Visual walkthrough — Blackbody radiation — Planck's quantum hypothesis

2.3.1 · D2 · Physics › Modern Physics › Blackbody radiation — Planck's quantum hypothesis

Hum ek number chase kar rahe hain: ek hot box mein har colour par kitni light energy hoti hai. "Colour" yahan matlab hai frequency $\nu$ (Greek "nu") — har second mein kitne wave wiggles hote hain. Zyada $\nu$ = bluer/zyada energetic light. Shuru karne ke liye bas yahi ek jargon chahiye.

Step 1 — Light ko ek box mein band karo aur uske "modes" gino

KYA. Light ko ek hollow metal box ("cavity") ke andar daalo. Light ek wave hai, isliye sirf woh waves survive karti hain jo walls ke beech perfectly fit hoti hain — jaise ek guitar string jo dono ends par still honi chahiye. Har ek allowed wave pattern ko mode kehte hain.

KYUN. Hum "saari light" track nahi kar sakte — yeh ek mess hai. Lekin hum allowed standing-wave patterns ki list bana sakte hain aur pooch sakte hain ki har frequency ke paas kitne hain. Yeh ek vague sawaal ko counting mein convert kar deta hai.

PICTURE. Neeche box dekho. Sabse chhota mode half a wave fit karta hai; agli ek poori wave; phir dedh, aur isi tarah. Har pattern ko har direction mein half-waves ki ek whole number se label kiya gaya hai.

Step 2 — Mode count ${\nu }^2$ ki tarah kyun badhta hai ($k$-space picture)

KYA. 3D mein, har allowed mode teen whole numbers $(n_x,n_y,n_z)$ se fix hota hai — box ki har edge ke along kitne half-waves fit hote hain. Har mode ko grid mein position $(n_x,n_y,n_z)$ par ek dot ki tarah plot karo. Dots ek perfect grid banate hain, ek unit cube mein ek dot.

KYUN. Ek mode ki frequency sirf is grid mein origin se uski distance par depend karti hai, $R=\sqrt{n_x^2+n_y^2+n_z^2}$ (kyunki $\nu \propto R$). Toh "frequency $\nu$ se neeche kitne modes hain?" ka sawaal ban jaata hai "radius $R$ ke sphere ke andar kitne grid dots hain?" — ek visual counting sawaal, shape dekhne ke liye koi algebra ki zaroorat nahi.

PICTURE. Neeche, eighth-of-a-sphere ke andar ke dots (hum sirf positive $n_x,n_y,n_z$ rakhte hain) frequency $\nu$ tak ke modes hain. Unki sankhya ≈ shell ki volume hai. $\nu$ aur $\nu +d\nu$ ke beech ek patli shell mein modes ki count shell ki surface area $\times$ thickness hai. Ek sphere ki surface $R^2$ ki tarah badhti hai, aur $R\propto \nu$, isliye shell count ${\nu }^2$ ki tarah badhta hai. Yahi woh jagah hai jahan se ${\nu }^2$ aata hai — yeh literally ek sphere ki surface area hai.

2 ka factor — do wiggle directions. In spatial patterns mein se har ek ke liye, light ka electric field do independent sideways directions mein wave kar sakta hai (imagine karo field aapar-neeche vibrate kar raha hai versus left-right jabki wave aage travel karti hai). Yeh do independent shake-directions do polarizations hain; har ek genuinely alag mode hai, isliye hum count ko 2 se multiply karte hain. Neeche picture same wave ko uske do perpendicular shake-directions ke saath dikhati hai.

Is shape ko yaad rakho: modes ${\nu }^2$ ki tarah pile up hote hain. Woh ${\nu }^2$ is kahani ka villain hai.

Step 3 — Poochho: ek MODE kitni energy rakhta hai?

KYA. Har mode ek chhota oscillator hai (yeh ek spring ki tarah aage-peechhe swing karta hai). Hum chahte hain uski average energy $⟨E⟩$ jab box temperature $T$ par baitha ho.

KYUN. Total light energy = (modes ki sankhya) × (har mode kitni energy carry karta hai). Steps 1–2 ne pehla factor diya. Yeh step poori ladaai hai: doosra factor kya hai?

PICTURE. Classical physics kehti hai: temperature har oscillator ko equally shake karti hai, aur har ek same average energy $k_{B}T$ rakhta hai — ek flat line, har colour ke liye same. ($k_B$ = Boltzmann's constant, temperature aur energy ke beech exchange rate.)

Equal-slice rule Equipartition theorem hai — ek rock-solid classical result jo yahan hume ek cliff se girata hai.

Step 4 — Ultraviolet Catastrophe (classical answer absurd kyun hai)

KYA. Steps 1–2 × Step 3 multiply karo: modes ($\propto {\nu }^2$) × energy each ($k_{B}T$, flat).

KYUN. Hum classical recipe ko imaandaari se uske end tak follow karte hain aur dekhte hain ki woh toot jaati hai — yeh hume exactly batata hai ki kaun si assumption fix karni hai.

PICTURE. Product ek aisa curve hai jo sirf badhta rehta hai: ${\nu }^2\cdot k_{B}T$. Jaise-jaise hum bluer hote hain yeh upar ki taraf rocket karta hai. Iske neeche ka area (total energy) infinite hai. Ek garam oven infinite X-rays blast karega. Aisa nahi hota. Nature kehti hai NAHI.

Step 5 — Planck ka rule: energy whole lumps mein aati hai

KYA. Planck ek oscillator ko koi bhi amount of energy rakhne se mana karta hai. Woh sirf ek smallest lump $h\nu$ ke whole multiples rakh sakta hai: $E_n=nh\nu ,n=0,1,2,3,\dots$

KYUN. Agar high-frequency modes ko excite karna kisi tarah expensive hai, toh woh empty rahenge aur runaway ${\nu }^2$ growth choke ho jaayegi. Lump size $h\nu$ ko frequency ke proportional banana crucial move hai: bluer light = mehnge tickets.

PICTURE. Do energy ladders compare karo. Low-$\nu$ mode ke liye rungs tiny aur close hain — thermal jiggling easily kaafi rungs climb kar leti hai, toh woh "continuously" behave karta hai, classical case ki tarah. High-$\nu$ mode ke liye rungs enormous hain — pehla rung akela itna cost karta hai jo box usually afford nahi kar sakta, toh mode ground floor par stuck rehta hai ($n=0$, zero energy). Woh frozen out hai.

Yahi lump idea Photoelectric effect aur Quantum harmonic oscillator ko bhi power deta hai.

Step 6 — Ladder weigho: raw weights (abhi tak true probabilities nahi)

KYA. Temperature $T$ par, har rung $n$ ko ek Boltzmann weight $w_n=e^{-E_n/k_{B}T}$ milta hai: high rungs ko exponentially chhote weights milte hain. Yeh raw weights hain, abhi finished probabilities nahi — yeh 1 tak add nahi hote. Inhe true probabilities mein convert karne ke liye hum inhe unke total (the "partition sum") se divide karenge Step 7 mein: $P(E_n)=\frac{w_n}{{\sum }_m{w}_m}=\frac{e^{-nx}}{{\sum }_{m=0}^{\infty }{e}^{-mx}}.$

KYUN. Average energy nikaalने ke liye hum har rung ki energy ko us rung ki likelihood se weight karte hain. Saste rungs (low $E$) ko bada weight milta hai; costly rungs (high $E$) ko tiny weight milta hai. Average ek tug-of-war hai "upar aur rungs hain" aur "har ek exponentially rarer hai" ke beech. Hum weights abhi unnormalized rakhte hain kyunki normalizing total hi woh do sums mein se ek hai jo hum aage compute karte hain.

PICTURE. Bars jinki heights raw weights $e^{-nx}$ hain (normalize karne se pehle), jahan $x=h\nu /k_{B}T$ price-to-budget ratio hai — ek lump ki cost divided by available thermal energy. Chhota $x$ (saste lumps): bars slowly fall off hote hain, kaafi rungs occupied. Bada $x$ (mehnge lumps): sirf $n=0$ bar tall hai — mode frozen baitha hai.

Step 7 — Average nikalo (do tidy sums)

KYA. Compute karo $⟨E⟩=\frac{{\sum }_{n=0}^{\infty }(nh\nu )e^{-nx}}{{\sum }_{n=0}^{\infty }{e}^{-nx}}.$ Top = har rung ki energy × uska raw weight, summed. Bottom = total weight (partition sum); isse divide karna exactly wahi hai jo Step 6 ke raw weights ko true probabilities mein convert karta hai jo 1 tak add hote hain.

KYUN har tool.

• Geometric series bottom ke liye: $\sum e^{-nx}$ hai $1+r+r^2+\dots$ jahan $r=e^{-x}<1$, ek aisi sum jo $\frac{1}{1-r}$ par close hoti hai. Hum ise isliye use karte hain kyunki yeh infinite ladder ka exact closed form hai.
• Derivative trick top ke liye: $\sum n{e}^{-nx}=-\frac{d}{dx}\sum e^{-nx}$. Hum differentiation isliye use karte hain kyunki $n$ ka factor neeche kheenchna exactly wahi hai jo $-\frac{d}{dx}$ karta hai $e^{-nx}$ ko — yeh ek mushkil weighted sum ko ek easy wali ki derivative mein convert kar deta hai.

PICTURE. Messy weighted sum ek clean fraction mein collapse ho jaata hai. $x$ ko $⟨E⟩$ ke curve ke along chhote se bade slide hote dekho: chhote $x$ par yeh classical line $k_{B}T$ se chipka rehta hai; bade $x$ par zero ki taraf girta hai.

$\text{Bottom}={\sum }_{n=0}^{\infty }{e}^{-nx}=\frac{1}{1-e^{-x}}$ $\text{Top}=h\nu {\sum }_{n=0}^{\infty }n{e}^{-nx}=h\nu \left(-\frac{d}{dx}\frac{1}{1-e^{-x}}\right)=h\nu \frac{e^{-x}}{(1-e^{-x}{)}^2}$

Step 8 — Dono edges check karo (koi scenario unchecked nahi)

KYA. Formula ko wahan test karo jahan hume pehle se answer pata hai: bahut low aur bahut high frequency.

KYUN. Ek correct formula ko classical physics par reduce karna chahiye jahan lumps negligible hain, aur wahan vanish karna chahiye jahan lumps ruinous hain. Agar nahi hota, toh humne koi galti ki hogi.

PICTURE. Poora $⟨E⟩$ curve uske do behaviours highlight karke — left mein flat-topped classical plateau, right mein exponential cliff.

Left edge isliye hai ki kuch blast nahi hua: $k_{B}T$ ka flat march infinite colours tak jaane ki bajay, curve ka right side mar jaata hai. Step 4 ke runaway se compare karo — yahi rescue hai.

Step 9 — Planck's law assemble karo

KYA. Steps 1–2 (mode density) ko Step 7 (energy per mode) se multiply karo — same recipe Step 4 jaisi, lekin sahi doosre factor ke saath.

KYUN. Ab ${\nu }^2$ pile-up exponential freeze-out se milta hai. Unka product badhta hai, peak karta hai, phir girta hai — exactly measured glow curve.

PICTURE. Catastrophe curve (grey, exploding) aur Planck's curve (coral, peaking and falling) saath mein draw ki gayi hain. Freeze-out factor rising ${\nu }^2$ ko zero ki taraf wapas mod deta hai.

Ek picture summary

Poora safar ek canvas par: (1–2) modes gino (${\nu }^2$, rising, sphere ki surface se) → (3) classical har ek ko $k_{B}T$ deta hai (flat) → (4) product explode karta hai: catastrophe → (5) Planck $h\nu$ per lump charge karta hai → (6–7) weighted average ban jaata hai $\frac{h\nu }{e^{h\nu /k_{B}T}-1}$, jo (8) high colours ko freeze karta hai → (9) product ab peak karta hai aur girta hai: real glow.

Flashcards

Neeche har card us figure ki taraf point karta hai jo uska answer prove karta hai — pehle picture yaad karo, phir words.