The Virial Theorem In the transition from classical to statistical mechanics, are there familiar qua

The Virial Theorem In the transition from classical to statistical mechanics, are there familiar quantities that remain constant? The Virial theorem defines a law for how the total kinetic energy of a system behaves under the right conditions, and is equally valid for a one particle system or a mole of particles. Rudolf Clausius, the man responsible for the first mathematical treatment of entropy and for one of the classic statements of the second law of thermodynamics, defined a quantity G (now called the Virial of Clausius): G ≡ Σi(pi · ri) Where the sum is taken over all the particles in a system. You may want to satisfy yourself (it’s a short derivation) that taking the time derivative gives: dG/dt = 2T + Σi(Fi · ri) Where T is the total kinetic energy of the system (Σ  ½mv2) and dp/dt = F. Now for the theorem: the Virial Theorem states that if the time average of dG/dt is zero, then the following holds (we use angle brackets ⟨·⟩ to denote time averages): 2⟨T⟩ = - Σi(Fi · ri) Which may not be surprising. If, however, all the forces can be written as power laws so that the potential is V=arn (with r the inter-particle separation), then 2⟨T⟩ = n⟨V⟩ Which is pretty good to know! (Here, V is the total kinetic energy of the particles in the system, not the potential function V=arn.) For an inverse square law (like the gravitational or Coulomb forces), F∝1/r2 ⇒ V∝1/r, so 2⟨T⟩ = -⟨V⟩. Try it out on a simple harmonic oscillator (like a mass on a spring with no gravity) to see for yourself. The potential V ∝ kx², so it should be the case that the time average of the potential energy is equal to the time average of the kinetic energy (n=2 matches the coefficient in 2⟨T⟩). Indeed, if x = A sin( √[k/m] · t ), then v = A√[k/m] cos( √[k/m] · t ); then x2 ∝ sin² and v² ∝ cos², and the time averages (over an integral number of periods) of sine squared and cosine squared are both ½. Thus the Virial theorem reduces to 2 · ½m·(A²k/2m) = 2 · ½k(A²/2) Which is easily verified. This doesn’t tell us much about the simple harmonic oscillator; in fact, we had to find the equations of motion before we could even use the theorem! (Try plugging in the force term F=-kx in the first form of the Virial theorem, without assuming that the potential is polynomial, and verify that the result is the same). But the theorem scales to much larger systems where finding the equations of motion is impossible (unless you want to solve an Avogadro’s number of differential equations!), and just knowing the potential energy of particle interactions in such systems can tell us a lot about the total energy or temperature of the ensemble. -- source link

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