles with equal mass m {\displaystyle m} , acted upon by mutually attractive forces. Suppose the particles are at diametrically opposite points of a circular orbit with radius r {\displaystyle r} . The velocities are v 1 ( t ) {\displaystyle \mathbf {v} _{1}(t)} and v 2 ( t ) = − v 1 ( t ) {\displaystyle \mathbf {v} _{2}(t)=-\mathbf {v} _{1}(t)} , which are normal to forces F 1 ( t ) {\displaystyle \mathbf {F} _{1}(t)} and F 2 ( t ) = − F 1 ( t ) {\displaystyle \mathbf {F} _{2}(t)=-\mathbf {F} _{1}(t)} . The respective magnitudes are fixed at v {\displaystyle v} and F {\displaystyle F} . The average kinetic energy of the system in an interval of time from t 1 {\displaystyle t_{1}} to t 2 {\displaystyle t_{2}} is
⟨ T ⟩ = 1 t 2 − t 1 ∫ t 1 t 2 ∑ k = 1 N 1 2 m k | v k ( t ) | 2 d t = 1 t 2 − t 1 ∫ t 1 t 2 ( 1 2 m | v 1 ( t ) | 2 + 1 2 m | v 2 ( t ) | 2 ) d t = m v 2 . {\displaystyle \langle T\rangle ={\frac {1}{t_{2}-t_{1}}}\int _{t_{1}}^{t_{2}}\sum _{k=1}^{N}{\frac {1}{2}}m_{k}|\mathbf {v} _{k}(t)|^{2}\,dt={\frac {1}{t_{2}-t_{1}}}\int _{t_{1}}^{t_{2}}\left({\frac {1}{2}}m|\mathbf {v} _{1}(t)|^{2}+{\frac {1}{2}}m|\mathbf {v} _{2}(t)|^{2}\right)\,dt=mv^{2}.}
Taking center of mass as the origin, the particles have positions r 1 ( t ) {\displaystyle \mathbf {r} _{1}(t)} and r 2 ( t ) = − r 1 ( t ) {\displaystyle \mathbf {r} _{2}(t)=-\mathbf {r} _{1}(t)} with fixed magnitude r {\displaystyle r} . The attractive forces act in opposite directions as positions, so F 1 ( t ) ⋅ r 1 ( t ) = F 2 ( t ) r 2 ( t ) = − F r {\displaystyle \mathbf {F} _{1}(t)\cdot \mathbf {r} _{1}(t)=\mathbf {F} _{2}(t)\mathbf {r} _{2}(t)=-Fr} . Applying the centripetal force formula F = m v 2 / r {\displaystyle F=mv^{2}/r} results in
− 1 2 ∑ k = 1 N ⟨ F k ⋅ r k ⟩ = − 1 2 ( − F r − F r ) = F r = m v 2 r ⋅ r = m v 2 = ⟨ T ⟩ , {\displaystyle -{\frac {1}{2}}\sum _{k=1}^{N}\langle \mathbf {F} _{k}\cdot \mathbf {r} _{k}\rangle =-{\frac {1}{2}}(-Fr-Fr)=Fr={\frac {mv^{2}}{r}}\cdot r=mv^{2}=\langle T\rangle ,}
as required. Note: If the origin is displaced, then we'd obtain the same result. This is because the dot product of the displacement with equal and opposite forces F 1 ( t ) {\displaystyle \mathbf {F} _{1}(t)} , F 2 ( t ) {\displaystyle \mathbf {F} _{2}(t)} results in net cancellation.
Statement and derivation Although the virial theorem depends on averaging the total kinetic and potential energies, the presentation here postpones the averaging to the last step. For a collection of N {\displaystyle N} point particles, the scalar moment of inertia I {\displaystyle I} about the origin is
I = ∑ k = 1 N m k | r k | 2 = ∑ k = 1 N m k r k 2 , {\displaystyle I=\sum _{k=1}^{N}m_{k}|\mathbf {r} _{k}|^{2}=\sum _{k=1}^{N}m_{k}r_{k}^{2},}
where m k {\displaystyle m_{k}} and r k {\displaystyle \mathbf {r} _{k}} represent the mass and position of the k {\displaystyle k} th particle and r k = | r k | {\displaystyle r_{k}=|\mathbf {r} _{k}|} is the position vector magnitude. Consider the scalar
G = ∑ k = 1 N p k ⋅ r k , {\displaystyle G=\sum _{k=1}^{N}\mathbf {p} _{k}\cdot \mathbf {r} _{k},}
where p k {\displaystyle \mathbf {p} _{k}} is the momentum vector of the k {\displaystyle k} th particle. Assuming that the masses are constant, G {\displaystyle G} is one-half the time derivative of this moment of inertia:
1 2 d I d t = 1 2 d d t ∑ k = 1 N m k r k ⋅ r k = ∑ k = 1 N m k d r k d t ⋅ r k = ∑ k = 1 N p k ⋅ r k = G . {\displaystyle {\begin{aligned}{\frac {1}{2}}{\frac {dI}{dt}}&={\frac {1}{2}}{\frac {d}{dt}}\sum _{k=1}^{N}m_{k}\mathbf {r} _{k}\cdot \mathbf {r} _{k}\\&=\sum _{k=1}^{N}m_{k}\,{\frac {d\mathbf {r} _{k}}{dt}}\cdot \mathbf {r} _{k}\\&=\sum _{k=1}^{N}\mathbf {p} _{k}\cdot \mathbf {r} _{k}=G.\end{aligned}}}
In turn, the time derivative of G {\displaystyle G} is
d G d t = ∑ k = 1 N p k ⋅ d r k d t + ∑ k = 1 N d p k d t ⋅ r k = ∑ k = 1 N m k d r k d t ⋅ d r k d t + ∑ k = 1 N F k ⋅ r k = 2 T + ∑ k = 1 N F k ⋅ r k , {\displaystyle {\begin{aligned}{\frac {dG}{dt}}&=\sum _{k=1}^{N}\mathbf {p} _{k}\cdot {\frac {d\mathbf {r} _{k}}{dt}}+\sum _{k=1}^{N}{\frac {d\mathbf {p} _{k}}{dt}}\cdot \mathbf {r} _{k}\\&=\sum _{k=1}^{N}m_{k}{\frac {d\mathbf {r} _{k}}{dt}}\cdot {\frac {d\mathbf {r} _{k}}{dt}}+\sum _{k=1}^{N}\mathbf {F} _{k}\cdot \mathbf {r} _{k}\\&=2T+\sum _{k=1}^{N}\mathbf {F} _{k}\cdot \mathbf {r} _{k},\end{aligned}}}
where m k {\displaystyle m_{k}} is the mass of the k {\displaystyle k} th particle, F k = d p k d t {\displaystyle \mathbf {F} _{k}={\frac {d\mathbf {p} _{k}}{dt}}} is the net force on that particle, and T {\displaystyle T} is the total kinetic energy of the system according to the v k = d r k d t {\displaystyle \mathbf {v} _{k}={\frac {d\mathbf {r} _{k}}{dt}}} velocity of each particle,
T = 1 2 ∑ k = 1 N m k v k 2 = 1 2 ∑ k = 1 N m k d r k d t ⋅ d r k d t . {\displaystyle T={\frac {1}{2}}\sum _{k=1}^{N}m_{k}v_{k}^{2}={\frac {1}{2}}\sum _{k=1}^{N}m_{k}{\frac {d\mathbf {r} _{k}}{dt}}\cdot {\frac {d\mathbf {r} _{k}}{dt}}.}
Connection with the potential energy between particles The total force F k {\displaystyle \mathbf {F} _{k}} on particle k {\displaystyle k} is the sum of all the forces from the other particles j {\displaystyle j} in the system:
F k = ∑ j = 1 N F j k , {\displaystyle \mathbf {F} _{k}=\sum _{j=1}^{N}\mathbf {F} _{jk},}
where F j k {\displaystyle \mathbf {F} _{jk}} is the force applied by particle j {\displaystyle j} on particle k {\displaystyle k} . Hence, the virial can be written as
− 1 2 ∑ k = 1 N F k ⋅ r k = − 1 2 ∑ k = 1 N ∑ j = 1 N F j k ⋅ r k . {\displaystyle -{\frac {1}{2}}\,\sum _{k=1}^{N}\mathbf {F} _{k}\cdot \mathbf {r} _{k}=-{\frac {1}{2}}\,\sum _{k=1}^{N}\sum _{j=1}^{N}\mathbf {F} _{jk}\cdot \mathbf {r} _{k}.}
Since no particle acts on itself (i.e., F j j = 0 {\displaystyle \mathbf {F} _{jj}=0} for 1 ≤ j ≤ N {\displaystyle 1\leq j\leq N} ), we split the sum in terms below and above this diagonal and add them together in pairs:
∑ k = 1 N F k ⋅ r k = ∑ k = 1 N ∑ j = 1 N F j k ⋅ r k = ∑ k = 2 N ∑ j = 1 k − 1 F j k ⋅ r k + ∑ k = 1 N − 1 ∑ j = k + 1 N F j k ⋅ r k = ∑ k = 2 N ∑ j = 1 k − 1 F j k ⋅ r k + ∑ j = 2 N ∑ k = 1 j − 1 F j k ⋅ r k = ∑ k = 2 N ∑ j = 1 k − 1 ( F j k ⋅ r k + F k j ⋅ r j ) = ∑ k = 2 N ∑ j = 1 k − 1 ( F j k ⋅ r k − F j k ⋅ r j ) = ∑ k = 2 N ∑ j = 1 k − 1