Exercises — Blackbody radiation from statistical mechanics — Planck distribution

$u(\nu ,T)=\underset{g(\nu )\text{= modes per vol per Hz}}{\underbrace{\frac{8\pi {\nu }^2}{c^3}}}\cdot \underset{⟨ϵ⟩\text{= energy per mode}}{\underbrace{\frac{h\nu }{e^{h\nu /k_{B}T}-1}}}$ Pehla factor $g(\nu )$ (upar define kiya hua) ${\nu }^2$ ki tarah badhta hai (high frequency par zyada modes); doosra Bose–Einstein average energy hai jo exponentially crash kar jaati hai jab $h\nu \gg k_{B}T$. Dono ka product ke units hain $({\text{m}}^{-3}{\text{Hz}}^{-1})\times (\text{J})={\text{J⋅s⋅m}}^{-3}$, jo $u$ se match karta hai, aur $\propto {\nu }^3/(e^x-1)$ ki tarah scale karta hai.

$\bar{n}=\frac{1}{e^x-1}=\frac{1}{e^1-1}=\frac{1}{2.71828-1}=\frac{1}{1.71828}\approx 0.582$ Toh average mein ek se bhi kam photon hai — ek mode jo "ek thermal kick ki price" par hai, woh bhi aksar empty ya sirf ek se occupied hota hai. Isliye classical $k_{B}T$ (jo hamesha ek poori unit energy maangta hai) overcount karta hai.

${\nu }_{\max }=2.821\cdot \frac{(1.381\times 10^{-23})(5772)}{6.626\times 10^{-34}}$ Numerator: $1.381\times 10^{-23}\times 5772=7.971\times 10^{-20}$; times $2.821=2.249\times 10^{-19}$. $h$ se divide karo: ${\nu }_{\max }=\frac{2.249\times 10^{-19}}{6.626\times 10^{-34}}\approx 3.39\times 10^{14}\text{Hz}$. Yeh near-infrared mein hai. Neeche figure mein is $T$ ke liye poori Planck curve $u(\nu ,T)$ plot ki gayi hai.

Figure (self-contained caption / alt-text): Suraj ki normalized blackbody spectral energy density $u(\nu ,T)$ $T=5772$ K par, frequency $\nu$ ke against plot ki gayi hai jiske units $10^{14}$ Hz hain (horizontal axis); vertical axis $u$ ko uske apne maximum se divide karke hai, toh peak height 1 par baitta hai. Blue curve zero ke paas se shuru hoti hai, ${\nu }^3$ ki tarah badhti hai (green label, left), ek single interior maximum tak pahunchi hai red dashed line par ${\nu }_{\max }\approx 3.39\times 10^{14}$ Hz, phir $e^{-x}$ ki tarah girती hai (orange label, right). Curve ke neeche shaded blue area total energy density hai.

Figure mein kya dekhna hai: peak single aur interior hai — na $\nu =0$ par, na $\nu =\infty$ par — yahi ultraviolet catastrophe ki maut visible form mein hai: $g(\nu )$ mein ${\nu }^2$ ka rise, $⟨ϵ⟩$ mein $e^{-x}$ ki fall se haara hua hai.

${\nu }_{\max }$ se: $\lambda =c/{\nu }_{\max }=\frac{2.998\times 10^8}{3.39\times 10^{14}}\approx 8.84\times 10^{-7}\text{m}=884\text{nm}$. Direct wavelength peak: ${\lambda }_{\max }=\frac{(6.626\times 10^{-34})(2.998\times 10^8)}{4.965(1.381\times 10^{-23})(5772)}$. Numerator $=1.986\times 10^{-25}$; denominator $=4.965\times 7.971\times 10^{-20}=3.958\times 10^{-19}$. Toh ${\lambda }_{\max }=5.02\times 10^{-7}\text{m}=502\text{nm}$ (greenish). Yeh alag hain (884 nm vs 502 nm). Yahan kyun hai, directly derive karke, bas assert nahi karke.

Spectral density ka MATLAB: $u(\nu )d\nu$ ek frequency slice ki energy hai width $d\nu$ ki; $u(\lambda )d\lambda$ ek wavelength slice ki energy hai width $d\lambda$ ki. Dono same physical energy describe karte hain, toh $u(\lambda )|d\lambda |=u(\nu )|d\nu |\Rightarrow u(\lambda )=u(\nu )\left|\frac{d\nu }{d\lambda }\right|.$ Factor kyun aata hai: $\nu =c/\lambda$ se differentiate karo, $\frac{d\nu }{d\lambda }=\frac{d}{d\lambda }\left(\frac{c}{\lambda }\right)=-\frac{c}{{\lambda }^2}\Rightarrow \left|\frac{d\nu }{d\lambda }\right|=\frac{c}{{\lambda }^2}.$ Yeh factor $c/{\lambda }^2$ Jacobian hai: ek fixed frequency slice $d\nu$ ek aisi wavelength slice se map hoti hai jo chhoti $\lambda$ par aur aur narrow hoti jaati hai. Toh $\lambda$ mein convert karne par curve $1/{\lambda }^2$ (equivalently ${\nu }^2$) se multiply ho jaati hai, jo density ko higher frequency / shorter wavelength ki taraf reweight kar deti hai — aur yahi reweighting maximum ko nai jagah khench le jaati hai. Isliye $x^3/(e^x-1)$ maximize karne par ($\nu$ ke liye) $x\approx 2.821$ milta hai lekin reweighted shape maximize karne par ($\lambda$ ke liye) $x\approx 4.965$ milta hai: same spectrum, do alag peaks.

KYA: shape function $f(x)=x^3/(e^x-1)$ differentiate karo. KYUN: spectral peak wahan hoti hai jahan slope zero ho. $f^{\prime }(x)=\frac{3x^2(e^x-1)-x^3{e}^x}{(e^x-1{)}^2}=0\Rightarrow 3x^2(e^x-1)=x^3{e}^x$ $x^2{e}^x$ se divide karo (valid hai, $x>0$): $3(1-e^{-x})=x$. Check $x=2.821$: $3(1-e^{-2.821})=3(1-0.0595)=3(0.9405)=2.822\approx x$. ✓

KYA: chhote $x=h\nu /k_{B}T$ ke liye $e^x=1+x+\frac{x^2}{2}+\dots$ expand karo. KYUN: low $\nu$ matlab saste photon, toh $x\to 0$. $⟨ϵ⟩=\frac{h\nu }{e^x-1}=\frac{h\nu }{x+\frac{x^2}{2}+\dots }\stackrel{\text{keep }x}{\to }\frac{h\nu }{x}=\frac{h\nu }{h\nu /k_{B}T}=k_{B}T$ Humne linear term $x$ rakha aur $x^2/2$ aur usse bade terms chhodh diye. Phir $u\approx \frac{8\pi {\nu }^2}{c^3}{k}_{B}T$ — Rayleigh-Jeans law, jo $\nu \to \infty$ par diverge karta hai, yahi ultraviolet catastrophe hai. Quantization tabhi matter karta hai jab $x^2/2$ negligible nahi rehta, yaani $h\nu \sim k_{B}T$.

Integrate karo: $U={\int }_0^{\infty }\frac{8\pi h{\nu }^3}{c^3}\frac{d\nu }{e^{h\nu /k_{B}T}-1}$. Substitute karo $x=h\nu /k_{B}T\Rightarrow \nu =\frac{k_{B}T}{h}x,d\nu =\frac{k_{B}T}{h}dx$ (kyun: dimensionless ban jaata hai): $U=\frac{8\pi h}{c^3}(\frac{k_{B}T}{h}{)}^4{\int }_0^{\infty }\frac{x^3}{e^x-1}dx,{\int }_0^{\infty }\frac{x^3}{e^x-1}dx=\frac{{\pi }^4}{15}$ $U=\frac{8{\pi }^5{k}_B^4}{15c^3{h}^3}{T}^4$ Numerically $a=\frac{8{\pi }^5(1.381\times 10^{-23}{)}^4}{15(2.998\times 10^8{)}^3(6.626\times 10^{-34}{)}^3}\approx 7.57\times 10^{-16}{\text{J⋅m}}^{-3}{\text{K}}^{-4}$. Yeh Stefan-Boltzmann law se connect hota hai; emitted flux hai $\sigma T^4$ jahan $\sigma =ac/4$.

$\sigma =\frac{ac}{4}=\frac{(7.57\times 10^{-16})(2.998\times 10^8)}{4}=\frac{2.269\times 10^{-7}}{4}\approx 5.67\times 10^{-8}$ Accepted Stefan–Boltzmann constant se match karta hai. modes count karo → quantize karo → integrate karo — yeh chain pure statistical mechanics se ek lab-measurable constant produce karta hai.

(a) $\bar{n}=1/(e^4-1)=1/(54.598-1)=1/53.598\approx 0.01866$. (b) $⟨ϵ⟩=h\nu \bar{n}=x{k}_{B}T\bar{n}=4\times 0.01866k_{B}T=0.0746k_{B}T$. Toh is mode mein classical equipartition value $k_{B}T$ ka sirf 7.5% hi hai. (c) Classical physics (Equipartition theorem) har aise mode ko poora $k_{B}T$ deta; unki sankhya $\propto {\nu }^2$ hai, toh sum diverge karta. Quantization ki wajah se har high-$x$ mode mein sirf $\approx x{e}^{-x}{k}_{B}T$ hota hai, ek exponential suppression jo polynomial ${\nu }^2$ growth ko overwhelm kar deta hai — total finite rehta hai. Link: Bose-Einstein statistics.

(a) Power $\propto T^4$, toh ratio $=(6000/3000{)}^4=2^4=16$. Total power solah guna ho jaati hai. (b) ${\nu }_{\max }\propto T$, toh ratio $=6000/3000=2$. Peak frequency double ho jaati hai, color blue ki taraf shift hota hai — Wien's law ke consistent.

KYA: photons integrate karo (energy nahi) — extra $h\nu$ ki jagah 1 rakho. $n={\int }_0^{\infty }\frac{8\pi {\nu }^2}{c^3}\frac{d\nu }{e^{h\nu /k_{B}T}-1}\stackrel{x=h\nu /k_{B}T}{\to }\frac{8\pi }{c^3}(\frac{k_{B}T}{h}{)}^3{\int }_0^{\infty }\frac{x^2}{e^x-1}dx$ Integral hai $\Gamma (3)\zeta (3)=2!\zeta (3)=2\zeta (3)\approx 2.404$. Isliye $n=\frac{8\pi k_B^3}{c^3{h}^3}2\zeta (3)T^3\propto T^3$ Numeric check $T=300\text{K}$ par: $\frac{8\pi (1.381\times 10^{-23}{)}^3}{(2.998\times 10^8{)}^3(6.626\times 10^{-34}{)}^3}(2)(1.202)(300{)}^3\approx 5.5\times 10^{14}{\text{photons/m}}^3$.