PHY304 - Statistical Mechanics
Spring Semester 2021 Dr. Anosh Joseph, IISER Mohali
LECTURE 18
Wednesday, February 17, 2021 (Note: This is an online lecture due to COVID-19 interruption.)
Contents
1 Virial Theorem 1
1 Virial Theorem
Let us make a statement on the mean energy U = hUi of a system at a given temperature T .
We begin with the Hamiltonian H(qν, pν) of the system.
We will denote all phase space coordinates qν and pν by xi, where xi runs over all momenta and coordinates (i = 1, 2, ··· , 6N). Let us compute the mean value of the quantity
∂H xi . ∂xk
If xi and xk are two arbitrary coordinates or momenta, then we have Z ∂H 1 6N ∂H xi = 3N d x ρ(~x) xi . (1) ∂xk h ∂xk
In the above we can consider ρ to be canonical or microcanonical phase-space density. The results should be identical in both cases. Let us look at the microcanonical case. We will look at the canonical case later. We have 1 Ω when E ≤ H(~x) ≤ E + ∆E ρmc(~x) = (2) 0 otherwise PHY304 - Statistical Mechanics Spring Semester 2021
Inserting this into Eq. (1), we get
Z ∂H 1 6N ∂H xi = 3N d x xi ∂xk Ωh E≤H(~x)≤E+∆E ∂xk Z 1 6N ∂(H − E) = 3N d x xi . (3) Ωh E≤H(~x)≤E+∆E ∂xk
In the above we used the fact that ∂E = 0 ∂xk since E is fixed. Now to this expression we apply the same trick which we used before to calculate Ω in the case of microcanonical ensemble. We have the phase space volume between two energy surfaces E and E + ∆E
∂ω ∆ω = ∆E . (4) ∂E V,N
We integrate over all phase space points below the energy surface and then obtain the result for ∂ the thin shell of thickness ∆E by application of ∆E ∂E . That is, Z ∂H 1 ∂ 6N ∂(H − E) xi = 3N ∆E d x xi . (5) ∂xk Ωh ∂E 0≤H(~x)≤E ∂xk
Let us apply integration by parts for the term xi ∂(H − E)/∂xk. This gives
∂H 1 ∂ Z xk max x = ∆E d6N−1x [x (H − E)] i 3N i ∂xk Ωh ∂E 0≤H≤E xk min Z ∂x − d6N x (H − E) i . (6) 0≤H≤E ∂xk
Note that in the first term we are left with an integral over 6N − 1 variables (without xk).
The extrema xk max and xk min need not be specified. The first integral is nothing but a surface integral, over the boundary of the region defined by H ≤ E, and on this boundary H − E = 0.
In other words, if xk assumes its extreme value, the corresponding point lies on the energy surface, which includes the integration region.
Thus, the first term vanishes, no matter whether xk is a coordinate or a momentum. Now, upon using ∂xi = δik, (7) ∂xk
2 / 6 PHY304 - Statistical Mechanics Spring Semester 2021 we get Z ∂H δik ∂ 6N xi = − 3N ∆E d x (H − E). (8) ∂xk Ωh ∂E 0≤H≤E Note that the integration limits also depend on E. We can perform the differentiation in the above, if we take care of this fact. We have the following general formula for the integral, whose integrand and limits depend on a parameter α
∂ Z x=g(α) Z x=g(α) ∂F (α, x) dx F (α, x) = dx ∂α x=f(α) x=f(α) ∂α ∂g ∂f + F (α, g(α)) − F (α, f(α)) . (9) ∂α ∂α
Applying this formula to Eq. (8), with F → (H − E) we get
Z ∂H δik 6N xi = − 3N ∆E d x(−1) (10) ∂xk Ωh 0≤H≤E ∂E + (E − E) − 0(0 − E) . (11) ∂E
That is, Z ∂H ∆E 1 6N xi = δik 3N d x . (12) ∂xk Ω h 0≤H≤E
The integral in the curly brackets (with the factor h−3N ) just gives the total number of all states Σ in the interior of the energy surface. Thus we have ∂H ∆E xi = δik Σ. (13) ∂xk Ω Note that Ω was the number of states in a shell of thickness ∆E. Thus we have Ω ∂Σ ≈ g(E) = (14) ∆E ∂E is the density of states. Note that g(E) ∼ EN for most of the systems. We have seen before that for large particle numbers (N → ∞) we can approximate
ln Σ ≈ ln Ω (15) since EN ≈ EN−1. (16)
Thus kB ln Σ may be replaced by the entropy, to very good approximation.
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We then get ∂H δik xi = kB = δikkBT. (17) ∂xk ∂S ∂E N,V This is the desired result. Note that this result is independent of the precise form of the Hamiltonian H. First, it means that the expression ∂H xi ∂xk has a non-zero mean value only for i = k.
Take xi as the coordinate qν. Then according to Hamilton’s equations of motion
∂H ∂H = = −p˙i. (18) ∂xi ∂qi
We have ∂H xi = −hqip˙ii = −hqiFii = kBT. (19) ∂xi
In this case, ∂H/∂xi is just generalized force Fi on the particle.
If we take xi to be pi then we have
∂H ∂H = =q ˙i, ∂xi ∂pi giving the result ∂H xi = hpiq˙ii = kBT. (20) ∂xi
The quantity piq˙i is noting but twice the kinetic energy in a certain direction. For a particle i moving in three dimensions, Eq. (20) becomes
3 hT i = k T, (21) i 2 B where Ti denotes the kinetic energy of particle i. Eq. (21) is a measure of the mean kinetic energy. We can rewrite Eq. (19) for vectors. Then we get
−h~ri · F~ii = 3kBT. (22)
The mean value given in Eq. (22) is called Clausius’ virial. For N number of particles we get the virial theorem
* N + 1 X 3 hT i = − ~r · F~ = Nk T. (23) 2 i i 2 B i=1
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Just as Eq. (21) is a measure of the mean kinetic energy, Clausius’ virial Eq. (22) is a measure of the mean potential energy. Let us understand this.
Consider the case that the force Fi can be written as the gradient of a potential V ,
* N + * N + X X − ~ri · F~i = ~ri · ∇Vi . (24) i=1 i=1
Assume a power function for the potential
V ∝ rα. (25)
Then, we obtain ∂V h~r · ∇V i = r i = αhV i. (26) i i ∂r i According to Eq. (23), α 3 hT i = hV i = k T. (27) i 2 i 2 B Thus the virial is indeed proportional to the mean potential energy. For quadratic potentials (α = 2) in the mean, the kinetic and potential energy are of equal 1 magnitude, and have the value 2 kBT per spatial direction.
2 Equipartition Theorem
We can formulate this statement more generally for a Hamiltonian that contains only quadratic terms 3N X 2 2 H = Aνpν + Bνqν . (28) ν=1 We can easily show that for such a Hamiltonian it holds that
3N X ∂H ∂H 2H = p + q . (29) ν ∂p ν ∂q ν=1 ν ν
The mean value of the total energy is then
( 3N 3N ) 1 X ∂H X ∂H hHi = p + q . (30) 2 ν ∂p ν ∂q ν=1 ν ν=1 ν
Let us denote f as the number of quadratic terms in the Hamiltonian. (Here f is equal to 6N.) Then, with the help of Eq. (17) we get
1 hHi = fk T. (31) 2 B
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In the literature f is often referred to as the number of degrees of freedom of the system. However, strictly speaking, f counts the number of quadratic terms in Eq. (28). If we keep the notation f, then Eq. (31) can be read as In the mean each degree of freedom of the system at a temperature T has the thermal energy 1 2 kBT . This is called the equipartition theorem (or equal distribution theorem). It tells us that the thermal energy is uniformly distributed over all degrees of freedom of the system. Note that the equipartition theorem is a special case of the virial theorem for quadratic poten- tials. We have obtained the virial theorem Eq. (23) with the help of Eq. (17) by taking a mean value over the ensembles. We have taken the mean value over all possible microstates of the microcanonical energy surface.
References
[1] W. Greiner, L. Neise, H. Stocker, and D. Rischke, Thermodynamics and Statistical Mechanics, Springer (2001).
[2] R. K. Pathria and Paul D. Beale, Statistical Mechanics, Elsevier; Third edition (2011).
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