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Next: 4.2 Canonical ensemble Up: 4. Statistical Thermodynamics Previous: 4. Statistical Thermodynamics

4.1 Microcanonical ensemble

We recall the definition of this ensemble - it is that set of microstates which for given $N,V$ have an energy in the interval $[E, \Delta E]$. The number of such microstates is proportional to the phase space volume they inhabit. And we found some reason to suspect that this volume - its logarithm, rather - may be identified as that property which the thermodynamicists have dubbed entropy and denoted by $S(N,V,E)$. This can hold only if $S$ has the two essential properties of entropy:
If the system is divided into two subsystems that may freely exchange energy, then the equilibrium state is the one in which the available energy $E = E_{1}+E_{2}$ is distributed such that
\left. \frac{d S(N_{1},V_{1},E_{1})}{dE_{1}}\right\vert _{E_...
...S(N_{2},V_{2},E_{2})}{dE_{2}}\right\vert _{E_{2}= E-E_{1}^{*}}
\end{displaymath} (4.1)

The quantity $T \equiv [d S(N,V,E)/dE]^{-1}$ is commonly called temperature.
In equilibrium the entropy $S$ is additive:
S(N_{1},V_{1},E_{1}) + S(N_{2},V_{2},E_{2}) \stackrel{?}{=}
\end{displaymath} (4.2)

Figure 4.1: Isolated system: $\rightarrow $ microcanonical ensemble

Now consider two systems ( $N_{1},V_{1},E_{1}$) and ( $N_{2},V_{2},E_{2}$) in thermal contact, meaning that they can exchange energy but keeping the total energy $E \equiv E_{1}+E_{2}$ constant; their ``private'' volumes and particles remain separated. We may determine the phase space volume of the combined system.

Ad 1. The optimal partial energy $E_{1}^{*}$ fulfills $\Gamma_{2}(E_{2})  \partial \Gamma_{1}(E_{1})/ \partial E_{1}$ $ =\Gamma_{1}(E_{1})  \partial \Gamma_{2}(E_{2})/ \partial E_{2}$, or
\frac{\partial}{\partial E_{1}}
\left. \ln \Gamma_{1}(E_{1}...
...\left. \ln \Gamma_{2}(E_{2}) \right\vert _{E_{2}=E-E_{1}^{*}}
\end{displaymath} (4.3)

This, however, is nothing else but the well-known thermodynamic equation
\left. \frac{\partial S_{1}(E_{1})}{\partial E_{1}}
...{2}(E_{2})}{\partial E_{2}}
\right\vert _{E_{2}=E-E_{1}^{*}}
\end{displaymath} (4.4)

or, with $\partial S / \partial E \equiv 1/T$,
T_{1}=T_{2}  .
\end{displaymath} (4.5)

Ad 2. Let the total energy be divided up according to $E \equiv E_{1}+E_{2}$. Then we have, by equ. 3.30,
\Gamma_{1+2}(E) = \Gamma_{1}(E_{1}^{*})   \Gamma_{2}(E-E_{1}^{*})
\end{displaymath} (4.6)

and thus
$\displaystyle S_{1+2}$ $\textstyle =$ $\displaystyle k   \ln \frac{\Gamma_{1+2}(E)}{g^{3N}}
= k \ln \left[ \frac{\Gam...
\frac{\Gamma_{2}(E-E_{1}^{*})}{g^{3N_{2}}} \right]$  
  $\textstyle =$ $\displaystyle k \ln \frac{\Gamma_{1}(E_{1}^{*})}{g^{3N_{1}}}   +  
k \ln \frac{\Gamma_{2}(E-E_{1}^{*})}{g^{3N_{2}}}$  
  $\textstyle =$ $\displaystyle S_{1}(E_{1}^{*}) + S_{2}(E-E_{1}^{*})$ (4.7)

where $E_{1}^{*}$ is that partial energy of system $1$ which maximizes the product $\Gamma_{1}(E_{1})\Gamma_{2}(E-E_{1})$.

In other words, at thermal contact between two systems isolated from the outside world there will be a regular flow of energy until the quantity $ \partial S / \partial E$ ($\equiv 1/T$) is equal in both systems. Since the combined system has then the largest extension in phase space, this will be the most probable distribution of energies upon the two systems. There may be fluctuations around the optimal energy distribution, but due to the extreme sharpness of the maximum of $\Gamma_{1}(E_{1})\Gamma_{2}(E-E_{1})$ these deviations remain very small.

It should be noted that these conclusions, although of eminent physical significance, may be derived quite simply from the geometrical properties of high-dimensional spheres.

Example: Consider two systems with $N_{1}=N_{2}=36$ and initial energies $E_{1}=36$, $E_{2}=72$. Now bring the systems in thermal contact. The maximum value of the product $\Gamma_{1}(E_{1})\Gamma_{2}(E-E_{1})$ occurs at $E_{1}^{*}= E_{2}^{*}=54$, and the respective phase space volume is

= \left[ \frac{1}{36} \left(\pi e 10^{4}\right)^{3/2}
\end{displaymath} (4.8)

How does another partitioning of the total energy - say, $(46,62)$ instead of $(54, 54)$ - compare to the optimal one, in terms of phase space volume and thus probability?

\frac{\Gamma_{1}(46) \Gamma_{2}(62)}{ \Gamma_{1}(54) \Gamma_...
= \left( \frac{46 \cdot 62}{54 \cdot 54}\right)^{54}
= 0.30
\end{displaymath} (4.9)

We can see that the energy fluctuations in these small systems are relatively large: $\delta E_{1}/E_{1}^{*} \approx 8/54 = 0.15$. However, for larger particle numbers $\delta E_{1}/E_{1}^{*}$ decreases as $1/\sqrt{N}$: $(515 \cdot 565 / 540 \cdot 540 )^{540} = 0.31 $; thus $\delta E_{1}/E_{1}^{*} \approx 25/540 = 0.05$.

Let the system under consideration be in mechanical or thermal contact with other systems. The macroscopic conditions ($V,E$) may then undergo changes, but we assume that this happens in a quasistatic way, meaning that the changes are slow enough to permit the system always to effectively perambulate the microensemble pertaining to the momentary macroconditions. To take the $N$-particle gas as an example, we require that its energy and volume change so slowly that the system may visit all regions of the the phase space shell $(E(t),V(t))$ before it moves to a new shell. Under these conditions the imported or exported differential energy $dE$ is related to the differential volume change according to
dS = \left( \frac{\partial S}{\partial E}\right)_{V}   dE
+ \left( \frac{\partial S}{\partial V}\right)_{E}   dV
\end{displaymath} (4.10)

Defining (in addition to $T \equiv (\partial S / \partial E)_{V}^{-1}$) the pressure by
P \equiv T   \left( \frac{\partial S}{\partial V}\right)_{E}
\end{displaymath} (4.11)

then equ. 4.10 is identical to the thermodynamic relation
dS = \frac{1}{T} (dE + P   dV) \;\; \;
{\rm or} \;\;\;
dE = T   dS - P   dV \;\; {\rm\bf (First \;\; Law)}
\end{displaymath} (4.12)

Example: Classical ideal gas with

S(N,V,E) = Nk \ln \left[ \frac{V}{N}
\left( \frac{4 \pi E e}{3 N m g^{2}} \right)^{3/2} \right]
\end{displaymath} (4.13)

Solving this equation for $E$ we find for the internal energy $U(S,V) \equiv E$ the explicit formula

U(S,V) = \left(\frac{N}{V}\right)^{2/3} 
\frac{3Nmg^{2}}{4 \pi}   e^{2S/3Nk - 1}
\end{displaymath} (4.14)

From thermodynamics we know that $ T =(\partial U / \partial S)_{V}$; therefore

T = \frac{2}{3} \frac{U}{Nk}
\end{displaymath} (4.15)

From this we conclude, in agreement with experiment, that the specific heat of the ideal gas is

C_{V}=\frac{3}{2} Nk  .
\end{displaymath} (4.16)

The pressure may be found from $P = - (\partial U / \partial V)_{S}$:

P = \frac{2}{3} \frac{U}{V} = \frac{NkT}{V}
\end{displaymath} (4.17)

We found it a simple matter to derive the entropy $S$ of an ideal gas as a function of $N$, $V$ and $E$; and once we had $S(N,V,E)$ the subsequent derivation of thermodynamics was easy. To keep things so simple we had to do some cheating, in that we assumed no interaction between the particles. Statistical mechanics is powerful enough to yield solutions even if there are such interactions. A discussion of the pertinent methods - virial expansion, integral equation theories etc. - is beyond the scope of this tutorial. However, there is a more pragmatic method of investigating the thermodynamic properties of an arbitrary model system: computer simulation. The classical equations of motion of mass points interacting via a physically plausible pair potential $u(r)$ such as the one introduced by Lennard-Jones read
\frac{d \vec{v}_{i}}{dt} = - \frac{1}{m} \sum_{j \neq i}
\frac{\partial u(r_{ij})}{\partial
\end{displaymath} (4.18)

If at some time $t$ the microstate $\left\{ \vec{r}_{i} , \vec{v}_{i}  ;\; i=1, \dots N \right\}$ is given we can solve these equations of motion for a short time step $\Delta t$ by numerical approximation; the new positions and velocities at time $t+\Delta t$ are used as starting values for the next integration step and so forth. This procedure is known as molecular dynamics simulation.

Since we assume no external forces but only forces between the particles, the total energy of the system remains constant: the trajectory in phase space is confined to the energy surface $E=const$. If the systems is chaotic it will visit all states on this hypersurface with the same frequency. An average over the trajectory is therefore equivalent to an average over the microcanonical ensemble. For example, the internal energy may be calculated according to $U_{i} \equiv \langle E_{kin}\rangle + \langle E_{pot} \rangle$, where $E_{kin}$ may be determined at any time from the particle velocities, and $E_{pot} \equiv (1/2)\sum_{i,j} u(r_{ij})$ from the positions. By the same token the temperature may be computed via $3NkT/2 \equiv \langle E_{kin}\rangle$, while the pressure is the average of the so-called ``virial''; that is the quantity $W \equiv (1/2)\sum_{i}\sum_{j \neq i} \vec{K}_{ij} \cdot \vec{r}_{ij}$. In particular we have $P=NkT/V + \langle W \rangle / 3V$.

In the case of hard spheres the particle trajectories are computed in a different manner. For given $\vec{r}_{i}, \vec{v}_{i}$ the time span $t_{0}$ to the next collision between any two particles in the system is determined. Calling these prospective collision partners $i_{0}$ and $j_{0}$ we first move all spheres along their specific flight directions by $\Delta \vec{r}_{i}=\vec{v}_{i} t_{0}$ and then simulate the collision ($i_{0},j_{0}$), computing the new directions and speeds of the two partners according to the laws of elastic collisions. Now we have gone full circle and can do the next $t_{0}$ and $i_{0},j_{0}$.

Further details of the MD method may be found in [VESELY 94] or [ALLEN 90]

next up previous
Next: 4.2 Canonical ensemble Up: 4. Statistical Thermodynamics Previous: 4. Statistical Thermodynamics
Franz Vesely