Visual walkthrough — Blackbody radiation from statistical mechanics — Planck distribution

2.4.19 · D2 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Blackbody radiation from statistical mechanics — Planck dist

Hum destination formula likhne se pehle, do nature ke constants ke naam jaanna zaroori hai jo us formula ke andar aayenge. Unhe hum properly Steps 6 aur 7 mein (pictures ke saath) milenge — yahan sirf unka plain-word meaning dete hain taaki koi symbol unexplained na rahe.

Hamari destination — woh cheez jiske liye har step build ho rahi hai:

$u(\nu ,T)d\nu =\underset{\text{how many modes}}{\underbrace{\frac{8\pi {\nu }^2}{c^3}}}\times \underset{\text{energy per mode}}{\underbrace{\frac{h\nu }{e^{h\nu /k_{B}T}-1}}}d\nu$

Yahan $u(\nu ,T)$ hai energy per unit volume per unit frequency — box mein kitni light-energy colour $\nu$ aur colour $\nu +d\nu$ ke beech baithti hai. Chhota $d\nu$ frequency ka ek patla slice hai. $\nu$ (Greek "nu") frequency hai, $T$ temperature hai, $c$ speed of light hai, aur $h,k_B$ woh do constants hain jo abhi introduce kiye. Baaki sab hum chalte chalte define karenge.

Step 1 — Light ko ek box mein daalo

KYA. Imagine karo ek hollow metal cube, side length $L$. Iske walls ko temperature $T$ tak heat karo. Walls glow karte hain, andar electromagnetic (light) waves se bharte hue. Equilibrium mein andar bounce karti light ko standing waves banana padta hai — aisi patterns jo drift nahi karti, jaise ek plucked guitar string.

KYUN. Ek wave jo standing wave nahi hai woh energy ko hamesha idhar-udhar slosh karti rahegi aur kabhi settle nahi karegi. Sirf standing patterns stable hain, isliye hum sirf unhi ko count karte hain. Ek standing wave ko dono walls par zero par pinned hona chahiye (metal wahan field ko short out kar deta hai), bilkul aise jaise ek rope dono ends par bandhi ho.

PICTURE. Figure dekho. Pehla pattern box ke across aadha wavelength fit karta hai, agla ek pura wavelength fit karta hai, phir dedh, aur aise aage. Sirf half-wavelengths ke whole numbers fit hote hain.

Kyunki $x$ axis ke saath half-wavelengths ka ek whole number fit hona chahiye, wave ki "tightness" $x$ ke saath sirf special values le sakti hai:

$k_x=\frac{n_x\pi }{L}$

• $k_x$ — $x$ ke saath wavenumber: har metre mein kitne radians of wave. Bada $k_x$ = tighter ripples.
• $n_x$ — ek counting integer $1,2,3,\dots$: kaun sa harmonic (kitne bumps).
• $\pi /L$ — sabse chhota step; har extra bump $k_x$ mein itna add karta hai.

Yahi $y$ aur $z$ ke liye bhi apne integers $n_y,n_z$ ke saath hold karta hai.

Step 2 — Har mode ek "mode-space" mein ek dot hai

KYA. Ek mode teen integers $(n_x,n_y,n_z)$ se fix hota hai. Har triple ko 3D grid par ek point ki tarah plot karo. Har mode = ek dot, aur dots exactly ek unit apart har direction mein baithte hain.

KYUN. Waves count karna mushkil hai; evenly-spaced dots count karna aasaan hai. "Kitne modes" ko "kitne grid points" mein badal ke, hum ek-ek add karne ki jagah volume use karke count kar sakte hain.

PICTURE. Figure pehle octant mein dots ka grid dikhata hai (teeno integers positive). Ek dot exactly ek unit cube of space occupy karta hai, isliye kisi region ke andar dots ki sankhya ≈ us region ka volume.

Hum measure karte hain ki ek dot corner se kitni door baitha hai:

$R=\sqrt{n_x^2+n_y^2+n_z^2}$

• $R$ — mode-space mein radius, origin se dot tak ki seedhi-line distance. (Hum $R$ use karte hain, $n$ nahi, taaki woh counting integer $n$ se kabhi clash na ho jise hum Step 7 mein milenge.)
• Sum-of-squares ka square-root sirf 3D mein Pythagoras hai: teen counts ko ek distance mein badalta hai.

Same $R$ wale saare dots radius $R$ ke sphere par lie karte hain. Yahi key hai: same "size" ke modes ek shell banate hain.

Step 3 — Ek sphere se modes count karo (dhyan se: ek-aathwan, aur sirf jab $R\gg 1$)

KYA. Radius $R$ tak ke saare modes count karne ke liye, radius $R$ ke sphere ka volume lo — lekin sirf woh corner rakhno jahan teeno integers positive hain.

KYUN. Kyunki har dot ek unit cube bharta hai, count hai hi volume. Lekin $n_x,n_y,n_z$ positive hone chahiye (koi "minus-first harmonic" nahi hota), isliye sphere ka sirf $\frac{1}{8}$ count hota hai — corner ke sabse kareeb wala single octant.

PICTURE. Figure full sphere ko faded dikhata hai, aur solid amber wedge woh ek-aathwan hai jo hum rakhte hain.

$N(R)=\underset{\text{keep one octant}}{\underbrace{\frac{1}{8}}}\times \underset{\text{full sphere volume}}{\underbrace{\frac{4}{3}\pi R^3}}=\frac{\pi R^3}{6}$

• $N(R)$ — radius $R$ tak ke saare modes ki total sankhya.
• $\frac{1}{8}$ — positive-integer restriction; ise drop karo toh aath guna overcount ho jaega.
• $\frac{4}{3}\pi R^3$ — radius $R$ ki ball ka ordinary volume.

Step 4 — Mode-count ko frequency mein translate karo

KYA. Physicists rang (frequency $\nu$) chahte hain, abstract radius $R$ nahi. Convert karo.

KYUN. Do facts $R$ ko $\nu$ se link karte hain. Wavevector magnitude hai $|\vec{k}|=R\pi /L$ (Step 1 ke $k$'s par Pythagoras). Aur koi bhi light wave satisfy karta hai $|\vec{k}|=2\pi \nu /c$ (wave speed = frequency × wavelength). Dono ko equal set karo.

PICTURE. Figure ek "dial converter" hai: $R$ knob ghuma, $\nu$ padho — ek straight-line (linear) relationship.

$\frac{R\pi }{L}=\frac{2\pi \nu }{c}⟹R=\frac{2L\nu }{c}$

• Left side: box geometry se wave tightness.
• Right side: frequency $\nu$ par light hone se wave tightness.
• $\pi$'s cancel ho jaate hain; $L$ aur $c$ scale set karte hain. Bada box ya zyada $\nu$ → bada $R$.

$N=\pi R^3/6$ mein substitute karo:

$N(\nu )=\frac{\pi }{6}{\left(\frac{2L\nu }{c}\right)}^3=\frac{4\pi L^3{\nu }^3}{3c^3}$

Step 5 — Do polarizations, phir differentiate → density of states

KYA. Count double karo (light har direction mein do independent polarizations rakhti hai), box volume $L^3$ se divide karo taaki per-volume number mile, phir differentiate karo taaki ek thin frequency slice mein count mile.

KYUN.

• ×2: har wavevector ke liye, light do perpendicular directions mein wiggle kar sakti hai — do alag modes.
• ÷ $L^3$: hum material ki ek property chahte hain, apne particular box size ki nahi.
• Differentiate: hum "sab modes up to $\nu$" nahi chahte, hum chahte hain "kitne modes $\nu$ par ek sliver $d\nu$ ke andar." Derivative exactly hai "count kitni tezi se per unit frequency grow karta hai."

PICTURE. Figure $N(\nu )$ ko ek rising curve ki tarah dikhata hai; har point par uski slope $g(\nu )$ hai, woh sankhya jo ek patli vertical strip mein rehti hai.

$g(\nu )=\frac{1}{L^3}\frac{d}{d\nu }[2N(\nu )]=\frac{1}{L^3}\frac{d}{d\nu }\left[\frac{8\pi L^3{\nu }^3}{3c^3}\right]=\boxed{\frac{8\pi {\nu }^2}{c^3}}$

• $g(\nu )$ — density of states: modes per unit volume per unit frequency. Dekho Density of states.
• $L^3$ cancel ho jaata hai — acha hai, answer ko box size ki parwah nahi karni chahiye.
• ${\nu }^3$ ka derivative $3{\nu }^2$ hai; woh 3, denominator ke 3 ko cancel kar deta hai.

Step 6 — Classical energy per mode (aur yeh kyun explode karta hai)

KYA. Classical physics (Equipartition theorem) kehti hai ki har oscillator same average energy $k_{B}T$ hold karta hai, chahe uski frequency kuch bhi ho. Yahan $k_B$ Boltzmann's constant hai — temperature aur energy ke beech ka exchange rate — isliye $k_{B}T$ temperature $T$ par typical thermal energy hai.

KYUN yeh ek catastrophe hai. "Kitne modes" ko "har ek ki energy" se multiply karo:

$u(\nu )=g(\nu )\cdot k_{B}T=\frac{8\pi {\nu }^2}{c^3}{k}_{B}T$

Yeh Rayleigh-Jeans law hai. Jaise $\nu \to \infty$ woh ${\nu }^2$ hamesha chadhta rahta hai — box infinite energy hold kar leta. Woh impossible result hai ultraviolet catastrophe.

PICTURE. Figure true (measured) spectrum ko Rayleigh–Jeans curve ke against overlay karta hai jo infinity ki taraf rocket karti hai. Woh sirf low $\nu$ par agree karti hain; classical curve phir page ke top se baahar nikal jaati hai.

• $k_B$ — Boltzmann's constant, temperature aur energy ke beech conversion.
• $k_{B}T$ — classical energy per oscillator; har mode ke liye same — yahi fatal assumption hai.

Hume flat $k_{B}T$ ko kuch aisa replace karna hai jo high $\nu$ par shrink kare.

Step 7 — Energy quantize karo: staircase, ramp nahi

KYA. Planck ki leap: frequency $\nu$ ka ek mode sirf $h\nu$ size ke whole lumps mein energy hold kar sakta hai, jahan $h$ Planck's constant hai:

${ϵ}_m=mh\nu ,m=0,1,2,\dots$

KYUN. Agar energy smooth ramp mein aati, toh har mode ek arbitrarily tiny sip accept kar sakta aur equipartition jeet jaata. Energy ko staircase par force karo aur ek high-$\nu$ mode ko ek pehli step $h\nu$ itni unchi milti hai ki $k_{B}T$ size ki thermal kick us par chadh nahi sakti. Toh woh step 0 par rehta hai, zero energy hold karte hue.

PICTURE. Figure classical smooth ramp ko quantum staircase se contrast karta hai; amber arrow $k_{B}T$ height ki thermal kick dikhata hai — woh aasani se bahut saare low steps tak pahunchi hai lekin ek tall step ke liye short pad jaati hai.

• $h$ — Planck's constant, per unit frequency ek energy lump ka size.
• $h\nu$ — step height; frequency ke saath badhta hai, isliye high-$\nu$ steps unaffordable hain.
• $m$ — is mode mein abhi kitne lumps hain (ek occupation number, ek alag integer Steps 2–4 ke geometric radius $R$ se).

Ab true average compute karo. Pehle humein ek tarika chahiye "allowed energies ki list" ko "temperature $T$ par average energy" mein turn karne ka. Woh machinery hai Partition function.

Har energy level ${ϵ}_m$ ek Boltzmann weight $e^{-\beta {ϵ}_m}$ carry karta hai — ek relative probability jo kehti hai "mode step $m$ par baithne kitna likely hai." Partition function bas un saare weights ka sum hai (ek normaliser):

$Z={\sum }_{m=0}^{\infty }{e}^{-\beta mh\nu }={\sum }_{m=0}^{\infty }(e^{-\beta h\nu }{)}^m=\frac{1}{1-e^{-\beta h\nu }}$

• Sum ek geometric series hai ratio $e^{-\beta h\nu }<1$ ke saath, isliye yeh $1/(1-e^{-\beta h\nu })$ mein close ho jaata hai.

Apply karo:

$⟨ϵ⟩=-\frac{\partial }{\partial \beta }\ln \left[\frac{1}{1-e^{-\beta h\nu }}\right]=\frac{h\nu e^{-\beta h\nu }}{1-e^{-\beta h\nu }}=\boxed{\frac{h\nu }{e^{h\nu /k_{B}T}-1}}$

(last step: upar aur neeche $e^{\beta h\nu }$ se multiply karo aur $\beta =1/k_{B}T$ wapas daalo). Yeh $h\nu$ times Bose–Einstein occupation $\bar{n}=1/(e^{h\nu /k_{B}T}-1)$ hai — mode mein lumps ki average sankhya.

Step 8 — Assemble karo, aur catastrophe ko marte dekho

KYA. Step 5 (count) ko Step 7 (energy) se multiply karo:

$u(\nu ,T)=\frac{8\pi {\nu }^2}{c^3}\cdot \frac{h\nu }{e^{h\nu /k_{B}T}-1}=\frac{8\pi h{\nu }^3}{c^3}\frac{1}{e^{h\nu /k_{B}T}-1}$

KYUN yeh ab converge karta hai. ${\nu }^3$ upar abhi bhi explode karna chahta hai — lekin high $\nu$ par denominator $e^{h\nu /k_{B}T}$ exponentially badhta hai, aur exponential decay kisi bhi power ko hara deta hai. Toh $u$ badhta hai, peak karta hai, aur zero par gir jaata hai.

PICTURE. Figure finished Planck curve plot karta hai: left par ${\nu }^3$ ki tarah badhta, peak par bend karta, right par exponentially marta. Disaster contained ho gaya.

Edge cases — dono ends check karo:

• Low frequency ($x\to 0$, jahan $x=h\nu /k_{B}T$): use karo $e^x-1\approx x$, toh $⟨ϵ⟩\to k_{B}T$ aur hum Rayleigh–Jeans recover karte hain. Classical physics Planck's law ka cheap-photon corner hai.
• High frequency ($x\to \infty$): $e^x-1\approx e^x$, toh $u\to \frac{8\pi h{\nu }^3}{c^3}{e}^{-h\nu /k_{B}T}$ — Wien's exponential tail.
• Degenerate $T\to 0$: har step infinitely expensive hai, $⟨ϵ⟩\to 0$ sab $\nu$ par — thanda body glow nahi karta. ✓
• Degenerate $\nu \to 0$: $u\to 0$ kyunki $g(\nu )\to 0$ (zero frequency par koi modes nahi), chahe har ek $k_{B}T$ hold kare. ✓

Ek-picture summary

Sab ek canvas par: count ${\nu }^2$ ki tarah badhta hai (green, climbing), energy per mode fall off hoti hai (cyan, exponential ke zariye girta), aur unka product Planck bump hai (amber). "${\nu }^3$ upar, $e^x$ neeche."

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