1.Derivation of the entropy formula

for the microcanonical ensemble

1.

We first show by straightforward calculation that the phase volume $\Gamma (E,V,N)$ of the microcanonical ensemble of a system upon an infinitesimal variation of energy $E$ and volume $V$, at fixed particle number $N$, changes by

$$d\Gamma = (\partial \Gamma/\partial E)_{V,N} (dE + PdV) ,$$ (1)

where $P$ is the equilibrium pressure. The phase volume is defined by

$$\Gamma (E,V, N) = \int d^{3N} p d^{3N} x \theta (E- {\cal{H}}_{N,V})$$ (2)

where $\theta (.)$ is the step function. The Hamilton ${\cal{H}}_{N,V}$ of the $N$-particle system is the sum of the kinetic energy, the potential energy of interparticle interactions and the wall potential:

$${\cal{H}}_{N,V} = {\cal{H}}_{\rm {kin}} + {\cal{H}}_{\rm {pot}} + V_{\rm {wall}} .$$ (3)

For simplicity we assume that the volume has the form of a cube of side length $L=V^{1/3}$, and that the volume is changed by shifting the position $L_x$ of the wall facing in $x$-direction from $L$ by a small amount $d L_x$.

Figure 1:

The potential of the particles near this wall is given by

$$V_{{\rm {wall}}, x} = \sum_n v(x_n - L_x) ,$$ (4)

where $v(x)$ is the wall potential for a particle at a distance $x$ from the wall, and the sum goes over all particles of the system. Since

$$-v' (x_n -L_x)=F_{n,x}$$ (5)

is the force on particle $n$ from the wall facing in $x$-direction, and, by Newton's third law,

$$-\sum_n F_{n,x} = F_x$$ (6)

is the force on that wall from all particles near it, we obtain for the derivative of the wall potential

$$\frac{\partial V_{\rm {wall}}}{\partial L_x} = - F_x ,$$ (7)

the microcanonical average of which is related to the pressure $P$ by

$$\frac{\overline{\partial V_{\rm {wall}}}}{\partial L_x} = - \overline F_x = -P \cdot L^2 .$$ (8)

The partial derivative of $\Gamma$ with respect to $L_x$ yields

$$\left( \frac{\partial \Gamma}{\partial L_x}\right)_{E,N} = -\int ... ...\delta (E-{\cal{H}}_{N,V} ) \frac{\partial V_{\rm {wall}}} {\partial L_x} .$$ (9)

Since the distribution function $f$ (= probability density in phase space) for the microcanonical ensemble can be chosen as

$$f=\delta (E-{\cal{H}}_{N,V})/(\partial \Gamma/\partial E)_{V,N}$$ (10)

(by letting the infinitesimal energy spread $\Delta E$ go to zero) the expression on the r. h. s. of eqn. (9) can be identified as

$$\left( \frac{\partial \Gamma}{\partial L_x}\right)_{E,N} = - \fra... ...partial L_x} \cdot \left( \frac{\partial \Gamma}{\partial E}\right)_{V,N} .$$ (11)

For the total differential of $\Gamma$ we thus obtain, using (8),

$$d\Gamma = \left( \frac{\partial \Gamma}{\partial E}\right)_{V,N} dE + \left( \frac{\partial \Gamma}{\partial L_x}\right)_{E,N} dL_x$$ (12)

$$= \left( \frac{\partial \Gamma}{\partial E}\right)_{V,N} \left\{ dE + P \underbrace{L^2 dL_x}_{=dV} \right\} .$$

The last expression is the desired result eqn. (1).

2.

How do we interpret this result? How can we relate it to real physical processes?
We recall that the microcanonical ensemble is used to describe the equilibrium state of a closed system, with a certain fixed energy $E$ and volume $V$. If we change this energy and volume, it is not, of course, a closed system any more. However, we have in mind that the procedure by which the energy and volume of the system is changed is such that at any moment the system can still be described by a microcanonical ensemble, with varying values of $E$ and $V$. We call such a process "quasistatic". Correspondingly, our result (1) expresses the change of the phase volume of a microcanonical ensemble subject to a quasistatic process.
If in the procedure energy is transferred to the system only in the form of mechanical work done against the internal pressure $P$, i.e. if

$$dE=-PdV$$ (13)

holds, our result (1) tells us that the phase volume remains unchanged:

$$d \Gamma = 0 .$$ (14)

Since this implies that no energy is transferred in the form of heat, the result (14) is referred to as ''adiabatic invariance of the phase volume''. However, if (13) does not hold, i.e. if the amount of transferred energy differs from the mechanical work, we attribute the difference to a transfer of heat $\delta Q$

$$dE + PdV = \delta Q ,$$ (15)

according to the law of energy conservation. (For simplicity we omit the possibility of energy transfer by electrical or magnetical work.) Comparison with eqn. (1) leads to the following relation between heat transfer and change of phase volume:

$$\delta Q = d\Gamma / (\partial \Gamma/\partial E)_{V,N} .$$ (16)

This result may seem somewhat surprising since in the derivation of eqn. (1) any consideration of heat is absent. However, for that derivation it is irrelevant in which way the energy of the system is changed, if only the process is quasistatic as defined above. This condition must also hold for a transfer of heat, which requires that the temperature differences driving heat flow must be infinitesimal. We know from thermodynamics that in this case, by the second law of thermodynamics, we may write

$$\delta Q = TdS ,$$ (17)

where $dS$ is the corresponding change of the entropy of the system, and the thermodynamic temperature $T$ is defined by

$$T^{-1} = (\partial S/\partial E)_{V,N} .$$ (18)

Combining (16), (17) and (18) we finally obtain

$$\frac{dS}{(\partial S/\partial E)_{V,N}} = \frac{d\Gamma}{(\partial \Gamma/\partial E)_{V,N}} .$$ (19)

It follows¹ from this relation that the entropy $S$ is determined by the phase volume $\Gamma$ alone. In other words, $S$ varies with $E$ and $V$ only via its dependence on $\Gamma$:

$$S(E,V,N) = \varphi (\Gamma(E, V,N))$$ (20)

with some function $\varphi (\Gamma)$. However, this function cannot be deduced from (19), which is fulfilled for any differentiable function $\varphi (\Gamma)$:

$$\frac{dS}{(\partial S/\partial E)_{V,N}} = \frac{ \varphi' \cdot ... .../\partial E)_{V,N} } = \frac{d\Gamma}{(\partial \Gamma/\partial E)_{V,N} } .$$ (21)

        The derivation of relation (20) is the main step in our derivation of the entropy formula. The function $\varphi$ can be inferred from the requirement that the entropy is an extensive quantity, using our knowledge of the function $\Gamma (E,V,N)$. A quantity $X(E,V,N)$ is extensive if it can be written as

$$X(E,V,N) = N \cdot x(E/N, V/N) ,$$ (22)

where $x(.,.)$ is some function of two variables. For ideal gases $\Gamma (E,V,N)$ can be calculated without great difficulty. It is found that $\ln \Gamma (E,V,N)$ is an extensive quantity. Therefore the desired function $\varphi (\Gamma)$ must be given, apart from constants, by $\ln \Gamma$. This leads to the result

$$S(E,V,N) = a \ln \Gamma (E,V,N) + b .$$ (23)

Without a proof that $\ln \Gamma$ has this property generally, we have to assume that $\varphi (\Gamma)$ is a universal function. The constant $a$ must be positive since the entropy is non-negative. Its magnitude fixes the thermodynamic temperature scale. The additive constant $b$ must be chosen to match the quantum-mechanical case, which defines the zero of entropy. With the proper choice of the constants $a$ and $b$, the entropy formula (23) becomes

$$S(E,V,N) = k_B \ln \Omega (E,V,N)$$ (24)

with

$$\Omega (E,V,N) = \frac{\Gamma (E,V,N)}{h^{3N} N!} .$$ (25)

$\Omega (E,V,N)$ is the semiclassical integrated density of states of the system. The entropy formula (24) thus has been derived by interpreting the statistical mechanics result (1) in terms of thermodynamics processes, taking thermodynamic laws into account.