Definition: A macroscopic state is described by a set of partial information. A microscopic state can be described by a set of either classical information or quantum information that is a complete set of detailed information at the finest level including boundary and initial conditions.

Example: Consider $N$ coupled harmonic oscillators with Hamiltonian H given by

(1.1) \begin{equation} H=\sum\limits ^N_{i=1}{{{\tfrac{1}{2}}}}\!\left (p^2_i+{\omega }^2_iq^2_i\right )+\lambda \sum\limits _{ijk}{c_{ijk}q_i}q_jq_k \end{equation}

with parameters $N, \lambda , \{c_{ijk}\},\{{\omega }_i\}$ . In a finite-sized box, the energy eigenstates for an uncoupled system of harmonic oscillators are discretized and representative of quantum systems. Because H in this example has no explicit time dependence, H is a constant of the motion; and the macrostate can be characterized by its energy without knowledge of the initial conditions. There is only partial information available in this example.

Postulate (Fundamental Postulate – R.C. Tolman) All microstates consistent with the given partial information (macrostate) are equally probable.

Definition: $\varGamma \!\left (E_0\right )\equiv$ number of microstates with $E\lt E_0$

At this point we drop the classical picture for a little while for pedagogic reasons, chiefly because counting and summing microstates over a discretized phase space resolves certain mathematical measure complications encountered in classical systems.

Definition: In the discretized quantum picture the probability of one given microstate is

(1.2) \begin{equation} \textrm {probability} \equiv w_n=\left \{ \begin{array}{l@{\quad}l} \frac {1}{\varGamma \!\left (E_0\right )},& {E}_n\lt E_0, \\[4pt] 0,& {E}_n\gt E_0. \end{array} \right . \end{equation}

In the limit of a very large number $\varGamma (E_0)$ we employ the correspondence principle, and in the volume $\Omega$ of allowed phase space one obtains

(1.3) \begin{equation} \varGamma \!\left (\Omega \right )=\mathop {\textrm {lim}}_{h\to 0}\frac {\int\nolimits _{\Omega }{\mathrm{d}q\mathrm{d}p}}{h^n}. \end{equation}

For large $\varGamma$ there are many accessible microstates, and the probability of a given microstate with energy ${E}_n$ is a relatively smooth function of energy because the granularity of the energy is such fine scale.

Example: N uncoupled simple harmonic oscillators. Let ${N} = 3$ and use a canonical transformation to action-angle variables:

(1.4) \begin{equation} \left . \begin{array}{c} q_i\equiv \sqrt {\frac {2J_i}{{\omega }_i}}{\sin {\theta }_i} \\[5pt] p_i\equiv \sqrt {2J_i{\omega }_i}{\cos {\theta }_i} \end{array} \right \}\to H=\sum\limits ^3_{i=1}{{\omega }_i}J_i. \end{equation}

Further simplify by requiring by requiring ${\omega }_1={\omega }_2={\omega }_3={\omega }_0$ . We note that $\{{\theta }_1,{\theta }_2,{\theta }_3\}$ are ignorable in H; hence, the actions $J_i$ are constants of the motion; and the energy is given by $E={\omega }_0\!\left (J_1+J_2+J_3\right )$ . The volume occupied by the system of three oscillators in phase space is a triangular solid in J-space, with vertices $\{J_0,0,0\}$ , $\{0,J_0,0\}$ , and $\left \{0,0,J_0\right \}$ where $J_0=E_0/{\omega }_0$ and a rectangular solid in $\theta$ -space spanning $[0,2\pi ]$ in each of the three $\theta$ ${}_{ }$ coordinates. The product volume in the $\{J,\theta \}$ phase space divided by h ${}^{3}$ yields the number of states:

(1.5) \begin{equation} \varGamma \!\left (E_0\right )=\frac {\textrm{volume}}{h^3}=\frac {{(2\pi )}^3}{h^3}{{\frac{1}{2}}}{\bigg({{\frac {E_0}{{\omega }_0}}}\bigg)}^2\frac {1}{3}\frac {E_0}{{\omega }_0}={{\frac {1}{3!}}}{\bigg(\frac {E_0}{{\hslash \omega }_0}\bigg)}^3{.} \end{equation}

In (1.5) we are assuming that typically the number of states is large, i.e., $E_0\gg {\hslash \omega }_0N$ . For the general case of N oscillators,

(1.6) \begin{equation} \varGamma \!\left (E_0,3\right )\to \varGamma \!\left (E_0,N\right )=\frac {1}{N!}{\bigg(\frac {E_0}{{\hslash \omega }_0}\bigg)}^N{.} \end{equation}

If we allow each of N oscillators to have any one of M possible energy states, where

(1.7) \begin{equation} M\equiv \frac {E-{{\tfrac{1}{2}}}{\hslash \omega }_0}{{\hslash \omega }_0}, \end{equation}

then Kubo (Reference Kubo1965, p. 38) shows that the number of distinguishable states is given by

(1.8) \begin{equation} \varGamma \!\left (E,{\omega }_0\right )=\sum\limits _M{\frac {\!\left (N+M-1\right )!}{\!\left (N-1\right )!M!}.} \end{equation}

Example: To illustrate (1.8) consider $N = 3$ oscillators and $M = 4$ energy levels, for which there are 15 states {(4,0,0), (3,1,0), (3,0,1), (2,2,0), (2,0,2), (2,1,1), (1,3,0), (1,0,3), (1,2,1), (1,1,2), (0,4,0), (0,0,4), (0,3,1), (0,1,3), (0,2,2)} in agreement with $(N + M - 1)$ !/[ $(N - 1)$ ! $M$ !] = 6!/(2! 4!) = 15.

Definition: The specific oscillator energy is ${\mathcal E}\equiv {E}/{N}$ .

Returning to (1.6) and using the definition of the specific oscillator energy one obtains

(1.9) \begin{equation} \varGamma \!\left (E_0,N\right )=\frac {1}{N!}{\bigg(\frac {E_0}{{\hslash \omega }_0}\bigg)}^N=\frac {N^N}{N!}{\bigg(\frac {{{\mathcal E}}_0}{{\hslash \omega }_0}\bigg)}^N\approx \frac {e^N}{\sqrt {2\pi N}}{\bigg(\frac {{{\mathcal E}}_0}{{\hslash \omega }_0}\bigg)}^N{,} \end{equation}

where in (1.9) we have made use of Stirling’s approximation $N!\approx \ \sqrt {2\pi N}\,{\!\left (N/e\right )}^N$ for large N. We identify ${{{\mathcal E}}_0}/{{\hslash \omega }_0}$ as a basic quantum number. We note that if the basic quantum number is O(10) and N ∼ 10 ${}^{10{-}20}$ , $\varGamma$ is rather large.

Definition: Take the natural logarithm of the number of states and introduce the concept of entropy. If all states in $\varGamma$ are equally probable, then define the entropy as

(1.10) \begin{equation} S\!\left (E_0,N\right )\equiv {\text{ ln } \varGamma }\to N\ \textrm {ln}\bigg(\frac {{{\mathcal E}}_0}{{\hslash \omega }_0}\bigg)+N-\frac{1}{2}{\text{ ln}\!\left (2\pi N\right )}\approx N{\text{ ln} \bigg(\frac {{{\mathcal E}}_0}{{\hslash \omega }_0}\bigg)}+N \end{equation}

for large N. Thus, S ∼ O(N).

Definition: Introduce the specific entropy ${\mathcal S}\equiv S/N$ . Hence,

(1.11) \begin{equation} {\mathcal S}={\text{ ln} \bigg(\frac {{{\mathcal E}}_0}{{\hslash \omega }_0}\bigg)}+1, \end{equation}

which has no N dependence (‘normal dependence’) and is a number of order unity.

[Editor’s Note: Prof. Kaufman remarked at this point that Problem 2-33 in Kubo (Reference Kubo1965) addressing the correspondence principle was interesting and not at all obvious.]

Example: Consider an ensemble of N atoms or molecules with a harmonic oscillator Hamiltonian. We will derive the specific entropy for an ideal gas.

Definition: The specific energy is ${\mathcal E}=E/N.$ The particle density is then $n\equiv N/V$ .

Definitions: An ensemble of states for energies $E_{N}\lt E$ is a subcanonical ensemble, and we denote the number of states by ${\varGamma }_{\sigma }$ . The ensemble of states for energies $E-\delta E\lt E_n\lt E+\delta E$ is defined as a microcanonical ensemble, and its number of states is denoted ${\varGamma }_{\mu }$ .

Example: Calculate $S_{\mu ,\textrm{class}}$ for the harmonic oscillator model of the ideal gas and compare it with the quantum entropy expression. We anticipate that if the two expressions are different, it is only due to constants. The classical microcanonical entropy is given in (1.21), while the quantum entropy is given by

(1.23) \begin{equation} S_{qm}\equiv {\text{ ln } \varGamma } \to \varGamma =e^{S_{qm}} \to \frac {\textrm{d}\varGamma }{\textrm{d}E}=e^{S_{qm}}\frac {\textrm{d}S_{qm}}{\textrm{d}E}\equiv \beta e^{S_{qm}} \end{equation}

Using ${\textrm{d}\varGamma }/{\textrm{d}E}=\beta e^{S_{qm}}$ in the last term in (1.21),

(1.24) \begin{equation} S_{\mu ,\textrm{class}}={\text{ ln } \delta E}+{\text{ ln } \beta e^{S_{qm}}={{\text{ ln } \delta E}+\text{ ln } \beta +S_{qm}}}=O(1)+O(1)+O(N)\approx S_{qm_{_{_{}}}} \end{equation}

We note that we should introduce h ${}^{N}$ in the denominator in the expressions for $\varGamma$ to give the correct dimensionless units for the phase-space normalized volume. However, this results in no change in the final formulas due to taking the logarithm of a product expands into the sum of logarithms; and the $S_{qm}\approx$ O(N) term remains dominant.

Next consider the phase space of a system with many degrees of freedom whose trajectory in phase space is constrained by a Hamiltonian. Define a subdomain in this phase space as a shell with thickness $\delta \ell$ and volume defined by $\textrm{d}p\textrm{d}q=\textrm{d}A\delta \ell$ , and the thickness of the shell is parametrized by a variation in the total energy:

(1.25) \begin{equation} H\!\left (p,q\right )=E-\delta E,\quad \frac {\delta E}{\delta \ell }= |{\nabla H}|,\quad \delta \ell =\frac {\delta E} {|{\nabla H(p,q)}|}, \end{equation}

which varies as a function of p and q in phase space. If the probability of the system occupying a given subdomain in phase space is proportional to the volume of the subdomain, then

(1.26) \begin{equation} \textrm {Probability}\propto \mathrm{d}A\ \delta \ell =\mathrm{d}A\frac {\delta E}{|{\nabla H(p,q)}|}. \end{equation}

Definition: C appears in (1.39) and is the derivative of the specific energy with respect to the temperature and depends explicitly on the degrees of freedom,

(1.40) \begin{equation} C\equiv \frac {\mathrm{d}{\mathcal E}}{\textrm{d}T}=\left \{ \begin{array}{l} \frac {3}{2}\ \ \textrm {ideal gas,} \\ 1\ \ \textrm {harmonic oscillator.}\ \end{array} \right . \end{equation}

One can read off the standard deviation $\sigma$ around the peak of the probability distribution $\rho \!\left ({\delta E}_{\textrm{I}}\right )$ from the right-hand side of (1.39)

(1.41) \begin{equation} \sigma =T{\left (\frac {1}{{C_{\textrm{I}}N}_{\textrm{I}}}+\frac {1}{C_{\textrm{II}}N_{\textrm{II}}}\right )}^{-{1}/{2}}\sim T\sqrt {N}. \end{equation}

We note that for $N_{\textrm{I}}\sim N_{\textrm{II}},$ (1.41) determines that $\sigma \sim T\sqrt {N}$ and ${\sigma }/{E_{\textrm{I}}}\sim {1}/{\sqrt {N}}\ll 1$ . In the limit that $N_{\textrm{I}}{\ll N}_{\textrm{II}}$ , e.g., a heat bath, then $\sigma =\sqrt {{C_{\textrm{I}}N}_{\textrm{I}}}T$ and ${\sigma }/{E_{\textrm{I}}}\sim {1}/{\sqrt {N_{\textrm{I}}}}\ll 1$ .

Let’s compare S(E) with S(E $(E^*_{\textrm{I}},E^*_{\textrm{II}})$ :

(1.42) \begin{equation} \varGamma \!\left (E\right )=\int\nolimits ^E_0{\textrm{d}E_{\textrm{I}}\varGamma \!\left (E_{\textrm{I}},E_{\textrm{II}}\right )=}\int\nolimits ^E_0{\textrm{d}E_{\textrm{I}}\varGamma \!\left (E^*_{\textrm{I}},E^*_{\textrm{II}}\right )e^{-\ \frac {{\!\left ({\delta E}_{\textrm{I}}\right )}^2}{2{\sigma }^2}}\approx \varGamma \!\left (E^*_{\textrm{I}},E^*_{\textrm{II}}\right )\sqrt {2\pi {\sigma }^2}} \end{equation}

aside from units. Using the relation $S={\text{ln}\,\varGamma}$ , (1.42) leads to

(1.43) \begin{equation} S\!\left (E\right )=S\!\left (E^*_{\textrm{I}},E^*_{\textrm{II}}\right )+\tfrac{1}{2}{\text{ ln}\!\left (2\pi {\sigma }^2\right )}\approx S\!\left (E^*_{\textrm{I}},E^*_{\textrm{II}}\right ) \end{equation}

because $S\!\left (E\right ),S\!\left (E^*_{\textrm{I}},E^*_{\textrm{II}}\right )\sim O(N)$ $\gg$ $({1}/{2}){\text{ ln}(2\pi {\sigma }^2)}\sim O({\text{ln } N)}$ . The conclusion is that the entropy is somewhat invariant relative to the system constraints involved.

Example: Consider $N_{\textrm{II}}\gg N_{\textrm{I}}$ and $E_{\textrm{II}}\gg E_{\textrm{I}}$ , a heat bath if you will. Then

(1.44) \begin{equation} \varGamma \!\left (E_{\textrm{I}},E\right )={\varGamma }_{\textrm{I}}\!\left (E_{\textrm{I}}\right ){\varGamma }_{\textrm{II}}\!\left ({E-E}_{\textrm{I}}\right )={\varGamma }_{\textrm{I}}\!\left (E_{\textrm{I}}\right )e^{S_{\textrm{II}}\left (E\right )-E_{\textrm{I}}\frac {\textrm{d}S_{\textrm{II}}}{\textrm{d}E_{\textrm{II}}}+\frac{1}{2}E^2_{\textrm{I}}\frac {\textrm{d}^2S_{\textrm{II}}}{\textrm{d}E^2_{\textrm{II}}}+\dots . } \end{equation}

We note that ${\textrm{d}^2S_{\textrm{II}}}/{\textrm{d}E^2_{\textrm{II}}}={\textrm{d}\beta }/{\textrm{d}E_{\textrm{II}}}\sim -{N^2_{\textrm{I}}}/{N_{\textrm{II}}}$ in (1.44), and we further impose that $N^2_{\textrm{I}}\ll N_{\textrm{II}}$ so that this term is small. Hence, (1.44) becomes

(1.45) \begin{equation} \varGamma \!\left (E_{\textrm{I}},E\right )\cong {\varGamma }_{\textrm{II}}\!\left (E\right )e^{-{\beta }_{\textrm{II}}\!\left (E_{\textrm{I}}-T_{\textrm{II}}S_{\textrm{I}}(E_{\textrm{I}}\right ))}\equiv {\varGamma }_{\textrm{II}}\!\left (E\right )e^{-{\beta }_{\textrm{II}}F_{\textrm{I}}(E_{\textrm{I}},T_{\textrm{II}})}, \end{equation}

where we have introduced the definition of the free energy $F_{\textrm{I}}(E_{\textrm{I}},T_{\textrm{II}})\equiv \ E_{\textrm{I}}-T_{\textrm{II}}S_{\textrm{I}}(E_{\textrm{I}})$ , so called because this is the energy available to do work. We recall that for an ideal gas $T_{\textrm{I}}=({2}/{3})({E_{\textrm{I}}}/{N_{\textrm{I}}})$

The probability is proportional to $\varGamma \!\left (E_{\textrm{I}},E\right )$ , i.e.,

(1.46) \begin{equation} \rho (E_{\textrm{I}})\propto e^{-{\beta }_{\textrm{II}}F_{\textrm{I}}(E_{\textrm{I}},T_{\textrm{II}})}. \end{equation}

The peak of the probability distribution determines the most probable value of E ${}_{\textrm{I}}$ which corresponds to the minimum of the free energy F ${}_{\textrm{I}}$ . Minimizing the free energy of the microstate is equivalent to maximizing the total system entropy.

Example: The free energy of an ideal gas is given by

(1.47) \begin{equation} F_{\textrm{I}}\!\left (E_{\textrm{I}},T_{\textrm{II}}\right )\equiv \ E_{\textrm{I}}-T_{\textrm{II}}S_{\textrm{I}}\!\left (E_{\textrm{I}}\right )=N_{\textrm{I}}\frac {3}{2}T_{\textrm{I}}-T_{\textrm{II}}N_{\textrm{I}}\left [\frac {5}{2}-{\text{ln}\!\left (n_{\textrm{I}}{\Lambda}^3_{\textrm{I}}(T_{\textrm{I}})\right )}\right ]. \end{equation}

Exercise: Show that the most probable temperature is $T^*_{\textrm{I}}=T_{\textrm{II}}$ .

Example: Assume a slowly varying Hamiltonian for a harmonic oscillator system with N = 1 modeled by

(1.50) \begin{equation} H\!\left (p,q;t\right )=\tfrac{1}{2}p^2+\tfrac{1}{2}{\omega }^2_0(t)q^2 \end{equation}

and we assume ${\textrm{d}{\omega }_0}/{\textrm{d}t}\ll {\omega }^2_0$ . Energy is not conserved here because the Hamiltonian is time dependent due to ${\omega }_0\!\left (t\right ).$ The elliptical orbit of the system in the (p,q) phase space evolves, but the area of the ellipse is conserved, i.e., there is an adiabatic invariant. The time derivative of the Hamiltonian can be calculated from

(1.51) \begin{equation} \frac {\textrm{d}H}{\textrm{d}t}=\frac {\partial H}{\partial t}={\omega }_0{\dot {\omega }}_0q^2 \end{equation}

From (1.51) we calculate the time-integrated change $\Delta H$ from (1.51):

(1.52) \begin{equation} \Delta H=\int\nolimits {\textrm{d}t\dot {H}=\int\nolimits {\textrm{d}t\frac {{\dot {\omega }}_0}{{\omega }_0}}}{\omega }^2_0q^2. \end{equation}

For purposes of calculating $\Delta H$ over time durations long compared with the oscillation period, we can assume that ${{\dot {\omega }}_0}/{{\omega }_0}$ is approximately constant over the oscillation period; and we can average ${\omega }^2_0q^2$ over the oscillation period. Noting that $({1}/{2})\langle {\omega }^2_0q^2\rangle =({1}/{2})\langle H\rangle$ , we conclude that

(1.53) \begin{equation} \Delta H\approx \int\nolimits {\textrm{d}t\frac {{\dot {\omega }}_0}{{\omega }_0}}\langle H\rangle \quad \textrm {and}\quad \frac {\langle \dot {H}\rangle }{\langle H\rangle }\approx \frac {{\dot {\omega }}_0}{{\omega }_0}. \end{equation}

Definition: Introduce the action

(1.54) \begin{equation} J\equiv \frac {1}{2\pi }\int\nolimits {p\mathrm{d}q=\frac {\langle H\rangle }{{\omega }_0}}. \end{equation}

Exercise: Calculate the time derivative of J in (1.54) and use (1.53) to deduce

(1.55) \begin{equation} \frac {\dot {J}}{J}=\frac {\dot {\langle H\rangle }}{\langle H\rangle }-\frac {{\dot {\omega }}_0}{{\omega }_0}\approx 0. \end{equation}

Hence, the action is ‘conserved.’

Example: $N \gt 1$ and generalize the time dependence of the Hamiltonian: $H\!\left (p,q;\lambda \!\left (t\right )\right )$ where $\lambda \to \lambda +\Delta \lambda $ in a time interval $\Delta t$ that is large, i.e., $\lambda$ is assumed to change at a very slow rate compared with ${\omega }_0$ :

(1.56) \begin{align} H&=H(p,q;\lambda \!\left (t\right )),\ \ \ \ \frac {\mathrm{d}\textrm {ln}\ \lambda }{\textrm{d}t}\ll \frac {{\dot {\omega }}_0}{{\omega }_0}\to \nonumber \\[4pt] \Delta H&\equiv \int\nolimits ^{\Delta t}_0{\dot {H}\textrm{d}t=\int\nolimits ^{\Delta t}_0{\frac {\partial H}{\partial t}\textrm{d}t=}}\int\nolimits ^{\Delta t}_0{\frac {\partial H}{\partial \lambda }\frac {\textrm{d}\lambda }{\textrm{d}t}\textrm{d}t\approx \dot {\lambda }}\int\nolimits ^{\Delta t}_0{\frac {\partial H}{\partial \lambda }\textrm{d}t\equiv }\dot {\lambda }\Delta t{\left\langle \frac {\partial H}{\partial \lambda }\right\rangle }_t\nonumber \\[4pt] &=\dot {\lambda }\Delta t{\left\langle \frac {\partial H}{\partial \lambda }\right\rangle }_{{\varGamma }_{\mu }(E)}, \end{align}

where the time average has been replaced by an average over the energy surface in phase-space ${\varGamma }_{\!\mu }\!\left (E\right ).$

Hence, $\Delta H=\Delta \lambda {\langle {\partial H}/{\partial \lambda }\rangle }_E$ or ${\Delta E}/{\Delta \lambda }={\langle {\partial H}/{\partial \lambda }\rangle }_E$ where the energy E characterizes the microcanonical ensemble.

Example: For an ideal gas $\varGamma (E,V)\sim V^NE^{3N/2}$ and an adiabatic change in V, then $E\sim V^{-2/3}$ in order that $\Delta \varGamma = \textrm {0.}$ Hence, the pressure is $P\sim E/V\sim 1/V^{5/3}$ , i.e., $PV^{5/3}=$ const, the usual adiabatic law for an ideal gas.

Example: Particles in their own electric field and in an externally applied electric field with electric potential $\phi _0$ have the Hamiltonian:

(1.59) \begin{equation} H=\sum\limits _i{\frac {p^2_i}{2m_i}+\sum\limits _{i\lt j}{\frac {e_ie_j}{r_{ij}}}+\sum\limits _i{{e_i\phi }_0({{\boldsymbol{r}}}_i)}}, \end{equation}

where e ${}_{i}$ are the particle charges, m ${}_{i }$ are the particle charges, p ${}_{i}$ are the momenta, and r ${}_{ij}$ are the distances between the i and j particles. We note that

(1.60) \begin{equation} \sum\limits _i{{e_i\phi }_0({{\boldsymbol{r}}}_i)}=\int\nolimits {\textrm{d}^3\textit {x}\ \rho ({\boldsymbol{x}},{\{{\boldsymbol{r}}}_i\})\phi _0({\boldsymbol{x}})} \end{equation}

and define the charge density as

(1.61) \begin{equation} \rho \!\left ({\boldsymbol{x}},{{\boldsymbol{r}}}_i\right )=\sum\limits _i{e_i\delta (}{\boldsymbol{x}}-{{\boldsymbol{r}}}_i). \end{equation}

We can choose $\lambda$ to be whatever attribute of the Hamiltonian is of interest, e.g., $\boldsymbol\lambda=\left \{e_i\right \}$ or $\left \{{{\boldsymbol{r}}}_i\right \}$ or other.

Definition: The generalized force is

(1.62) \begin{equation} {\boldsymbol {\Lambda} }(p,q;{\boldsymbol {\lambda }}{\mathbf )}\equiv \frac {\partial H(p,q;\boldsymbol\lambda)}{\partial\lambda} \quad\textrm{and}\quad {\boldsymbol {\Lambda} }(E;{\boldsymbol {\lambda }}{\mathbf )}\equiv {\langle {\boldsymbol {\Lambda} }(p,q;{\boldsymbol {\lambda }}{\mathbf )}\rangle }_{E,{\boldsymbol {\lambda } }} \end{equation}

Example: The functional derivative $({\partial H})/({\partial \phi _0(x)})= \rho \!\left ({\boldsymbol{x}},\{{{\boldsymbol{r}}}_i\}\right )$ (Goldstein Reference Goldstein1950; Schiff Reference Schiff1968).

Example: The macroscopic charge density averaged over the phase space constrained by constant energy E and fixed entropy is

(1.64) \begin{equation} \langle \rho \rangle \!\left (x\right )={\left .\frac {\partial E(S,\phi _0({\boldsymbol{x}}))}{\partial \phi _0(\textit{x})}\right .}\bigg \vert_S{.} \end{equation}

We next introduce the concept of pressure. Let $\lambda =V$ where V is the volume. Then using (1.63) the pressure P is

(1.65) \begin{equation} P\!\left (p,q;V\right )\equiv -\Lambda =-\frac {\partial H(p,q;V)}{\partial V}. \end{equation}

Exercise: Take the model Hamiltonian for an ideal gas or a charged particle plasma and show the consistency of (1.65) with the elementary definition $P\equiv F/\textrm {area}$ where F is the macroscopic force. We note from (1.63) and (1.65) that $P=-({\partial E(S,V)})/({\partial V})\big \vert _S=-({\partial E\!\left (S,V\right )})/({\textrm {area}\partial \ell })\big \vert _S$ and one can identify the force from $-({\partial E\!\left (S,V\right )})/({\partial \ell })\big \vert _S$ noting that $\textrm{d}V={\textbf{area}} \cdot {\textbf{d}}\boldsymbol\ell$ . It is also helpful to note $T\equiv (\partial E(S,V))/({\partial S})\big \vert _V$ and generally $\textrm{d}E\!\left (S,\lambda \right )=T\mathrm{d}S+{\boldsymbol {\Lambda} }\cdot \textrm{d}{\boldsymbol {\lambda}}$ At constant entropy, $\textrm{d}E\!\left (\lambda \right )\vert_S={\boldsymbol {\Lambda} }\cdot \textrm{d}{\boldsymbol {\lambda } }=-P\mathrm{d}V=-P{{\textbf{area}}} \cdot {\textbf{d}}\boldsymbol\ell =-{\boldsymbol{F}}\cdot {\textbf{d}}\boldsymbol\ell =-\textrm{d}E$ . Hence, the pressure at constant entropy is the force divided by the area. We will return to consideration of the pressure subsequently.

[Editor’s Note: Kaufman made the cryptic remark that this exercise is not trivial and alluded to the Ergodic Theorem (§ 1.1.3) without further explanation.]

Next we introduce the concept of heat. Again consider a physical system composed of two subsystems I and II. The composite Hamiltonian is $H=H_{\textrm{I}}+\ H_{\textrm{II}}+H_{\textrm{int}}$ where $H_{\textrm{int}}$ is the interaction Hamiltonian. The energy gained or lost by subsystem I is then

(1.66) \begin{equation} \Delta E_{\textrm{I}}=\int\nolimits {\textrm{d}t}{\dot {H}}_{\textrm{I}}=\int\nolimits {\textrm{d}t}\left \{H_{\textrm{I}},H\right \}=\int\nolimits {\textrm{d}t}\left \{H_{\textrm{I}},H_{\textrm{int}}\right \}\equiv Q_{\textrm{I}}. \end{equation}

The Poisson bracket is $\left \{A\!\left (p,q\right ),B(p,q)\right \}=-\Sigma _i(({\partial A})/({\partial p_i})({\partial B})/({\partial q_i})- ({\partial A})/({\partial q_i}) ({\partial B})/({\partial p_i})),$ and $Q_{\textrm{I}}$ is the heat transfer. The total time derivative of any quantity can be shown to be

(1.67) \begin{equation} \dot {A}=\frac {\partial A}{\partial t}+\frac {\partial A}{\partial p}\dot {p}+\frac {\partial A}{\partial q}\dot {q}=\frac {\partial A}{\partial t}+\left \{A,H\right \}. \end{equation}

Example: $\langle K\rangle =({3}/{2})NT$ which is valid for an ideal or nonideal gas (with interactions), and (1.80) becomes

(1.81) \begin{equation} \langle P\rangle =nT+\frac {1}{6V}\left\langle \sum\limits _{i\ne j}{\sum\limits _j{{{\boldsymbol{f}}}_{j,i}\cdot ({{\boldsymbol{r}}}_i-}}{{\boldsymbol{r}}}_j)\right\rangle , \end{equation}

where $n=N/V.$

Postulate: Consider a general force law of particle j on particle i represented by

(1.82) \begin{equation} {{\boldsymbol{f}}}_{ij}=-{\hat {{\boldsymbol{r}}}}_{ij}\frac {\partial }{\partial r_i}\phi ({{\boldsymbol{r}}}_{ij}), {{\boldsymbol{r}}}_{ij}\equiv {{\boldsymbol{r}}}_i-{{\boldsymbol{r}}}_j{,} \end{equation}

where $\left ({2}/{3}\right )\langle {K}/{V}\rangle =\left ({N}/{V}\right )T\equiv nT$ . We justify (1.82) based on Newton’s third law and symmetry. Then the equilibrium pressure deduced from (1.81) is

(1.83) \begin{equation} \langle P\rangle =nT-\frac {1}{6V}\left\langle \sum\limits _{i\ne j}{{\sum\limits _{j}}{\boldsymbol{r}}_{ij}\cdot }{\hat {{\boldsymbol{r}}}}_{ij}\frac {\textrm{d}\phi }{\textrm{d}r_{ij}}\right\rangle = nT-\frac {1}{6V}\sum\limits _{i\ne j}{\sum\limits _j \left\langle {{\boldsymbol{r}}}_{ij}\cdot {\hat {{\boldsymbol{r}}}}_{ij}\frac {\textrm{d}\phi }{\textrm{d}r_{ij}}\right\rangle },\ \end{equation}

commuting the sum over N(N−1)∼N ${}^{2 }$ pairs of interacting particles with the averaging bracket. Hence,

(1.84) \begin{equation} P= nT-\frac {N^2}{6V}\langle r_{12}\phi^{\prime} (r_{12})\rangle . \end{equation}

The nT term is the kinetic pressure and the second term in (1.84) is the interaction pressure. The average in the interaction pressure is

(1.85) \begin{equation} \langle r_{12}\phi^{\prime} (r_{12})\rangle \equiv \int\nolimits {\textrm{d}^3r_{12}\rho (}{{\boldsymbol{r}}}_{ij}\to r_{12})\ r_{12}\phi^{\prime} (r_{12}) \end{equation}

due to isotropy in the probability density and because the interaction force depends only on the scalar separation distance. The probability density can be represented as

(1.86) \begin{equation} \rho \!\left (r_{12}\right )=\frac {g(r_{12})}{V}, \end{equation}

where $g(r_{12})$ is the pair correlation function such that $g(\infty )\to 1$ and $\smallint {g\ \textrm {dvol}=V}$ . Then (1.84) becomes

(1.87) \begin{equation} P= nT-\frac {N^2}{6V}\langle r_{12}\phi^{\prime} \!\left (r_{12}\right )\rangle =nT-\frac {n^2}{6}\int\nolimits {4\pi r^2_{12}\textrm{d}r_{12}g(r_{12}){r_{12}\phi }^{\prime} (r_{12})}. \end{equation}

If we formulate the energy of a particle (atom or molecule) from first principles by summing the kinetic energy and the potential energy due to interactions over the volume, one obtains

(1.88) \begin{equation} {\mathcal E}= \frac {3}{2}T+\frac {n}{2}\int\nolimits {\textrm{d}^3r_{12}g(r_{12})\phi (r_{12})}. \end{equation}

We can make some qualitative remarks regarding the dependencies of the pair correlation function g and the interaction potential $\phi$ on $r_{12}$ so that the integral in (1.88) remains well behaved, and the results for P and $\mathcal E$ are physical. Given the constraints g $(\infty )\to 1$ and $\smallint {g\ \textrm {dvol}=V}$ , $\phi (r_{12})$ must fall off faster than $1/r^3_{12}$ as $r_{12}\to \infty$ . For $r_{12}\to 0$ , $g(r_{12})\phi (r_{12})$ cannot diverge as fast as $1/r^3_{12}$ . As a result, excluded are the Coulomb and gravitational potentials $\sim 1/r$ and the dipole–dipole interaction potential $\sim 1/r^3$ .

Example: Van der Waals force + hard sphere – Consider the schematic for the electric potential shown in figure 1.

Exercise: (i) Show that ${\mathcal E}=({3}/{2})T-n\alpha$ and there is no contribution from the $V_0$ constant term. (ii) Show from $T\textrm{d}{\mathcal S}=\textrm{d}{\mathcal E}+P\textrm{d}{\mathcal V}$ where ${\mathcal V}\equiv {1}/{n}$ that ${\mathcal S}=({5}/{2})-{\text{ln } n{\Lambda}^3-4\!\left ({V_0}/{{\mathcal V}}\right )}$ , where $\Lambda ={h}/{\sqrt {2\pi mT}}$ (iii) Convert to standard van der Waals form:

(1.93) \begin{equation} \!\left (P+\frac {\alpha }{{{\mathcal V}}^2}\right )\!\left ({\mathcal V}-4V_0\right )=T \end{equation}

Example: Let ${n} = 1,2,3$ and ${E} = E_{1}$ , ${E}_{2}$ , ${E}_{3}$ , and make $M(M\to \infty )$ measurements. As a matter of definition what we mean by ${w}_i$ is that ${n} = i$ occurs ${w}_{i}$ M times. The number of states for a given number of measurements M is

(1.95) \begin{equation} {\varGamma }_M=\frac {M!}{\prod\limits _n{\!\left (Mw_n\right )}!}=\frac {M!}{\!\left (w_1M\right )!\!\left (w_2M\right )!\!\left (w_3M\right )!} \end{equation}

and the corresponding entropy is

(1.96) \begin{align} S_M\equiv \text{ ln } {\varGamma }_M&=M\text{ ln } M-M-\sum\limits _n\left [Mw_n{\text{ ln}(Mw_n})-Mw_n\right ]\nonumber\\&=-M\sum\limits _n{w_n{\text{ ln}(w_n)}}, \end{align}

where we have used $\Sigma _n{w_n=1}$ . From (1.96) we define the entropy associated with making a single measurement on the ensemble of three states in equilibrium with a heat bath:

(1.97) \begin{equation} S\equiv {\mathop {\lim }_{M\to \infty } \!\left (\frac {S_M}{M}\right )= }-\sum\limits _n{w_n{\text{ ln}(w_n})}. \end{equation}

Example: Suppose all $\varGamma$ states are accessible with equal probability, such that

(1.98) \begin{equation} w_n=\frac {1}{\varGamma }\quad \textrm {and}\quad S=-\sum\limits ^{\varGamma }_{n=1}{\frac {1}{\varGamma }{\text{ ln}\frac {1}{\varGamma }={\text{ ln } \varGamma }}}. \end{equation}

More generally, the entropy for the canonical ensemble using (1.49) is

(1.99) \begin{equation} S_{can}=-\sum\limits _n{\frac {1}{Z}e^{-\beta E_n}{\text{ ln}\frac {1}{Z}e^{-\beta E_n}= }}{\text{ ln } Z+\beta \sum\limits _n{w_nE_n={\text{ ln } Z+\beta \langle E\rangle }}}. \end{equation}

For a canonical macroensemble we can convert the sum in the partition function into an integral:

(1.100) \begin{equation} Z\!\left (\beta \right )=\int\nolimits {\textrm{d}\varGamma e^{-\beta E}=\int\nolimits {\textrm{d}E\frac {\textrm{d}\varGamma }{\textrm{d}E}}} e^{-\beta E}. \end{equation}

Recall that $S_{\mu }\sim S_{\sigma }={\text{ ln } \varGamma }$ from which it follows that $\varGamma =e^{S_{\sigma }}$ and $({\textrm{d}\varGamma }/{\textrm{d}E})=e^{S_{\sigma }}({\textrm{d}S_{\sigma }})/({\textrm{d}E})=\beta e^{S_{\sigma }}$ ; then we obtain

(1.101) \begin{equation} Z\!\left (\beta \right )= \beta \int\nolimits {\mathrm{d}Ee^{\left (-\beta E+S_{\sigma }\right )}=\beta \int\nolimits {\mathrm{d}Ee^{-\beta \left (E-TS_{\sigma }(E)\right )}}}= \beta \int\nolimits {\mathrm{d}Ee^{-\beta F(E,T)}}, \end{equation}

where F is the free energy and T is the temperature of the heat bath independent of the energy of the system. That energy for which F is a minimum will maximize the partition function. We require the most probable energy E ${}^{ *}$ (T) is that which determines $({\partial F(E\cdot T)})/({\partial E})=0$ . Then we expand the free energy around E ${}^{ *}$ :

(1.102) \begin{equation} F\!\left (E,T\right )=F\!\left (E^*,T\right )+\frac{1}{2}\delta E^2\frac {{\partial }^2F}{\partial E^2}+\dots . \end{equation}

From (1.101) and (1.102):

(1.103) \begin{align} &Z\!\left (\beta \right )\approx \beta e^{-F\left (E^*,T\right )}\int\nolimits ^{\infty }_{-\infty }{\mathrm{d}\delta E}e^{-\beta \frac{1}{2}\delta E^2\frac {{\partial }^2F}{\partial E^2}} =\beta e^{-F\left (E^*,T\right )}\sqrt {2\pi }{\sigma }_E\equiv \beta e^{-F\left (E^*,T\right )}\sqrt {2\pi T/F^{\prime\prime}}, \end{align}

where ${\sigma }_E=\sqrt {T/F^{\prime\prime}},$ and from (1.99) and (1.103)

(1.104) \begin{align} S_{can}&={\text{ ln } Z+\beta \langle E\rangle }={\text{ ln } \beta }-\beta F\!\left (E^*,T\right )+\tfrac {1}{2}{\text{ ln}(2\pi T/F^{\prime\prime})+}\beta \langle E\rangle \nonumber \\[4pt] &\approx -\beta F\!\left (E^*,T\right )+\ \beta \langle E\rangle =-\beta \!\left (E^*-TS_{\sigma }\!\left (E^*\right )\right )+\beta \langle E\rangle , \end{align}

where ${\text{ ln } \beta =O\!\left (1\right ),\ \ \ \beta F\!\left (E^*,T\right )=O\!\left (N\right ),\ \textrm {and}\ }({1}/{2}){\text{ ln}(2\pi T/F^{\prime\prime})=O(1).}$ Note that $F\sim O\!\left (E\right )\sim O\!\left (N\right ),\ \ F^{\prime\prime}\sim O(1/E)\sim O(1/N)$ and ${\text{ ln}\!\left (N\right )}\sim O(1)$ to justify $({1}/{2}){\text{ ln}(2\pi T/F^{\prime\prime})=O(1).}$

One rearranges terms in (1.104) to obtain

(1.105) \begin{equation} S_{can}=S_{\sigma }\!\left (E^*\right )+\beta \!\left (\langle E\rangle -E^*\right ). \end{equation}

For the canonical ensemble one can calculate $\langle E\rangle$ as a function of $\beta$

(1.106) \begin{equation} \langle E\rangle =\sum\limits _n{w_nE_n=\frac {1}{Z}\sum\limits _n{e^{-\beta E_n}E_n=-}}\frac {1}{Z}\frac {\partial Z}{\partial \beta }=-\frac {\partial {\text{ ln } Z}}{\partial \beta }. \end{equation}

We note that $F\!\left (E^*,T\right )=F(T)$ because fluctuations about E ${}^{*}$ are small. To O(N) from (1.104)

(1.107) \begin{equation} {\text{ ln } Z}=-\beta F\!\left (E^*,T\right )=-\beta F\!\left (T\right )\quad \text{and}\quad F\!\left (T\right )=-T{\text{ ln } Z(\beta )}. \end{equation}

For ${\text{ ln } Z(\beta )}\gt 1,$ then $F\!\left (T\right )\lt 0$ . From (1.106) and (1.107) we deduce

(1.108) \begin{equation} \langle E\rangle =-\frac {\partial {\text{ ln } Z}}{\partial \beta }=\frac {\partial \beta F}{\partial \beta } \quad \textrm{and}\quad \langle E\rangle (T)=E^*(T) \;\textrm{to}\; \textit {O}(\textit {N}). \end{equation}

It also follows that

(1.109) \begin{equation} S_{\text{can}}\!\left (T\right )={\text{ ln } Z-}\beta \frac {\partial {\text{ ln } Z}}{\partial \beta }=-\frac {\partial F(T)}{\partial T}, \textrm {and hence}\; \textrm{d}F=-S\textrm{d}T. \end{equation}

Definition: Let $R\equiv \Delta E-Q=\Lambda \Delta \lambda$ where $\Lambda =(\partial E)/(\partial \lambda) \vert _S.$

From the definition of the free energy

(1.114) \begin{align} F\equiv E-TS \to \Delta F&=\Delta E-T\Delta S-S\Delta T= T\Delta S+\Lambda \Delta \lambda -T\Delta S-S\Delta T\nonumber\\[4pt] &=\Lambda \Delta \lambda =R,\ \end{align}

because $\Delta T=0$ due to the contact with the heat bath. We can now make some general remarks regarding systems that are either thermally isolated or in contact with a heat bath (table 1). We note that if there are no changes in the system parameters, a system in contact with a heat bath experiences no change in S and E; and the heat bath is superfluous.

Example: N identifiable microsystems, e.g., N weakly interacting harmonic oscillators

(1.115) \begin{equation} Z_N\!\left (\beta \right )=\sum\limits _{n=\left \{n_i\right \}}{e^{-\beta E_n}\ \ \ \ \textrm {and}\ \ \ \ E_{\left \{n_i\right \}}=}\sum\limits ^N_{n_i=1}{{{\mathcal E}}^{(i)}_{n_i}}, \end{equation}

where $n_i$ can be thought of as the quantum numbers for the energy levels. The partition function becomes

(1.116) \begin{equation} Z=\sum\limits _{n_1}{\sum\limits _{n_2}{\dots \sum\limits _{n_N}{e^{-\beta \left ({{\mathcal E}}^{(1)}_{n_1}+{{\mathcal E}}^{(2)}_{n_2}+\dots +{{\mathcal E}}^{(N)}_{n_N}\right )}=\prod\limits^N_{i=1}{\sum\limits _{n_i}{e^{-\beta {{\mathcal E}}^{(i)}_{n_i}}=\prod\limits^N_{i=1}{Z_i}}}}}}. \end{equation}

Hence, ${\text{ ln } Z=\Sigma _i{{\textrm {ln}\ Z}_i}}.$ In the special case where all N subsystems have the same properties, $Z={(Z_1)}^N$ and $\text{ ln } Z=N{\text{ ln } Z_1}$ .

Example: A classical gas or fluid consisting of N indistinguishable particles

(1.117) \begin{equation} Z_N(\beta ,V)\equiv \int\nolimits _V{\frac {\textrm{d}^{3N}p\textrm{d}^{3N}q}{h^{3N}N!}e^{-\beta H(p,q)}}, \end{equation}

(1.118) \begin{equation} H\!\left (p,q\right )\to \sum\limits ^N_{i=1}{\frac {p^2_i}{2m}+\Phi (\left \{r_i\right \})}. \end{equation}

Using (1.118) in (1.117) one obtains

(1.119) \begin{equation} Z_N\!\left (\beta ,V\right )=\frac {1}{N!}{\left [V\int\nolimits {\frac {\textrm{d}^3p}{h^3}e^{-\beta \frac {p^2}{2m}}}\right ]}^N\left [\int\nolimits {\frac {\textrm{d}^3r^{(N)}}{V^N}e^{-\beta \Phi (\left \{r\right \})}}\right ] \equiv \frac {1}{N!}{\left (\frac {V}{{\Lambda}^3\!\left (\beta \right )}\right )}^NQ_N(\beta ), \end{equation}

where $Q_N\!\left (\beta \right )\equiv [(\smallint {\textrm{d}q^N}/{V^N})e^{-\beta \Phi (\{r\})}]$ is the configurational partition function independent of volume.

Table 1. Adiabatically evolving systems.

For an ideal gas $\Phi (\{r\})\to 0$ and $Q_N\!\left (\beta \right )=1$ ; hence, $Z_N\!\left (\beta ,V\right )=$ $({1}/{N!}){\left ({V}/{{\Lambda}^3\!\left (\beta \right )}\right )}^N.$

Exercise: Show the specific free energy is given by $f={F}/{N}=T\!\left ({\text{ln } n{\Lambda}^3-1}\right )$ , ${\mathcal E}=({3}/{2})T$ , and ${\mathcal S}=({5}/{2})-{\text{ ln } n{\Lambda}^3}$ . (Recall (1.47) and (1.92), and the exercise following (1.92) in the limit $\Phi \to 0$ .)

Example: Single harmonic oscillator ( $\ell )$ with quantized energy levels

Energy levels:

(1.120) \begin{equation} E^{\ell }_n=\hslash {\omega }_{\ell }n+E_0,\ \ E_0=\tfrac{1}{2}\hslash {\omega }_{\ell }, \end{equation}

(1.121) \begin{equation} Z_{\ell }\!\left (\beta \right )=\sum\limits ^{\infty }_{n=0}{e^{-\beta \hslash \omega_{\ell }n}=1+e^{-x}+e^{-2x}+\dots =\frac {1}{1-e^{-x}}=\frac {1}{1-e^{-\beta \hslash {\omega }_{\ell }}}}. \end{equation}

We could compare the energy spectrum for the quantum harmonic oscillator in (1.120) to a few continuous medium systems.

1. 1. Vibrating string: $\lambda ={2L}/{\ell },\ \ \ell =1,2,3,\dots .$

2. 2. Drumhead.

3. 3. Water waves, e.g., one-dimensional gravity waves in a narrow channel, three-dimensional ocean waves (surface gravity waves, internal waves, etc.).

4. 4. Electromagnetic waves, e.g., free space ( $\omega =kc)$ , wave-guide modes, cavity modes.

5. 5. Plasma waves, e.g., electromagnetic waves ( ${\omega }_k=\sqrt {k^2c^2+{\omega }^2_p}),$ longitudinal waves, etc.

6. 6. Fluid sound waves: ${\omega }_k=kc_s, c_s=\sqrt {{\gamma P}/{\rho }}$ .

7. 7. Waves in a solid: longitudinal sound wave ${\omega }_k=kc_{\ell }$ , transverse shear wave ${\omega }_k=kc_t$ .

In order to calculate the partition function and the statistical properties of any of these systems, one must properly count the distinct modes. Here are two illustrative examples.

1. (a) One-dimensional standing waves with nodes at x = 0 and x = L: $U\!\left (x,t\right )=$ $A\,\textrm {sin}(kx){\sin(\omega t)},\, \omega \gt 0,\, {\lambda }/{2}={L}/{\ell },\, k={2\pi }/{\lambda }$ and $\Sigma ^{\infty }_{\ell =1}{\to}\smallint\nolimits ^{\infty }_0\mathrm{d}\ell =$ $({L}/{\pi})\smallint\nolimits ^{\infty }_0{\textrm{d}k}$ .

2. (b) One-dimensional traveling waves with periodic boundary conditions at x = 0 and x = L: $U\!\left (x,t\right )=A{\sin \!\left (kx-\omega t\right )},\ \omega \gt 0,\ \lambda ={L}/{\ell },\ k={2\pi }/{\lambda }$ and $\Sigma ^{\infty }_{\ell =-\infty }{\to }\smallint\nolimits ^{\infty }_{-\infty }{\mathrm{d}\ell =({L}/{2\pi })}\smallint\nolimits ^{\infty }_{-\infty }{\textrm{d}k}$ . If the traveling wave spectrum is symmetric with respect to positive and negative k, then $({L}/{2\pi })\smallint\nolimits ^{\infty }_{-\infty }{\textrm{d}k}\to ({L}/{\pi})\smallint\nolimits ^{\infty }_0{\textrm{d}k}$ .

In three spatial dimensions $\Sigma _{\text{modes}}{\to {(2/\pi)}^3}\smallint {\textrm{d}^3k}$ .

Example: Classical noninteracting oscillators with $H_{\ell }\!\left (J_{\ell }\right )=J_{\ell }{\omega }_{\ell }$

(1.122) \begin{equation} Z_{\ell }\!\left (\beta \right )=\int\nolimits {\frac {\textrm{d}p\textrm{d}q}{h}}e^{-\beta H(p,q)}=\frac {2\pi }{h}\int\nolimits ^{\infty }_0{\textrm{d}J_{\ell }e^{-\beta J_{\ell }{\omega }_{\ell }}=\frac {2\pi }{h{\beta \omega }_{\ell }}}=\frac {T}{\hslash {\omega }_{\ell }}. \end{equation}

We note that the result (1.121) for the partition function for the quantized harmonic oscillator recovers the classical limit in (1.122) in the limit $\hslash {\omega }_{\ell }\ll T$ :

(1.123) \begin{equation} Z_{\ell }{(\beta )}_{\textrm{quantum}}={\mathop {\lim }_{\hslash {\omega }_{\ell }\ll T} \ \frac {1}{1-e^{-\beta \hslash {\omega }_{\ell }}}\to }\frac {T}{\hslash {\omega }_{\ell }}. \end{equation}

From (1.123) we can calculate the average energy per mode:

(1.124) \begin{equation} \langle E_{\ell }\rangle \!\left (T\right )=-\frac {\partial {\text{ ln } Z}}{\partial \beta }=\frac {\hslash {\omega }_{\ell }}{e^{\beta \hslash {\omega }_{\ell }}-1}, \end{equation}

which yields $\langle E_{\ell }\rangle \!\left (T\right )\to T$ in the classical limit and recovers the Rayleigh–Jeans classical result. For black-body radiation, each of the infinite number of modes has energy T in the classical limit which leads to an infinite total energy when summing over all of the modes, i.e., the ultraviolet catastrophe!

Definition: The Wien wavelength and its inverse $k_w$ are defined by $\overline {\lambda }={1}/{k_w}={\hslash c}/{T}$ .

Exercise: For a d-dimensional medium supporting normal modes with ${\omega }_k\sim k^p$ with $p\gt 0$ , e.g., ${p}= 1/2$ for water waves, ${p} = 1$ for sound waves, and ${p }= 2$ for a de Broglie matter wave, find the specific heat C ∼ T ${}^{ q}$ . The specific heat capacity is defined as $C=T{\partial {\mathcal S}}/{\partial T}$ and recall that $\mathcal S$ is the specific entropy. Use (1.99) to evaluate the entropy in terms of the partition function and the examples in § 1.2.5 as a template to calculate the partition function.

Exercise: Work out the Euler–Lagrange equations for the particles using (1.137) to find

(1.140) \begin{align} {m}_i{\dot {{{{\boldsymbol v}}}}}_i&=e_i\bigg [{{\boldsymbol{E}}}_0\!\left ({{\boldsymbol{r}}}_i,t\right )+\frac {1}{c}{{{{\boldsymbol v}}}}_i\times {{\boldsymbol{B}}}_0\!\left ({{\boldsymbol{r}}}_i,t\right )\bigg ]+e_i\sum\limits _j{e_j\!\left (-{\nabla }_i\frac {1}{r_{ij}}\right )}\nonumber \\[4pt] &\quad +e_i\bigg [{{\boldsymbol{E}}}_{\mathrm{rad}}\!\left ({{\boldsymbol{r}}}_i,t\right )+\frac {1}{c}{{{{\boldsymbol v}}}}_i\times {{\boldsymbol{B}}}_{\mathrm{rad}}\!\left ({{\boldsymbol{r}}}_i,t\right )\bigg ]. \end{align}

Definition: Define the functional derivatives (introducing bars over the partial signs $\overline {\boldsymbol\partial }$ )

(1.141) \begin{equation} {\boldsymbol \Pi }\equiv \frac {\overline {\boldsymbol\partial }{\boldsymbol{L}}}{\overline {\boldsymbol\partial }\dot {{\boldsymbol{A}}}{\mathbf (}{\boldsymbol{x}}{\mathbf )}}\equiv \frac {1}{4\pi c^2}\dot {{\boldsymbol{A}}}\!\left ({\boldsymbol{x}}\right )=-\frac {1}{4\pi }{{\boldsymbol{E}}}_{\mathrm{rad}}\quad \textrm {and}\quad \dot {{\boldsymbol \Pi }}\!\left ({\boldsymbol{x}}\right )\equiv \frac {\overline {\boldsymbol\partial }{\boldsymbol{L}}}{\overline {\boldsymbol\partial }{\boldsymbol{A}}\!\left ({\boldsymbol{x}}\right )}=-\frac {1}{4\pi c}{\dot {{\boldsymbol{E}}}}_{\mathrm{rad}}\!\left ({\boldsymbol{x}}\right )\!. \end{equation}

These are used in recovering Maxwell’s equations from the Euler–Lagrange equations applied to (1.137). For example, from the term $\smallint {\textrm{d}^3\textit {x}}{\vert \nabla \times {\boldsymbol{A}}\vert }^2$ one forms $2\smallint {\textrm{d}^3\textit {x}}{\nabla \times {\boldsymbol{A}}\cdot \nabla \times \delta {\boldsymbol{A}}}\to 2\smallint {\textrm{d}^3\textit{x}}\;\delta {\boldsymbol{A}}\cdot {\nabla \times \!\left (\nabla \times {\boldsymbol{A}}\right )=}2\smallint {\textrm{d}^3\textit {x}}\;\delta {\boldsymbol{A}}\cdot ({4\pi }/{c}){{\boldsymbol{j}}}$ from which $\dot {{\boldsymbol \Pi }}\!\left ({\boldsymbol{x}}\right )=-({1}/{4\pi c}){\dot {{\boldsymbol{E}}}}_{\mathrm{rad}}\!\left ({\boldsymbol{x}}\right )=-({1}/{4\pi })\nabla \times {{\boldsymbol{B}}}_{{\textrm{rad}}}{\mathbf +}({{1}}/{{{{c}}}}){{\boldsymbol{j}}}^{\textit {t}}$ . In summary, the Lagrangian in (1.137) recovers the correct Maxwell equations:

(1.142) \begin{eqnarray} {{\boldsymbol{E}}}_{\mathrm{rad}}=-\frac {1}{c}{\dot {{\boldsymbol{A}}}}_{\mathrm{rad}}, \nabla \times {{\boldsymbol{B}}}_{\mathrm{rad}}-\frac {1}{c}{\dot {{\boldsymbol{E}}}}_{\mathrm{rad}}=\frac {4\pi }{c}{{\boldsymbol{j}}}^t\quad \textrm{and}\qquad\quad \nonumber \\[4pt] \nabla \cdot \!\left (\nabla \times {{\boldsymbol{B}}}-\frac {1}{c}{\dot {{\boldsymbol{E}}}}=\frac {4\pi }{c}{{\boldsymbol{j}}}\right ) \to 0= {\dot {{\boldsymbol{E}}}}+4\pi {{\boldsymbol{j}}}\to \nabla \cdot {\boldsymbol{E}}=4\pi \rho , \end{eqnarray}

where we have made use of charge continuity: $\dot {\rho }+\nabla \cdot {\boldsymbol{j}}=0$ .

Definition: The canonical momentum in an electromagnetic field is defined by

(1.143) \begin{equation} {{\boldsymbol{p}}}_i\equiv \frac {\partial L}{\partial {{{{\boldsymbol v}}}}_{{\boldsymbol{i}}}}=m_i{{{{\boldsymbol v}}}}_i+\frac {e_i}{c}\left [{{\boldsymbol{A}}}_0\!\left ({{\boldsymbol{r}}}_i,t\right )+{\boldsymbol{A}}\!\left ({{\boldsymbol{r}}}_i,t\right )\right ] \end{equation}

and as noted in (1.141)

(1.144) \begin{equation} {\boldsymbol \Pi }\!\left ({\boldsymbol{x}}\right )\equiv \frac {\partial L}{\partial {\boldsymbol{A}}\!\left ({\boldsymbol{x}}\right )}=\frac {1}{4\pi c^2}\dot {{\boldsymbol{A}}}\!\left ({\boldsymbol{x}}\right )=-\frac {1}{4\pi c}{\boldsymbol{E}}\!\left ({\boldsymbol{x}}\right ). \end{equation}

Recalling the definitions in (1.141), the Hamiltonian implied by the Lagrangian in (1.137) is

(1.145) \begin{align} H&=\sum\limits _i{{{\boldsymbol{p}}}_i\cdot }{{{{\boldsymbol v}}}}_i+\int\nolimits {\textrm{d}^3\textit {x}{\boldsymbol \Pi }\!\left ({\boldsymbol{x}}\right )}\cdot \dot {{\boldsymbol{A}}}\!\left ({\boldsymbol{x}}\right )-L\nonumber\\[4pt] &= \sum\limits _i{\frac{1}{2}}m_i{{{{v}}}}^2_i+\sum\limits _{i\lt j}{\frac {e_ie_j}{r_{ij}}}+\sum\limits _i{e_i}\phi _0\!\left ({{\boldsymbol{r}}}_i,t\right )+\int\nolimits {\frac {\textrm{d}^3{\boldsymbol{x}}}{8\pi }\ \!\left (E^2_{\mathrm{rad}}+B^2_{\mathrm{rad}}\right )}\nonumber \\[4pt]&=K+C+R, \end{align}

where $K\equiv$ $\Sigma _i({{1}/{2}})m_i{{{{v}}}}^2_i,\, C\equiv \Sigma _{i\lt j}({ {e_ie_j}/{r_{ij}}})+\Sigma _i{e_i}\phi _0({{\boldsymbol{r}}}_i,t), R\equiv \smallint \textrm{d}^3 \boldsymbol{x} (E_{\textrm{rad}}^2 + B_{\textrm{rad}}^2)/$ $(8\pi).$ We note that there is no magnetic interaction energy in the Hamiltonian.

Exercise: Calculate $\dot {p}=-{\partial H}/{\partial q}$ , $\dot {q}={\partial H}/{\partial p}$ , and recover Maxwell’s equations. The generalized momentum and Maxwell’s equation in (1.141), (1.142), and (1.143) have been calculated already from the Lagrangian in (1.137).

Definition: ${\boldsymbol{j}}({\boldsymbol{x}}\!\left ({{\boldsymbol{r}}}_i,{{{{\boldsymbol v}}}}_i\right ))\equiv \Sigma _i{e_i{{{{\boldsymbol v}}}}_i\delta ({\boldsymbol{x}}-}{{\boldsymbol{r}}}_i).$

Exercise: In a system with a uniform constant applied magnetic field B ${}_{0}$ and ${{\boldsymbol{A}}}_0=({1}/{2}){{\boldsymbol{B}}}_0 \times {\boldsymbol{x}}$ with $L_{\textrm{int}}={\boldsymbol \mu }\cdot {{\boldsymbol{B}}}_0$ and ${\boldsymbol \mu }\equiv (2c)^{-1}\smallint {\textrm{d}^3\textit{x}}{\boldsymbol{x}}\times {\boldsymbol{j}}\!\left ({\boldsymbol{x}}\right ),$ show that $\langle {\boldsymbol \mu }\rangle =0=\langle {\partial L}/{\partial {{\boldsymbol{B}}}_0}\rangle$ analogous to the arguments and results in (1.160)–(1.165).

Definition: ${{\boldsymbol \mu }}_s\equiv -T{{\boldsymbol \gamma }}_s$ is a set of chemical potentials. Equivalently, ${{\boldsymbol \mu }}_s\equiv {\partial E(S,{{\boldsymbol{N}}}_s,\lambda )}/{\partial {{\boldsymbol{N}}}_s}$ .

Recalling the definition of ${{\boldsymbol \gamma }}_s$ , we can express the chemical potential as follows:

(1.172) \begin{equation} {\mu }_s=T{\text{ ln } n_s{\Lambda}^3_s}. \end{equation}

Keep in mind that N ${}_{s}$ is variable.

Example: The grand partition function for an ideal gas (of identical particles) is

(1.173) \begin{equation} \mathbb{Z} = \sum\limits _N{e^{-\gamma N}\frac {{(Z_1)}^N}{N!}\approx \sum\limits ^{\infty }_{N=0}{\frac {{\!\left (Z_1e^{-\gamma }\right )}^N}{N!}}}= e^{Z_1e^{-\gamma }}, \end{equation}

where Z ${}_{1}$ is the one-particle partition function as defined after (1.116), Z ${}_{1}$ $\equiv$ $\Sigma _{n_i}{e^{-\beta {{\mathcal E}}_{n_i}}}\to {V_{\textrm{I}}}/{{\Lambda}^3}$ using (1.119). We note that (1.170) and (1.171) yield the simple identities

(1.174) \begin{equation} \langle {{\boldsymbol{N}}}_{\textrm{I}}\rangle =-\frac {\partial {\text{ ln } \mathbb {Z}\textrm {(}{\beta }_{\textrm{II}},{{\boldsymbol \gamma }}_{\textrm{II}})}}{\partial {{\boldsymbol \gamma }}_{\textrm{II}}}\quad \textrm {and}\quad \langle E_{\textrm{I}}\rangle =-\frac {\partial {\text{ ln } \mathbb {Z}\textrm {(}{\beta }_{\textrm{II}},{{\boldsymbol \gamma }}_{\textrm{II}})}}{\partial {\beta }_{\textrm{II}}}. \end{equation}

Example: In an ideal gas $\langle N_{\textrm{I}}\rangle =-({\partial {\text{ ln } \textrm {Z}\!\left ({\beta }_{\textrm{II}},{{\boldsymbol \gamma }}_{\textrm{II}}\right )}}/{\partial {{\boldsymbol \gamma }}_{\textrm{II}}})=Z_1e^{-\gamma }=$ ${V_{\textrm{I}}e^{-{\gamma }_{\textrm{II}}}}/{{\Lambda}^3{(\beta }_{\textrm{II}})}=(V_{\textrm{I}} n_{\textrm{II}} \Lambda^3(\beta_{\textrm{II}})/ \Lambda^3(\beta_{\textrm{II}})) = n_{\textrm{II}}V_{\textrm{I}}$ and the number densities $n_{\textrm{II}}=n_{\textrm{I}}$ using (1.171), (1.172), and the definitions, and assuming that the bath and system I share the same mix of species and can freely exchange particles without disturbing the physics.

Exercise: Show that $\langle {\left (\delta N_{\textrm{I}}\right )}^2\rangle \sim {{\partial }^2{\text{ ln } \textrm {Z}}}/{\partial {\gamma }^2_{\textrm{II}}}$ is small if system I is macroscopic, i.e., $N_{\textrm{I}}\sim O\!\left ({10}^{23}\right ).$

Consider a macroscopic system I whose probability is given by

(1.175) \begin{equation} w_{{\{N}^I\}}\equiv \sum\limits _n{w_{n,\{N^I\}}=\frac {1}{\mathbb {Z}}e^{-{\boldsymbol \gamma }\cdot {\boldsymbol{N}}}Z\!\left (\beta ,{\boldsymbol{N}}\right )=}\frac {1}{\mathbb {Z}}e^{-\beta (F-{\boldsymbol \mu }\cdot {\boldsymbol{N}})}\equiv \frac {1}{\mathbb { Z}}e^{-\beta \Omega }, \end{equation}

where F is the Helmholtz free energy, ${\boldsymbol \mu }=-T{\boldsymbol \gamma }$ , and the grand potential is defined as

(1.176) \begin{equation} \Omega \equiv \Omega \!\left (\beta ,{\boldsymbol \mu };{\boldsymbol{N}}\right )\equiv F\!\left (\beta ,{\boldsymbol{N}}\right )-{\boldsymbol \mu }\cdot {\boldsymbol{N}}. \end{equation}

Recall that $\boldsymbol{N}$ is the vector representing the set of occupation numbers in different states.

The probability of having a given set of occupation numbers is

(1.177) \begin{equation} w_{{\boldsymbol{N}}}=\frac {1}{\textrm {Z}\textrm {(}\beta \textrm {,}\mu \textrm {)}}e^{-\beta \Omega \textrm {(}\beta \textrm {,}\mu ;{\boldsymbol{N}}{\mathbf )}}. \end{equation}

The maximum of the probability $w_{{\boldsymbol{N}}}$ with respect to $\boldsymbol{N}$ occurs at ${{\boldsymbol{N}}}^*$ , i.e., $\langle {\boldsymbol{N}}\rangle ={{\boldsymbol{N}}}^*$ . Given (1.177), the maximum of $w_{{\boldsymbol{N}}}$ corresponds to minimizing $\Omega$ with respect to ${\boldsymbol{N}}.$ At the most probable set of occupation numbers ${{\boldsymbol{N}}}^*$ the grand potential satisfies the relation

(1.178) \begin{equation} \Omega \cong -\frac {1}{\beta }{\text{ln }\textrm {Z}} \end{equation}

plus a constant of smaller order.

Using the relations $F\equiv E-TS,$ $\Omega =F-{\boldsymbol \mu }\cdot {\boldsymbol{N}}$ , (1.171), and (1.172), we have

(1.179) \begin{equation} \textrm{d}E=T\textrm{d}S+{\boldsymbol \mu }\cdot \textrm{d}{\boldsymbol{N}}+\Lambda \textrm{d}\lambda \quad \textrm {and}\quad \textrm{d}F=-S\textrm{d}T+{\boldsymbol \mu }\cdot \textrm{d}{\boldsymbol{N}}+\Lambda \textrm{d}\lambda \, \end{equation}

where, for example, we might choose $\lambda =V$ and $\Lambda =-P$ from (1.65); and we arrive at

(1.180) \begin{equation} \textrm{d}\Omega =-S\textrm{d}T-{\boldsymbol{N}}\cdot \textrm{d}{\boldsymbol \mu }+\Lambda \textrm{d}\lambda \quad \textrm {and}\quad {\boldsymbol{N}}=-\frac {\partial \Omega }{\partial {\boldsymbol \mu }} =-\frac {\partial \textrm {ln}\mathbb {Z}}{\partial {\boldsymbol \gamma }} \to \langle {\boldsymbol{N}}\rangle ={{\boldsymbol{N}}}^{{\mathbf *}} \end{equation}

for a macroscopic system. From (1.180) with $\Lambda =-P$ and $\lambda =V$

(1.181) \begin{equation} \textrm{d}\Omega =-S\textrm{d}T-{\boldsymbol{N}}\cdot \textrm{d}{\boldsymbol \mu }-P\textrm{d}V \to P=-\frac {\partial \Omega }{\partial V}\!\left (T,{\boldsymbol \mu },V\right )=-\frac {\Omega }{V}. \end{equation}

In deriving $P=-{\Omega }/{V}$ we argue that $\Omega$ is extensive (should scale with volume), while T and ${\boldsymbol \mu }$ are intensive. Hence, $\Omega =V\ \textrm {fn}(T,{\boldsymbol \mu }\textrm {)}$ and

(1.182) \begin{equation} -PV=\Omega = F-{\boldsymbol \mu }\cdot {\boldsymbol{N}}. \end{equation}

Example: Grand partition function and grand potential for an ideal classical gas

(1.183) \begin{equation} \Omega \!\left (T,\mu ,V\right )=-T{\text{ ln } \mathbb {Z} = -\frac {VT}{{\Lambda}^3(\beta )}e^{\beta \mu }}, \end{equation}

(1.184) \begin{equation} P=-\frac {\Omega }{V}= \frac {T}{{\Lambda}^3(\beta )}e^{\beta \mu },\quad N=\frac {V}{{\Lambda}^3(\beta )}e^{\beta \mu },\quad \frac {P}{N}=\frac {T}{V} \to P=nT. \end{equation}

Example: Quantum ideal gas.

Consider a subsystem consisting of a single-particle quantum state k for a simple noninteracting electron gas. The energy of a single electron is

(1.187) \begin{equation} {{\mathcal E}}_k=\left \{ \begin{array}{c@{\quad}l} \frac {p^2}{2m} & (\textrm {non-relativistic}), \\[4pt] \sqrt {p^2c^2+m^2c^4} & (\textrm {relativistic).} \end{array} \right . \end{equation}

Note: If the particles are photons (bosons) instead, one must be careful because they are not conserved. Ions, molecules, fermions, molecules, etc., are conserved if noninteracting (no ionization, no recombination, no chemistry). Including a magnetic field B and spin, the subsystem energy is

(1.188) \begin{equation} E=\sum\limits _k{{{\mathcal E}}_kN_k\pm \hat {\mu }B}, \end{equation}

where $N_k$ is an occupation number ( $N_k=0,1$ for fermions due to the Pauli principle; and $N_k=0,1,2,\ldots ,\infty$ for bosons); and $\hat {\mu }$ is the magnetic moment associated with the spin. The probability of the macroscopic state k with occupation number $N_k$ is then

(1.189) \begin{equation} w_{N_k}=\frac {e^{-\gamma N_k-\beta {{\mathcal E}}_kN_k}}{{\textrm {Z}}_k}\equiv \frac {e^{-\beta ({{\mathcal E}}_k-{{\mu }_s})N_k}}{\sum\nolimits ^{1\ or\ \infty }_{N_k=0}{e^{-\beta ({{\mathcal E}}_k-{{\mu }_s})N_k}}}, \end{equation}

where the sum in the denominator for the grand partition function is just two terms for fermions and a geometric series for bosons.

Example: Bosons.

In order that ${\textrm {Z}}_k$ converges, ${{\mathcal E}}_k\gt {\mu }$ for all k. We also note that ${{\mathcal E}}_0=0$ , which implies that ${\mu }\lt 0.$ Hence, $w_{N_k}\propto e^{-\beta ({{\mathcal E}}_k+\vert \mu \vert )N_k}$ , i.e., $w_{N_k}$ is a monotonic and exponentially decreasing function of $N_k$ . The most probable state is the state $N_k=0.$ From (1.189) one concludes

(1.190) \begin{equation} \langle N_k\rangle \equiv \sum\limits ^{\infty }_{N_k=0}{w_{N_k}}N_k=\frac {1}{e^{\beta ({{\mathcal E}}_k+|{\mu }|)}-1}. \end{equation}

Here $\langle N_k\rangle$ is a monotonically decreasing function of ${{\mathcal E}}_k,$ and Einstein condensation can occur when $\langle N_k\rangle$ becomes macroscopically large which is possible at ${{\mathcal E}}_k=0$ .

Example: Fermions.

Because of the Pauli principle the occupation number $N_k$ = 0 or 1 for fermions. The probability $w_{N_k}$ is proportional to

(1.191) \begin{equation} w_{N_k}\propto e^{-\beta ({{\mathcal E}}_k-{\mu })N_k}. \end{equation}

For ${\mu }\gt 0$ the argument of the exponential is positive for ${{\mathcal E}}_k\lt {\mu }_s$ and is negative for ${{\mathcal E}}_k\gt {\mu }$ . $w_{N_k}$ takes on just two values as a function of $N_k$ , at $N_k$ = 0 and 1. The argument of the exponential vanishes for ${{\mathcal E}}_k={\mu }$ the Fermi level. Figure 2 plots $\langle N_k\rangle =\Sigma ^1_{N_k=0}{w_{N_k}}N_k$ as a function of energy ${{\mathcal E}}_k$ (Fermi–Dirac distribution function).

Figure 2. Fermi–Dirac distribution function $\langle$ N ${}_{k}$ $\rangle$ =n( ${{\mathcal E}})$ (Riebesell Reference Riebesell2022).

We define ${{\mathcal E}}^{\prime} \equiv {{\mathcal E}}_k-{\mu }$ and the partition function ${\textrm {Z}}_k$ is then

(1.192) \begin{equation} {\textrm {Z}}_k={\!\left (1+\sigma e^{-\beta {{\mathcal E}}^{\prime} }\right )}^{\sigma } \sigma \equiv \left \{ \begin{array}{l@{\quad}l} +1 & (\textrm {fermions}), \\[4pt] -1 & (\textrm {bosons}). \end{array} \right . \end{equation}

Then

(1.193) \begin{equation} {\text{ ln } {\mathbb {Z}}_k}=\sigma {\text{ ln}(1+\sigma e^{-\beta {{\mathcal E}}^{\prime} })}\quad \textrm {and}\quad \mathbb {Z} = \prod\limits _k{{\mathbb {Z}}_k}\to {\text{ ln } \mathbb {Z} = \sum\limits _k{{\text{ ln } {\mathbb {Z}}_k}}} \end{equation}

We recall the analysis leading to (1.182) and obtain

(1.194) \begin{equation} P=-\frac {\Omega }{V}=\frac {T}{V}{\text{ ln } \mathbb {Z} = \frac {T}{V}\sigma \sum\limits _k{{\text{ ln}\!\left (1+\sigma e^{-\beta ({{\mathcal E}}_k-{\mu })}\right )}}}. \end{equation}

We introduce the de Broglie wavenumber k and evaluate ${{\mathcal E}}_k={{\left (\hslash k\right )}^2}/({2m})$ . In (1.194) we replace $\Sigma _k{ \to gV\smallint {({\textrm{d}^3{\boldsymbol{k}}})/{{(2\pi )}^3}}}$ where the factor $g=2{\mathcal S}+1$ and the spin factor is ${\mathcal S}={1}/{2}$ or an integer. Then (1.194) becomes using $\text{ ln}\!\left (1+x\right )=\Sigma ^{\infty }_{\ell =1}{{(-1)}^{\ell -1}{x^{\ell }}/{\ell }}$

(1.195) \begin{align} P&={\sigma gT\int\nolimits {\frac {\textrm{d}^3k}{{(2\pi )}^3}}\textrm {ln}\!\left (1+\sigma e^{-\beta ({{\mathcal E}}_k-{\mu })}\right )}\nonumber \\[4pt] &= \sigma gT\sum\limits ^{\infty }_{\ell =1}{\frac {{\!\left (-1\right )}^{\ell -\textrm {1}}}{\ell }}{\sigma }^{\ell }\int\nolimits {\frac {\textrm{d}^3{\boldsymbol{k}}}{{\!\left (2\pi \right )}^3}}e^{-\beta \ell \left ({{\mathcal E}}_k-{\mu }\right )}\nonumber \\[4pt] &= gT\sum\limits ^{\infty }_{\ell =1}{\frac {{\!\left (-\sigma \right )}^{\ell -\textrm {1}}}{\ell }}\int\nolimits {\frac {\textrm{d}^3{\boldsymbol{k}}}{{\!\left (2\pi \right )}^3}}e^{-\beta \ell \left ({{\mathcal E}}_k-{\mu }\right )}. \end{align}

Recalling the definition $\Lambda ={h}/{\sqrt {2\pi mT}}$ and introducing the dimensionless fugacity or absolute activity $\xi \equiv e^{\beta {\mu }}$ , (1.195) leads to

(1.196) \begin{equation} P=\frac {gT}{{\Lambda}^{\textrm{d}}(\beta )}\sum\limits ^{\infty }_{\ell =1}{\frac {{\!\left (-\sigma \right )}^{\ell -\textrm {1}}}{{\ell }^{1+\frac {\textrm{d}}{2}}}}{\xi }^{\ell } \to P(T,\xi )=\frac {gT}{{\Lambda}^3(\beta )}\left [\xi -\frac {\sigma {\xi }^2}{2^{5/2}}+\frac {{\xi }^3}{3^{5/2}}+\dots \right ], \end{equation}

where the dimensionality d has been set to d = 3. We recall (1.186) which determines the particle density $n={\partial P}/{\partial {\mu }_s}$

(1.197) \begin{align} n\!\left (T,\xi \right )&=\frac {\partial P\!\left (T,{\mu }\right )}{\partial {\mu }}=\frac {\beta \xi \partial P\!\left (T,\xi \right )}{\partial \xi }=\frac {g}{{\Lambda}^{\text{d}}\!\left (\beta \right )}\left [\xi \!\left (1-\frac {\sigma }{2^{\frac {\textrm{d}}{2}}}\xi +\dots \right )\right ]\nonumber \\[4pt]&=\frac {g}{{\Lambda}^{\text{d}}(\beta )}\sum\limits ^{\infty }_{\ell =1}{\frac {{\!\left (-\sigma \right )}^{\ell -1}}{{\ell }^{\text{d}/2}}{\xi }^{\ell }}. \end{align}

We note that the convergence of the expressions in (1.196) and (1.197) for P and $n$ requires that the absolute activity $\xi \equiv e^{\beta {\mu }}\lt 1$ , i.e., ${\mu }\lt 0$ . (1.197) can be inverted and solved iteratively for

(1.198) \begin{equation} \xi \!\left (T,n\right )=\frac {n{\Lambda}^{\text{d}}(\beta )}{g}\!\left (1+\frac {\sigma }{2^{\text{d}/2}}\frac {n{\Lambda}^{\text{d}}(\beta )}{g}+\dots \right )\!. \end{equation}

The value of $n{\Lambda}^3$ yields a measure of how quantum mechanical the gas is: $n{\Lambda}^3\ll 1$ is the classical limit.

Using the definition $\xi \equiv e^{\beta {\mu }}$ and (1.198), one obtains

(1.199) \begin{equation} \mu =T{\text{ ln}\!\left (\frac {n{\Lambda}^{\text{d}}(\beta )}{g}\right )}+ \textrm{corrections} \end{equation}

and from (1.196) and (1.198)

(1.200) \begin{equation} P\!\left (n,T\right )=nT\!\left (1+\frac {\sigma }{2^{1+\text{d}/2}}\frac {n{\Lambda}^{\text{d}}}{g}+O{(n{\Lambda}^{\text{d}})}^2\right ) \end{equation}

where 3 is replaced by d.

The influence of $\sigma$ on the pressure is clear: $\sigma =+1$ for fermions has a repulsive effect, while $\sigma =-1$ for bosons has an attractive effect (symmetric wave function).

Example: Bose gas.

Consider a gas of bosons, $\sigma =-1.$ The pressure relation (1.196) becomes

(1.201) \begin{equation} P(T,\xi )=\frac {gT}{{\Lambda}^{\text{d}}(\beta )}\sum\limits ^{\infty }_{\ell =1}{\frac {1}{{\ell }^{1+\text{d}/2}}{\xi }^{\ell }} \end{equation}

and the density relation (1.197) becomes

(1.202) \begin{equation} n\!\left (T,\xi \right )=\frac {g}{{\Lambda}^{\text{d}}(\beta )}\sum\limits ^{\infty }_{\ell =1}{\frac {1}{{\ell }^{{\text{d}}/2}}{\xi }^{\ell }}. \end{equation}

The expression

(1.203) \begin{equation} \frac {{\Lambda}^{\text{d}}(\beta )}{g}n\!\left (T,\xi \right )=\sum\limits ^{\infty }_{\ell =1}{\frac {1}{{\ell }^{{\text{d}}/2}}{\xi }^{\ell }} \end{equation}

is a monotonic increasing function of $\xi$ on [0,1] and takes on larger values for d = 2 than for d = 3 (it diverges at $\xi =1$ for d = 2). The limiting value of $\xi$ is $\xi =1,$ and the right-hand side of (1.203) becomes the Riemann zeta function ${R}(x={d}/{2})$ where $R (x )=\Sigma ^{\infty }_{\ell =1}{{1}/{{\ell }^x}}$ . A few values of ${R}(x\big)$ are given in the following list in order of increasing x: ${R}\big({1}/{2})=\infty, {R} (1)=\infty, {R}({3}/{2})=2.612, {R} (2 )={{\pi}^2}/{6}=1.645, {R} (4 )={{\pi}^4}/{90}=1.082, {R} (10 )=1.001.\ \textrm {Here}\;R$ is a monotonic decreasing function of its argument and asymptotes to unity for large argument. Thus, for d = 3, ${ (n{\Lambda}^3 )}_{\text{max}}=2.612g, g=2{\mathcal S}+1$ . Because $\xi$ is less than one, the value of ${ (n{\Lambda}^3 )}_{\text{max}}$ is actually less than $2.612g$ . Recall the discussion accompanying (1.190) that $\langle$ $N_k\rangle$ is a monotonically decreasing function of energy ${{\mathcal E}}_k$ and that a condensate can occur in the ground state. From (1.193) the partition function satisfies

(1.204) \begin{equation} {\text{ ln } {\textrm {Z}}_0}=-{\text{ ln}(1-\xi )} \quad \textrm {and}\quad \text{ln } \textrm {Z} = \sum\limits _k{\sigma {\text{ ln}(1+\sigma e^{-\beta {{\mathcal E}}^{\prime} }).}} \end{equation}

Hence,

(1.205) \begin{equation} \langle N_0\rangle =-\frac{\partial {\text{ ln } {\textrm {Z}}_0}}{\partial \gamma }=\xi \frac {\partial {\text{ ln } {\textrm{Z}}_0}}{\partial \xi }=\frac {\xi }{1-\xi } \to {\mathop {\lim }_{\xi \to 1} \frac {\xi }{1-\xi }=\frac {1}{1-\xi }} \end{equation}

and this is volume independent. Thus, $\langle N_0\rangle \to \infty$ as $\xi \to 1,$ and $\xi =1-1/\langle N_0\rangle$ . To illustrate the onset of the Bose–Einstein condensation, set $\xi$ to its limiting value $\xi$ =1 in (1.203), set d = 3, use $R\!\left ({3}/{2}\right )=2.612$ , and evaluate $\Lambda$ , $g=1,$ and n in a specific experiment for helium II to obtain

(1.206) \begin{equation} T_0\!\left (n\right )=3.31\frac {{\hslash }^2}{m}n^{2/3}\; \to \; T_0\ \!\left (\textrm {theory}\right )=3.13^{\circ}\;\textrm{K}, \end{equation}

where $2\pi /{R\!\left ({3}/{2}\right )}^{{2}/{3}}=3.3128\dots$ . This compares with an experimental result of 2.19 $^{\circ}$ K.

[Editor’s Note: No reference to a specific experiment was given. The boiling point of He is 4.2 K, and He II becomes a superfluid at approximately 2.17 $^{\circ}$ K at 1 atmosphere pressure.]

There is a problem with applying the grand canonical ensemble to the description of the Bose–Einstein condensate. Consider (1.201) and (1.202) for the pressure P and the density n in the limits g = 1 and d = 3:

(1.207) \begin{equation} P(T,\xi )=\frac {T}{{\Lambda}^3(\beta )}\sum\limits ^{\infty }_{\ell =1}{\frac {1}{{\ell }^{5/2}}{\xi }^{\ell }},\quad n\!\left (T,\xi \right )=\frac {1}{{\Lambda}^3(\beta )}\sum\limits ^{\infty }_{\ell =1}{\frac {1}{{\ell }^{3/2}}{\xi }^{\ell }}, \end{equation}

where $\xi =e^{-\gamma }=e^{\beta \mu }\lt 1$ because $\mu \lt 0$ . In the limit that $\xi \to 0 \xi =n{\Lambda}^3$ is the number of particles in a de Broglie cube. However, from the expression for the number density in (1.207), as $\xi \to 1$

(1.208) \begin{equation} n{\Lambda}^3={\Lambda}^3\frac {\langle N\rangle }{V}=2.612\to \langle N\rangle =2.612\frac {V}{{\Lambda}^3}, \end{equation}

while in contrast (1.205) asserts $\langle N_0\rangle ={1}/({1-\xi })$ which diverges as $\xi \to 1$ and is volume independent, while $\langle N\rangle$ scales with V. Thus, there is a problem here. The difficulty is that in using the grand canonical ensemble, any particular energy state uses all the other states (systems) as the bath at a given temperature. However, when the particular state is the ground state, the bath is not so large in comparison with the number of states occupying the ground state at conditions such that $\xi \to 1$ are approached. The model of the grand canonical ensemble falls apart here for Bose statistics.

A solution for bosons is to use the Gibbs ensemble $Z(\beta ,V_{\textrm{I}},N_{\textrm{I}})$ in which the system (I) is in contact with a heat bath allowing exchange of energy but in which particle exchange is prohibited. In § 9.6 of Reif (Reference Reif1965) the authors present presented an analysis of Bose–Einstein statistics. There is a clever use of a Lagrange multiplier there. They show N ${}_{0}$ to be proportional to V, and the ground state is shown to support large fluctuations. Landau & Lifshitz (Reference Landau and LIfshitz1969) is another good reference on the Bose gas.

Figure 3. Phase diagram for Bose–Einstein condensate, density versus temperature.

A plot of the density versus temperature where the Bose–Einstein condensate onsets based on (1.206) is shown in figure 3. A schematic of the relative fraction in the ground state for the corrected theory is shown in figure 4, ${\langle N_0\rangle }/{N}\;\textrm {versus}\; T$ . Note that in the corrected theory both ${N}_{0}$ and N scale with volume. Regarding the pressure, the ground state has no energy; only the excited states contribute. Equation (1.207) gives the correct pressure. Figure 5 presents a schematic for the pressure versus density for various temperatures. For $n{\Lambda}^3\ll 1$ the system satisfies the ideal gas relation $P\sim nT$ while for $n{\Lambda}^3\gt 2.612$ the system begins to fill the ground state.

Figure 4. Schematic of ${\langle N_0\rangle }/{N}\,\textrm {versus}\ T$ for the bose condensate.

Figure 5. Schematic: P versus n for various temperatures.

Figure 6 presents a schematic of the pressure P versus the volume V = N/n for various isotherms. The critical pressure for a given volume above which volume there is no condensate scales as $P_c\sim V^{3/5}.$

Figure 6. Schematic: P versus V for various temperatures.

Example: Bosons in which the number N is not conserved, e.g., excitations and photons. In such situations N is not conserved, $N_k=0, 1, \dots ,\infty$ and $\mu =0.$ Calculate the properties using the grand canonical ensemble but with $\mu =0$ (should agree with canonical ensemble). For the special case of photons in vacuum with ${\omega }_k=kc,$ ${{\mathcal E}}_k=\hslash {\omega }_k=\hslash kc$ , summing over right-hand and left-hand circularly polarized waves in three dimensions, we can calculate the grand potential $\Omega$ from (1.183) and (1.203):

(1.209) \begin{equation} \Omega =-VT\ {R}\!\left (4\right )\frac {1}{{\pi}^2}\frac {1}{{\lambda }^3_w},\ \ \ \ \ \,{\lambda }_w\equiv \frac {\hslash c}{T}, \end{equation}

where ${R}\!\left (4\right )= {{\pi}^4}/{90}$ .

Exercise: Verify (1.209).

Example: Ideal Fermi gas. Consider an electron gas with $g=2{\mathcal S}+1=2$ and $\sigma =1.$ The pressure from (1.195) is

(1.210) \begin{equation} P(T, \mu )={\textrm {} 2T\int\nolimits {\frac {\textrm{d}^3\textit{k}}{{(2\pi )}^3}}\textrm {ln}\!\left (1+\sigma e^{-\beta ({{\mathcal E}}_k-{\mu })}\right )}. \end{equation}

For ${{\mathcal E}}_k\gt \ \mu :\ e^{-\beta ({{\mathcal E}}_k-\mu )}\ll 1$ in the limit that $T\to 0$ ( $\beta \to \infty$ ), so that $\text{ ln}\!\left (1+0\right )=0$ ; and the electrons are completely degenerate. Note that for ${{\mathcal E}}_k\equiv {p^2}/{2m}={{{\hslash }^2k}^2}/{2m}\lt \mu \ \textrm {:}\ \ e^{\beta \left ({\mu -{\mathcal E}}_k\right )}\gg 1$ and ${\text{ ln } e^{\beta (\mu -{{\mathcal E}}_k)}=\beta \!\left (\mu -{{\mathcal E}}_k\right )}\ \textrm{in }$ (1.210). The pressure receives finite contributions only for ${{\mathcal E}}_k\lt \mu$ when $T\to 0$ :

(1.211) \begin{equation} P(0, \mu )={2\int\nolimits ^{k_{\text{max}}\ ({{\mathcal E}}_k\lt \mu )}_0{\frac {\textrm{d}^3\textit{k}}{{(2\pi )}^3}}(\mu -{{\mathcal E}}_k)}. \end{equation}

The density satisfies

(1.212) \begin{equation} n\!\left (T=0,\mu \right )=\frac {\partial P}{\partial \mu }=2\int\nolimits ^{k_{\text{max}}\ ({{\mathcal E}}_k\lt \mu )}_0{\frac {\textrm{d}^3\textit{k}}{{(2\pi )}^3}}=2\frac {\frac {4\pi }{3}k^3_f}{{(2\pi )}^3}=\frac {8\pi }{3}{\left (\frac {p_F}{h}\right )}^3\equiv \frac {8\pi }{3}{{\Lambda}}^{-3}_F, \end{equation}

where we define the Fermi level

(1.213) \begin{equation} \mu ={{\mathcal E}}_F=\frac {p^2_F}{2m}=\frac {{\hslash }^2k^2_F}{2m}. \end{equation}

From (1.212) and (1.213)

(1.214) \begin{equation} k^3_F=\frac {3}{8\pi }\frac {N_0}{V}{(2\pi )}^3\Longrightarrow \,{\Lambda}_F=\frac {h}{p_F}\quad \textrm {and}\quad n{\Lambda}^3_F=\frac {8\pi }{3}. \end{equation}

Hence, as $T\to 0$

(1.215) \begin{equation} E=\frac {3}{5}{{\mathcal E}}_FN \quad \textrm {and}\quad P=\frac {2}{3}\frac {\frac {3}{5}{{\mathcal E}}_FN}{V}=\frac {2}{5}{n{\mathcal E}}_F \end{equation}

in the nonrelativistic limit. Beware that ${{\mathcal E}}_F$ depends on n via (1.213) and (1.214).

Example: A nonideal gas (‘real gas’) in the classical and quantum limits

Consider a simple one-species gas with partition function in the classical limit. From (1.119)

(1.216) \begin{equation} Z(\beta ,N,V)\equiv \sum\limits _n{e^{-\beta E_n(V,N)}=\frac {1}{N!}{\left (\frac {V}{{\Lambda}^3}\right )}^NQ_N(\beta ,V)}, \end{equation}

where $Q_N(\beta ,V)\equiv \ \smallint {{\left ({\textrm{d}^3{\boldsymbol{r}}}/{V}\right )}^N}e^{-\beta {\Phi \textrm {(}r_i)}}$ is the configurational partition function and $\Phi$ is the interaction potential. The grand partition function is then

(1.217) \begin{equation} \mathbb {Z}\!\left (\gamma ,\beta ,V\right )=\sum\limits _N{e^{-\gamma N}}Z(\beta ,N,V). \end{equation}

Substituting (1.216) into (1.217)

(1.218) \begin{align} \mathbb {Z}\!\left (\gamma ,\beta ,V\right )&=\sum\limits ^{\infty }_{N=0}{\frac {1}{N!}{\left (\frac {Ve^{-\gamma }}{{\Lambda}^3}\right )}^N}Q_N\!\left (\beta ,V\right ) =\sum\limits ^{\infty }_{N=0}{\frac {1}{N!}{\left (Vz\right )}^N}Q_N\!\left (\beta ,V\right )\nonumber\\[4pt] &=1+VzQ_1+\tfrac{1}{2}V^2z^2Q_2+\dots , \end{align}

where $z\equiv {\xi }/{{\Lambda}^3}={e^{-\gamma }}/{{\Lambda}^3}$ is the activity in density units. Note that in the Boltzmann gas limit $\xi \to n{\Lambda}^3.$ In (1.218), $Q_1=1$ (no self-interaction); and

(1.219) \begin{equation} Q_2=\int\nolimits {\frac {\textrm{d}^3{\textit {r}}_1\textrm{d}^3{\textit {r}}_2}{V^2}}e^{-\beta {\Phi }_{12}}=\int\nolimits {\frac {\textrm{d}^3{\textit{r}}_1\textrm{d}^3{\textit {r}}_2}{V^2}}e^{-\beta \phi (r_{12})}. \end{equation}

For large numbers of particles the thermodynamics is independent of the particular ensemble. However, for finite systems, e.g., N = 100, the grand canonical ensemble is invalid; N is not much larger than ln(N). Consider applications of using the grand canonical ensemble (1.217)–(1.219). For $z={e^{-\gamma }}/{{\Lambda}^3}\to n$ as $n\to 0$ with $\gamma ={\partial S}/{\partial N}(E,V,N)$ , we have (1.218) for $\textrm {Z}\!\left (\gamma ,\beta ,V\right )$ with

(1.220) \begin{equation} {Q}_N\!\left (T,V\right )=\int\nolimits {\frac {\textrm{d}^{3N}r_i}{V^N}}e^{-\beta {\Phi \left (r_i\right )}}\gt 0. \end{equation}

Consider (1.219) in more detail:

(1.221) \begin{equation} Q_2=\int\nolimits {\frac {\textrm{d}^3{\textit {r}}_1\textrm{d}^3{\textit {r}}_2}{V^2}}e^{-\beta \phi (r_{12})}=\int\nolimits {\frac {\textrm{d}^3{\textit {r}}_1}{V}}{\left \{\int\nolimits {\frac {\textrm{d}^3\textit {s}}{V}}\left [e^{-\beta \phi (s)}-1\right ]+1\right \}=1-\frac {2b_2(T)}{V}}, \end{equation}

where $b_2\!\left (T\right )\equiv -({1}/{2})\smallint {\textrm{d}^3\textrm {s}}\left [e^{-\beta \phi (s)}-1\right ]$ is defined in (1.91). We note that the term ${2b_2(T)}/{V}=O\!\left ({1}/{V}\right ).$ Then

(1.222) \begin{equation} \textrm {Z}\!\left (\gamma ,\beta ,V\right )=1+Vz+\tfrac{1}{2}V^2z^2Q_2+O\!\left (N^{*3}\right )+\dots +O{\left (N^*\right )}^{N^*}, \end{equation}

where $Vz={O(N}^*)$ , $({1}/{2})V^2z^2Q_2=O(N^{*2})$ , and so on. The terms increase successively until the $N^*$ term, after which the terms in the series fall off, and the series converges. The expansion in (1.222) is not as useless as one might think because the series is monotonic increasing in z, the activity. As a function of z, $\textrm {Z}\!\left (\gamma ,\beta ,V\right )$ increases from unity at z = 0; and ln $\textrm {Z}$ is greater than 0 and increases with z. There are some assumptions to keep in mind. For example, one requires that $\Phi \gt -\infty$ in order that Q ${}_{N }$ not go to $\infty ,$ which excludes point masses and point charges. For the hard core + van der Waals potential diagrammed in figure 1, there is a minimum volume for the hard-sphere particle $({4\pi}/{3}) r^3_0$ and a maximum number of particles in the volume: $N_{\text{max}}\sim {V}/{({4\pi}/{3}) r^3_0}$ For N $\gt$ N ${}_{\text{max}}$ $Q_N\to 0.$ In these circumstances we can terminate the series in (1.222) at N = $N_{\text{max}}$ . The grand partition function is analytic and finite. From (1.194)

(1.223) \begin{equation} P=\frac {1}{\beta V}{\text{ ln } \mathbb {Z}(z,T,V)} \end{equation}

and the physical pressure independent of volume satisfies

(1.224) \begin{equation} P{(z,T)}_{\mathrm{physical}}={\mathop {\lim }_{V\to \infty } \frac {1}{\beta V}{\text{ ln } \textrm {Z}(z,T,V)}}. \end{equation}

Figure 7. Schematic: $\beta P=P/T$ versus n equation of state and phase diagram.

Here $\beta P$ is a nonnegative and a monotonic increasing function of z. There is a possibility of a discontinuity in the slope of $\beta P$ with respect to z. Using (1.197)

(1.225) \begin{equation} n\!\left (z,T, V\right )=\frac {\partial P\!\left (\mu ,T,V\right )}{\partial {\mu }}=z\frac {\partial}{\partial z}\left [\beta P(z,T,V\right ]=z\frac {\partial}{\partial z}\frac {{\text{ ln}\!\left (1+Vz+\frac{1}{2}V^2z^2Q_2\right )}}{V}. \end{equation}

The density n is positive and so is ${\partial n}/{\partial z}\gt 0$ . Where $\beta P$ has a discontinuity in its slope at $z=z_T$ , for $V\to \infty$ , a jump discontinuity can develop in the density n. The equation of state for $P$ as a function of density n can then exhibit a phase transition at $z_T$ . Figure 7 depicts a schematic of an equation of state and a phase diagram. Phase I might represent a gas, while phase II might represent a liquid or a solid. In constructing the equation of state and phase diagram in figure 7, we note that $0\lt \ \langle {(\delta N)}^2\rangle \equiv \langle N^2\rangle -{\langle N\rangle }^2$ and, as a consequence of (1.225),

(1.226) \begin{equation} N-\langle N\rangle =\frac {{\partial }^2{\text{ ln } \textrm {Z}}}{{\partial }^2{\gamma }^2}=Vz\frac {\partial n}{\partial z}=T\langle N\rangle \frac {\partial n}{\partial P}\gt 0\ \ \Longrightarrow \ \ \frac {\partial n}{\partial P}\gt 0\ \textrm {and}\ \frac {\textrm{d}P}{\textrm{d}n}\ge 0 \end{equation}

using ${T}\gt 0$ , $\langle N\rangle \gt 0,$ and ${\partial n}/{\partial z}\gt 0$ . Hence, P either increases with respect to increasing n or has flat intervals.

There was fundamental work on phase transitions in physical systems in the thermodynamic limit based on the properties of small, model systems by Lee and Yang (Lee & Yang Reference Lee and Yang1952; Yang & Lee Reference Yang and Lee1952). The theory revolves around complex zeros of the partition function in finite-sized systems which may permit the possibility of phase transitions.

Definition: Introduce $f_{ij},$ $e^{-\beta \phi _{ij}}=1+f_{ij}$ where $\mathop {\lim }\limits_{r_{ij}\to \infty } f_{ij}\to 0$ . Then

(1.242) \begin{align} \!\left (1+f_{12}\right )\!\left (1+f_{23}\right )\!\left (1+f_{13}\right )&=1+f_{12}+f_{23}+f_{13}+f_{12}f_{23}+f_{12}f_{13}\nonumber\\[4pt] &\quad +{\ f}_{23}f_{13}+ f_{12}f_{23}f_{13}. \end{align}

Now start performing the integrals in (1.241), e.g., $({V}/{V^3})\smallint \textrm{d}^3r_1\textrm{d}^3r_2f_{12}=$ $({V^2}/{V^3})\smallint {\textrm{d}^3r_{21}f_{12}}$ ; all of the terms are like this. Hence,

(1.243) \begin{align} Q_3&=1+\frac {3}{V}\int\nolimits {\textrm{d}^3r_{21}f_{12}}+\frac {3}{V^2}\int\nolimits {\textrm{d}^3r_{21}{\int\nolimits {\textrm{d}^3r_{31}f_{12}}f_{13}}}\nonumber\\[4pt]&+\frac {1}{V^3}\int\nolimits {\textrm{d}^3r_{23}}\int\nolimits {\textrm{d}^3r_{21}{\int\nolimits {\textrm{d}^3r_{31}f_{12}}f_{13}f_{23}}}. \end{align}

The second through fourth terms on the right-hand side of (1.243) correspond pictorially to $-,\wedge, {\Delta }$ , i.e., two-particle interactions among two vertices (particles) or three vertices (particles). Here $Q_4$ has additional terms like $\boxtimes$ and other ways to connect four vertices depicting two-particle interactions among four vertices (particles). From (1.218) we can employ a Taylor-series expansion to calculate

(1.244) \begin{align} \text{ ln } \mathbb {Z}&=Vz+\frac{1}{2}V^2z^2\left [Q_2-1\right ]+\frac {1}{6}V^3z^3\left [Q_3-3Q_2+2\right ]\dots\nonumber\\[4pt] &= Vz+\frac{1}{2}V^2z^2\bigg[\frac {1}{V}(-)\bigg]+\frac {1}{6}V^3z^3\bigg[\frac {1}{V^2}(3\wedge {+\ \Delta })\bigg]+\dots \ \end{align}

in terms of the pictograms (Hill Reference Hill1960). Hence,

(1.245) \begin{equation} \frac {{\text{ ln } \mathbb {Z}}}{V}={z+}\frac{1}{2}z^2(-)+\frac {1}{6}z^3{(3\wedge {+\ \Delta })+}\dots =\beta P. \end{equation}

Thus, the pressure is volume independent. We can rewrite (1.245) in the following form:

(1.246) \begin{align} \beta P\!\left (z,T\right )&=\sum\limits ^{\infty }_{j=1}{z^{\,j}C_j(T)},\nonumber\\[4pt] C_1&=1,\nonumber \\[4pt] C_2&=\tfrac{1}{2}\!\left (-\right ),\nonumber\\[4pt] C_3&=\tfrac{1}{2}\!\left (\Lambda \right )+\frac {1}{6}\!\left (\Delta \right )=\tfrac{1}{2}{(-)}^2+\frac {1}{6}\!\left (\Delta \right ),\nonumber\\[4pt] C_4&=\dots . \end{align}

From (1.225)

(1.247) \begin{equation} n\!\left (z,T\right )=z{\partial }/{\partial z}\!\left (\beta P\right )=\sum\limits ^{\infty }_{j=1}{{jz}^jC_j(T)}. \end{equation}

Hence,

(1.248) \begin{equation} n=z+2z^2C_2+3z^3C_3+\dots \end{equation}

and

(1.249) \begin{equation} \beta P=z+z^2C_2+z^3C_3+\dots \end{equation}

from which it follows that

(1.250) \begin{equation} z=n-2n^2C_2+n^3\!\left (8{C_2}^2-3C_3\right )+\dots . \end{equation}

Substituting z from (1.250) into (1.249) and collecting terms in a power series in n, a virial expansion for $\beta P$ can be obtained:

(1.251) \begin{equation} \beta P\!\left (n,T\right )=\sum\limits ^{\infty }_{j=1}{n^{\,j}b_j(T)}, \end{equation}

where $b_1=1, b_2=-C_2=-({1}/{2})\!\left (-\right ), b_3=4{C_2}^2-2C_3={\!\left (-\right )}^2-{\!\left (-\right )}^2-({1}/{3})\!\left (\Delta \right )=-({1}/{3})\!\left (\Delta \right )$ , and $b_4$ contains irreducible forms involving ( $\boxtimes,$ …). We discussed the evaluation of $b_2$ earlier for the van der Waals model (1.92). For $b_3$ we have

(1.252) \begin{equation} b_3=-\frac {1}{3}\!\left (\Delta \right )=-\frac {1}{3}\int\nolimits {\textrm{d}^3r_{32}}\int\nolimits {\textrm{d}^3r_{21}{\int\nolimits {\textrm{d}^3r_{31}f_{12}}f_{13}f_{23}}}. \end{equation}

Example: Assume a hard-sphere model

(1.253) \begin{align} f_{12} = \left\{ \begin{array}{l@{\quad}l} -1,& r_{12}\le a, \\[4pt]0, & r_{12} \gt a, \end{array} \right. \end{align}

which corresponds to $\phi _{12}=\infty$ for $r_{12}\le a$ and $\phi _{12}=0$ for $r_{12}\gt a$ , where a is the diameter of the hard sphere or disk. We note that this model is highly simplified and artificial. Actual configurations of three hard spheres generally cannot satisfy $r_{ij}\le a$ for all three interacting pairs except for one orientation in which the three hard sphere centers correspond to vertices of an equilateral triangle. Configurations of four hard spheres cannot satisfy $r_{ij}\le a$ for all possible pairs. Nevertheless, use of the hard-sphere model in (1.253) and ignoring the reality that certain interacting pairs can never satisfy $r_{ij}\le a$ allow us to push through the calculations determining the virial coefficients (Frisch Reference Frisch1964).

The evaluation of the virial coefficients in two dimensions leads to

(1.254) \begin{equation} b_1=1,\ \ b_2=-C_2=-\tfrac{1}{2}\!\left (-\right )=\tfrac{1}{2}\pi a^2,\ \ b_3=0.782(b_2)^2, b_4=0.5327(b_2)^3. \end{equation}

Then using (1.251) it follows that

(1.255) \begin{equation} \beta P=n+b_2n^2+0.782{b_2}^2n^3+0.533{b_2}^3n^4+O(n^5). \end{equation}

We note that $({1}/{2})\pi a^2$ is the area of two hard disks, and $n\pi a^2/2$ is the average number of particles within two disks in two dimensions. The maximum two-dimensional density for close-packed hard disks is $n_{\text{max}}=2/a^2$ from which $n\pi a^2/2=\pi n/n_{\text{max}}$ and (1.255) becomes

(1.256) \begin{equation} \frac {P}{nT}=1+\frac {n}{n_{\text{max}}}+{\pi}^20.782{\left (\frac {n}{n_{\text{max}}}\right )}^2+{\pi}^30.533{\left (\frac {n}{n_{\text{max}}}\right )}^3+\dots , \end{equation}

where P/nT is an increasing function of n/n ${}_{\text{max}}$ , and increases faster as n/n ${}_{\text{max}}$ increases in this example; P/nT diverges at a value of n/n ${}_{\text{max}}$ equal to the radius of convergence which is less than or equal to unity. However, the virial coefficients are not necessarily positive in the nonfluid region; and the P/T versus n/n ${}_{\text{max}}$ plot can have a flat region as in figure 7 where there is a phase transition. In the flat region the local number density is the ratio of the sum of the number of particles or molecules in the two phases divided by the sum of the volumes of the two phases. Recall the arguments accompanying (1.224)–(1.226) regarding the positivity of both P/T and ${\textrm{d}P}/{\textrm{d}n}$ for the grand canonical ensemble. By including more physics in $\phi _{12}$ , e.g., attractive forces, P/T versus n can acquire more structure.

Example: Hard-disk interaction in a two-dimensional periodic domain (particles leaving the domain re-enter symmetrically). Alder and collaborators computed the pressure using the virial expansion. In two dimensions and taking the time average to simplify

(1.257) \begin{equation} P=\frac {\langle K\rangle }{V}+\frac {1}{2V}\bigg\langle \sum\limits _{\mathrm{collisions}}{\left ({{\boldsymbol{r}}}_i-{{\boldsymbol{r}}}_j\right )\cdot {{\boldsymbol{f}}}^{\,i}_j}\bigg\rangle . \end{equation}

The collisional interaction can be evaluated as ${{\boldsymbol{f}}}^{\,i}_j=({\textrm{d}}/{\textrm{d}t}){m{{{{\boldsymbol v}}}}_i\vert }_{\text{collision}}={\Delta m{{{{\boldsymbol v}}}}_i}/{\Delta t}$ and

(1.258) \begin{equation} \frac {1}{\Delta t}\sum\limits ^{in\ \Delta t}_{\mathrm{coll}}{r_{ij}\langle }\vert \Delta m{{{{\boldsymbol v}}}}_i\vert \rangle ={\nu }_{\mathrm{coll}}\frac {N}{2}a\langle \vert \Delta m{{{{\boldsymbol v}}}}_i\vert \rangle , \end{equation}

where $r_{ij}=a$ for hard disks and ${\nu }_{\mathrm{coll}}$ is the single-particle collision rate. We define an effective temperature although there is no heat bath, $K\equiv NT$ . Hence,

(1.259) \begin{equation} P=nT+\frac {na}{4}{\nu }_{\mathrm{coll}}\langle \vert \Delta m{{{{\boldsymbol v}}}}_i\vert \rangle \end{equation}

or using ${\nu }_{\mathrm{coll}}\equiv {\sqrt {\langle v^2_i\rangle }}/{\ell }$ where $\ell$ is the collisional mean-free-path

(1.260) \begin{equation} \frac {P}{nT}=1+\frac {a{\nu }_{\mathrm{coll}}}{4}\frac {\langle \vert \Delta m{{{\boldsymbol v}}}_i\vert \rangle }{\langle \frac{1}{2}mv^2_i\rangle }\equiv 1+\frac {a}{\ell }\frac {\langle {{\boldsymbol \vert }\Delta {{{\boldsymbol v}}}}_i\vert \rangle }{2\sqrt {\langle v^2_i\rangle }}. \end{equation}

The right-hand side of (1.260) has no explicit temperature dependence, only geometry and density dependence, at least for hard disks.

Hence, P/T versus n is a universal curve independent of isotherm for hard-sphere interactions.

We note that in a dilute medium the collisional mean-free-path scales as $\ell \sim 1/n\sigma \sim 1/na^2$ where $\sigma$ is the collision cross-section. In dense regimes the collisions are pretty much head on. As $n\to n_{\text{max}}$ , then $\ell \to 0$ in crystalline structures.

In 1957 Alder and Wainwright published the results of numerical Monte Carlo calculations based on a hard-sphere model for the interaction potential leading to equation-of-state results (Alder & Wainwright Reference Alder and Wainwright1957; Wood & Jacobson Reference Wood and Jacobson1957). Figure 8 taken from Wood & Jacobson (Reference Wood and Jacobson1957) shows Alder and Wainwright’s equation of state results for systems with 108 and 32 molecules, in which PV ${}_{0}$ /T is plotted versus the normalized volume V/V ${}_{0}$ , which is proportional to the number density n, and V ${}_{0}$ is the close-packed volume for the system. Alder and Wainwright’s results show an overlapping region wherein two distinct pressure states can coexist for the same volume. The system supports the possibility of a spontaneous transition. In the overlap region, i.e., the two-phase regime, there is a great amount of instability with large fluctuations.

Figure 8. Monte Carlo equation of state results from Wood and Jacobson (1957) showing their results and those of Alder & Wainwright (Reference Alder and Wainwright1957) (solid line for 108 molecules; + for 32 molecules).

If we were to plot P/T versus n for Alder and Wainwright’s results in figure 8, P/T would grow from 0 at n = 0 in the fluid region for increasing n until it reaches a critical density n ${}_{c}$ , above which P/T might continue to increase if the system remains a fluid, but the system may instead jump to a crystalline solid branch at a lower value of P/T whose P/T then increases with n. Note that we earlier proved ${\textrm{d}P}/{\textrm{d}n}\gt 0$ for the grand canonical ensemble, whereas the Alder and Wainwright systems do not satisfy this constraint because their systems are microcanonical ensembles with a relatively small number of particles/molecules rather than an infinite number.

Theories of the liquid state are difficult. Liquids are hard-sphere systems with a weak interaction between particles/molecules considered as a perturbation.

(1.261) \begin{equation} \frac {P}{T}=\frac {\partial S(E,V,N)}{\partial V}=-n^2\frac {\partial {\mathcal S}\!\left ({\mathcal E},n\right )}{\partial n}\to \frac {P}{nT}=1+h(n). \end{equation}

From (1.167) and (1.261) one obtains

(1.262) \begin{equation} {\mathcal S}\!\left ({\mathcal E},n\right )=\frac {5}{2}-{\text{ ln } n{\Lambda}^3\!\left ({\mathcal E}\right )-\int\nolimits ^n_0{\frac {\textrm{d}n^{\prime} }{n^{\prime} }h(n').}} \end{equation}

Why is the entropy contribution negative as n increases? As n increases for N fixed, V decreases and configuration space is limited so $\varGamma$ decreases and the entropy must decrease.

Consider a solid and its own vapor with strong enough forces such that even at $P\to 0$ there is a solid. Assume there is a solid surrounded by a gas. Further assume that the energy of the solid can be represented by

(1.263) \begin{equation} E_{\mathrm{sol}}=-N_{\mathrm{sol}}{{\mathcal E}}_b+f(T)\!\left (\textrm {neglible term}\right )+g(V)\!\left (\textrm {neglible term}\right ), \frac {1}{V}\frac {\textrm{d}V}{\textrm{d}P}P_0\ll 1. \end{equation}

The specific energy of the solid is ${{\mathcal E}}_{\mathrm{sol}}=-{{\mathcal E}}_b$ , the binding energy. Fixed constants are

(1.264) \begin{equation} N=N_{\mathrm{gas}}+N_{\mathrm{sol}} \quad\textrm{and}\quad V=V_{\mathrm{gas}}+V_{\mathrm{sol}}, \;\textrm{where}\; V_{\mathrm{sol}}=N_{\mathrm{sol}}{{\mathcal V}}_0, \end{equation}

where ${{\mathcal V}}_0$ is the volume per solid molecule. At equilibrium

(1.265) \begin{equation} {\mu }_{\mathrm{sol}}={\mu }_{\mathrm{gas}} \to {\mu }_{\mathrm{sol}}=\frac {\partial E_{\mathrm{sol}}}{\partial N_{\mathrm{sol}}}=-{{\mathcal E}}_b={\mu }_{\mathrm{gas}}=T\ \textrm {ln}\ n_{\mathrm{gas}}{\Lambda}^3\!\left (T\right ) \Longrightarrow \ n_{\mathrm{gas}}{\Lambda}^3=e^{-{{\mathcal E}}_b/T}. \end{equation}

Equation (1.265) is a nice formula for the vapor pressure. The classical limit in (1.265) is $n_{\mathrm{gas}}{\Lambda}^3\ll 1$ , which implies that for ${{\mathcal E}}_b\sim 1$ eV, $T\ll$ 1 eV = 12,000 $^{\circ}{} \textrm{K}$ .

Consider increasing N ${}_{\mathrm{sol}}$ and concomitantly $V_{\mathrm{sol}}=N_{\mathrm{sol}}{{\mathcal V}}_0$ , while holding T fixed. The volume $V_{\mathrm{gas}}$ must decrease for the total volume $V$ fixed. In this case, $N_{\mathrm{gas}}{\Lambda}^3=V_{\mathrm{gas}}e^{-{{\mathcal E}}_b/T}$ decreases with decreasing $V_{\mathrm{gas}}$ . Here $n_{\mathrm{gas}}=N_{\mathrm{gas}}/V_{\mathrm{gas}}$ is just a function of temperature and remains fixed. Hence, $P_{\mathrm{gas}}=n_{\mathrm{gas}}T=({T}/{{\Lambda}^3})e^{-{{\mathcal E}}_b/T}$ remains fixed, i.e., the vapor pressure remains constant as we increase N ${}_{\mathrm{sol}}$ .

We next include volume dependence in (1.263) so that $V_{\mathrm{sol}}$ is allowed to vary around its optimum equilibrium value:

(1.266) \begin{equation} E_{\mathrm{sol}}=-N_{\mathrm{sol}}{{\mathcal E}}_b+\alpha {{\mathcal E}}_b\frac {N_{\mathrm{sol}}}{2}\frac {{\!\left ({{\mathcal V}}_s-{{\mathcal V}}_0\right )}^2}{{{\mathcal V}}^2_0}, \end{equation}

where ${N_{\mathrm{sol}}}/{2}$ represents the number of interacting pairs and ${{\mathcal V}}_s=V_{\mathrm{sol}}/N_{\mathrm{sol}}$ . From (1.266) is follows that

(1.267) \begin{equation} P_{\mathrm{sol}}=-{\frac {\partial E_{\mathrm{sol}}}{\partial V_{\mathrm{sol}}}\bigg\vert }_{N_{\mathrm{sol}}}=-\alpha {{\mathcal E}}_b\frac {{\!\left ({{\mathcal V}}_s-{{\mathcal V}}_0\right )}}{{{\mathcal V}}^2_0}, \end{equation}

which has the form of a Hooke’s force law. We can divide through by $N_{\mathrm{sol}}$ in (1.266) to obtain the specific energy per solid molecule. We note that the system has the following attributes:

(1.268) \begin{equation} P_{\mathrm{sol}}=P_{\mathrm{gas}},\quad {\mu }_{\mathrm{sol}}={\mu }_{\mathrm{gas}},\quad V=V_{\mathrm{sol}}+V_{\mathrm{gas}},\quad N=N_{\mathrm{sol}}+N_{\mathrm{gas}},\quad S=S_{\mathrm{sol}}+S_{\mathrm{gas}}.\ \end{equation}

The total entropy is the sum of the gas and solid entropies:

(1.269) \begin{align} S&=S_{\mathrm{sol}}\!\left (E_{\mathrm{sol}},V_{\mathrm{sol}},N_{\mathrm{sol}}\right )+S_{\mathrm{gas}}\!\left (E_{\mathrm{gas}},V_{\mathrm{gas}},N_{\mathrm{gas}}\right )\nonumber\\[4pt]& = N_{\mathrm{sol}}{{\mathcal S}}_{\mathrm{sol}}\!\left ({{\mathcal E}}_{\mathrm{sol}},n_{\mathrm{sol}}\right ) + N_{\mathrm{gas}}{{\mathcal S}}_{\mathrm{gas}}\!\left ({{\mathcal E}}_{\mathrm{gas}},n_{\mathrm{gas}}\right ). \end{align}

Given that the solid and gas are in equilibrium with one another at the same temperature, then $P_{\mathrm{sol}}=P_{\mathrm{gas}}$ and ${\mu }_{\mathrm{sol}}={\mu }_{\mathrm{gas}}$ in (1.268). From ${\mu }_{\mathrm{sol}}={{\partial E_{\mathrm{sol}}}/{\partial N_{\mathrm{sol}}}\vert }_{S_{\mathrm{sol}},V_{\mathrm{sol}}}=-{{\mathcal E}}_b+({1}/{2})\alpha {{\mathcal E}}_b{{\left ({{\mathcal V}}_s-{{\mathcal V}}_0\right )}^2}/{{{\mathcal V}}^2_0}$ , ${\mu }_{\mathrm{gas}}=T\ \textrm {ln}\ n_{\mathrm{gas}}{\Lambda}^3\!\left (T\right )$ , ${\mu }_{\mathrm{sol}}={\mu }_{\mathrm{gas}}$ , $P_{\mathrm{gas}}=n_{\mathrm{gas}}T=({T}/{{\Lambda}^3})e^{-{{\mathcal E}}_b/T}=P_{\mathrm{sol}}=-\alpha {{\mathcal E}}_b{{\left ({{\mathcal V}}_s-{{\mathcal V}}_0\right )}}/{{{\mathcal V}}^2_0}$ , it then follows that

(1.270) \begin{equation} \frac {{\!\left ({{\mathcal V}}_s-{{\mathcal V}}_0\right )}}{{{\mathcal V}}_0}=-\frac {1}{\alpha }\frac {T}{{{\mathcal E}}_b}\frac {v_0}{{\Lambda}^3} e^{-{{\mathcal E}}_b/T}, \end{equation}

where ${1}/{\alpha }\sim \ O\!\left (1\right ),\ \ {T}/{{{\mathcal E}}_b}\lt O\!\left (1\right ),$ ${v_0}/{{\Lambda}^3}$ $\ \sim O\!\left (1\right ),\ \textrm {and }e^{-{{\mathcal E}}_b/T}\ll 1$ .

From (1.270) we conclude that only negligible deviations (compression or expansion) of ${{\mathcal V}}_s$ from ${{\mathcal V}}_0$ are allowed. From (1.268) and (1.270) we can also derive

(1.271) \begin{equation} N_{\mathrm{gas}}\cong \frac {V-N_{\mathrm{sol}}{{\mathcal V}}_0}{{\Lambda}^3}e^{-{{\mathcal E}}_b/T}. \end{equation}

Figure 9 presents of schematic diagram for the P versus V relation for the gas–solid system with N and T fixed. For small values of V greater than the minimum value Nv ${}_{0 }$ the system is a solid and the pressure is largest. As V increases, P decreases while ${N_{\mathrm{sol}}}/{N}\sim 1$ for a while, until ${N_{\mathrm{sol}}}/{N}$ begins to decrease and ${N_{\mathrm{gas}}}/{N}$ increases. Both gas and solid phases occupy the flattish intermediate region of the P versus V relation. At the largest values of V there is only the gas phase. Not shown in figure 9 are trajectories followed from either end of P versus V where we progress along curves in the complete absence of the other phase, namely, beginning at largest V gas only and beginning at smaller V solid phase only. Along these curves we can have metastable states which require either nonuniformities, e.g., for the supersaturated vapor to collect and precipitate upon, or over a long time cavities will appear (when the solid is subjected to too little pressure) in which there is vapor.

Figure 9. Schematic for P versus V phase diagram for the gas–solid system.

Example: For an ideal gas we employ classical counting for the possible states and conclude that $z_{2}=z_{1}^{2}/2$ and $b_{2}=0$ .

Example: For quantum systems we carefully count the possible states. Consider bosons with no internal structure, e.g., the helium atom with two states: $n=(k_1,k_2).$ Then with $E_n={{\mathcal E}}_{k_1}+{{\mathcal E}}_{k_2}$

(1.277) \begin{align} Z_2&=\sum\limits _n{e^{-E_n}=\left (\sum\limits _{k_1,k_2,k_1\lt k_2}{+\sum\limits _{k_1=k_2}{}}\right )e^{-\beta \left ({{\mathcal E}}_{k_1}+{{\mathcal E}}_{k_2}\right )}}\nonumber\\[4pt]&=\left (\tfrac{1}{2}\sum\limits _{k_1,k_2}{+\tfrac{1}{2}\sum\limits _{k_1=k_2}{}}\right )e^{-\beta \left ({{\mathcal E}}_{k_1}+{{\mathcal E}}_{k_2}\right )}\nonumber\\[4pt]&= \tfrac{1}{2}\left (Z^2_1\!\left (\beta \right )+\sum\limits _{k_1}{e^{-2\beta {{\mathcal E}}_k}}\right )=\tfrac{1}{2}\left (Z^2_1\!\left (\beta \right )+Z_1\!\left (2\beta \right )\right ), \end{align}

where, from (1.119) and (1.274)–(1.276)

(1.278) \begin{equation} Z_1\!\left (\beta \right )=V/{\Lambda}^3(\beta ) \quad \textrm{and}\quad b_2\!\left (T\right )=\frac {-V}{2Z^2_1\!\left (\beta \right )}Z_1\!\left (2\beta \right )=-\frac {{\Lambda}^3(\beta )}{4\sqrt {2}}. \end{equation}

Example: For Fermions we exclude the state with $k_1=k_2$ and derive

(1.279) \begin{equation} b_2\!\left (T\right )=\frac {{\Lambda}^3(\beta )}{4\sqrt {2}}, \end{equation}

because Fermions have repulsive interactions.

Example: Consider hard spheres for which $\phi =0$ or $\infty$ (not accessible). Load N hard spheres in a defined volume V for a given temperature T. Randomly displace one sphere in the list. If this results in no overlap with another hard sphere, then this is a successful new configuration and weight it by $e^{-V_N/kT}$ . If, instead, this results in overlapping another hard sphere, then return the sphere to its original position and count the old configuration in the sum over states weighted by $e^{-V_N/kT}$ . Continue through the list. This kind of method works in practice, but in a certain sense it is theoretically incapable of coming up with all configurations for hard spheres. The simulation examples shown in figure 8 illustrate phase transition phenomena for relatively small ensembles of simulation test particles. For insufficient numbers of particles it becomes difficult to distinguish the distinct phases, and the relative size of the statistical fluctuations becomes problematic.

Example: Bosons. Bosons have symmetric wave functions, and the angular momentum quantum number $\ell$ is even. Note that Fermions have antisymmetric wave functions. If we assume that the bosons have repulsive interactions then there are no bound states, and the first sum inside the square bracket in (1.287) vanishes. Furthermore, from quantum mechanics ${\delta }_{\ell }\lt 0$ , which then leads directly to $b^{\text{int}}_2\!\left (T\right )\gt 0$ ; and the pressure has increased due to the interaction.

Example: Bose gas with hard-sphere repulsive interaction. From the quantum mechanical treatment for the phase shift ${\delta }_{\ell }$ in Schiff (Reference Schiff1968, § 19), there is a treatment for a spherically symmetric interaction potential in the special limit of a hard-sphere interaction, Schiff’s (19.20), which in the low-energy limit simplifies to Schiff’s (19.21). As $T\to 0$ only the $\ell =0$ quantum number significantly contributes, in which limit

(1.288) \begin{equation} {\delta }_0\!\left (ka\right )=-ka \quad\mathrm{and}\quad b^{\text{int}}_2\!\left (T\to 0\right )={\Lambda}^2\!\left (\beta ,\frac {m}{2}\right )a, \end{equation}

where $r_{ij}=a$ defines the distance between the hard-sphere centers. The pressure increases because b ${}_{2}$ $\gt$ 0. The pressure at low temperatures for the Bose gas with hard-sphere interaction then follows from (1.200), (1.276), and (1.288), and recalling $b_2=b^{(0)}_2+b^{\text{int}}_2$ :

(1.289) \begin{equation} \frac {P}{nT}=1-\frac {n{\Lambda}^3\!\left (\frac {m}{2}\right )}{16}+n{\Lambda}^2a+O(n^2). \end{equation}

Example: Bose gas with an attractive interaction potential. In this case, bound states are possible and $b^{\text{int}}_2\lt 0$ , so the pressure is reduced. The term involving $\Sigma _{{{{\mathcal E}}}^{\ell }_k\lt 0}{{[e}^{+\beta \vert {{{\mathcal E}}}^{\ell }_k\vert }-1]}$ in (1.287) contributes a net negative contribution to $b^{\text{int}}_2,$ and from quantum mechanics ${\delta }_{\ell }\gt 0$ so the second term in the bracket in (1.287) also contributes a net negative contribution. As $T\to 0\ (\beta \to \infty )$ just a single bound state and only $\ell =0$ are important. Hence, at low temperatures

(1.290) \begin{equation} b^{\text{int}}_2\!\left (T\to 0\right )=-{\Lambda}^3e^{\beta {{\mathcal E}}_b} \;\textrm{and}\; \frac {P}{nT}=1-ne^{\beta {{\mathcal E}}_b}{\Lambda}^3\!\left (\frac {m}{2}\right ), \end{equation}

where ${{\mathcal E}}_b$ is the disassociation energy, i.e., the binding energy of the molecule.

Definition:

(1.298) \begin{equation} {\mu }_A\equiv -T{\gamma }_a,\,{\ \ \mu }_B\equiv -T{\gamma }_b,\quad {\mu }_c\equiv -T({\gamma }_a+{\gamma }_b) \end{equation}

We note that the definitions in (1.298) dictate ${\mu }_c={\mu }_A+{\mu }_B$ . The probability in (1.297) can then be rewritten as

(1.299) \begin{equation} \rho\! \left(N^I_A,N^I_B,N^I_C\right)\sim \ \sum\limits _{E^I}{e^{-{\beta }F^I+\beta {\boldsymbol \mu }\cdot {\boldsymbol{N}}}= \sum\limits _{{E^I}}{e^{-{\beta }(E^I-TS\!\left (E^I,N\right ))+\beta {\boldsymbol \mu }\cdot {\boldsymbol{N}}}}}, \end{equation}

where ${\boldsymbol \mu }\equiv \{{\mu }_A,{\mu }_B,{\mu }_C\}$ and ${\boldsymbol{N}}\equiv \{N_A,N_B,N_C\}$ .

Definition: The degree of ionization is

(1.316) \begin{equation} f\equiv \frac {[H^+]}{\left [H^+\right ]+[H]}=\frac {n_e}{n_e+n_H}\equiv \frac {n_e}{n_0}. \end{equation}

The total density is defined $n=n_H+n_e+n_p=n_0+n_e=n_0(1+f)$ and P = nT. We use these definitions and divide (1.315) by $n_0=n/(1+f)$ to obtain the following.

From Landau & Lifshitz (Reference Landau and LIfshitz1969, equation (106.5)),

(1.317) \begin{equation} \frac {f^2}{1-f^2}=\frac {1}{n{\Lambda}^3_e}e^{-\beta I} \to f^2\!\left (n,T\right )=\frac {1}{1+n{\Lambda}^3_ee^{\beta I}}. \end{equation}

Definition: We define the ionization temperature $T_{\textrm{I}}$ by setting $n{\Lambda}^3_e=e^{-\beta I}\ll 1$ so that $I\gg T_{\textrm{I}}$ and

(1.318) \begin{equation} f(n,T_{\textrm{I}})\equiv \frac {1}{\sqrt {2}}. \end{equation}

From the ionization temperature relation

(1.319) \begin{equation} {\text{ ln } \frac {1}{n{\Lambda}^3_e}=\frac {I}{T_{\textrm{I}}}}\;\textrm{or}\; T_{\textrm{I}}\!\left (n\right )=\frac {1}{{\text{ ln } \frac {1}{n{\Lambda}^3_e(T_{\textrm{I}})}}} \end{equation}

and

(1.320) \begin{equation} {\text{ ln } \frac {1}{n{\Lambda}^3_e}=49.3-2.3\left [{\textrm {log}}_{\textrm {10}}n_{{cc}^{-1}}-\frac {3}{2}{\textrm{log}}_{\textrm {10}}T_{eV}\right ].} \end{equation}

Example: For the interstellar gas, $n\sim 1\ \textrm {cm}^{-3},$ $T_{\textrm{I}}={13.6\,\textrm{eV}}/(49.3+ 3.5\text{ ln} (T_{\textrm{I}}\sim {1}/ {4}))=({13.6}/{47.2})\,\textrm {eV}=0.29\,\textrm {eV}.$

Example: For atmospheric densities, $n\sim {10}^{19}\ \textrm {cm}^{-3}$ , $T_{\textrm{I}}=({13.6}/{6.6})\,\textrm{eV}\sim 2\,\textrm {eV}.$

Note: The recombination temperature for hydrogen is $1000^{\circ}-2000^{\circ\textrm{K}}\sim 0.1-0.2\ \,\textrm {eV}$ , which sets a lower limit for a physically realistic value of $T_{\textrm{I}}.$

Equation (1.317) can be expressed using (1.319)

(1.321) \begin{equation} f\!\left (n,T\right )=\frac {1}{\sqrt {1+{\left (\frac {T_{\textrm{I}}(n)}{T}\right )}^{3/2}e^{T_{\textrm{I}}(n)/T}}}. \end{equation}

The fractional ionization in (1.321) is plotted in figure 10 as a function of ${T/T}_{\textrm{I}}(n)$ ; we note that most of the change in f occurs in the range $1/2\le \,{T/T}_{\textrm{I}}\le 3/2$ .

Figure 10. Fractional ionization versus ${T/T}_{\textrm{I}}(n)$ based on (1.321).