---
title: Thermodynamics of the Bose Gas and Superfluidity
module: Bosonic Systems
moduleNumber: 8
lessonNumber: 6
order: 806
summary: >
  The energy and pressure of the ideal Bose gas follow from the Bose function at
  the order above the density, and below the critical temperature the pressure
  depends on temperature alone because the condensate carries none. The heat
  capacity rises to a cusp at the transition. Real superfluid helium departs from
  the ideal gas because interactions matter: the Landau criterion ties frictionless
  flow to the phonon-roton excitation spectrum, and the two-fluid model carries a
  second sound.
topics: [Bosonic Systems]
sources:
  - book: Pathria & Beale
    ref: "Ch. 7 — Ideal Bose Systems; §7.1–7.2"
  - book: Schroeder
    ref: "Ch. 7 — Quantum Statistics; §7.6 Bose-Einstein Condensation"
  - book: Kardar (Statistical Physics of Particles)
    ref: "Ch. 7 — Quantum Statistical Mechanics; §7.4"
  - book: Tipler & Llewellyn
    ref: "Ch. 8 — Statistical Physics; §8-3 (Liquid Helium)"
draft: false
---

The [condensation](/statistical-mechanics/bose-systems/bose-einstein-condensation-derived)
of an ideal Bose gas fixes the fugacity below $T_c$ at $z = 1$. That single fact
determines the energy, the pressure, and the heat capacity in the condensed
phase, because every thermodynamic function reduces to a Bose function evaluated
at unit fugacity. The ideal-gas results are exact and show the transition as a
cusp in the heat capacity. Real superfluid helium condenses with the same
statistics but differs quantitatively: the interactions between atoms reshape the
excitation spectrum, and it is that spectrum, through the Landau criterion, that
makes the flow frictionless.

## Energy and pressure

The energy integrates $\varepsilon g(\varepsilon)\langle n\rangle$ over the
$\varepsilon^{1/2}$ density of states, one power of $\varepsilon$ higher than the
particle count, giving the next Bose function:

$$
U = \int_0^\infty \frac{\varepsilon\, g(\varepsilon)\,\d\varepsilon}{z^{-1}e^{\beta\varepsilon} - 1}
= \frac{3}{2}\,k_B T\,\frac{V}{\lambda^3}\,g_{5/2}(z),
\qquad
g_{5/2}(z) = \sum_{\ell=1}^\infty \frac{z^\ell}{\ell^{5/2}}.
$$

The grand potential is $\Phi = -PV = -k_B T (V/\lambda^3) g_{5/2}(z)$, so the
pressure is

$$
P = \frac{k_B T}{\lambda^3}\,g_{5/2}(z),
\qquad
PV = \frac{2}{3}U.
$$

The relation $PV = \tfrac{2}{3}U$ is the nonrelativistic result, unchanged by
statistics; the statistics enter only through $g_{5/2}(z)$. The condensate — the
particles in the zero-energy ground state — contributes nothing to either sum: it
carries no energy, no momentum, and no pressure.

## The condensed phase and the isotherm plateau

Below $T_c$ the fugacity is pinned at $z = 1$, so $g_{5/2}(z) \to \zeta(5/2) =
1.341$ and both $U$ and $P$ lose all dependence on the volume:[^zeta]

$$
P = \frac{k_B T}{\lambda^3}\,\zeta\!\left(\tfrac{5}{2}\right) \propto T^{5/2},
\qquad
U = \frac{3}{2}\,\zeta\!\left(\tfrac{5}{2}\right) k_B T\,\frac{V}{\lambda^3} \propto V T^{5/2}.
$$

The pressure of the condensed gas is a function of temperature alone. Compressing
the gas at fixed $T$ below $T_c$ pushes more particles into the condensate without
raising the pressure, exactly as compressing a vapor along a coexistence line
converts it to liquid at fixed pressure. On a $P$–$V$ isotherm the condensed
region is a horizontal plateau, and the locus of transition points across
isotherms is the curve $P_c \propto V_c^{-5/3}$.

$$
% caption: Isotherms of the ideal Bose gas. Compressing at fixed temperature raises the pressure until condensation begins, after which the pressure stays flat because the condensate absorbs added particles without contributing pressure.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.4,0) node[right] {V};
\draw[->, black] (0,0) -- (0,3.6) node[above] {P};
% isotherm: plateau then falling curve
\draw[acc, very thick] (0.5,2.5) -- (2.3,2.5);
\draw[black, very thick, smooth] plot[domain=2.3:6.0, samples=60] (\x,{2.5*pow(2.3/\x,1.4)});
\fill[acc] (2.3,2.5) circle (1.6pt);
\node[text=acc, anchor=south] at (1.2,2.55) {condensed plateau};
\node[text=black, anchor=west] at (3.6,1.5) {normal gas};
% transition locus P ~ V^{-5/3} through the plateau corner
\draw[black, thick, dashed, smooth] plot[domain=1.9:5.4, samples=50] (\x,{9.85*pow(\x,-1.667)});
\node[text=black, anchor=west] at (3.1,2.75) {transition locus};
\end{tikzpicture}
$$

## The heat-capacity cusp

The heat capacity below $T_c$ follows from $U \propto V T^{5/2}$ at fixed $V,N$:

$$
\frac{C_V}{Nk_B} = \frac{15}{4}\,\frac{\zeta(5/2)}{\zeta(3/2)}\left(\frac{T}{T_c}\right)^{3/2},
\qquad T < T_c,
$$

which rises as $T^{3/2}$ and reaches $\tfrac{15}{4}\,\zeta(5/2)/\zeta(3/2) = 1.925$
at $T_c$. Above $T_c$ the fugacity falls below $1$ and

$$
\frac{C_V}{Nk_B} = \frac{15}{4}\,\frac{g_{5/2}(z)}{g_{3/2}(z)}
- \frac{9}{4}\,\frac{g_{3/2}(z)}{g_{1/2}(z)}.
$$

As $T \to T_c^+$ the fugacity approaches $1$, where $g_{1/2}(z)$ diverges, so the
second term vanishes and $C_V/Nk_B \to 1.925$ from above — the same value as from
below. The heat capacity is therefore continuous at $T_c$, but its slope is not:
the derivative $\d C_V/\d T$ jumps, producing a cusp. Far above the transition the
gas becomes classical and $C_V \to \tfrac{3}{2}Nk_B$.

> **Theorem (Bose-gas heat-capacity cusp).** The ideal Bose gas has a continuous
> heat capacity with a peak value $C_V/Nk_B = 1.925$ at $T_c$ and a discontinuous
> temperature derivative there. The transition is continuous (no latent heat), with
> the cusp the sharpest ideal-gas signature of condensation.

$$
% caption: The ideal Bose-gas heat capacity rises as temperature to the three-halves below the transition to a peak, then relaxes toward the classical value; the peak is a cusp, continuous in value but with a broken slope.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.4,0) node[right] {T};
\draw[->, black] (0,0) -- (0,3.4) node[above] {$C_V$};
% classical level 3/2 Nk
\draw[black, dashed] (0,1.7) -- (6.0,1.7);
\node[black, anchor=east] at (-0.05,1.7) {$\tfrac{3}{2}Nk_B$};
% rising T^{3/2} to peak at Tc
\draw[acc, very thick, smooth] plot coordinates {(0,0)(0.7,0.32)(1.4,0.85)(2.1,1.5)(2.7,2.05)(3.1,2.45)};
% falling above Tc toward 3/2
\draw[acc, very thick, smooth] plot coordinates {(3.1,2.45)(3.6,2.15)(4.3,1.92)(5.1,1.8)(6.0,1.73)};
\draw[black, dashed] (3.1,0) -- (3.1,2.47);
\node[black, anchor=north] at (3.1,-0.05) {$T_c$};
\node[text=acc, anchor=south east] at (3.05,2.5) {cusp};
\end{tikzpicture}
$$

Superfluid $^4$He shows a related but sharper feature. Its heat capacity diverges
logarithmically at the lambda point $T_\lambda = 2.17\ \mathrm{K}$, the shape that
named the transition, rather than forming the finite ideal-gas cusp. The
difference is the interactions: liquid helium is dense, its atoms strongly
coupled, and the ideal-gas model captures only that a transition occurs near the
right temperature. The ideal $T_c$ for helium's density is $3.1\ \mathrm{K}$,
close enough to identify the mechanism, wrong in detail.[^he]

## The Landau criterion for superfluidity

Frictionless flow is a statement about excitations. Consider superfluid moving at
velocity $\vec v$ through a stationary tube. Dissipation requires the fluid to
create an excitation, which draws momentum and energy from the flow. Transforming
to the fluid rest frame, an excitation of momentum $\vec p$ and energy
$\varepsilon(p)$ can be created only if it lowers the total energy, which happens
only when the flow speed exceeds

$$
v_c = \min_{p}\frac{\varepsilon(p)}{p}.
$$

Below $v_c$ no excitation can form and the flow persists without loss.

> **Theorem (Landau critical velocity).** A fluid flows without dissipation up to
> the speed
> $$
> v_c = \min_p \frac{\varepsilon(p)}{p},
> $$
> the smallest slope of a line from the origin to the excitation spectrum. If the
> spectrum has $\varepsilon/p$ bounded below by a positive value, superfluidity
> persists up to that critical velocity.

For a free particle $\varepsilon = p^2/2m$, the ratio $\varepsilon/p = p/2m \to 0$
as $p \to 0$, so $v_c = 0$: an ideal Bose gas is not, strictly, a superfluid — any
flow can decay by exciting a slow long-wavelength particle. Superfluidity is a
property of the interacting system, whose spectrum is linear at small $p$ and has
a positive minimum slope.

$$
% caption: The Landau construction. The critical velocity is the slope of the lowest line from the origin that touches the excitation spectrum; a parabolic free-particle spectrum has zero minimum slope, so it gives no superfluidity.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.2,0) node[right] {momentum p};
\draw[->, black] (0,0) -- (0,3.4) node[above] {energy};
% phonon-roton spectrum
\draw[acc, very thick, smooth] plot coordinates {(0,0)(0.6,0.85)(1.2,1.6)(1.9,2.3)(2.6,2.0)(3.2,1.35)(3.7,1.15)(4.3,1.5)(5.0,2.25)(5.6,2.9)};
\node[text=acc, anchor=south west] at (4.6,2.5) {excitation spectrum};
% tangent line from origin to roton minimum
\draw[black, thick] (0,0) -- (5.2,1.9);
\node[text=black, anchor=north] at (4.7,0.6) {critical slope};
\fill[black] (3.55,1.16) circle (1.6pt);
\node[black, anchor=north] at (3.2,0.95) {roton minimum};
\end{tikzpicture}
$$

## The phonon-roton spectrum and two sounds

The measured excitation spectrum of superfluid $^4$He has two regions:

- **Phonons.** At small momentum the excitations are quantized sound waves,
  $\varepsilon = c\,p$ with $c \approx 240\ \mathrm{m/s}$. The slope $\varepsilon/p
  = c$ is finite, the first ingredient of a positive critical velocity.
- **Rotons.** At larger momentum the spectrum dips to a local minimum,
  parametrized by Landau as $\varepsilon(p) = \Delta + (p - p_0)^2/2m^\ast$, with a
  gap $\Delta/k_B \approx 8.6\ \mathrm{K}$ at momentum $p_0/\hbar \approx 19\
  \ \mathrm{nm}^{-1}$. The roton minimum, not the phonon slope, sets the critical
  velocity: $v_c \approx \Delta/p_0 \approx 60\ \mathrm{m/s}$.

The tangent from the origin touches the spectrum at the roton minimum, so the
Landau critical velocity is the roton gap divided by the roton momentum. Measured
critical velocities in wide channels are lower, limited by vortex formation rather
than roton creation, but the roton value is recovered for ions moving fast through
the bulk.[^roton]

$$
% caption: The phonon-roton dispersion of superfluid helium. The spectrum is linear at small momentum (phonons), peaks at the maxon, and dips to the roton minimum before rising again; the roton gap and momentum set the critical velocity.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (6.4,0) node[right] {momentum};
\draw[->, black] (0,0) -- (0,3.4) node[above] {energy};
\draw[acc, very thick, smooth] plot coordinates {(0,0)(0.5,0.75)(1.0,1.45)(1.6,2.35)(2.1,2.7)(2.7,2.35)(3.3,1.6)(3.8,1.25)(4.3,1.35)(4.9,1.95)(5.6,2.9)};
% phonon linear guide
\draw[black, dashed] (0,0) -- (1.55,2.3);
\node[text=acc, anchor=east] at (1.25,0.7) {phonon};
\node[black, anchor=south] at (2.1,2.75) {maxon};
\node[black, anchor=north] at (3.6,1.15) {roton};
\end{tikzpicture}
$$

The two-fluid model that describes helium II — a normal component of density
$\rho_n$ and a superfluid component $\rho_s$ — supports two distinct sound modes:

- **First sound.** An ordinary pressure and density wave, in which the normal and
  superfluid components oscillate together. Its speed is the usual
  $\sqrt{(\partial P/\partial\rho)_S}$.
- **Second sound.** A temperature and entropy wave, in which the two components
  oscillate in antiphase at nearly constant total density, carrying heat as a wave
  rather than by diffusion. It exists only because the superfluid carries no
  entropy, and it has no counterpart in an ordinary fluid.

## Summary

- The ideal Bose gas has $U = \tfrac{3}{2}k_B T (V/\lambda^3) g_{5/2}(z)$ and
  $P = (k_B T/\lambda^3) g_{5/2}(z)$ with $PV = \tfrac{2}{3}U$; the condensate
  contributes nothing to either.
- Below $T_c$ the fugacity is $1$, so $P = \zeta(5/2) k_B T/\lambda^3 \propto
  T^{5/2}$ depends on temperature alone and the $P$–$V$ isotherm is flat.
- The heat capacity rises as $T^{3/2}$ to a peak $C_V/Nk_B = 1.925$ at $T_c$ and
  is continuous with a discontinuous slope — a cusp; real $^4$He diverges
  logarithmically because interactions matter.
- Superfluidity requires an excitation spectrum with positive minimum slope: the
  Landau critical velocity is $v_c = \min_p \varepsilon(p)/p$, set in helium by
  the roton minimum at about $60\ \mathrm{m/s}$; the two-fluid model carries first
  sound (density) and second sound (temperature).

[^he]: Tipler & Llewellyn, §8-3, gives the lambda point and the ideal-gas $T_c
\approx 3.1\ \mathrm{K}$ for helium's density; Pathria & Beale, §7.1, contrasts
the ideal cusp with the observed logarithmic divergence.
[^roton]: Pathria & Beale, §7.2, presents the Landau criterion and the roton
parametrization; the roton gap $\Delta/k_B \approx 8.6\ \mathrm{K}$ and momentum
$p_0/\hbar \approx 19\ \mathrm{nm}^{-1}$ are the neutron-scattering values quoted
there and by Kardar, §7.4.
[^zeta]: The **Riemann zeta function** is $\zeta(s)=\sum_{n=1}^{\infty} n^{-s}=1+2^{-s}+3^{-s}+\cdots$. It enters these gas integrals through the standard Bose result $\int_0^\infty \frac{x^{s-1}}{e^x-1}\,\d x=\Gamma(s)\,\zeta(s)$. The half-integer values $\zeta(3/2)\approx 2.612$ and $\zeta(5/2)\approx 1.341$ used here arise for the nonrelativistic Bose gas.