2.4.16 · D3Bose-Einstein statistics — bosons
Yeh page ek shooting range hai: har tarah ka sawal jo Bose-Einstein parent note pooch sakta hai, zero se worked out. Examples se pehle, hum har case class ka ek matrix banate hain taaki tum dekh sako ki koi gap nahi hai.
Neeche sab kuch ek hi formula use karta hai, Bose-Einstein mean occupancy:
n ˉ ( ε ) = e ( ε − μ ) / k B T − 1 1 .
Pehle hum har symbol ko naam dete hain taaki koi bhi symbol bina samjhe use na ho.
Definition Symbols, seedhi zubaan mein
ε energy (socho: ek seat ki price ). Units: joules (J) ya, suvidha ke liye, k B T μ chemical potential (discount ): reservoir kitni energy deta hai jab tum use ek particle dete ho. Dekho Chemical potential .T temperature ; k B 1.381 × 1 0 − 23 k B T thermal energy scale , woh natural yardstick jiske against hum ε − μ Hum ek dimensionless number ko abbreviate karte hain jo poora show chalata hai:
y ≡ k B T ε − μ . n ˉ = e y − 1 1 Sab kuch neeche bas itna hai: y n ˉ
y ε μ T y = ( ε − μ ) / k B T seat price hai thermal units mein, discount ke baad . Chota y y
Curve dekho: yeh infinity ki taraf dive karta hai jab y → 0 + e − y y
n ˉ case classes mein se exactly ek mein aata hai. Neeche ke examples mein se har ek ek ya zyada cells cover karta hai.
Cell
Case class
Defining condition
Covered by
A
Photon gas (no μ μ = 0 y = ε / k B T Ex 1, Ex 6
B
Intermediate y (quantum bunching)y ∼ 0.1 1 Ex 2
C
Classical tail (dilute limit)y ≫ 1 Ex 3
D
Degenerate: y → 0 + (near-condensation)μ → ε m i n − Ex 4
E
Forbidden input (y < 0 μ > ε n ˉ Ex 5
F
Real-world word problem T ε Ex 6
G
Exam twist : ratio / BE-vs-FDsame y
Ex 7
H
Temperature sweep (limiting behaviour)fixed ε − μ T
Ex 8
Recall Poora matrix ek sentence meinSawal ::: Har BE problem reduce hoti hai "kya hai y = ( ε − μ ) / k B T n ˉ = 1/ ( e y − 1 ) y > 0 y → 0 y → ∞
Worked example Ex 1 · Ek light mode jis ki energyε = k B T Statement. Ek cavity temperature T ε = k B T
Forecast: compute karne se pehle andaza lagao — 1 se zyada, ya 1 se kam? (Photons social hote hain, lekin yeh seat sasti nahi hai…)
Step 1 — μ = 0
Yeh step kyun? Photons cavity walls se create aur destroy ho sakte hain, toh unka number conserved nahi hai ; ek add karne ki thermodynamic cost zero hai, jisse μ = 0 Planck radiation law derivation). Tab y = k B T ε − 0 = k B T ε
Step 2 — ε = k B T
Yeh step kyun? Humein bataya gaya hai ki numerator denominator ke barabar hai, toh y = 1 n ˉ = e 1 − 1 1 = 2.71828 − 1 1 = 1.71828 1 .
Step 3 — Evaluate karo.
n ˉ ≈ 0.582.
Verify: n ˉ < 1 0.582 × ( e − 1 ) = 0.582 × 1.718 = 1.000 k B T wajah hai blackbody intensity high frequency par kyun girती है.
Worked example Ex 2 · Bosonsμ Statement. Ek boson level ka y = ( ε − μ ) / k B T = 0.1 n ˉ
Forecast: net seat bahut sasti hai (y
Step 1 — Small-y
Yeh step kyun? y = 0.1 ≪ 1 n ˉ
Step 2 — Exponential compute karo phir ek subtract karo.
Yeh step kyun? Poora quantum effect "− 1 e 0.1 = 1.10517 , e 0.1 − 1 = 0.10517.
Step 3 — Invert karo.
n ˉ = 0.10517 1 ≈ 9.51.
Verify: Chote y e y − 1 ≈ y n ˉ ≈ 1/ y = 1/0.1 = 10 9.51 y n ˉ ≈ k B T / ( ε − μ ) e − 0.1 = 0.905 10× chota hai: yeh 10 ka factor boson bunching hai, lasers aur Bose-Einstein condensation ka beej.
Worked example Ex 3 · High-energy state Maxwell–Boltzmann recover karta haiStatement. Ek boson level ka y = 5 n ˉ BE n ˉ MB = e − y
Forecast: s01 par bilkul right — kya BE aur classical yahan agree karte hain, ya clash?
Step 1 — BE value.
Yeh step kyun? Direct substitution; humein expect hai ki "− 1 e 5 e 5 = 148.413 , n ˉ BE = 148.413 − 1 1 = 147.413 1 = 0.006784.
Step 2 — Classical value.
Yeh step kyun? Yeh Maxwell-Boltzmann statistics estimate hai — woh limit jis ke paas BE ko aana chahiye.
n ˉ MB = e − 5 = 0.006738.
Step 3 — Percent difference.
0.006738 0.006784 − 0.006738 × 100% = 0.68%.
Verify: Dono 1% ε − μ ≫ k B T − 1 n ˉ
Worked example Ex 4 · Ground state jabμ → ε m i n − Statement. Ground state ε m i n n ˉ 0 μ y = 0.01 y = 0.001
Forecast: discount μ
Step 1 — y = 0.01
Yeh step kyun? Hum forbidden point y = 0 n ˉ 0 = e 0.01 − 1 1 = 0.0100502 1 ≈ 99.5.
Step 2 — y = 0.001
Yeh step kyun? y 1/ y n ˉ 0 = e 0.001 − 1 1 = 0.00100050 1 ≈ 999.5.
Step 3 — Trend padho.
Yeh step kyun? Step 1 se Step 2 tak, y × 10 n ˉ 0 × ∼ 10 n ˉ 0 → ∞ y → 0 +
Verify: Law n ˉ 0 ≈ 1/ y 100 1000 99.5 999.5 divergence Bose-Einstein condensation ka mathematical chehra hai: μ = ε m i n 1/ ( 1 − x ) μ ε m i n
Worked example Ex 5 · Agarμ > ε Statement. Ek student ε = k B T μ = 2 k B T y = − 1 n ˉ
Forecast: n ˉ
Step 1 — y
y = k B T ε − μ = k B T k B T − 2 k B T = − 1.
Step 2 — Substitute karo (blindly, trap expose karne ke liye).
Yeh step kyun? Hum plug in karte hain taaki dekh sakein kyun negative y e − 1 = 0.3679 , e − 1 − 1 = − 0.6321 , n ˉ = − 0.6321 1 = − 1.582.
Step 3 — Physical grounds par reject karo.
Yeh step kyun? Particles ki count kabhi negative nahi ho sakti. Formula ne n ˉ < 0 μ > ε forbidden hai. Condition μ < ε m i n Grand canonical ensemble geometric-sum convergence x < 1 har level ke liye hold karni chahiye.
Verify: − 1.582 × ( e − 1 − 1 ) = − 1.582 × ( − 0.6321 ) = 1.000 physical hai, arithmetic nahi. ✓ Rule lock in: n ˉ y > 0
Common mistake Negative occupancy par trust karnaKyun sahi lagta hai: formula ne "ek answer diya," toh zaroor kuch matlab hoga. Fix: BE sirf y > 0 n ˉ cheekh ke bolna hai ki tumne μ < ε m i n
Worked example Ex 6 · Cosmic Microwave Background photon modeStatement. CMB space ko T = 2.725 ν = 160.2 ε = h ν h = 6.626 × 1 0 − 34 n ˉ
Forecast: spectral peak par, kya n ˉ k B T
Step 1 — Photon ⇒ μ = 0
Yeh step kyun? Same wajah jaise Ex 1 mein: photon number conserved nahi hai, μ = 0 y = h ν / k B T
Step 2 — ε = h ν
ε = 6.626 × 1 0 − 34 × 1.602 × 1 0 11 = 1.0615 × 1 0 − 22 J .
Step 3 — k B T
k B T = 1.381 × 1 0 − 23 × 2.725 = 3.763 × 1 0 − 23 J .
Step 4 — y
Yeh step kyun? y y = 3.763 × 1 0 − 23 1.0615 × 1 0 − 22 = 2.821 , n ˉ = e 2.821 − 1 1 = 16.79 − 1 1 = 15.79 1 ≈ 0.0633.
Verify: y ≈ 2.82 ≈ 2.82 radiance peak ke liye — Planck radiation law ke saath ek sundar consistency. ✓ Units pure number mein cancel hue. ✓ n ˉ ≈ 0.063
Worked example Ex 7 · Samey Statement. y = 0.5 − 1 + 1 e − y n ˉ
Forecast: kaun sabse bada hai — social boson, antisocial fermion, ya classical middle?
Step 1 — Common exponential.
Yeh step kyun? Teeno e 0.5 = 1.64872
Step 2 — BE.
n ˉ BE = 1.64872 − 1 1 = 0.64872 1 = 1.5415. Step 3 — FD (dekho Fermi-Dirac statistics — fermions ).
Yeh step kyun? Sirf sign flip hota hai — "+ 1 n ˉ FD = 1.64872 + 1 1 = 2.64872 1 = 0.3775. Step 4 — Classical.
n ˉ MB = e − 0.5 = 0.60653.
Step 5 — Rank karo. n ˉ BE ( 1.54 ) > n ˉ MB ( 0.61 ) > n ˉ FD ( 0.38 ) .
Verify: Bosons classical se upar bunch karte hain , fermions neeche anti-bunch karte hain — classical value hamesha dono ke beech rehti hai, bilkul jaise theory demand karti hai. ✓ Note karo n ˉ FD < 1 n ˉ BE
Bar chart ordering ko visual banata hai: cyan boson tower sabse upar, amber fermion sabse neeche, white classical beech mein.
ε − μ T Statement. Ek level ka fixed ε − μ = Δ n ˉ k B T = Δ y = 1 k B T = 10Δ y = 0.1 k B T = Δ/5 y = 5 T → 0 T → ∞
Forecast: boson gas ko heat karo — har level ki bheed badhti hai ya ghatti hai?
Step 1 — (a) y = 1
Yeh step kyun? Baseline, same arithmetic jaise Ex 1 mein.
n ˉ = e 1 − 1 1 = 0.5820. Step 2 — (b) hot, y = 0.1
Yeh step kyun? T y n ˉ = e 0.1 − 1 1 = 9.508. Step 3 — (c) cold, y = 5
Yeh step kyun? T y n ˉ = e 5 − 1 1 = 0.006784.
Step 4 — Limits padho.
Yeh step kyun? Trend ko endpoints tak extrapolate karo.
T → ∞ y → 0 + n ˉ → ∞ T → 0 y → ∞ n ˉ → 0 Bose-Einstein condensation ).
Verify: (c)→(a)→(b) se, heating (T n ˉ 0.0068 → 0.582 → 9.51 y = Δ/ k B T
Mnemonic One-number workflowy y y = ( ε − μ ) / k B T y > 0 n ˉ = 1/ ( e y − 1 ) y y B oson = B elow-by-one = woh − 1
