Exercises — Blackbody radiation from statistical mechanics — Planck distribution

$u(\nu ,T)=\underset{g(\nu )\text{= modes per vol per Hz}}{\underbrace{\frac{8\pi {\nu }^2}{c^3}}}\cdot \underset{⟨ϵ⟩\text{= energy per mode}}{\underbrace{\frac{h\nu }{e^{h\nu /k_{B}T}-1}}}$ The first factor $g(\nu )$ (defined above) grows as ${\nu }^2$ (more modes at high frequency); the second is the Bose–Einstein average energy that crashes exponentially once $h\nu \gg k_{B}T$. Their product has units $({\text{m}}^{-3}{\text{Hz}}^{-1})\times (\text{J})={\text{J⋅s⋅m}}^{-3}$, matching $u$, and scales as $\propto {\nu }^3/(e^x-1)$.

$\bar{n}=\frac{1}{e^x-1}=\frac{1}{e^1-1}=\frac{1}{2.71828-1}=\frac{1}{1.71828}\approx 0.582$ So less than one photon on average — even a mode "priced at one thermal kick" is usually empty or singly occupied. This is why classical $k_{B}T$ (which would demand a full unit of energy always) overcounts.

${\nu }_{\max }=2.821\cdot \frac{(1.381\times 10^{-23})(5772)}{6.626\times 10^{-34}}$ Numerator: $1.381\times 10^{-23}\times 5772=7.971\times 10^{-20}$; times $2.821=2.249\times 10^{-19}$. Divide by $h$: ${\nu }_{\max }=\frac{2.249\times 10^{-19}}{6.626\times 10^{-34}}\approx 3.39\times 10^{14}\text{Hz}$. That is in the near-infrared. The figure below plots the full Planck curve $u(\nu ,T)$ for this $T$.

Figure (self-contained caption / alt-text): Normalized blackbody spectral energy density $u(\nu ,T)$ of the Sun at $T=5772$ K, plotted against frequency $\nu$ in units of $10^{14}$ Hz (horizontal axis); vertical axis is $u$ divided by its own maximum, so the peak sits at height 1. The blue curve starts near zero, rises like ${\nu }^3$ (green label, left), reaches a single interior maximum at the red dashed line ${\nu }_{\max }\approx 3.39\times 10^{14}$ Hz, then falls off as $e^{-x}$ (orange label, right). The shaded blue area under the curve is the total energy density.

What to notice on the figure: the peak is single and interior — not at $\nu =0$ nor $\nu =\infty$ — which is exactly the death of the ultraviolet catastrophe made visible: the ${\nu }^2$-rise in $g(\nu )$ is overpowered by the $e^{-x}$-fall in $⟨ϵ⟩$.

From ${\nu }_{\max }$: $\lambda =c/{\nu }_{\max }=\frac{2.998\times 10^8}{3.39\times 10^{14}}\approx 8.84\times 10^{-7}\text{m}=884\text{nm}$. 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 $=4.965\times 7.971\times 10^{-20}=3.958\times 10^{-19}$. So ${\lambda }_{\max }=5.02\times 10^{-7}\text{m}=502\text{nm}$ (green-ish). They differ (884 nm vs 502 nm). Here is why, derived rather than asserted.

WHAT a spectral density means: $u(\nu )d\nu$ is the energy in a frequency slice of width $d\nu$; $u(\lambda )d\lambda$ is the energy in a wavelength slice of width $d\lambda$. They must describe the same physical energy, so $u(\lambda )|d\lambda |=u(\nu )|d\nu |\Rightarrow u(\lambda )=u(\nu )\left|\frac{d\nu }{d\lambda }\right|.$ WHY a factor appears: from $\nu =c/\lambda$ we differentiate, $\frac{d\nu }{d\lambda }=\frac{d}{d\lambda }\left(\frac{c}{\lambda }\right)=-\frac{c}{{\lambda }^2}\Rightarrow \left|\frac{d\nu }{d\lambda }\right|=\frac{c}{{\lambda }^2}.$ This factor $c/{\lambda }^2$ is the Jacobian: a fixed frequency slice $d\nu$ maps to a wavelength slice that gets narrower and narrower at short $\lambda$. So converting to $\lambda$ multiplies the curve by an extra $1/{\lambda }^2$ (equivalently ${\nu }^2$), which reweights the density toward higher frequency / shorter wavelength — that reweighting drags the maximum to a new place. That is why maximizing $x^3/(e^x-1)$ (for $\nu$) gives $x\approx 2.821$ but maximizing the reweighted shape (for $\lambda$) gives $x\approx 4.965$: same spectrum, two different peaks.

WHAT: differentiate the shape function $f(x)=x^3/(e^x-1)$. WHY: the spectral peak is where the slope is zero. $f^{\prime }(x)=\frac{3x^2(e^x-1)-x^3{e}^x}{(e^x-1{)}^2}=0\Rightarrow 3x^2(e^x-1)=x^3{e}^x$ Divide by $x^2{e}^x$ (valid, $x>0$): $3(1-e^{-x})=x$. Check $x=2.821$: $3(1-e^{-2.821})=3(1-0.0595)=3(0.9405)=2.822\approx x$. ✓

WHAT: expand $e^x=1+x+\frac{x^2}{2}+\dots$ for small $x=h\nu /k_{B}T$. WHY: low $\nu$ means cheap photons, so $x\to 0$. $⟨ϵ⟩=\frac{h\nu }{e^x-1}=\frac{h\nu }{x+\frac{x^2}{2}+\dots }\stackrel{\text{keep }x}{\to }\frac{h\nu }{x}=\frac{h\nu }{h\nu /k_{B}T}=k_{B}T$ We kept the linear term $x$ and dropped $x^2/2$ and higher. Then $u\approx \frac{8\pi {\nu }^2}{c^3}{k}_{B}T$ — Rayleigh-Jeans law, which diverges as $\nu \to \infty$, the ultraviolet catastrophe. Quantization only matters once $x^2/2$ is no longer negligible, i.e. $h\nu \sim k_{B}T$.

Integrate: $U={\int }_0^{\infty }\frac{8\pi h{\nu }^3}{c^3}\frac{d\nu }{e^{h\nu /k_{B}T}-1}$. Substitute $x=h\nu /k_{B}T\Rightarrow \nu =\frac{k_{B}T}{h}x,d\nu =\frac{k_{B}T}{h}dx$ (why: makes it dimensionless): $U=\frac{8\pi h}{c^3}(\frac{k_{B}T}{h}{)}^4{\int }_0^{\infty }\frac{x^3}{e^x-1}dx,{\int }_0^{\infty }\frac{x^3}{e^x-1}dx=\frac{{\pi }^4}{15}$ $U=\frac{8{\pi }^5{k}_B^4}{15c^3{h}^3}{T}^4$ Numerically $a=\frac{8{\pi }^5(1.381\times 10^{-23}{)}^4}{15(2.998\times 10^8{)}^3(6.626\times 10^{-34}{)}^3}\approx 7.57\times 10^{-16}{\text{J⋅m}}^{-3}{\text{K}}^{-4}$. This connects to the Stefan-Boltzmann law; the emitted flux is $\sigma T^4$ with $\sigma =ac/4$.

$\sigma =\frac{ac}{4}=\frac{(7.57\times 10^{-16})(2.998\times 10^8)}{4}=\frac{2.269\times 10^{-7}}{4}\approx 5.67\times 10^{-8}$ Matches the accepted Stefan–Boltzmann constant. The chain count modes → quantize → integrate produces a lab-measurable constant from pure statistical mechanics.

(a) $\bar{n}=1/(e^4-1)=1/(54.598-1)=1/53.598\approx 0.01866$. (b) $⟨ϵ⟩=h\nu \bar{n}=x{k}_{B}T\bar{n}=4\times 0.01866k_{B}T=0.0746k_{B}T$. So this mode holds only 7.5% of the classical equipartition value $k_{B}T$. (c) Classical physics (Equipartition theorem) would give every such mode a full $k_{B}T$; there are $\propto {\nu }^2$ of them, so the sum diverges. Quantization instead makes each high-$x$ mode hold $\approx x{e}^{-x}{k}_{B}T$, an exponential suppression that overwhelms the polynomial ${\nu }^2$ growth — the total stays finite. Link: Bose-Einstein statistics.

(a) Power $\propto T^4$, so ratio $=(6000/3000{)}^4=2^4=16$. Sixteen times the total power. (b) ${\nu }_{\max }\propto T$, so ratio $=6000/3000=2$. The peak frequency doubles, shifting the color toward the blue — consistent with Wien's law.

WHAT: integrate photons (not energy) — replace the extra $h\nu$ by 1. $n={\int }_0^{\infty }\frac{8\pi {\nu }^2}{c^3}\frac{d\nu }{e^{h\nu /k_{B}T}-1}\stackrel{x=h\nu /k_{B}T}{\to }\frac{8\pi }{c^3}(\frac{k_{B}T}{h}{)}^3{\int }_0^{\infty }\frac{x^2}{e^x-1}dx$ The integral is $\Gamma (3)\zeta (3)=2!\zeta (3)=2\zeta (3)\approx 2.404$. Hence $n=\frac{8\pi k_B^3}{c^3{h}^3}2\zeta (3)T^3\propto T^3$ Numeric check at $T=300\text{K}$: $\frac{8\pi (1.381\times 10^{-23}{)}^3}{(2.998\times 10^8{)}^3(6.626\times 10^{-34}{)}^3}(2)(1.202)(300{)}^3\approx 5.5\times 10^{14}{\text{photons/m}}^3$.