Bose-Einstein condensation — concept: Worked examples

2.4.18 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Bose-Einstein condensation — concept

Shuru karne se pehle, ek reminder un symbols ka jo hume chahiye — sab parent mein define hain, yahan restate kiye hain taaki yeh page akele bhi kaam kar sake:

Definition Jo quantities hum reuse karte hain (har symbol ka matlab)

$N$ = box mein atoms ki total sankhya.
$N_{0}$ = atoms ki sankhya jo single lowest-energy state mein baithe hain (condensate). Toh $N_{0} / N$ hai "kitne fraction atoms condense hue hain," ek pure number $0$ aur $1$ ke beech.
$n$ = particles per unit volume (density). Picture karo: box kitna crowded hai.
$T$ = temperature; $T_{c}$ = critical temperature jahan condensation shuru hoti hai.
$h$ = Planck's constant; $ℏ = h / (2 π)$ wahi constant hai $2 π$ se divided ("h-bar"). Dono neeche aate hain — yeh same physics hai, bas $2 π$ ke factor se alag hain.
$λ_{T} = \frac{h}{2 π m k _{B} T}$ = thermal de Broglie wavelength — roughly ek atom ki matter-wave ka "quantum size." Thanda $\Rightarrow$ bada $λ_{T}$ . ( $k_{B}$ = Boltzmann's constant, $m$ = atom mass.)
$μ$ = chemical potential — ek aur atom add karne ki energy cost; bosons ke liye yeh $\leq 0$ rehna chahiye aur condensate ban'ne par $0^{-}$ pe pin ho jaata hai.
$g$ = spin degeneracy (kitne spin states har energy level share karte hain). Spin-0 atom ke liye, $g = 1$ .
$ζ (3/2) \approx 2.612$ = Riemann zeta function se ek fixed number (parent mein density of states integral dekho).

Do boxed results jinpar hum rely karte hain: $k_{B} T_{c} = \frac{2 π ℏ ^{2}}{m} (\frac{n}{g ζ ( 3/2 )})^{2/3}, \frac{N _{0}}{N} = 1 - (\frac{T}{T _{c}})^{3/2} (T \leq T_{c}) .$

Scenario matrix

Is topic ke har problem ka ek cell hota hai. Neeche ke examples tag karte hain ki kaun sa cell hit hua.

#	Cell class	Kya special hai	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 $1$ (sab condense)	Ex 2
D	Forbidden region $T > T_{c}$	formula negative jaayegi — clamp karna hoga	Ex 3
E	Density se $T_{c}$ nikalo (real lab)	$n, m$ ko $T_{c}$ mein plug karo	Ex 4
F	Inverse: kaun sa $n$ ya $T$ target fraction hit kare	ulta solve karo	Ex 5
G	Degenerate: $g$ badlo	spin degeneracy $T_{c}$ ko kaise shift karti hai	Ex 6
H	Degenerate: dimension badlo (2D)	integral diverge — BEC nahi hogi	Ex 7
I	Limiting behaviour / $T_{c}$ ke paas slope	steep, faster-than-linear onset	Ex 8
J	Word-problem / scaling twist	density double karo, $T_{c}$ ka kya hoga?	Ex 9

Cell A — generic condensate fraction

Worked example Ex 1 ·

$T = 0.8 T_{c}$ par Fraction (cell A) Ek gas $T = 0.8 T_{c}$ par rakhi hai. Kitne fraction atoms condensate mein hain?

Forecast: guess karo — kya yeh $10%$ ke paas hai ya $40%$ ke? (Zyaadatar log overshoot karte hain; $3/2$ power $T_{c}$ ke paas utni brutal nahi lagti.)

Boxed fraction likho. Yeh step kyun? Hum $T \leq T_{c}$ ke andar hain, toh formula valid hai — clamping ki zaroorat nahi. $\frac{N _{0}}{N} = 1 - (\frac{T}{T _{c}})^{3/2} = 1 - (0.8)^{3/2} .$
$(0.8)^{3/2}$ evaluate karo. Yeh step kyun? $x^{3/2} = x x$ , toh yeh hai $0.8 \times 0.8 = 0.8 \times 0.8944 = 0.7155$ .
Subtract karo. Yeh step kyun? Jo excited states mein nahi hai woh sab condensate hai. $\frac{N _{0}}{N} = 1 - 0.7155 = 0.2845.$

Answer: atoms ka lagbhag $28.5%$ condense hua hai.

Verify: sanity — $0.8 T_{c}$ par hum transition ke paas hain, toh ek chota fraction condense hona sahi lagta hai (forecast ke $10%$ side ke karib). Dimensionless in, dimensionless out. ✓

Cells B & C — do exact boundaries

Worked example Ex 2 · Endpoints

$T = T_{c}$ aur $T = 0$ (cells B, C) Do extremes par condensate fraction evaluate karo.

Forecast: exactly $T_{c}$ par, kya condensate khaali hai ya already macroscopic? $T = 0$ par, kya yeh sab hai ya sirf zyaadatar?

$T = T_{c}$ par ratio ko $1$ set karo. Yeh step kyun? $T_{c}$ define hi woh point hai jahan excited states sirf full hain aur condensate ban'ne wala hi hai. $\frac{N _{0}}{N} = 1 - (1)^{3/2} = 1 - 1 = 0.$ Toh condensate transition par bilkul khaali hai — yeh zero se shuru hota hai.
$T = 0$ par ratio ko $0$ set karo. Yeh step kyun? Absolute zero: excited states kuch hold nahi kar sakti. $\frac{N _{0}}{N} = 1 - (0)^{3/2} = 1 - 0 = 1.$ Har atom condense hota hai.

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

Verify: fraction $0$ aur $1$ ke beech rehni chahiye kisi bhi $0 \leq T \leq T_{c}$ ke liye; hamare do boundary values wahi extremes hain, toh curve sahi bounded hai. ✓

Figure neeche padho: magenta curve hai $N_{0} / N = 1 - (T / T_{c})^{3/2}$ . Ise left orange dot se follow karo $(0, 1)$ par — woh step 2 se " $T = 0$ , sab condense" endpoint hai — neeche right orange dot tak $(1, 0)$ par — step 1 se " $T = T_{c}$ , khaali condensate" endpoint. $T_{c}$ ke daayein taraf violet dashed line woh flat-zero forbidden region hai jo hum Ex 3 mein dekhenge. Notice karo ki curve top-left corner se steep jaati hai aur $T_{c}$ tak visible downward tilt ke saath pahunchti hai (Ex 8 woh tilt measure karta hai).

Figure — Bose-Einstein condensation — concept

Cell D — forbidden region (the trap)

Worked example Ex 3 ·

$T = 1.2 T_{c}$ par "fraction" kya hai? (cell D) Ek student blindly $T = 1.2 T_{c}$ ko boxed formula mein plug karta hai aur answer report karta hai. Kya galat hai?

Forecast: formula ek negative number deta hai. Kya woh physical hai?

Blindly plug in karo. Yeh step kyun? Trap ko expose karne ke liye. $1 - (1.2)^{3/2} = 1 - 1.3145 = - 0.3145.$ Ek negative condensate fraction — impossible; tumhare paas $- 31%$ atoms kahi nahi ho sakte.
Formula ka domain yaad karo. Yeh step kyun? Parent note ne box ke paas $(T \leq T_{c})$ likha tha purpose ke saath. Apni symbol list se yaad karo ki $μ$ (chemical potential, ek atom add karne ki energy cost) $0^{-}$ pe pinned tab rehta hai jab condensate exist karta hai. $T_{c}$ ke upar excited states full nahi hoti, toh $μ$ safely $0$ se neeche drop karta hai, ground state mein sirf gine-chune atoms hote hain, aur simply koi condensate nahi hota.
Answer clamp karo. Yeh step kyun? $T > T_{c}$ ke liye correct value hai $\frac{N _{0}}{N} = 0 (T > T_{c}) .$

Answer: $0$ — $T_{c}$ ke upar koi condensate nahi. Negative number ek sign hai ki tum formula ki valid range se bahar chale gaye.

Verify: $N_{0} / N$ must be $\geq 0$ ; raw formula ne $< 0$ diya, jo built-in alarm hai ki $T > T_{c}$ . ✓

Cell E — real numbers se $T_{c}$ nikalo (lab problem)

Worked example Ex 4 · Rubidium cloud ka critical temperature (cell E)

Ek $^{87}$ Rb gas mein density $n = 1.0 \times 1 0^{19} m^{- 3}$ , atom mass $m = 1.44 \times 1 0^{- 25} kg$ , spin degeneracy $g = 1$ hai. $T_{c}$ nikalo. 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, ya millikelvin? (Parent ne hint diya tha "sub-microkelvin.")

Density ratio banao. Yeh step kyun? Formula ko apne core ke roop mein $n / (g ζ (3/2))$ chahiye. $\frac{n}{g ζ ( 3/2 )} = \frac{1.0 \times 1 0 ^{19}}{2.612} = 3.829 \times 1 0^{18} m^{- 3} .$
$2/3$ power tak raise karo. Yeh step kyun? Box mein $2/3$ exponent hai $λ_{T} \propto T^{- 1/2}$ ko invert karne se. $(3.829 \times 1 0^{18})^{2/3} = 2.436 \times 1 0^{12} m^{- 2} .$
Prefactor $\frac{2 π ℏ ^{2}}{m}$ multiply karo. Yeh step kyun? Yeh saari physics constants aur units carry karta hai. $\frac{2 π ( 1.055 \times 1 0 ^{- 34} ) ^{2}}{1.44 \times 1 0 ^{- 25}} = 4.855 \times 1 0^{- 43} J\cdotm^{2} .$
$k_{B} T_{c}$ paane ke liye combine karo, phir $k_{B}$ se divide karo. Yeh step kyun? Step 3 $\times$ step 2 ek energy deta hai; $k_{B}$ se divide karne par energy temperature ban jaati hai. $k_{B} T_{c} = 4.855 \times 1 0^{- 43} \times 2.436 \times 1 0^{12} = 1.183 \times 1 0^{- 30} J,$ $T_{c} = \frac{1.183 \times 1 0 ^{- 30}}{1.381 \times 1 0 ^{- 23}} = 8.56 \times 1 0^{- 8} K .$

Answer: $T_{c} \approx 8.6 \times 1 0^{- 8} K \approx 86 nK$ — sub-microkelvin regime mein deep, forecast se match karta hai.

Verify: units — $[J\cdotm^{2}] \times [m^{- 2}] = [J]$ ; phir $[J] / [J/K] = [K]$ . ✓ Aur value "hundreds of nK" par aata hai, exactly jo parent ke Example 1 ne predict kiya tha. ✓

Cell F — inverse problem (ulta solve karo)

Worked example Ex 5 · Kis

$T / T_{c}$ par exactly aadha gas condense hota hai? (cell F) Hum chahte hain $N_{0} / N = 0.5$ . $T / T_{c}$ nikalo.

Forecast: kya yeh $0.5 T_{c}$ se upar hai ya neeche? (Parent note ke worked Example 2 ne $0.5 T_{c}$ par $65%$ condense paaya, toh sirf aadha condensed hone ke liye hume $0.5 T_{c}$ se zyaada garm rehna hoga.)

Fraction ko target ke barabar set karo. Yeh step kyun? Hum boxed relation ko forward use karne ki jagah invert kar rahe hain. $0.5 = 1 - (\frac{T}{T _{c}})^{3/2} \Rightarrow (\frac{T}{T _{c}})^{3/2} = 0.5.$
$3/2$ power ko $2/3$ raise karke undo karo. Yeh step kyun? $(x^{3/2})^{2/3} = x$ , ratio isolate karne ka saaf tarika. $\frac{T}{T _{c}} = 0. 5^{2/3} = 0.6300.$

Answer: $T / T_{c} \approx 0.63$ . Aadhe atoms critical temperature ke lagbhag $63%$ par condense hote hain — $0.5 T_{c}$ se zyaada garm, jaise forecast mein tha.

Verify: plug back karo — $(0.6300)^{3/2} = 0.500$ , toh $N_{0} / N = 1 - 0.500 = 0.500$ . ✓

Cell G — degeneracy $g$ $T_{c}$ ko shift karta hai

Worked example Ex 6 · Spin degeneracy double karne par

$T_{c}$ kaise shift hoti hai? (cell G) Do identical gases (same $n, m$ ) sirf spin degeneracy mein alag hain: gas A mein $g = 1$ , gas B mein $g = 2$ . Ratio $T_{c}^{(B)} / T_{c}^{(A)}$ kya hai?

Forecast: zyaada spin channels matlab zyaada "cheap seats," toh $T_{c}$ upar jaayegi ya neeche?

$g$ -dependence isolate karo. Yeh step kyun? Sirf $g$ badal raha hai, toh baaki sab ko constant $C$ maano: $T_{c} = C (\frac{n}{g ζ ( 3/2 )})^{2/3} \Rightarrow T_{c} \propto g^{- 2/3} .$
Ratio lo. Yeh step kyun? Unknown constant $C$ cancel ho jaata hai. $\frac{T _{c}^{(B)}}{T _{c}^{(A)}} = (\frac{g _{A}}{g _{B}})^{2/3} = (\frac{1}{2})^{2/3} = 0.6300.$

Answer: $T_{c}^{(B)} = 0.63 T_{c}^{(A)}$ — zyaada degeneracy $T_{c}$ ko lower karti hai. Extra excited "seats" zyaada atoms absorb karte hain, toh overflow hone se pehle aur thanda karna padta hai.

Verify: dimensionless ratio, aur $g ↑\Rightarrow T_{c} ↓$ "excited states mein zyaada jagah" wali intuition se match karta hai. Numerically $2^{- 2/3} = 0.6300$ . ✓ (Fermi–Dirac statistics se contrast karo, jahan aisa koi condensation exist nahi karta — Pauli principle piling up block karta hai.)

Cell H — degenerate dimension: 2D mein BEC nahi

Worked example Ex 7 · Ek uniform 2D ideal Bose gas condense kyun

nahi karta? (cell H) Saturation argument ko do dimensions mein repeat karo aur dekho kya toot'ta hai.

Forecast: 3D mein density of states $ε$ ki tarah thi. 2D mein yeh constant hai. Guess karo ki saturation integral finite rehta hai ya nahi.

2D density of states likho. Yeh step kyun? $d$ dimensions mein $g (ε) \propto ε^{d /2 - 1}$ ; $d = 2$ ke liye woh exponent $0$ hai, yaani $ε$ mein constant. $g_{2 D} (ε) = const .$
$μ = 0$ par max excited population set up karo. Yeh step kyun? Saturation hamesha $μ = 0$ par test hoti hai (parent ka Step 2). Yahan $μ$ hamare symbol list se chemical potential hai. $N_{exc}^{m a x} \propto \int_{0}^{\infty} \frac{d ε}{e ^{ε / k_{B} T} - 1} x = ε / k_{B} T \int_{0}^{\infty} \frac{d x}{e ^{x} - 1} .$
Small- $x$ behaviour check karo. Yeh step kyun? Divergence $ε = 0$ ke paas hoti hai. Chhote $x$ ke liye, $e^{x} - 1 \approx x$ , toh integrand $\approx 1/ x$ , aur $\int_{0} \frac{d x}{x}$ diverge karta hai (logarithmically).

Answer: maximum excited population infinite hai — excited states kitne bhi atoms hold kar sakti hain, toh woh kabhi overflow nahi karti. Uniform 2D ideal gas mein BEC nahi hoti.

Verify: 3D integral $\int_{0}^{\infty} \frac{x d x}{e ^{x} - 1} = Γ (3/2) ζ (3/2)$ finite hai kyunki extra $x$ $1/ x$ blow-up ko khatam kar deta hai; ise hataane par (2D) divergence wapas aa jaati hai. Dimensionality decide karta hai. ✓

Cell I — limiting behaviour: $T_{c}$ ke paas steep onset

Worked example Ex 8 ·

$T_{c}$ ke bilkul neeche condensate kitni tezi se badhta hai? (cell I) $T = 0.99 T_{c}$ aur $T = 0.98 T_{c}$ par condensate fraction compare karo — kya top ke paas growth gentle hai ya steep?

Forecast: guess karo ki $T_{c}$ ke paas $1%$ cooling $1%$ se zyaada ya kam condensate deti hai.

Setup — ek shorthand. "Fraction thande hone par kaise badlti hai" ke baare mein baat karne ke liye, scaled temperature $u = T / T_{c}$ define karo. Toh fraction hai $f (u) = 1 - u^{3/2}$ , aur thanda hona matlab $u$ ko $1$ se neeche $0$ ki taraf le jaana. Hum $u$ neeche ke steps mein use karte hain.

$u = 0.99$ par evaluate karo. Yeh step kyun? Slope feel karne ke liye hume do nearby points chahiye. $f (0.99) = 1 - (0.99)^{3/2} = 1 - 0.98507 = 0.01493.$
$u = 0.98$ par evaluate karo. Yeh step kyun? Same, cooling ka ek aur step. $f (0.98) = 1 - (0.98)^{3/2} = 1 - 0.97015 = 0.02985.$
Cooling se compare karo. Yeh step kyun? Humne $T_{c}$ ka $1%$ thanda kiya (yaani $Δ u = 0.01$ ) aur $\approx 1.5%$ condensate gain kiya — temperature drop se zyaada condensate. Yeh ek slope ka fingerprint hai jo $1$ se steeper hai.
Calculus se confirm karo. Yeh step kyun? $f (u) = 1 - u^{3/2}$ ka derivative edge par exact steepness batata hai: $\frac{df}{d u} = - \frac{3}{2} u^{1/2}, \frac{df}{d u}_{u = 1} = - \frac{3}{2} .$ Toh $T_{c}$ ke paas thanda hone par fraction scaled variable $u$ mein slope $\frac{3}{2}$ se badhti hai — finite, lekin $1$ ke linear slope se steeper jis ki tum naively expect kar sakte ho, toh condensate faster than linearly switch on hoti hai. Yeh sharp "knee" ek phase transition aur superfluidity ke onset ka fingerprint hai.

Answer: condensate abruptly switch on hoti hai — $1%$ cooling edge par $\sim 1.5%$ condensate deta hai, slope $\frac{3}{2}$ se match karta hai.

Verify: $0.02985 - 0.01493 = 0.01492$ , aur $0.01492/0.01 = 1.49 \approx \frac{3}{2}$ — measured slope calculus se match karta hai. ✓

Cell J — word-problem scaling twist

Worked example Ex 9 · Gas compress karo density double karo —

$T_{c}$ kitna garm ho sakta hai? (cell J) Ek experimenter trap squeeze karta hai taaki density double ho jaaye, $n \to 2 n$ , fixed $m, g$ ke saath. $T_{c}$ kis factor se badlega?

Forecast: denser matlab atoms karib hain, matter-waves jaldi overlap karti hain — toh $T_{c}$ upar jaayega ya neeche?

$n$ -dependence extract karo. Yeh step kyun? Sirf $n$ badal raha hai; prefactor $\frac{2 π ℏ ^{2}}{m}$ aur $g, ζ (3/2)$ sab fixed hain, toh $T_{c} \propto n^{2/3} .$
$n \to 2 n$ ke liye ratio lo. Yeh step kyun? Constants cancel ho jaate hain, sirf ek pure number bachta hai. $\frac{T _{c}^{new}}{T _{c}^{old}} = (\frac{2 n}{n})^{2/3} = 2^{2/3} = 1.587.$
Words mein batao. Yeh step kyun? Number ko physics mein badlo. Density double karne par transition temperature lagbhag $59%$ badh jaati hai.

Answer: $T_{c}$ factor $2^{2/3} \approx 1.59$ se badh'ta hai — lagbhag $59%$ zyaada. Dense gas zyaada garm temperature par condense hoti hai, exactly forecast ki "jaldi overlap" wali intuition.

Verify: phase-space criterion $n λ_{T}^{3} = g ζ (3/2)$ fixed hai; agar $n$ double ho, $λ_{T}^{3}$ halve hona chahiye, toh $λ_{T} \propto T^{- 1/2}$ ke saath hume milta hai $T \propto λ_{T}^{- 2} \propto (n^{1/3})^{2} = n^{2/3}$ . Consistent, aur $2^{2/3} = 1.587$ . ✓

Active recall

Recall Kaun se cells students ko sabse zyaada trap karte hain?

Q: Do "silent" traps aur unka fix batao. A: (1) $T > T_{c}$ ko $1 - (T / T_{c})^{3/2}$ mein plug karne par negative "fraction" aati hai — $0$ par clamp karo (cell D). (2) Yeh assume karna ki BEC kisi bhi dimension mein hoti hai — 2D saturation integral diverge karta hai, toh condensation nahi hoti (cell H).

Recall Ek saanch mein scaling laws

Q: $T_{c}$ $n$ aur $g$ ke saath kaise scale karta hai? A: $T_{c} \propto n^{2/3}$ (denser $\Rightarrow$ garm transition) aur $T_{c} \propto g^{- 2/3}$ (zyaada degeneracy $\Rightarrow$ thandi transition).

Condensate-fraction curve $::: N_{0} / N = 1 - (T / T_{c})^{3/2}$ $T \leq T_{c}$ ke liye, aur $T_{c}$ ke upar flat $0$

T > T_{c}

par "fraction formula" ka sign tumhe kis cheez ki warning deta hai?

Tum uske domain se bahar aa gaye ho — sahi answer

0

hai, koi condensate nahi.