Exercises — Blackbody radiation — Planck's quantum hypothesis

WHAT we use: Wien's displacement law, ${\lambda }_{\text{peak}}T=b$. This law says: the wavelength where a blackbody glows brightest is inversely proportional to its temperature. Hotter → shorter (bluer) peak.

WHY inverse: if ${\lambda }_{\text{peak}}T$ is a fixed constant $b$, then pushing $T$ up must pull ${\lambda }_{\text{peak}}$ down to keep the product constant.

Because $b$ is the same for both bodies: ${\lambda }_A{T}_A={\lambda }_B{T}_B\Rightarrow \frac{T_B}{T_A}=\frac{{\lambda }_A}{{\lambda }_B}=\frac{1000}{500}=2.$ Body B is hotter, by a factor of exactly 2. The shorter the peak wavelength (look at the red curve in the figure), the hotter the body.

(Q) is the classical Rayleigh–Jeans law: mode density $\frac{8\pi {\nu }^2}{c^3}$ times the classical average energy $k_{B}T$ per mode.

(P) is Planck's law. The quantum signature is the "$-1$" in the denominator $e^{h\nu /k_{B}T}-1$. If you deleted that $-1$ you would get Wien's high-frequency tail; if instead $h\nu \ll k_{B}T$ you recover (Q). That lonely $-1$ is exactly what prevents the ultraviolet catastrophe.

WHAT: Wien's law again, rearranged for $T$. $T=\frac{b}{{\lambda }_{\text{peak}}}=\frac{2.898\times 10^{-3}\text{m⋅K}}{290\times 10^{-9}\text{m}}.$ WHY convert nm → m: $b$ is expressed in metre·kelvin, so both lengths must be in metres or the units won't cancel. $T=\frac{2.898\times 10^{-3}}{2.90\times 10^{-7}}=1.00\times 10^4\text{K}=10,000\text{K}.$ A hot blue-white star.

WHAT: Planck's postulate says energy comes in lumps of size $h\nu$. $E=h\nu =(6.626\times 10^{-34})(5.0\times 10^{14})=3.31\times 10^{-19}\text{J}.$ Convert to electron-volts (divide by the joules-per-eV): $E=\frac{3.31\times 10^{-19}}{1.602\times 10^{-19}}=2.07\text{eV}.$ This "a few eV" size is exactly the scale of visible-light photons and of the Photoelectric effect work functions — no coincidence, they are the same quanta.

WHICH tool and WHY: a Taylor expansion of the exponential. We use it because "$h\nu \ll k_{B}T$" means the exponent $x=h\nu /k_{B}T$ is a small number near zero, and Taylor expansion is precisely the tool that approximates a function near a chosen point using its first few terms. Here the point is $x=0$.

For small $x$: $e^x=1+x+\frac{x^2}{2}+\cdots \approx 1+x.$ WHY keep only $1+x$: the next term $x^2/2$ is the square of a tiny number — negligible compared to $x$ itself.

Then $e^x-1\approx x$, so $⟨E⟩=\frac{h\nu }{e^x-1}\approx \frac{h\nu }{x}=\frac{h\nu }{h\nu /k_{B}T}=k_{B}T.✓$ This is the Equipartition theorem result — the quantum formula smoothly contains the classical one as its low-frequency limit.

WHAT: when $x$ is large, $e^x$ is enormous, so $e^x-1\approx e^x$ (the $-1$ is a tiny correction to a huge number). $⟨E⟩=\frac{h\nu }{e^x-1}\approx \frac{h\nu }{e^x}=h\nu e^{-h\nu /k_{B}T}.$ WHY this matters: the factor $e^{-h\nu /k_{B}T}$ crashes toward zero far faster than the mode count $\frac{8\pi {\nu }^2}{c^3}$ grows. So even though there are more and more high-frequency modes, each one carries almost no energy.

Physical picture: a single quantum costs $h\nu$. At high $\nu$ that lump costs more energy than is thermally available ($\sim k_{B}T$). The bath simply cannot afford to excite even one quantum, so the mode stays in its ground state — "frozen out." No infinity. This is the whole point of quantisation.

WHICH tool and WHY: we set the derivative $\frac{\partial u}{\partial \nu }=0$. A derivative measures the slope of the curve; at a peak the curve is momentarily flat, so its slope is zero. That is what "find the peak" means mathematically.

Write $x=h\nu /k_{B}T$, so $\nu \propto x$ and $u\propto \frac{x^3}{e^x-1}\equiv f(x)$. Maximising over $\nu$ is the same as maximising $f(x)$.

Step — quotient rule. With numerator $x^3$ and denominator $g=e^x-1$: $f^{\prime }(x)=\frac{3x^2(e^x-1)-x^3{e}^x}{(e^x-1{)}^2}.$ Set the numerator to zero (a fraction is zero when its top is zero): $3x^2(e^x-1)-x^3{e}^x=0.$ Divide through by $x^2$ (valid since the peak has $x\ne 0$): $3(e^x-1)-x{e}^x=0.$ Divide by $e^x$ and rearrange: $3(1-e^{-x})=x.✓$ Solve numerically: the root is $x\approx 2.821$. (You can see this in the figure: the two curves $y=x$ and $y=3(1-e^{-x})$ cross near $x\approx 2.82$.)

So the frequency peak sits at ${\nu }_{\text{peak}}=\frac{2.821k_{B}T}{h}$ — proportional to $T$.

Naïvely converting: ${\lambda }_{\nu \text{-peak}}=c/{\nu }_{\text{peak}}=\frac{hc}{2.821k_{B}T}$, whereas the true wavelength peak is ${\lambda }_{\text{peak}}=\frac{hc}{4.965k_{B}T}$. $\frac{{\lambda }_{\text{peak}}}{{\lambda }_{\nu \text{-peak}}}=\frac{2.821}{4.965}=0.568\ne 1.$ They differ by about $43\%$.

WHY they differ: the spectrum measured "per unit frequency" and "per unit wavelength" are different functions, because converting requires the factor $|d\nu /d\lambda |=c/{\lambda }^2$. That extra $\lambda$-dependent factor reshapes the curve and shifts where the maximum lands. A maximum is not preserved under a nonlinear change of variable. The peak of $u_{\nu }$ and the peak of $u_{\lambda }$ are genuinely different physical statements — both correct, just answering different questions ("brightest per unit frequency" vs "per unit wavelength").

WHAT: just substitute the constants into the closed form. $\sigma =\frac{2{\pi }^5{k}_B^4}{15h^3{c}^2}.$ Piece by piece:

• $k_B^4=(1.381\times 10^{-23}{)}^4=3.638\times 10^{-92}$
• $h^3=(6.626\times 10^{-34}{)}^3=2.909\times 10^{-100}$
• $c^2=(3.00\times 10^8{)}^2=9.00\times 10^{16}$
• $2{\pi }^5=612.0$

Combine: $\sigma =\frac{612.0\times 3.638\times 10^{-92}}{15\times 2.909\times 10^{-100}\times 9.00\times 10^{16}}\approx 5.67\times 10^{-8}{\text{W⋅m}}^{-2}{\text{K}}^{-4}.✓$ This matches the measured $\sigma =5.670\times 10^{-8}$. The Stefan–Boltzmann law is not an independent empirical law — it falls straight out of Planck's spectrum, with $\sigma$ built entirely from $h,k_B,c$.

Temperature via Wien: $T=\frac{b}{{\lambda }_{\text{peak}}}=\frac{2.898\times 10^{-3}}{1.06\times 10^{-3}}=2.73\text{K}.$ The famous $2.7\text{K}$ of the CMB — the whole sky is a blackbody at nearly absolute zero.

Peak frequency (frequency spectrum): ${\nu }_{\text{peak}}=\frac{2.821k_{B}T}{h}=\frac{2.821\times 1.381\times 10^{-23}\times 2.73}{6.626\times 10^{-34}}=1.60\times 10^{11}\text{Hz}\approx 160\text{GHz}.$ Note this is a genuinely different peak than the wavelength one (recall L4.2): $c/{\nu }_{\text{peak}}=1.87\text{mm}\ne 1.06\text{mm}$ — the Jacobian shift in action, in a real experiment.

Set $x=h\nu /k_{B}T=1$: $\bar{n}=\frac{1}{e^1-1}=\frac{1}{2.71828-1}=\frac{1}{1.71828}=\mathrm{0.582.}$ So a mode whose quantum costs exactly one $k_{B}T$ holds, on average, about 0.58 photons — less than one. Modes with $h\nu \gg k_{B}T$ hold far fewer (the freeze-out of L3.2); modes with $h\nu \ll k_{B}T$ hold $\bar{n}\approx k_{B}T/h\nu \gg 1$ (many photons, the classical crowd). Each such mode behaves like a Quantum harmonic oscillator in thermal equilibrium.

Peak shifts how with $T$? ::: Inversely — ${\lambda }_{\text{peak}}=b/T$, hotter is bluer. What single feature of Planck's law kills the UV catastrophe? ::: The $-1$ in $e^{h\nu /k_{B}T}-1$, causing exponential freeze-out at high $\nu$. Why do the $\nu$-peak and $\lambda$-peak disagree? ::: A spectral density picks up a Jacobian $|d\nu /d\lambda |$; peaks aren't preserved under nonlinear variable change. $\sigma$ is built from which three constants? ::: $h$, $k_B$, $c$ (via $\sigma =2{\pi }^5{k}_B^4/15h^3{c}^2$).