2.3.1 · Physics › Modern Physics
Intuition Ek-sentence ki kahani
Ek garam object glow karta hai, aur kaise glow karta hai (colours ka spectrum) classical physics ko follow karne se mana kar diya. Planck ne ise ek wild guess se fix kiya: ==energy of oscillators is quantized in lumps of h ν ==. Isi ek guess ne quantum mechanics shuru ki.
Ek blackbody ek idealized object hai jo absorbs all radiation falling on it (kuch reflect nahi karta) aur, jab garam hota hai, ==emits a characteristic spectrum that depends only on its temperature T ==, na ki uske material ya shape par.
YEH kyun important hai: Kyunki emit kiya hua spectrum sirf T par depend karta hai, yeh ek universal function hai — fundamental physics ke liye ek perfect testbed. Ek cavity mein chhota sa hole real-world ka sabse accha blackbody hai: andar jaane wali koi bhi ray bounce karti rahti hai aur bahar nikalne se pehle absorb ho jaati hai.
Intuition Hole = blackbody kyun
Ek photon jo kisi cave mein ghus jaata hai, use waapas bahar aane ka raasta lagbhag kabhi nahi milta — woh walls par absorb ho jaata hai. Isliye hole bilkul black dikhta hai. Jab cavity garm hoti hai, toh hole andar trapped equilibrium radiation emit karta hai.
Spectral curve u ( ν , T ) badhti hai, peak karti hai, phir high frequency par zero ho jaati hai.
Wien's displacement law: peak wavelength temperature ke saath shift hoti hai: λ peak T = b , jahan b ≈ 2.898 × 1 0 − 3 m⋅K .
Stefan–Boltzmann law: total emitted power ∝ T 4 .
Definition Rayleigh–Jeans law (classical)
Ek cavity mein electromagnetic standing-wave modes count karke aur har ek ko classical average energy k B T (equipartition) dene par spectral energy density milti hai:
u ( ν , T ) d ν = c 3 8 π ν 2 k B T d ν
YEH kaise milta hai (first principles se):
[ ν , ν + d ν ] mein modes ka number per unit volume c 3 8 π ν 2 d ν hai (standing waves ki geometry, ×2 polarizations).
Classical equipartition: har mode (ek harmonic oscillator) ki average energy ⟨ E ⟩ = k B T hoti hai.
Multiply karo: energy density = (modes)×(energy per mode).
Common mistake Steel-man: "Har mode ko
k B T milta hai — yeh toh sahi hi hai na?"
Kyun sahi lagta hai: Equipartition ek rock-solid classical theorem hai — har quadratic degree of freedom ko 2 1 k B T milta hai, ek oscillator mein 2 hote hain (KE+PE) → k B T . Gases ke liye toh kaam karta hai!
Catastrophe: Modes ν 2 ke saath scale karti hain, isliye u → ∞ jab ν → ∞ . Total energy ∫ 0 ∞ ν 2 d ν = ∞ . Tumhara oven infinite UV/X-ray energy emit karta. Bilkul bakwaas.
Fix: Galti mode-counting mein nahi hai — galti yeh assume karne mein hai ki har mode koi bhi (continuous) amount of energy rakh sakti hai. Planck ne energy ko quantize kiya.
Definition Planck's postulate
Frequency ν ka ek oscillator sirf discrete steps mein energy rakh sakta hai:
E n = nh ν , n = 0 , 1 , 2 , …
jahan h = 6.626 × 1 0 − 34 J⋅s Planck's constant hai. Energy h ν ke size ke lumps ("quanta") mein exchange hoti hai.
Intuition Yeh catastrophe ko kyun khatam karta hai
Low ν par (h ν ≪ k B T ): ek quantum chhota hota hai, energy continuous lagti hai → ⟨ E ⟩ → k B T (classical recover hota hai, achha!). High ν par (h ν ≫ k B T ): ek quantum itni energy maangta hai jo thermally available nahi hai, isliye woh modes "frozen out" ho jaate hain — e − h ν / k B T se exponentially suppress ho jaate hain. Koi infinity nahi.
Worked example Stefan–Boltzmann (
∝ T 4 ) — Yeh step kyun?
Total energy density U = ∫ 0 ∞ u ( ν , T ) d ν . Substitute x = h ν / k B T ⟹ ν = h k B T x , d ν = h k B T d x .
Kyun substitute karte hain? Saari T dependence ko saamne nikalne ke liye.
U = c 3 8 π h ( h k B T ) 4 ∫ 0 ∞ e x − 1 x 3 d x
Integral ek pure number hai = π 4 /15 . Isliye U ∝ T 4 . 4th power numerator mein ν 3 + d ν + substitution se 1/ h 3 aane se aata hai = T ke chaar factors. ✓
Worked example Wien's displacement law — Yeh step kyun?
u ( λ , T ) maximize karo: ∂ λ ∂ u = 0 set karo. y = h c / λ k B T maano.
Kyun? Peak wahan hoti hai jahan slope zero ho. Differentiate karne par transcendental equation milti hai:
5 ( 1 − e − y ) = y ⇒ y ≈ 4.965
Isliye λ peak T = 4.965 k B h c = b ≈ 2.898 × 1 0 − 3 m⋅K . ✓
Worked example Numerical: Sun ka temperature
Sun ka spectrum λ peak ≈ 500 nm ke paas peak karta hai. Wien use karke:
T = λ peak b = 500 × 1 0 − 9 2.898 × 1 0 − 3 ≈ 5800 K
Yeh kyun kaam karta hai: Wien's law ek measurable peak colour ko surface temperature se link karta hai — astronomers exactly yahi use karte hain.
Intuition Agar sirf yahi yaad rakhna hai
Classical = k B T per mode ⟹ UV catastrophe.
Planck: energy h ν ke lumps mein aati hai ⟹ high-ν modes frozen out ⟹ ⟨ E ⟩ = e h ν / k B T − 1 h ν .
Sab kuch (Planck law, Stefan T 4 , Wien λ p e ak T = b ) usi ek average energy se nikalta hai.
Recall Feynman: ek 12-saal ke bachche ko explain karo
Socho ek piano hai jahan tum sirf poore keys daba sakte ho, kabhi "aadha key" nahi. Low notes ke liye keys easy aur sasti hain, isliye warm room unhe bahut dabaata hai — yahi visible glow hai. Lekin bahut high notes dabane ke liye ek baar mein bhi bahut zyada energy chahiye, isliye ek garam room bhi unhe afford nahi kar sakta — woh chup rahte hain. Old physics sochti thi tum key ka koi bhi chhota sa fraction daba sakte ho, isliye usne predict kiya ki room infinitely many ultra-high notes chilla raha hai ("ultraviolet catastrophe"). Planck ne kaha: nahi, energy h ν ke poore keys mein aati hai. Us ek rule ne silly infinity ko gayab kar diya — aur accidentally poori quantum physics launch kar di.
Mnemonic Average energy yaad karo
"H-nu over EXP minus one" → e h ν / k B T − 1 h ν . Woh "− 1 " quantum signature hai; ise hata do aur classical k B T behaviour waapis aa jaayega. Woh akela "− 1 " hi universe ko UV catastrophe se bachata hai.
Blackbody kya hota hai? Ek ideal body jo saari incident radiation absorb karta hai aur sirf apne temperature par depend karta spectrum emit karta hai.
Ek garam cavity mein chhota sa hole ideal blackbody kyun hai? Andar jaane wali radiation bounce karti hai aur absorb ho jaati hai escape se pehle (absorptivity ≈ 1), isliye yeh equilibrium thermal radiation emit karta hai.
Rayleigh–Jeans law kya hai? u ( ν , T ) d ν = c 3 8 π ν 2 k B T d ν — classical mode density × k B T .
Ultraviolet catastrophe kya hai? Rayleigh–Jeans predict karta hai u → ∞ jab ν → ∞ (aur infinite total energy), jo physically impossible hai.
Planck ki hypothesis kya thi? Oscillators sirf E n = nh ν energies rakh sakte hain (energy h ν ke lumps mein quantized hai).
Planck oscillator ki average energy? ⟨ E ⟩ = e h ν / k B T − 1 h ν .
Quantization catastrophe ko kyun rokta hai? High-ν modes ko ek quantum h ν ≫ k B T chahiye hota hai, isliye woh exponentially suppress ho jaate hain (frozen out).
Planck's radiation law (frequency form) batao. u ( ν , T ) d ν = c 3 8 π ν 2 ⋅ e h ν / k B T − 1 h ν d ν .
Planck's law ka low-frequency limit? Rayleigh–Jeans reduce ho jaata hai, c 3 8 π ν 2 k B T .
Planck's law ka high-frequency limit? Wien's law reduce ho jaata hai, c 3 8 π h ν 3 e − h ν / k B T .
Wien's displacement law? λ peak T = b ≈ 2.898 × 1 0 − 3 m·K.
Planck se Stefan–Boltzmann law? Total energy density U ∝ T 4 (from ∫ x 3 / ( e x − 1 ) d x = π 4 /15 ).
Planck's constant ki value? h ≈ 6.626 × 1 0 − 34 J·s.
Photoelectric effect — E = h ν quanta real photons ban jaate hain.
Wien's displacement law · Stefan–Boltzmann law — limiting cases yahan derive hue.
Equipartition theorem — classical assumption jo fail hoti hai.
Bose–Einstein statistics — e x − 1 factor ki modern derivation.
Cosmic Microwave Background — sabse perfect blackbody known (T = 2.725 K).
Quantum harmonic oscillator — jahan E n = nh ν ban jaata hai E n = ( n + 2 1 ) h ν .
flaw is continuous energy
Blackbody absorbs all emits by T only
Mode counting 8pi nu^2 / c^3
Classical equipartition kBT
Planck postulate E=n h nu