Blackbody radiation from statistical mechanics — Planck distribution

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1. What are we counting? (WHAT)

Inside a cubic cavity of side $L$, the EM field is a superposition of standing-wave modes. Each mode is labeled by a wavevector $\vec k$ and a polarization. We need two ingredients:

1. How many modes exist at each frequency — the density of states $g(\nu)$.
2. How much energy each mode carries on average at temperature $T$ — $\langle \epsilon \rangle$.

The spectrum is the product: $u(\nu)\,d\nu = g(\nu)\,\langle\epsilon(\nu)\rangle\,d\nu$

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2. Density of states (HOW — derive it)

Why these modes? Standing waves must vanish at the walls, so the allowed wavevector components are quantized: $k_x = \frac{n_x \pi}{L},\quad k_y=\frac{n_y\pi}{L},\quad k_z=\frac{n_z\pi}{L},\quad n_i = 1,2,3,\dots$ Why this step? A wave with a node at $x=0$ and $x=L$ must fit a half-integer number of wavelengths.

In $\vec n$-space the modes form a cubic lattice with spacing 1. The number of modes with magnitude up to $n$ is the volume of one octant of a sphere (since $n_i>0$): $N(n) = \frac{1}{8}\cdot\frac{4}{3}\pi n^3 = \frac{\pi n^3}{6}$ Why one-eighth? Only positive $n_i$ count — that's $1/8$ of the full sphere.

Now convert $n$ to frequency. With $|\vec k| = n\pi/L$ and $|\vec k| = 2\pi\nu/c$: $n = \frac{2L\nu}{c}\quad\Rightarrow\quad N(\nu) = \frac{\pi}{6}\left(\frac{2L\nu}{c}\right)^3 = \frac{4\pi L^3 \nu^3}{3c^3}$

Multiply by 2 for the two polarizations of light, divide by volume $L^3$, and differentiate: $g(\nu) = \frac{1}{L^3}\frac{d}{d\nu}\big[2N(\nu)\big] = \boxed{\dfrac{8\pi \nu^2}{c^3}}$

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3. Average energy per mode (HOW — the quantum step)

Classical answer (wrong): equipartition gives each oscillator $\langle\epsilon\rangle = k_BT$ regardless of $\nu$. Then $u(\nu) = \frac{8\pi\nu^2}{c^3}k_BT$ — the Rayleigh–Jeans law, which diverges as $\nu\to\infty$.

Planck's quantization (right): a mode of frequency $\nu$ has only discrete energies $\epsilon_n = n h\nu,\qquad n = 0,1,2,\dots$

Use the canonical ensemble. Define $\beta = 1/(k_BT)$ and $x = \beta h\nu$. The partition function: $Z = \sum_{n=0}^{\infty} e^{-n h\nu\beta} = \sum_{n=0}^\infty (e^{-x})^n = \frac{1}{1-e^{-x}}$ Why this step? Geometric series with ratio $e^{-x}<1$.

The average energy:

= \frac{h\nu\, e^{-\beta h\nu}}{1-e^{-\beta h\nu}}$$ Multiply top and bottom by $e^{\beta h\nu}$: $$\boxed{\langle\epsilon\rangle = \dfrac{h\nu}{e^{h\nu/k_BT}-1}}$$ This is the average energy of a quantum oscillator (minus zero-point — equilibrium *photon* picture: it's the Bose–Einstein occupation $\bar n = 1/(e^{h\nu/k_BT}-1)$ times $h\nu$). --- ## 4. The Planck distribution (assemble) $$\boxed{u(\nu,T)\,d\nu = \frac{8\pi\nu^2}{c^3}\cdot\frac{h\nu}{e^{h\nu/k_BT}-1}\,d\nu = \frac{8\pi h\nu^3}{c^3}\frac{1}{e^{h\nu/k_BT}-1}\,d\nu}$$ ![[2.4.19-Blackbody-radiation-from-statistical-mechanics-—-Planck-distribution.png]] > [!intuition] How quantization kills the catastrophe > When $h\nu \gg k_BT$, $e^{h\nu/k_BT}$ is huge, so $\langle\epsilon\rangle \approx h\nu\,e^{-h\nu/k_BT}\to 0$. > High-frequency modes are *exponentially* suppressed — they need a whole quantum $h\nu$ to even turn on, > and thermal kicks of size $k_BT$ can't pay. The $\nu^2$ growth is defeated by exponential decay. --- ## 5. The two limits (Forecast-then-Verify) > [!example] Forecast: what should $u(\nu)$ do at low and high $\nu$? > **Low $\nu$** ($h\nu \ll k_BT$): photons cheap → expect classical $k_BT$ per mode → recover Rayleigh–Jeans. > **High $\nu$** ($h\nu \gg k_BT$): photons expensive → expect exponential cutoff (Wien). > Now verify by expanding. **Low frequency** ($x = h\nu/k_BT \to 0$): $e^x - 1 \approx x$, so $$\langle\epsilon\rangle \approx \frac{h\nu}{h\nu/k_BT} = k_BT \quad\Rightarrow\quad u \approx \frac{8\pi\nu^2}{c^3}k_BT$$ ✓ Rayleigh–Jeans recovered. *Why this step?* Taylor: $e^x = 1+x+\dots$. **High frequency** ($x\to\infty$): $e^x - 1 \approx e^x$, so $$u \approx \frac{8\pi h\nu^3}{c^3}e^{-h\nu/k_BT}$$ ✓ Wien's law — the exponential tail. --- ## 6. Worked consequences > [!example] Stefan–Boltzmann from Planck (Why this step? at each line) > Total energy density: integrate over all $\nu$. > $$U = \int_0^\infty \frac{8\pi h\nu^3}{c^3}\frac{d\nu}{e^{h\nu/k_BT}-1}$$ > Substitute $x = h\nu/k_BT$, so $\nu = \frac{k_BT}{h}x$, $d\nu = \frac{k_BT}{h}dx$ *(why? makes integral dimensionless)*: > $$U = \frac{8\pi h}{c^3}\left(\frac{k_BT}{h}\right)^4 \int_0^\infty \frac{x^3\,dx}{e^x-1}$$ > The integral is a standard result $\displaystyle\int_0^\infty\frac{x^3}{e^x-1}dx = \frac{\pi^4}{15}$ *(why? it equals $\Gamma(4)\zeta(4)=6\cdot\pi^4/90$)*. > $$U = \frac{8\pi^5 k_B^4}{15 c^3 h^3}T^4 \propto T^4$$ > So total radiated power $\propto T^4$ — **Stefan–Boltzmann**. Doubling $T$ multiplies brightness 16×. > [!example] Wien displacement (peak frequency) > Maximize $u(\nu)$: set $\frac{d}{dx}\left(\frac{x^3}{e^x-1}\right)=0$ with $x=h\nu/k_BT$. > This gives the transcendental equation $3(1-e^{-x}) = x$, solved numerically by $x \approx 2.821$. > Hence $\nu_{\max} = 2.821\,\frac{k_BT}{h} \propto T$: hotter bodies peak at higher frequency (why a star looks > blue-white when very hot, red when cooler). --- ## 7. Common mistakes (Steel-man + fix) > [!mistake] "Equipartition must hold — every mode gets $k_BT$." > **Why it feels right:** equipartition is rigorously true for *continuous* classical energies. **The flaw:** > it secretly assumes energy is continuous. With discrete levels $nh\nu$, a mode whose gap $h\nu \gg k_BT$ > is essentially stuck in its ground state and carries far less than $k_BT$. **Fix:** compute $\langle\epsilon\rangle$ > from the *quantized* partition function — it gives $k_BT$ only in the $h\nu\ll k_BT$ limit. > [!mistake] Forgetting the factor of 2 for polarization. > **Why it feels right:** you counted all wavevectors carefully, seems complete. **The flaw:** each $\vec k$ > supports two independent transverse polarizations. **Fix:** multiply $g(\nu)$ by 2 → $8\pi\nu^2/c^3$ (a 4 would be wrong, a missing 2 halves the Stefan constant). > [!mistake] Confusing $u(\nu)$ peak with $u(\lambda)$ peak. > **Why it feels right:** "the peak is the peak." **The flaw:** $u(\nu)d\nu = u(\lambda)d\lambda$ and the > Jacobian $|d\nu/d\lambda| = c/\lambda^2$ shifts the maximum. **Fix:** $x_\nu\approx2.82$ but $x_\lambda\approx4.97$; > they describe the *same* spectrum in different variables. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine a room full of jump-ropes of different lengths. Slow, long ropes are easy to get swinging — a > little energy starts them. Super-fast tiny ropes need a big shove all at once to even move; tiny taps do > nothing. Heat gives random little taps. So slow ropes swing happily, but the fast ones mostly sit still. > That's why a warm object glows mostly at certain colors and never at infinitely-blue light — the fast > "ropes" are too hard to start, so they stay quiet. Make the room hotter (bigger taps) and faster ropes > finally join in, so the glow shifts toward blue. > [!mnemonic] **"Count modes, quantize energy, the catastrophe dies."** > $g\sim\nu^2$ (count) × $\langle\epsilon\rangle\sim h\nu/(e^{x}-1)$ (quantize) ⇒ exponential tail beats $\nu^2$. > Remember the combo as **"$\nu^3$ up, $e^{x}$ down."** --- ## Flashcards #flashcards/physics What physical system is a blackbody modeled as, statistically? ::: A cavity of EM standing-wave modes (each a quantum oscillator) in thermal equilibrium with the walls. Write the density of states for EM modes. ::: $g(\nu)=8\pi\nu^2/c^3$ (includes ×2 for polarization). Why the factor $1/8$ in mode counting? ::: Allowed $n_x,n_y,n_z>0$, so only one octant of the sphere in $\vec n$-space counts. What quantization did Planck impose? ::: A mode of frequency $\nu$ has energies $\epsilon_n=nh\nu$, $n=0,1,2,\dots$ Average energy per mode (Planck)? ::: $\langle\epsilon\rangle=\dfrac{h\nu}{e^{h\nu/k_BT}-1}$. State the full Planck distribution $u(\nu,T)$. ::: $u=\dfrac{8\pi h\nu^3}{c^3}\dfrac{1}{e^{h\nu/k_BT}-1}$. Low-frequency limit of $\langle\epsilon\rangle$? ::: $\to k_BT$ (Rayleigh–Jeans), via $e^x-1\approx x$. High-frequency limit of $u(\nu)$? ::: $\propto \nu^3 e^{-h\nu/k_BT}$ (Wien's law). What causes the ultraviolet catastrophe classically? ::: Equipartition gives every mode $k_BT$; with $g\sim\nu^2$ the integral diverges. How does quantization remove the catastrophe? ::: For $h\nu\gg k_BT$ modes need a full quantum $h\nu$, so they freeze out exponentially. Stefan–Boltzmann temperature dependence? ::: $U\propto T^4$ (from $\int x^3/(e^x-1)dx=\pi^4/15$). Wien displacement: peak frequency vs T? ::: $\nu_{\max}\propto T$, with $x=h\nu/k_BT\approx2.82$. --- ## Connections - [[Bose-Einstein statistics]] — $\langle n\rangle = 1/(e^{\beta h\nu}-1)$ is the photon occupation number. - [[Partition function]] — $Z=\sum e^{-\beta\epsilon_n}$ drives the whole derivation. - [[Equipartition theorem]] — the classical limit and *why* it fails here. - [[Rayleigh-Jeans law]] and [[Wien's law]] — the two limits unified by Planck. - [[Stefan-Boltzmann law]] — integral of the Planck spectrum. - [[Density of states]] — same mode-counting reused for phonons (Debye model) and gases. - [[Photoelectric effect]] — independent evidence for $E=h\nu$ quanta. ## 🖼️ Concept Map ```mermaid flowchart TD BB[Blackbody cavity] -->|filled with| MODES[EM standing wave modes] MODES -->|each is| OSC[Harmonic oscillator] MODES -->|count gives| GDOS[Density of states g of nu] GDOS -->|grows as nu squared| SEED[Many high-freq modes] OSC -->|classical energy kBT| RJ[Rayleigh-Jeans law] SEED -->|combined with kBT| RJ RJ -->|diverges at high nu| UV[Ultraviolet catastrophe] OSC -->|energy in lumps h nu| PLANCK[Planck quantization] PLANCK -->|freezes high modes| AVGE[Average energy per mode] GDOS -->|times avg energy| SPEC[Spectrum u of nu] AVGE -->|times g of nu| SPEC PLANCK -->|resolves| UV ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, blackbody radiation ka idea simple hai: ek band box (cavity) ke andar electromagnetic > standing waves hoti hain, aur har wave ek chhota harmonic oscillator ki tarah behave karta hai. > Classical physics kehti thi ki har oscillator ko equipartition se barabar $k_BT$ energy milegi — > chahe uski frequency kuch bhi ho. Problem ye hai ki high-frequency modes infinite hote hain > ($g(\nu)\sim\nu^2$ badhta hai), to total energy infinite ho jaati — isko **ultraviolet catastrophe** > bolte hain. Yeh clearly galat tha. > > Planck ne ek bold step liya: oscillator ki energy continuous nahi, balki packets mein aati hai — > $\epsilon = nh\nu$. Iska matlab high-frequency mode ko chalu karne ke liye ek poora bada quantum > $h\nu$ chahiye. Lekin thermal energy sirf $k_BT$ jitni hoti hai, to jab $h\nu \gg k_BT$ ho, woh > mode practically off rehta hai. Partition function $Z = 1/(1-e^{-x})$ se average energy nikalta > hai $\langle\epsilon\rangle = h\nu/(e^{h\nu/k_BT}-1)$. Yahi quantization catastrophe ko maar deta > hai — exponential decay $\nu^2$ growth ko hara deta hai. > > Final Planck formula: $u(\nu,T) = \frac{8\pi h\nu^3}{c^3}\frac{1}{e^{h\nu/k_BT}-1}$. Low frequency > par yeh Rayleigh–Jeans ($k_BT$ per mode) ban jaata hai, aur high frequency par Wien ka exponential > tail. Isko integrate karo to Stefan–Boltzmann $U\propto T^4$ milta hai, aur peak shift karne se > Wien displacement law ($\nu_{max}\propto T$). Yeh derivation isliye important hai kyunki yahi se > quantum mechanics ka janm hua — "energy quantized hoti hai" ka pehla pukka proof. ![[audio/2.4.19-Blackbody-radiation-from-statistical-mechanics-—-Planck-distribution.mp3]]