Bose-Einstein condensation — concept: Worked examples

Before we start, one reminder of the only symbols we need — all defined in the parent, restated here so this page stands alone:

Definition The quantities we reuse (every symbol earned)

$N$ = total number of atoms in the box.
$N_{0}$ = number of atoms sitting in the single lowest-energy state (the condensate). So $N_{0} / N$ is "what fraction of all atoms have condensed," a pure number between $0$ and $1$ .
$n$ = number of particles per unit volume (density). Picture: how crowded the box is.
$T$ = temperature; $T_{c}$ = the critical temperature where condensation begins.
$h$ = Planck's constant; $ℏ = h / (2 π)$ is the same constant divided by $2 π$ ("h-bar"). Both appear below — they are the same physics, just differing by the factor $2 π$ .
$λ_{T} = \frac{h}{2 π m k _{B} T}$ = the thermal de Broglie wavelength — roughly the "quantum size" of one atom's matter-wave. Colder $\Rightarrow$ bigger $λ_{T}$ . ( $k_{B}$ = Boltzmann's constant, $m$ = atom mass.)
$μ$ = the chemical potential — the energy cost of adding one more atom; for bosons it must stay $\leq 0$ and pins to $0^{-}$ once the condensate forms.
$g$ = spin degeneracy (how many spin states share each energy level). For a spin-0 atom, $g = 1$ .
$ζ (3/2) \approx 2.612$ = a fixed number from the Riemann zeta function (see density of states integral in the parent).

The two boxed results we lean on: $k_{B} T_{c} = \frac{2 π ℏ ^{2}}{m} (\frac{n}{g ζ ( 3/2 )})^{2/3}, \frac{N _{0}}{N} = 1 - (\frac{T}{T _{c}})^{3/2} (T \leq T_{c}) .$

The scenario matrix

Every problem this topic throws is one of these cells. The examples below tag which one they hit.

#	Cell class	What is special	Covered by
A	Generic finite $T < T_{c}$	ordinary condensate fraction	Ex 1
B	Boundary $T = T_{c}$	fraction exactly $0$	Ex 2
C	Zero input $T = 0$	fraction exactly $1$ (all condense)	Ex 2
D	Forbidden region $T > T_{c}$	formula would go negative — must clamp	Ex 3
E	Find $T_{c}$ from density (real lab)	plug $n, m$ into $T_{c}$	Ex 4
F	Inverse: what $n$ or $T$ to hit a target fraction	solve backwards	Ex 5
G	Degenerate: change $g$	how spin degeneracy shifts $T_{c}$	Ex 6
H	Degenerate: change dimension (2D)	integral diverges — no BEC	Ex 7
I	Limiting behaviour / slope near $T_{c}$	steep, faster-than-linear onset	Ex 8
J	Word-problem / scaling twist	double the density, what happens to $T_{c}$ ?	Ex 9

Cell A — generic condensate fraction

Worked example Ex 1 · Fraction at

$T = 0.8 T_{c}$ (cell A) A gas is held at $T = 0.8 T_{c}$ . What fraction of the atoms are in the condensate?

Forecast: guess — is it closer to $10%$ or $40%$ ? (Most people overshoot; the $3/2$ power is gentler than it looks near $T_{c}$ .)

Write the boxed fraction. Why this step? We are inside $T \leq T_{c}$ , so the formula is valid — no clamping needed. $\frac{N _{0}}{N} = 1 - (\frac{T}{T _{c}})^{3/2} = 1 - (0.8)^{3/2} .$
Evaluate $(0.8)^{3/2}$ . Why this step? $x^{3/2} = x x$ , so it is $0.8 \times 0.8 = 0.8 \times 0.8944 = 0.7155$ .
Subtract. Why this step? Everything not in excited states is condensate. $\frac{N _{0}}{N} = 1 - 0.7155 = 0.2845.$

Answer: about $28.5%$ of the atoms have condensed.

Verify: sanity — at $0.8 T_{c}$ we are close to the transition, so a small fraction condensed makes sense (closer to $10%$ side of the forecast). Dimensionless in, dimensionless out. ✓

Cells B & C — the two exact boundaries

Worked example Ex 2 · The endpoints

$T = T_{c}$ and $T = 0$ (cells B, C) Evaluate the condensate fraction at the two extremes.

Forecast: at exactly $T_{c}$ , is the condensate empty or already macroscopic? At $T = 0$ , is it all or just most?

At $T = T_{c}$ set the ratio to $1$ . Why this step? $T_{c}$ is defined as the point where excited states are just full and the condensate is about to be born. $\frac{N _{0}}{N} = 1 - (1)^{3/2} = 1 - 1 = 0.$ So the condensate is exactly empty at the transition — it starts from zero.
At $T = 0$ set the ratio to $0$ . Why this step? Absolute zero: excited states can hold nothing. $\frac{N _{0}}{N} = 1 - (0)^{3/2} = 1 - 0 = 1.$ Every atom condenses.

Answer: $N_{0} / N = 0$ at $T_{c}$ , and $N_{0} / N = 1$ at $T = 0$ .

Verify: the fraction must live between $0$ and $1$ for any $0 \leq T \leq T_{c}$ ; our two boundary values are those extremes, so the curve is bounded correctly. ✓

Read the figure below: the magenta curve is $N_{0} / N = 1 - (T / T_{c})^{3/2}$ . Follow it from the left orange dot at $(0, 1)$ — that is the " $T = 0$ , everything condensed" endpoint from step 2 — down to the right orange dot at $(1, 0)$ — the " $T = T_{c}$ , empty condensate" endpoint from step 1. The violet dashed line to the right of $T_{c}$ is the flat-zero forbidden region we tackle in Ex 3. Notice the curve leaves the top-left corner steeply and hits $T_{c}$ with a visible downward tilt (Ex 8 measures that tilt).

Figure — Bose-Einstein condensation — concept

Cell D — the forbidden region (the trap)

Worked example Ex 3 · What is the "fraction" at

$T = 1.2 T_{c}$ ? (cell D) A student blindly plugs $T = 1.2 T_{c}$ into the boxed formula and reports the answer. What is wrong?

Forecast: the formula gives a negative number. Is that physical?

Blindly plug in. Why this step? To expose the trap. $1 - (1.2)^{3/2} = 1 - 1.3145 = - 0.3145.$ A negative condensate fraction — impossible; you cannot have $- 31%$ of atoms anywhere.
Recall the domain of the formula. Why this step? The parent note wrote $(T \leq T_{c})$ next to the box on purpose. Remember from our symbol list that $μ$ (the chemical potential, the energy cost of adding one atom) sits pinned at $0^{-}$ only when the condensate exists. Above $T_{c}$ the excited states are not full, so $μ$ drops safely below $0$ , the ground state holds only a handful of atoms, and there is simply no condensate.
Clamp the answer. Why this step? The correct value for $T > T_{c}$ is $\frac{N _{0}}{N} = 0 (T > T_{c}) .$

Answer: $0$ — there is no condensate above $T_{c}$ . The negative number is a sign you left the formula's valid region.

Verify: $N_{0} / N$ must be $\geq 0$ ; the raw formula returned $< 0$ , which is the built-in alarm that $T > T_{c}$ . ✓

Cell E — find $T_{c}$ from real numbers (lab problem)

Worked example Ex 4 · Critical temperature of a rubidium cloud (cell E)

A $^{87}$ Rb gas has density $n = 1.0 \times 1 0^{19} m^{- 3}$ , atom mass $m = 1.44 \times 1 0^{- 25} kg$ , spin degeneracy $g = 1$ . Find $T_{c}$ . Constants: $ℏ = h / (2 π) = 1.055 \times 1 0^{- 34} J\cdots$ , $k_{B} = 1.381 \times 1 0^{- 23} J/K$ , $ζ (3/2) = 2.612$ .

Forecast: nanokelvin, microkelvin, or millikelvin? (The parent hinted "sub-microkelvin.")

Form the density ratio. Why this step? The formula needs $n / (g ζ (3/2))$ as its core. $\frac{n}{g ζ ( 3/2 )} = \frac{1.0 \times 1 0 ^{19}}{2.612} = 3.829 \times 1 0^{18} m^{- 3} .$
Raise to the $2/3$ power. Why this step? The box has an exponent $2/3$ from inverting $λ_{T} \propto T^{- 1/2}$ . $(3.829 \times 1 0^{18})^{2/3} = 2.436 \times 1 0^{12} m^{- 2} .$
Multiply the prefactor $\frac{2 π ℏ ^{2}}{m}$ . Why this step? This carries all the physics constants and units. $\frac{2 π ( 1.055 \times 1 0 ^{- 34} ) ^{2}}{1.44 \times 1 0 ^{- 25}} = 4.855 \times 1 0^{- 43} J\cdotm^{2} .$
Combine to get $k_{B} T_{c}$ , then divide by $k_{B}$ . Why this step? Step 3 $\times$ step 2 gives an energy; dividing by $k_{B}$ turns energy into temperature. $k_{B} T_{c} = 4.855 \times 1 0^{- 43} \times 2.436 \times 1 0^{12} = 1.183 \times 1 0^{- 30} J,$ $T_{c} = \frac{1.183 \times 1 0 ^{- 30}}{1.381 \times 1 0 ^{- 23}} = 8.56 \times 1 0^{- 8} K .$

Answer: $T_{c} \approx 8.6 \times 1 0^{- 8} K \approx 86 nK$ — deep in the sub-microkelvin regime, matching the forecast.

Verify: units — $[J\cdotm^{2}] \times [m^{- 2}] = [J]$ ; then $[J] / [J/K] = [K]$ . ✓ And the value lands at "hundreds of nK," exactly what the parent's Example 1 predicted. ✓

Cell F — inverse problem (solve backwards)

Cell G — degeneracy $g$ changes $T_{c}$

Worked example Ex 6 · How does doubling spin degeneracy shift

$T_{c}$ ? (cell G) Two identical gases (same $n, m$ ) differ only in spin degeneracy: gas A has $g = 1$ , gas B has $g = 2$ . What is the ratio $T_{c}^{(B)} / T_{c}^{(A)}$ ?

Forecast: more spin channels means more "cheap seats," so should $T_{c}$ go up or down?

Isolate the $g$ -dependence. Why this step? Only $g$ changes, so treat everything else as a constant $C$ : $T_{c} = C (\frac{n}{g ζ ( 3/2 )})^{2/3} \Rightarrow T_{c} \propto g^{- 2/3} .$
Take the ratio. Why this step? The unknown constant $C$ cancels. $\frac{T _{c}^{(B)}}{T _{c}^{(A)}} = (\frac{g _{A}}{g _{B}})^{2/3} = (\frac{1}{2})^{2/3} = 0.6300.$

Answer: $T_{c}^{(B)} = 0.63 T_{c}^{(A)}$ — more degeneracy lowers $T_{c}$ . Extra excited "seats" absorb more atoms, so you must cool further before they overflow.

Verify: dimensionless ratio, and $g ↑\Rightarrow T_{c} ↓$ matches the "more room in excited states" intuition. Numerically $2^{- 2/3} = 0.6300$ . ✓ (Contrast with Fermi–Dirac statistics, where no such condensation exists — the Pauli principle blocks piling up.)

Cell H — degenerate dimension: no BEC in 2D

Worked example Ex 7 · Why does a uniform 2D ideal Bose gas

not condense? (cell H) Repeat the saturation argument in two dimensions and see what breaks.

Forecast: in 3D the density of states went as $ε$ . In 2D it is constant. Guess whether the saturation integral stays finite.

Write the 2D density of states. Why this step? In $d$ dimensions $g (ε) \propto ε^{d /2 - 1}$ ; for $d = 2$ that exponent is $0$ , i.e. constant in $ε$ . $g_{2 D} (ε) = const .$
Set up the max excited population at $μ = 0$ . Why this step? Saturation is always tested at $μ = 0$ (the parent's Step 2). Here $μ$ is the chemical potential from our symbol list. $N_{exc}^{m a x} \propto \int_{0}^{\infty} \frac{d ε}{e ^{ε / k_{B} T} - 1} x = ε / k_{B} T \int_{0}^{\infty} \frac{d x}{e ^{x} - 1} .$
Check the small- $x$ behaviour. Why this step? Divergence lives near $ε = 0$ . For small $x$ , $e^{x} - 1 \approx x$ , so the integrand $\approx 1/ x$ , and $\int_{0} \frac{d x}{x}$ diverges (logarithmically).

Answer: the maximum excited population is infinite — excited states can hold any number of atoms, so they never overflow. No BEC in a uniform 2D ideal gas.

Verify: the 3D integral $\int_{0}^{\infty} \frac{x d x}{e ^{x} - 1} = Γ (3/2) ζ (3/2)$ is finite because the extra $x$ kills the $1/ x$ blow-up; removing it (2D) restores the divergence. Dimensionality decides. ✓

Cell I — limiting behaviour: the steep onset near $T_{c}$

Worked example Ex 8 · How fast does the condensate grow just below

$T_{c}$ ? (cell I) Compare the condensate fraction at $T = 0.99 T_{c}$ and $T = 0.98 T_{c}$ — is the growth gentle or steep near the top?

Forecast: guess whether $1%$ of cooling near $T_{c}$ gives more or less than $1%$ condensate.

Setup — a shorthand. To talk about "how the fraction changes as we cool," define the scaled temperature $u = T / T_{c}$ . Then the fraction is $f (u) = 1 - u^{3/2}$ , and cooling means moving $u$ from $1$ downward toward $0$ . We use $u$ throughout the steps below.

Evaluate at $u = 0.99$ . Why this step? We need two nearby points to feel the slope. $f (0.99) = 1 - (0.99)^{3/2} = 1 - 0.98507 = 0.01493.$
Evaluate at $u = 0.98$ . Why this step? Same, one more step of cooling. $f (0.98) = 1 - (0.98)^{3/2} = 1 - 0.97015 = 0.02985.$
Compare to the cooling. Why this step? We cooled by $1%$ of $T_{c}$ (i.e. $Δ u = 0.01$ ) and gained $\approx 1.5%$ condensate — more condensate than the temperature drop. That is the fingerprint of a slope steeper than $1$ .
Confirm with calculus. Why this step? The derivative of $f (u) = 1 - u^{3/2}$ tells us the exact steepness at the edge: $\frac{df}{d u} = - \frac{3}{2} u^{1/2}, \frac{df}{d u}_{u = 1} = - \frac{3}{2} .$ So as we cool past $T_{c}$ the fraction rises with slope $\frac{3}{2}$ in the scaled variable $u$ — finite, but steeper than the linear slope of $1$ you might naively expect, so the condensate switches on faster than linearly. This sharp "knee" is the fingerprint of a phase transition and the onset of superfluidity.

Answer: the condensate switches on abruptly — $1%$ of cooling yields $\sim 1.5%$ condensate right at the edge, matching the slope $\frac{3}{2}$ .

Verify: $0.02985 - 0.01493 = 0.01492$ , and $0.01492/0.01 = 1.49 \approx \frac{3}{2}$ — the measured slope matches the calculus. ✓

Cell J — word-problem scaling twist

Worked example Ex 9 · Compress the gas to double its density — how much warmer can

$T_{c}$ be? (cell J) An experimenter squeezes the trap so the density doubles, $n \to 2 n$ , at fixed $m, g$ . By what factor does $T_{c}$ change?

Forecast: denser means atoms are closer, matter-waves overlap sooner — so should $T_{c}$ rise or fall?

Extract the $n$ -dependence. Why this step? Only $n$ changes; the prefactor $\frac{2 π ℏ ^{2}}{m}$ and $g, ζ (3/2)$ are all fixed, so $T_{c} \propto n^{2/3} .$
Take the ratio for $n \to 2 n$ . Why this step? The constants cancel, leaving a pure number. $\frac{T _{c}^{new}}{T _{c}^{old}} = (\frac{2 n}{n})^{2/3} = 2^{2/3} = 1.587.$
State it in words. Why this step? Turn the number into physics. Doubling the density raises the transition temperature by about $59%$ .

Answer: $T_{c}$ rises by a factor $2^{2/3} \approx 1.59$ — about $59%$ higher. Denser gas condenses at a warmer temperature, exactly the "overlap sooner" intuition of the forecast.

Verify: the phase-space criterion $n λ_{T}^{3} = g ζ (3/2)$ is fixed; if $n$ doubles, $λ_{T}^{3}$ must halve, so with $λ_{T} \propto T^{- 1/2}$ we get $T \propto λ_{T}^{- 2} \propto (n^{1/3})^{2} = n^{2/3}$ . Consistent, and $2^{2/3} = 1.587$ . ✓

Active recall

Recall Which cells trap students most?

Q: Name the two "silent" traps and their fix. A: (1) Plugging $T > T_{c}$ into $1 - (T / T_{c})^{3/2}$ gives a negative "fraction" — clamp to $0$ (cell D). (2) Assuming BEC happens in any dimension — the 2D saturation integral diverges, so no condensation (cell H).

Recall Scaling laws in one breath

Q: How does $T_{c}$ scale with $n$ and with $g$ ? A: $T_{c} \propto n^{2/3}$ (denser $\Rightarrow$ warmer transition) and $T_{c} \propto g^{- 2/3}$ (more degeneracy $\Rightarrow$ colder transition).

The condensate-fraction curve $::: N_{0} / N = 1 - (T / T_{c})^{3/2}$ for $T \leq T_{c}$ , and flat $0$ above $T_{c}$

Sign of the "fraction formula" at

T > T_{c}

warns you of what?

You have left its domain — the true answer is

0

, no condensate.