Critical Temperature and Condensate Fraction

In 1924 Einstein predicted that a gas of ordinary atoms, cooled enough, would do something no classical gas ever does: a finite fraction of all the atoms would crowd into the single lowest quantum state and stay there, a macroscopic clump of matter occupying one wavefunction. For seventy years it was a paper curiosity. Then in June 1995, a cloud of two thousand rubidium atoms in Boulder crossed a temperature of about a hundred billionths of a kelvin and a sharp spike rose out of the thermal haze — the condensate, made visible. This article answers the two quantitative questions that turn the prediction into a number you can measure: at what temperature does condensation begin, and how much of the gas is condensed once it has. The companion article on Bose–Einstein condensation in the ideal gas (aq-sm-bose-einstein-condensation-ideal-gas) establishes the mechanism — that the excited states have a finite carrying capacity, and the surplus has nowhere to go but the ground state. Here we take that mechanism as given and extract the critical temperature $k_BT_c=\dfrac{2\pi\hbar^2}{m}\left(\dfrac{n}{\zeta(3/2)}\right)^{2/3}$ and the condensate fraction $N_0/N=1-(T/T_c)^{3/2}$ . Both results are pure quantum statistics — no interactions, no forces, only the indistinguishability of bosons and a density of states that vanishes at zero energy. The onset has a beautifully simple reading: condensation begins exactly when the thermal de Broglie wavelength grows to the interparticle spacing, so that neighbouring wavepackets overlap and the atoms can no longer be told apart. We will also see how trapping geometry rewrites the exponent, why lighter atoms condense more easily, and why the ideal-gas formula lands only a factor of order one from the superfluid transition of liquid helium.

Key takeaways

The critical temperature follows from setting the fully-degenerate excited-state count ( $\mu=0$ ) equal to $N$ : $n\lambda_{T_c}^3=\zeta(3/2)\approx 2.612$ , i.e. condensation begins when the thermal wavelength reaches the interparticle spacing.
Explicitly $k_BT_c=\dfrac{2\pi\hbar^2}{m}\left(\dfrac{n}{\zeta(3/2)}\right)^{2/3}\approx 3.31\,\dfrac{\hbar^2 n^{2/3}}{m}$ , scaling as $n^{2/3}/m$ — light atoms and high densities condense first.
Below $T_c$ the condensate fraction obeys the universal law $N_0/N=1-(T/T_c)^{3/2}$ , independent of every gas parameter — at $T=0.5\,T_c$ already 65 % is condensed.
Trapping changes the exponent: a 3D harmonic trap gives $g(\varepsilon)\propto\varepsilon^2$ , so $N_0/N=1-(T/T_c)^3$ and $k_BT_c=\hbar\bar\omega[N/\zeta(3)]^{1/3}$ — the formula behind the 1995 experiments.
Liquid ⁴He plugged into the ideal formula gives $T_c\approx 3.1$ K, tantalisingly near the observed λ-point of 2.17 K — but interactions cut the $T=0$ condensate fraction to only ≈ 8 %.
Key idea: $T_c$ is fixed by $n\lambda_{T_c}^3=\zeta(3/2)$ and scales as $n^{2/3}/m$ ; below it the ground-state occupation grows as $N_0/N=1-(T/T_c)^{3/2}$ , an order parameter set by pure Bose statistics.

The core insight

The critical temperature is fixed by $n\lambda_{T_c}^3=\zeta(3/2)\approx 2.612$ , giving $k_BT_c=\tfrac{2\pi\hbar^2}{m}(n/\zeta(3/2))^{2/3}\propto n^{2/3}/m$ ; below it the condensate fraction grows as $N_0/N=1-(T/T_c)^{3/2}$ , a universal order parameter with no free constants.

Where this lives today

The condensate fraction is now a routine laboratory observable. Releasing a trapped cloud and imaging its ballistic expansion (time-of-flight) produces a bimodal profile: a narrow, dense central peak — the condensate — sitting on a broad thermal Gaussian. The area ratio of the two components reads off $N_0/N$ directly, and watching the peak sharpen as temperature drops traces out the $1-(T/T_c)^\alpha$ curve, cubic for the harmonic traps used in practice. Beyond dilute atoms, the same critical-temperature reasoning governs exciton-polariton condensates in semiconductor microcavities (condensing near room temperature because the effective mass is tiny), magnon condensates in yttrium iron garnet, and — through the ideal-gas estimate's near-miss for helium — the entire modern understanding of superfluidity as a condensate stiffened by interactions.

The physical picture

Start with the mechanism established next door in aq-sm-bose-einstein-condensation-ideal-gas: the excited states of an ideal Bose gas can hold only a finite number of particles once the chemical potential has climbed to the ground energy. The question here is purely quantitative — when is that ceiling reached, and how fast does the ground state fill afterward. The cleanest way to see the onset is to picture each atom not as a point but as a wavepacket of size $\lambda_T=h/\sqrt{2\pi m k_BT}$ , the thermal de Broglie wavelength. At high temperature the wavepackets are tiny compared to the spacing $n^{-1/3}$ between atoms, and the gas behaves classically. As you cool, the wavepackets swell. Condensation begins at the moment they start to overlap — when $\lambda_T$ reaches $n^{-1/3}$ , up to the numerical factor $\zeta(3/2)^{1/3}$ . From that moment on, quantum indistinguishability governs the gas.

Figure

The wavepackets swell on cooling until they overlap — the onset of condensation when nλ³ reaches ζ(3/2)

Once past $T_c$ , the split is clean. The chemical potential locks at the ground energy and cannot rise further, so the number of atoms the thermal cloud can hold is set by temperature alone — and it falls as $T^{3/2}$ , because the excited-state count is $\propto\lambda_T^{-3}\propto T^{3/2}$ . Every atom that the shrinking cloud can no longer carry drops into the condensate. The result is the order parameter $N_0/N=1-(T/T_c)^{3/2}$ , an S-free curve that leaves zero at $T_c$ with infinite slope in the derivative and climbs to one at absolute zero.

Figure

The order parameter: condensate fraction N₀/N = 1 − (T/T_c)³ᐟ² rises steeply below T_c to reach 1 at T = 0

The core equations

The physics rests on one occupation rule, one number equation with the condensate reinstated by hand, and the two results — $T_c$ and the fraction — that follow. We recall the occupation from the quantum-statistics foundation and the mechanism from aq-sm-bose-einstein-condensation-ideal-gas; the derivations here own the two quantitative results. The mean occupation of a single-particle level in the grand-canonical ensemble is the Bose–Einstein form, and the chemical potential can never exceed the ground energy — for otherwise the ground occupation would go negative.

\langle n_\varepsilon\rangle=\frac{1}{e^{\beta(\varepsilon-\mu)}-1},\qquad \mu\le\varepsilon_0 \qquad (1)

where: $\langle n_\varepsilon\rangle$ — mean number of bosons in a level of energy $\varepsilon$ ; $\beta=1/k_BT$ ; $\mu$ — chemical potential, capped at the ground energy $\varepsilon_0$ (taken as 0). The cap is the entire origin of condensation: $\mu$ presses up against $\varepsilon_0$ and can rise no further, so the excited states run out of room. To count particles we must add the condensate explicitly, because the smooth density of states $g(\varepsilon)\propto\varepsilon^{1/2}$ assigns zero weight to $\varepsilon=0$ and silently loses the ground state. This is the single most important bookkeeping point in the subject.

N=N_0+\underbrace{\frac{V}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\int_0^\infty\frac{\varepsilon^{1/2}\,\mathrm{d}\varepsilon}{e^{\beta(\varepsilon-\mu)}-1}}_{N_{\rm exc}} \qquad (2)

where: $N_0$ — ground-state (condensate) occupation, reinstated by hand; $N_{\rm exc}$ — the number the excited states can hold; $V$ — volume; $m$ — atomic mass. Read it as a bucket ( $N_{\rm exc}$ ) plus an overflow reservoir ( $N_0$ ): once the bucket is full, everything else pours into the reservoir. Setting $\mu=0$ (the fully degenerate excited states) makes $N_{\rm exc}$ a function of $T$ alone. Equating it to $N$ defines the critical temperature — the point at which the bucket is exactly full and the condensate is about to appear.

k_BT_c=\frac{2\pi\hbar^2}{m}\left(\frac{n}{\zeta(3/2)}\right)^{2/3},\qquad n\lambda_{T_c}^3=\zeta(3/2)\approx 2.612 \qquad (3)

where: $T_c$ — critical temperature; $n=N/V$ — number density; $\lambda_T=h/\sqrt{2\pi m k_BT}$ — thermal de Broglie wavelength; $\zeta(3/2)\approx 2.612$ — the Riemann zeta value that emerges from the Bose integral. Numerically the prefactor $2\pi/\zeta(3/2)^{2/3}\approx 3.31$ , so $k_BT_c\approx 3.31\,\hbar^2 n^{2/3}/m$ . The two forms are the same statement — one names the temperature, the other names the degeneracy threshold. Below $T_c$ the chemical potential is pinned at zero, so the excited-state population is $\propto\lambda_T^{-3}\propto T^{3/2}$ and the remainder condenses. This gives the universal order parameter with no adjustable constants.

\frac{N_0}{N}=1-\left(\frac{T}{T_c}\right)^{3/2}\qquad (T<T_c) \qquad (4)

where: $N_0/N$ — condensed fraction, the BEC order parameter; the exponent $3/2$ is the fingerprint of a 3D uniform gas (it becomes 3 in a harmonic trap). At $T_c$ the fraction is 0 (continuous onset); at $T=0$ it is 1.

The physical picture, made quantitative

Derivation AThe critical temperature from chemical-potential saturation

We want the temperature at which the excited states, working at their absolute maximum capacity ( $\mu=0$ ), can exactly hold every particle. Above this temperature they hold all of $N$ with room to spare and $\mu<0$ ; below it they cannot, and the surplus condenses. The strategy is to evaluate the excited-state integral at $\mu=0$ , recognise it as a standard Bose integral proportional to $\lambda_T^{-3}$ , set it equal to $N$ , and invert for $T$ . The one place to be careful is the Bose integral's zeta value, which sets the numerical prefactor 3.31; a dropped $\Gamma(3/2)=\sqrt\pi/2$ is the usual slip.

Step 1. Write the excited-state count with $\mu=0$ , its maximum: $N_{\rm exc}^{\max}=\dfrac{V}{4\pi^2}\left(\dfrac{2m}{\hbar^2}\right)^{3/2}\displaystyle\int_0^\infty\dfrac{\varepsilon^{1/2}\,\mathrm{d}\varepsilon}{e^{\beta\varepsilon}-1}$ . Why: the integrand grows monotonically with $\mu$ , so the largest number the excited states can carry at a given $T$ is reached when $\mu$ hits the ground energy.

Step 2. Substitute $x=\beta\varepsilon$ and evaluate the standard Bose integral: $\displaystyle\int_0^\infty\frac{x^{1/2}\,\mathrm{d}x}{e^{x}-1}=\Gamma(\tfrac32)\,\zeta(\tfrac32)=\frac{\sqrt\pi}{2}\,\zeta(\tfrac32)$ . Why: $\int_0^\infty x^{s-1}/(e^x-1)\,\mathrm{d}x=\Gamma(s)\zeta(s)$ from expanding $1/(e^x-1)=\sum_{k\ge1}e^{-kx}$ and integrating term by term; with $s=3/2$ this is $\Gamma(3/2)\zeta(3/2)$ .

Step 3. Collect the constants. The prefactor and the integral combine to $N_{\rm exc}^{\max}=\dfrac{V}{\lambda_T^3}\,\zeta(\tfrac32)$ with $\lambda_T=h/\sqrt{2\pi m k_BT}$ . Why: the factors of $2m/\hbar^2$ , $\sqrt\pi/2$ and $(k_BT)^{3/2}$ assemble exactly into the inverse cube of the thermal de Broglie wavelength — the natural quantum volume per atom.

Step 4. Set $N_{\rm exc}^{\max}=N$ and invert. $n\lambda_{T_c}^3=\zeta(3/2)$ gives, using $\lambda_{T_c}\propto T_c^{-1/2}$ ,

k_BT_c=\frac{2\pi\hbar^2}{m}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}.

Why: solving $\lambda_{T_c}^{-3}=n/\zeta(3/2)$ for $T_c$ turns the degeneracy threshold into a temperature; the $n^{2/3}$ and $1/m$ dependences are read straight off. *Verified: reproduces Pathria & Beale eq. (7.1.20); the numerical prefactor $2\pi/\zeta(3/2)^{2/3}=3.31$ and the scalings $T_c\propto n^{2/3}$ , $T_c\propto 1/m$ checked in verify_bec-critical-temperature-condensate-fraction.py.*

Go deeper · Derivation B · 1 minThe condensate fraction below the transition

Derivation BThe condensate fraction below the transition

Once $T<T_c$ the chemical potential is stuck at $\mu=0$ to macroscopic accuracy — any tiny negative $\mu$ that a finite $N_0$ would require is of order $1/N$ and vanishes in the thermodynamic limit. The strategy is therefore to compute the thermal population at $\mu=0$ for any $T<T_c$ , notice that it is exactly the $T_c$ population rescaled by $(T/T_c)^{3/2}$ , and let particle conservation deliver the condensate. There are no free constants — the gas parameters $m$ , $n$ , $V$ all cancel into $T_c$ , which is the whole content of "universal law".

Step 1. For $T<T_c$ the excited-state count is $N_{\rm exc}(T)=\dfrac{V}{\lambda_T^3}\,\zeta(\tfrac32)$ , the same Bose integral as at $T_c$ but now valid for every $T$ below it. Why: with $\mu$ pinned at 0 the excited-state occupation depends only on $T$ , so the count is fixed by temperature alone.

Step 2. Divide by $N=V\zeta(3/2)/\lambda_{T_c}^3$ : $\dfrac{N_{\rm exc}(T)}{N}=\dfrac{\lambda_{T_c}^3}{\lambda_T^3}=\left(\dfrac{T}{T_c}\right)^{3/2}$ . Why: $\lambda_T\propto T^{-1/2}$ makes $\lambda_T^{-3}\propto T^{3/2}$ , so the ratio is a pure power of $T/T_c$ — every material constant has cancelled.

Step 3. Impose conservation $N=N_0+N_{\rm exc}$ :

\frac{N_0}{N}=1-\left(\frac{T}{T_c}\right)^{3/2}.

Why: the thermal cloud shrinks as $T^{3/2}$ ; whatever it can no longer hold is, by conservation, in the condensate. The exponent $3/2$ is inherited directly from the 3D density-of-states power $\varepsilon^{1/2}$ . *Verified: $N_0/N=0$ at $T=T_c$ (continuous onset) and $\to 1$ as $T\to 0$ ; the value $1-0.5^{3/2}=0.646$ at $T=0.5\,T_c$ checked in verify_bec-critical-temperature-condensate-fraction.py.*

Limiting cases

Everything — whether condensation happens at all, and with what exponent — is controlled by the power in the density of states $g(\varepsilon)\propto\varepsilon^{\alpha-1}$ . The 3D box has $\alpha=3/2$ ; a harmonic trap effectively has $\alpha=3$ ; low dimensions push $T_c$ to zero.

Case	Result	Meaning
3D box ( $\alpha=3/2$ )	$N_0/N=1-(T/T_c)^{3/2}$ , $k_BT_c=\tfrac{2\pi\hbar^2}{m}(n/\zeta(3/2))^{2/3}$	The standard uniform ideal gas; onset at $n\lambda_T^3=\zeta(3/2)$ .
3D harmonic trap ( $\alpha=3$ )	$N_0/N=1-(T/T_c)^3$ , $k_BT_c=\hbar\bar\omega\,[N/\zeta(3)]^{1/3}$	Trapping steepens $g(\varepsilon)\propto\varepsilon^2$ ; the exponent becomes 3 and $\zeta(3)\approx 1.202$ enters. This is the experimental formula.
Uniform 2D box	$T_c\to 0$	$g(\varepsilon)$ is constant; the excited-state integral diverges logarithmically as $\mu\to 0$ , so the excited states absorb any $N$ — no finite- $T$ BEC.
Uniform 1D box	$T_c\to 0$	$g(\varepsilon)\propto\varepsilon^{-1/2}$ diverges at the origin; again no surplus is forced out.
$T\to 0$	$N_0/N\to 1$ , $\mu\to\varepsilon_0$	Every atom in the ground state; the thermal cloud vanishes as $T^{3/2}$ .

Worked examples

1 · Rubidium-87 in a trap — the JILA number (confidence builder)

Step 1. Take the historic JILA parameters: $N\approx 2\times10^4$ atoms of ⁸⁷Rb ( $m=87\times1.6605\times10^{-27}=1.445\times10^{-25}$ kg) in a harmonic trap of geometric-mean frequency $\bar\omega/2\pi\approx 100$ Hz, so $\bar\omega=6.28\times10^2$ rad/s. Use the trap formula $k_BT_c=\hbar\bar\omega[N/\zeta(3)]^{1/3}$ with $\zeta(3)=1.202$ .

Step 2. Evaluate the bracket: $N/\zeta(3)=2\times10^4/1.202=1.66\times10^4$ , and $(1.66\times10^4)^{1/3}=25.6$ .

Step 3. Assemble: $k_BT_c=(1.0546\times10^{-34})(6.28\times10^2)(25.6)=1.70\times10^{-30}$ J, so $T_c=1.70\times10^{-30}/1.381\times10^{-23}=1.23\times10^{-7}$ K $\approx$ 123 nK.

Step 4. Sanity: this is in the 100–200 nK band the 1995 JILA experiment reported — the trap, not the uniform box, sets the observed scale. The uniform-box formula (3) at a comparable peak density $n\sim10^{13}\,\mathrm{cm}^{-3}$ gives $T_c\approx 86$ nK, the same order of magnitude, confirming that either route lands near a hundred nanokelvin. The condensate appears at about 100 nK for a few thousand rubidium atoms.

2 · Liquid helium-4 versus the observed λ-point (exam level)

Step 1. Model liquid ⁴He as an ideal Bose gas at its actual density. Mass $m=6.6465\times10^{-27}$ kg, mass density $\rho=145$ kg/m³, so $n=\rho/m=145/6.6465\times10^{-27}=2.18\times10^{28}$ m⁻³.

Step 2. Compute $n^{2/3}=(2.18\times10^{28})^{2/3}=7.86\times10^{18}$ m⁻². Then $k_BT_c=3.31\,\hbar^2 n^{2/3}/m=3.31\,(1.0546\times10^{-34})^2(7.86\times10^{18})/(6.6465\times10^{-27})$ .

Step 3. Numerator: $3.31\times(1.112\times10^{-68})\times7.86\times10^{18}=2.89\times10^{-49}$ . Divide by $m$ : $2.89\times10^{-49}/6.6465\times10^{-27}=4.35\times10^{-23}$ J. Then $T_c=4.35\times10^{-23}/1.381\times10^{-23}=$ 3.15 K.

Step 4. Compare to experiment: the observed superfluid λ-transition is at 2.17 K. The ideal-gas estimate lands within a factor 1.5 — close enough that Fritz London (1938) argued BEC underlies superfluidity, yet off enough to show interactions matter. The $T=0$ condensate fraction confirms the caveat: measured ≈ 8 %, not the ideal 100 %. BEC sets the scale of the λ-point but does not quantitatively describe the strongly interacting liquid.

Origins · 1 minFritz London and how this was discovered

Fritz London

1900–1954 · Berlin, Oxford, Paris, Duke

London made the leap nobody else dared: in 1938, when Einstein's condensation was widely dismissed as an artifact of the non-interacting model, he proposed that the newly discovered superfluidity of liquid helium-4 was Bose–Einstein condensation in disguise — pointing to the striking closeness of the ideal-gas $T_c\approx 3.1$ K to the observed λ-point at 2.17 K. He had earlier co-founded the quantum theory of the chemical bond (the London dispersion force bears his name) and, with his brother Heinz, written the phenomenological electrodynamics of superconductors (the London equations). Forced from Germany in 1933, he carried the condensation idea through Oxford and Paris to Duke University, where he built the conceptual bridge between an abstract 1924 prediction and a liquid you could pour.

History

1924

— Bose derives Planck's law by counting indistinguishable photon states, μ = 0

1924

— Einstein extends the counting to a gas of massive atoms with conserved number, μ < 0

1925

— Einstein predicts that below a critical temperature a macroscopic fraction condenses into the lowest state

1938

— London proposes that superfluid helium-4's λ-transition is Bose–Einstein condensation

1947

— Bogoliubov gives the first microscopic theory of the weakly interacting Bose gas

1995

— Cornell and Wieman condense rubidium-87 at JILA; Ketterle condenses sodium-23 at MIT

2001

— Cornell, Wieman, and Ketterle share the Nobel Prize for the first atomic condensates

Going further

Texts. Pathria & Beale, Statistical Mechanics, 3rd ed., ch. 7 (Ideal Bose Systems), eq. (7.1.20) · Kardar, Statistical Physics of Particles, ch. 7 · Reif, Fundamentals of Statistical and Thermal Physics, ch. 9 · Sethna, Statistical Mechanics, ch. 7 · Pethick & Smith, Bose–Einstein Condensation in Dilute Gases, ch. 2 · Landau & Lifshitz, Statistical Physics Part 1, §62.

Provenance. Content upgraded from the AllQFT base article (Pathria & Beale, Kardar, Reif). The critical-temperature formula $k_BT_c=2\pi\hbar^2(n/\zeta(3/2))^{2/3}/m$ , the numerical prefactor $2\pi/\zeta(3/2)^{2/3}=3.31$ , the condensate-fraction law $1-(T/T_c)^{3/2}$ and its trap analogue $1-(T/T_c)^3$ , the ⁸⁷Rb trap $T_c\approx 123$ nK and ⁴He $T_c\approx 3.1$ K numbers, and the $C_V$ peak $\tfrac{15}{4}\zeta(5/2)/\zeta(3/2)\approx 1.925$ (owned by aq-sm-thermodynamics-bec-transition) were python-verified in verify_bec-critical-temperature-condensate-fraction.py.

Prerequisites: Quantum Statistics of Ideal Gases: Occupation Numbers · Density of States for Free Particles · Bose-Einstein Condensation in the Ideal Gas.

Leads to: Thermodynamics of the BEC Transition · Photon Gas and Planck Radiation Law.

Why does the condensate fraction fall exactly as (T/T_c)^{3/2} and not some other power?

The exponent is inherited from the density of states. Below $T_c$ the thermal cloud holds $N_{\rm exc}\propto\lambda_T^{-3}\propto T^{3/2}$ particles, because the 3D free-particle density of states goes as $\varepsilon^{1/2}$ and the Bose integral converts that into a $T^{3/2}$ dependence. Dividing by the same expression at $T_c$ gives $(T/T_c)^{3/2}$ , and conservation turns it into $1-(T/T_c)^{3/2}$ . Change the density of states — for instance by trapping, which makes it $\varepsilon^2$ — and the exponent changes to 3. The power is a direct readout of the single-particle spectrum.

Is T_c a real thermodynamic phase transition or just a crossover?

It is a genuine phase transition in the thermodynamic limit. The condensate fraction is a bona fide order parameter that is exactly zero above $T_c$ and non-zero below, switching on continuously — the hallmark of a second-order transition. The heat capacity is continuous but its temperature derivative is discontinuous, producing the cusp at $T_c$ (owned in detail by aq-sm-thermodynamics-bec-transition). In a finite trapped cloud the transition is technically a smooth crossover smeared over a width $\sim N^{-1/3}$ , but for $10^4$ – $10^7$ atoms this smearing is tiny and the transition is sharp for all practical purposes.

Why do lighter atoms condense at higher temperatures?

Because $T_c\propto 1/m$ . The onset condition is $n\lambda_T^3=\zeta(3/2)$ , and the thermal de Broglie wavelength $\lambda_T=h/\sqrt{2\pi m k_BT}$ is larger for lighter atoms at a given temperature. A lighter atom therefore reaches the wavepacket-overlap condition — $\lambda_T\sim n^{-1/3}$ — at a higher temperature. This is why hydrogen and helium are the natural candidates, and why, at fixed density, ⁴He has a higher ideal $T_c$ than a heavy alkali. In traps the interplay is subtler because $T_c$ also depends on $\bar\omega$ , but the mass sensitivity is the same physics.

The formula predicts T_c ≈ 3.1 K for helium but the transition is at 2.17 K. Is the theory wrong?

Not wrong — incomplete. The ideal-gas formula ignores interactions entirely, and liquid helium is a dense, strongly interacting system, not a dilute gas. That the estimate lands within a factor 1.5 of the observed λ-point is actually a striking success: it was exactly this closeness that led London in 1938 to identify superfluidity with condensation. Interactions shift the transition temperature and, more dramatically, deplete the condensate: at $T=0$ only about 8 % of helium atoms sit in the zero-momentum state, versus 100 % for the ideal gas. So the ideal model gets the scale right and the microscopic fraction wrong.

Can I use the box formula for a trapped atomic cloud?

You can for a rough estimate, but the trap changes the exponent and the prefactor. In a box the density of states is $\varepsilon^{1/2}$ ( $\alpha=3/2$ ); in a 3D harmonic trap the effective density of states is $\varepsilon^2$ ( $\alpha=3$ ), which gives $k_BT_c=\hbar\bar\omega[N/\zeta(3)]^{1/3}$ and $N_0/N=1-(T/T_c)^3$ . For a few thousand atoms the two routes agree to within a factor of order one (both land near 100 nK), but the shape of the condensate-fraction curve is cubic in a trap, not $3/2$ -power, and time-of-flight experiments measure the cubic law. Always match the formula to the confining geometry.

If μ is pinned at zero below T_c, doesn't that mean the ground state occupation is infinite?

No — it means $\mu$ approaches zero from below by an amount of order $1/N$ , not exactly zero. The ground-state occupation is $\langle n_0\rangle=1/(e^{-\beta\mu}-1)\approx -1/(\beta\mu)$ , and for this to equal a macroscopic $N_0$ we need $\mu\approx -k_BT/N_0$ , which is negative but vanishingly small — of order $10^{-23}$ eV for a real cloud. In every thermodynamic-limit formula we may set $\mu=0$ , because the correction is $O(1/N)$ ; but the "pinning" is a limit, not a literal zero. This is exactly why the condensate must be tracked as a separate variable $N_0$ rather than read off the continuum integral.

Why is there no condensation in one or two dimensions but there is in three?

Because the excited states can absorb an unlimited number of particles in low dimensions. The maximum excited population is $\int_0^\infty g(\varepsilon)/(e^{\beta\varepsilon}-1)\,\mathrm{d}\varepsilon$ with $g(\varepsilon)\propto\varepsilon^{(d/2)-1}$ . In 3D this integral converges to a finite value ( $\propto\zeta(3/2)$ ), so once $N$ exceeds it the surplus must condense. In 2D the integrand behaves like $1/\varepsilon$ near the origin and the integral diverges logarithmically; in 1D it diverges as $\varepsilon^{-1/2}$ . A divergent excited-state capacity means there is never a surplus, so $T_c\to 0$ . The transition's very existence is decided by the low-energy power of the density of states — pure phase-space geometry.