Contents 1 Entropy of Ideal

Home , Entropy (astrophysics), Gibbs paradox, Ideal gas

PHY304 - Statistical Mechanics

Spring Semester 2021 Dr. Anosh Joseph, IISER Mohali

LECTURE 11

Wednesday, January 27, 2021 (Note: This is an online lecture due to COVID-19 interruption.)

Contents

1 Entropy of Ideal Gas - Statistical Calculation 1

2 Gibbs’ Paradox 3

3 Counting Microstates using Quantum Mechanics 6

1 Entropy of Ideal Gas - Statistical Calculation

Let us begin with the Hamiltonian of the ideal gas. We have N 3N X ~p2 X p2 H(q , p ) = ν = ν . (1) ν ν 2m 2m ν=1 ν=1 We have numbered the coordinates and momenta from 1 to 3N. Let us evaluate ω(E,V,N) according to the equation we encountered in the previous lecture Z ω(E,V,N) = d3N qd3N p. H(qν ,pν )≤E

The integral over the coordinates of the particles can be performed trivially, since H does not depend on them. We have Z ω(E,V,N) = V N d3N p. (2) H(pν )≤E √ The above integral is just the volume of a 3N-dimensional sphere of radius 2mE, since we PHY304 - Statistical Mechanics Spring Semester 2021

have H(pν) ≤ E implying 3N √ 2 X 2 pν ≤ 2mE . (3) ν=1 It is given by N/2 π N VN (R) = N N R . (4) 2 Γ( 2 ) For ω(E,V,N) this gives Z ω(E,V,N) = V N d3N qd3N p H(pν ,qν ) 3N/2 π 3N/2 N = 3N 3N (2mE) V . (5) 2 Γ( 2 )

We have

1 ∂ω Ω(E,V,N) = σ0 ∂E 1 π3N/2 = V N (2m)3N/2E(3N/2)−1. (6) σ 3N 0 Γ( 2 )

Once we have the number of microstates, we can ﬁnd the entropy. We have " # 1 π3N/2 S(E,V,N) = k ln V N (2m)3N/2E(3N/2)−1 (7) B σ 3N 0 Γ( 2 ) Note that the argument of logarithm is indeed dimensionless. We can considerably simplify the above equation when N 1. In that limit we have E3N/2−1 ≈ E3N/2. (8)

The logarithm of a Gamma function can be approximated according to Stirling’s formula

ln Γ(n) ≈ (n − 1) ln(n − 1) − (n − 1) ≈ n ln n − n. (9)

Let us also deﬁne a new constant 1/N σ ≡ σ0 . (10) Then the formula for the entropy of the ideal gas becomes " ( )# 3 V 4πmE 3/2 S(E,V,N) = Nk + ln . (11) B 2 σ 3N

We can conﬁrm from above that we get the right equations of state for the ideal gas.

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1 ∂S 3 1 3 = = Nk or E = Nk T. (12) T ∂E 2 B E 2 B V,N

p ∂S Nk = = B or pV = Nk T. (13) T ∂V V B E,N

The above expressions are independent of σ0. To our surprise, we can see that the above expression for entropy cannot be the correct entropy of an ideal gas. According to Eq. (11), S is not a purely extensive quantity. In the case of enlargement of the system by a factor α, each extensive factor in the logarithm causes the total S to be magniﬁed by a factor α. We see that in the case at hand we will get an additional ln α factor. Our calculation above looks formally correct. There must be a principal error in the calculation of entropy. Let us clarify this problem by means of the well-known Gibbs paradox.

2 Gibbs’ Paradox

Let us write down the expression for S we derived above as a function of T , V , and N

3 V S(T,V,N) = Nk + ln (2πmk T )3/2 . (14) B 2 σ B

Consider a closed system consisting of two containers, which are separated by a wall and which contain two diﬀerent ideal gases A and B. They are under the same pressure and temperature. See Fig. 1.

gas A gas B

T, p T, p VA, NA VB, NB

Figure 1: Gibbs’ paradox.

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If we remove the wall, then both gases will spread out over the whole of the container until a new equilibrium situation is reached. We know that the internal energy of the ideal gas does not depend on the volume, but on temperature. Since the internal energy remains constant during the whole process, the temperature and pressure do not change either. The entropy however, increases. This is called the mixing entropy. We can easily calculate this using Eq. (14). Before we remove the separating wall, we have

(0) (0) (0) Stotal = SA (T,VA,NA) + SB (T,VB,NB). (15)

After removing the wall we have

(1) (1) (1) Stotal = SA (T,VA + VB,NA) + SB (T,VA + VB,NB). (16)

Now let us insert Eq. (14) for the entropy in the above equation. We get

(1) (0) ∆S = Stotal − Stotal VA + VB VA + VB = NAkB ln + NBkB ln . (17) VA VB

This looks promising. ∆S > 0, as it should be for this irreversible process. Now, let us perform the same analysis for two identical gasses in the two sub-systems. The initial entropy equation

(0) (0) (0) Stotal = SA (T,VA,NA) + SB (T,VB,NB) is still right. However, in the ﬁnal situation, we have

(1) Stotal = S(T,VA + VB,NA + NB), (18) since NA + NB particles of the same gas are now distributed over the total volume VA + VB. So in this case also, we get exactly same entropy diﬀerence ∆S > 0 (Eq. (17)). However, this cannot be correct, since after the removal of the separating wall no macroscopically observable processes are happening at all. We can simply re-insert the separating wall and recover the initial situation. In the case of identical gases, the removal of the separating wall is a reversible process and we must have ∆S = 0.

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Let us try to understand what might be going on. In classical mechanics, the particles are distinguishable.

We can label them as 1, 2, ··· ,NA and NA + 1,NA + 2, ··· ,NA + NB. Once we remove the wall, the gas particles (treat them as microscopic billiard balls) freely move around in the volume VA + VB. This is an irreversible process. We can mix the particles again and again, but we will never (almost never, in the case of finite number of particles) recover the initial situation. However, when we apply quantum mechanics, this argument does not work. We cannot attach a label to identical particles. They are all indistinguishable. But classical particles are distinguishable and we can identify them. This paradigm leads to the Gibbs’ paradox. We must note that two microstates should be considered different, if they differ in more than the enumeration of the particles. (Having other quantum numbers, for example.) For N particles, we can enumerate them in N! ways. Thus instead of σ(E,V,N) Ω(E,V,N) = , (19) σ0 we now try a new definition of Ω

1 σ(E,V,N) Ω(E,V,N) = . (20) N! σ0

∂ω The calculation of σ(E) by σ(E) = ∂E is still unchanged. The factor 1/N! in Eq. (20) is called the Gibbs’ correction factor. After incorporating the Gibbs’ correction factor we have the expression for the entropy of an ideal gas " ( )# 3 V 4πmE 3/2 S(E,V,N) = Nk + ln − k ln N!. (21) B 2 σ 3N B

For the case of very large N, we can apply the Stirling’s formula (ln N! ≈ N ln N − N), and we get " ( )# 5 V 4πmE 3/2 S(E,V,N) = Nk + ln . (22) B 2 Nσ 3N

Thus we get the anticipated result that entropy is indeed an extensive quantity. 5 A factor 2 also appears. Let us now go back to our mixing experiment. We have 5 V S(E,V,N) = Nk + ln (2πmk T )3/2 . (23) B 2 Nσ B

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In the above we used 3 E = Nk T. 2 B (0) (1) Upon using the expression we found for Stotal and Stotal for diﬀerent gases we obtain VA + VB VA + VB ∆S = NAkB ln + NBkB ln . (24) VA VB

This is nothing but the expression we found for ∆S in Eq. (17). For identical gases, we get 5 VA + VB 3/2 δS = (NA + NB)kB + ln (2πmkBT ) 2 (NA + NB)σ 5 VA 3/2 −NAkB + ln (2πmkBT ) 2 NAσ 5 VB 3/2 −NBkB + ln (2πmkBT ) . (25) 2 NBσ

As mentioned before we have thermal and mechanical equilibrium in the mixing process. Thus we have V V V + V A = B = A B . (26) NA NB NA + NB This implies, ∆S = 0. (27)

Thus for two identical gases we get the entropy change ∆S = 0, as it should be.

3 Counting Microstates using Quantum Mechanics

Let us determine Ω by counting the quantum mechanical states of distinguishable particles. This will give us an absolute number for Ω and thus an absolute entropy.

We will also get a value for the unit surface (unit volume) σ0 in phase space. Let us revisit the quantum mechanical problem of a particle in a box of length L. See Fig. 2. The one-particle states have the wave function

Ψnx,ny,nz = A sin(kxx) sin(kyy) sin(kzz) n πx n πy n πz = A sin x sin y sin z , (28) L L L where

nx, ny, nz = 1, 2, ··· . (29)

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E ⋮

0 L Figure 2: Quantum mechanical particle in a cubic box.

The single particle energy levels are

( k)2 2 = ~ ~ (k2 + k2 + k2) nx,ny,nz 2m 2m x y z h2 = (n2 + n2 + n2). (30) 8mL2 x y z

In classical mechanics, the state of motion (microstate of a particle) is ﬁxed by ~q and ~p.

In quantum mechanics, it is ﬁxed by the quantum numbers nx, ny, nz.

Each single particle state corresponds exactly to a point in the 3-dimensional (nx, ny, nz) space. See Fig. 3.

h2 h2 L√ = r2 =⇒ = r2 =⇒ r = 8m. (31) nx,ny,nz 8mL2 x,y,z 8mL2 h In this space, a given energy of a particle corresponds to a spherical shell with radius

L√ 8m. (32) h

We note that only integer points which lie on this sphere are possible single particle states for

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Figure 3: The space of quantum numbers ni. this energy . The N-particle problem is now solved by occupation of the single-particle states with N particles. The total energy is then determined by the 3N quantum numbers of the occupied states

3N h2 X E = n2. (33) 8mL2 i i=1

For the case of a 3N-dimensional space, we will have a (3N − 1)-dimensional energy sphere. The number of microstates Ω for a given macrostate (E, V , and N, with V = L3) is just the number of integer grid points on the energy surface. Let us consider the case N = 3.

We then have 9 quantum numbers n1, n2, ··· , n9. We have 9 h2 X E = n2. (34) 8mL2 i i=1

The lowest energy state is obtained when all ni = 1. That is 9 h2 X 9h2 E = 12 = . (35) lowest 8mL2 8mL2 i=1 Let us denote a given conﬁguration by (state)occupation. Then, the corresponding dimensionless energy is E∗ = E/(8mL2E/9h2) = E/∗ = 9. We have E∗ = 9 = 9 × 12 =⇒ Ω = 1, conﬁguration (1)9. (36)

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For the next higher state we have

E∗ = 12 = 8 × 12 + 1 × 22 =⇒ Ω = 9, conﬁguration (1)8(2)1. (37)

Each of the 9 quantum numbers can assume the value 2. Thus we have Ω = 9 possibilities. The next higher states of the 3-particle system are given in Table. 1.

State Ω Conﬁguration

∗ 2 2 9 7 2 E = 15 = 7 × 1 + 2 × 2 C2 = 36 (1) (2) ∗ 2 2 9 8 0 1 E = 17 = 8 × 1 + 1 × 3 C1 = 9 (1) (2) (3) ∗ 2 2 9 6 3 E = 18 = 6 × 1 + 3 × 2 C3 = 84 (1) (2) ∗ 2 2 2 9 8 7 1 1 E = 20 = 7 × 1 + 1 × 2 + 1 × 3 C1 C1 = 72 (1) (2) (3) ∗ 2 2 9 5 4 E = 21 = 5 × 1 + 4 × 2 C4 = 126 (1) (2)

Table 1: Higher states of the 3-particle system.

Let us plot E∗ against Ω. It is shown in Fig. 4.

Ω 100

0 10 15 20 E*

Figure 4: Plot of Ω against E∗. The irregularities are coming from quantum eﬀects.

We see that Ω is indeed a very irregular function of E∗. On average, it strongly increases with energy.

If Ω strongly ﬂuctuate with energy, the thermodynamic properties of the system (S = kB ln Ω) will also strongly ﬂuctuate, but this is in contradiction to experience.

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The irregularities tell us that the system cannot assume arbitrary energy states, but can absorb or emit energy only in discrete quanta. This property of the system becomes especially important when we consider very small systems or very low temperatures.

There, the typical energy of a particle ≈ kBT at temperature T is lower than or equal to the energy diﬀerences which can be estimated by ∆E∗ = 1.

References

[1] W. Greiner, L. Neise, H. Stocker, and D. Rischke, Thermodynamics and Statistical Mechanics, Springer (2001).

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