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Behavior of Density of States

Consider an isolated system in equilibrium whose volume is $ V$ , and whose energy lies in the range $ E$ to $ E+\delta E$ . Let $ {\mit\Omega}(E, V)$ be the total number of microscopic states that satisfy these constraints. It would be useful if we could estimate how this number typically varies with the macroscopic parameters of the system. The easiest way to do this is to consider a specific example. For instance, an ideal gas made up of spinless monatomic particles. This is a particularly simple example, because, for such a gas, the particles possess translational, but no internal (e.g., vibrational, rotational, or spin), degrees of freedom. By definition, interatomic forces are negligible in an ideal gas. In other words, the individual particles move in an approximately uniform potential. It follows that the energy of the gas is just the total translational kinetic energy of its constituent particles. In other words,

$\displaystyle E = \frac{1}{2 m}\sum_{i=1,N}{\bf p}_i^{ 2},$ (3.20)

where $ m$ is the particle mass, $ N$ the total number of particles, and $ {\bf p}_i$ the vector momentum of the $ i$ th particle.

Consider the system in the limit in which the energy, $ E$ , of the gas is much greater than the ground-state energy, so that all of the quantum numbers are large. The classical version of statistical mechanics, in which we divide up phase-space into cells of equal volume, is valid in this limit. The number of states, $ {\mit\Omega}(E, V)$ , lying between the energies $ E$ and $ E+\delta E$ is simply equal to the number of cells in phase-space contained between these energies. In other words, $ {\mit\Omega}(E, V)$ is proportional to the volume of phase-space between these two energies:

$\displaystyle {\mit\Omega}(E, V) \propto \int^{E+\delta E}_Ed^{ 3}{\bf r}_1\cdots d^{ 3}{\bf r}_N  d^{ 3} {\bf p_1}\cdots d^{ 3} {\bf p_N}.$ (3.21)

Here, the integrand is the element of volume of phase-space, with

$\displaystyle d^{ 3}{\bf r}_i$ $\displaystyle \equiv dx_i dy_i dz_i,$ (3.22)
$\displaystyle d^{ 3}{\bf p}_i$ $\displaystyle \equiv dp_{x i} dp_{y i}  dp_{z i},$ (3.23)

where $ (x_i$ , $ y_i$ , $ z_i)$ and $ (p_{x i}$ , $ p_{y i}$ , $ p_{z i})$ are the Cartesian coordinates and momentum components of the $ i$ th particle, respectively. The integration is over all coordinates and momenta such that the total energy of the system lies between $ E$ and $ E+\delta E$ .

For an ideal gas, the total energy $ E$ does not depend on the positions of the particles. [See Equation (3.20).] This implies that the integration over the position vectors, $ {\bf r}_i$ , can be performed immediately. Because each integral over $ {\bf r}_i$ extends over the volume of the container (the particles are, of course, not allowed to stray outside the container), $ \int d^{ 3}{\bf r}_i= V$ . There are $ N$ such integrals, so Equation (3.21) reduces to

$\displaystyle {\mit\Omega}(E, V) \propto V^{ N} \chi(E),$ (3.24)

where

$\displaystyle \chi(E) \propto \int^{E+\delta E}_E d^{ 3} {\bf p_1}\cdots d^{ 3} {\bf p_N}$ (3.25)

is a momentum-space integral that is independent of the volume.

The energy of the system can be written

$\displaystyle E = \frac{1}{2 m} \sum_{i=1,N} \sum_{\alpha=1,3} p_{\alpha i}^{ 2},$ (3.26)

because $ {\bf p}_i^{ 2} = p_{1 i}^{ 2}+p_{2 i}^{ 2}+p_{3 i}^{ 2}$ , denoting the $ (x$ , $ y$ , $ z)$ components by (1, 2, 3), respectively. The previous sum contains $ 3N$ square terms. For $ E=$ constant, Equation (3.26) describes the locus of a sphere of radius $ R(E) = (2 m  E)^{1/2}$ in the $ 3N$ -dimensional space of the momentum components. Hence, $ \chi(E)$ is proportional to the volume of momentum phase-space contained in the region lying between the sphere of radius $ R(E)$ , and that of slightly larger radius $ R(E+\delta E)$ . This volume is proportional to the area of the inner sphere multiplied by $ \delta R \equiv
R(E+\delta E)- R(E)$ . Because the area varies like $ R^{ 3N-1}$ , and $ \delta R \propto \delta E/ E^{ 1/2}$ , we have

$\displaystyle \chi(E) \propto R^{ 3N-1}/E^{ 1/2} \propto E^{ 3N/2-1}.$ (3.27)

Combining this result with Equation (3.24), we obtain

$\displaystyle {\mit\Omega}(E, V) \simeq B V^{ N} E^{ 3N/2},$ (3.28)

where $ B$ is a constant independent of $ V$ or $ E$ , and we have also made use of the fact that $ N\gg 1$ . Note that, because the number of degrees of freedom of the system is $ f=3N$ , the previous relation can be very approximately written

$\displaystyle {\mit\Omega}(E, V) \propto V^{ f} E^{ f}.$ (3.29)

In other words, the density of states varies like the extensive macroscopic parameters of the system raised to the power of the number of degrees of freedom. An extensive parameter is one that scales with the size of the system (e.g., the volume). (See Section 7.8.) Because thermodynamic systems generally possess a very large number of degrees of freedom, this result implies that the density of states is an exceptionally rapidly increasing function of the energy and volume. This result, which turns out to be quite general, is of great significance in statistical thermodynamics.


next up previous
Next: Exercises Up: Statistical Mechanics Previous: Probability Calculations
Richard Fitzpatrick 2016-01-25