advanced32 min studyLesson 327 in Physics

Locate the ideal Bose condensation threshold

Separate the ground state from the continuum and state the dimensional and thermodynamic assumptions behind a condensation temperature.

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What you’ll understand

  • Derive saturation of excited states for a uniform three-dimensional ideal Bose gas.
  • Compute the condensate fraction below the thermodynamic critical temperature.
  • Distinguish finite-size occupation, chemical-potential limits, and superfluidity.

Bosons may occupy one state in arbitrarily large numbers, but that fact alone does not guarantee condensation at every density and temperature. For a uniform ideal gas in three dimensions, the continuum of excited states has a finite maximum thermal population at fixed temperature. Additional particles then occupy the lowest state macroscopically. The derivation depends on the low-energy density of states, on separating the discrete ground state, and on taking an appropriate thermodynamic limit.

The chemical potential approaches the lowest energy

Choose the one-particle ground energy as zero and assume one internal state. At positive temperature, each ideal Bose occupation contains energy minus chemical potential in an exponential. The chemical potential must be negative in a finite grand-canonical system for the ground-state geometric sum to converge. Lowering temperature at fixed density raises fugacity toward one. The zero-energy ground occupation can then grow large even though the continuum density of states vanishes at zero energy.

That last observation explains why a naive continuum integral loses the condensate: multiplying by a square-root density of states assigns zero measure to one special discrete level. Keep its occupation separate. The excited-state integral is controlled by a dimensionless Bose function of order three halves. Its largest value occurs as fugacity approaches one from below and is the Riemann zeta value about 2.612. This finite upper bound is the source of the threshold, not an attractive force between ideal particles.

N=N0+VλT3g3/2(z),N0=z1−z,gs(z)=∑ℓ=1∞zℓℓs,z=eβμ<1N=N_0+\frac{V}{\lambda_T^3}g_{3/2}(z),\quad N_0=\frac{z}{1-z},\quad g_s(z)=\sum_{\ell=1}^{\infty}\frac{z^\ell}{\ell^s},\quad z=e^{\beta\mu}<1
N₀ is the mean discrete ground occupation, λT=h/√(2πmkBT) in metres, and gₛ(z) is dimensionless. The formula uses a continuum excited spectrum for a uniform three-dimensional ideal gas with one internal state.↔ Scroll wide equations horizontally.

Excited states saturate

Set the total density equal to the largest excited density to define the thermodynamic critical temperature. At fixed density, lowering temperature below that value makes the maximum excited population fall as temperature to the three-halves power. The remaining fraction occupies the ground state. In the infinite-volume fixed-density limit the chemical potential approaches the ground energy, while the condensate population remains extensive. Keeping chemical potential exactly at zero in a finite grand sum would instead make that sum diverge.

The low-energy convergence depends on dimension. In a uniform ideal gas with quadratic dispersion, the excited-number integral at zero chemical potential converges only above two spatial dimensions. Thus the same finite-temperature homogeneous ideal-gas condensation mechanism fails in one and two dimensions in the thermodynamic limit. Trapping potentials change the density of states and therefore the conclusion. One cannot transfer the three-dimensional box formula to every laboratory trap merely by inserting its particle number.

nλTc3=ζ(3/2),Tc=2πℏ2mkB(nζ(3/2))2/3,N0N=1−(TTc)3/2(T<Tc)n\lambda_{T_c}^3=\zeta(3/2),\qquad T_c=\frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3},\qquad \frac{N_0}{N}=1-\left(\frac{T}{T_c}\right)^{3/2}\quad(T<T_c)
Tc and T are in kelvin, n=N/V in m⁻³, and ζ(3/2)≈2.612. These are leading thermodynamic-limit results for a nonrelativistic uniform three-dimensional noninteracting Bose gas with a single internal state.↔ Scroll wide equations horizontally.

A macroscopic occupation is not the whole fluid story

Finite systems show a smooth crossover rather than the exact thermodynamic nonanalyticity. When the ground occupation is large, its finite-system fugacity is approximately one minus the reciprocal of that occupation, not exactly one. Moreover, the ideal grand-canonical ground-state number variance can be of order the mean squared. Matching a mean condensate fraction therefore does not prove that every ensemble predicts its fluctuations equally; exact number constraints remain significant for that global observable.

Condensation also does not by itself establish superfluid response. An ideal Bose gas's occupation counting omits the interaction-dependent excitation structure relevant to flow stability and collective modes. The lab can test the quadratic relation between a normalized thermal wavelength and its square, then infer the temperature scaling on paper. It does not calculate a Bose function, detect a condensate, or demonstrate superfluidity, and its clock has no meaning as a cooling protocol.

z=N0N0+1,μ=kBTln⁡z<0,Var⁡grand(N0)=N0(1+N0)z=\frac{N_0}{N_0+1},\qquad \mu=k_BT\ln z<0,\qquad \operatorname{Var}_{\mathrm{grand}}(N_0)=N_0(1+N_0)
These finite grand-canonical one-ground-state relations use a zero ground energy and ideal Bose counting. They explain why finite μ and condensate fluctuations require care even when the leading thermodynamic mean is accurate.↔ Scroll wide equations horizontally.

Count particles below the threshold

A uniform ideal three-dimensional gas has N=27000 and is held at T=4Tc/9. Within the thermodynamic condensate-fraction approximation, find the excited and ground populations. Estimate how close finite-system fugacity is to one if the ground mean has that value.

  1. The excited fraction is (4/9)^(3/2). Taking the square root gives 2/3 and cubing gives 8/27.

  2. The excited population is 27000×8/27=8000. Subtracting from the total leaves N₀=19000 in the ground state, a macroscopic fraction 19/27.

  3. For a finite grand-canonical ground mean of 19000, z=19000/19001. Thus 1−z=1/19001≈5.26×10⁻⁵. It is close to one but remains strictly below it.

The result

The leading populations are 8000 excited and 19000 condensed. A corresponding finite ground-state fugacity differs from one by about 5.26×10⁻⁵; the exact finite-size crossover needs a discrete spectrum.

Explore the model

Interactive lab

Interactive lab · function

Track the thermal-wavelength scaling

For fresh runs Reset and set BOTH coefficient a=1, position x=1; then Reset and set a=1, x=2. Interpret x as λT divided by a reference wavelength and y as its square. Predict y=1 and 4; infer that T falls by four and the saturated excited density by eight. The parabola is not a Bose-gas solver.

Track the thermal-wavelength scalingThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThermal wavelength scaling helps explain the T^(3/2) excited population, while condensation needs the state-counting argument.

Check your understanding

Why must the ground state be kept separately from the three-dimensional continuum number integral below Tc?

Try it yourself

At fixed particle mass and internal state, a uniform ideal Bose gas density increases by a factor of eight. Find the factor change in Tc. At a fixed temperature equal to the old Tc, what is the new condensate fraction in the thermodynamic approximation?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Tc scales as density to the power two thirds; then use the new temperature ratio.

Compare with the solution

The new Tc is 8^(2/3)=4 times the old one. The unchanged temperature is therefore one quarter of the new Tc, giving condensate fraction 1−(1/4)^(3/2)=1−1/8=7/8. The result assumes the uniform three-dimensional ideal-gas thermodynamic limit and does not describe an arbitrary trap.

Take it with you

  • Uniform three-dimensional excited states have finite thermal capacity at μ approaching the ground energy.
  • The condensate is a separate discrete level with macroscopic occupation.
  • Finite-size fluctuations and superfluid response require more than the leading ideal condensate fraction.

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