Blackbody radiation from statistical mechanics — Planck distribution: Question bank

For the full derivations these questions poke at, keep the parent note open, plus Density of states, Partition function, Bose-Einstein statistics, Equipartition theorem, Rayleigh-Jeans law, Wien's law and Stefan-Boltzmann law.

Definition Symbols used on this page (defined here so you never chase the parent)

$u (ν, T) d ν$ = spectral energy density: the electromagnetic energy per unit volume carried by modes with frequency in $[ν, ν + d ν]$ . It is a density in frequency — you must multiply by $d ν$ to get an energy.
$U$ = total energy density: all of it summed over frequency, $U = \int_{0}^{\infty} u (ν, T) d ν$ .
$β \equiv \frac{1}{k _{B} T}$ = inverse temperature: the natural "coldness" variable in statistical mechanics; large $β$ means cold. Boltzmann weights are $e^{- β ϵ}$ .
$x \equiv β h ν = \frac{h ν}{k _{B} T}$ = the dimensionless energy ratio: how big one photon lump $h ν$ is compared with the thermal budget $k_{B} T$ . Small $x$ = cheap mode, large $x$ = frozen mode.
$μ$ = chemical potential: the energy cost of adding one particle to the system. For photons $μ = 0$ (explained in WQ7).

The three figures below are the visual backbone — glance at them before answering.

Figure — Blackbody radiation from statistical mechanics — Planck distribution

True or false — justify

Every answer must give the reason, not just the verdict. (Look at figure s01 for the shape of the curves.)

TF1. "The ultraviolet catastrophe is caused by there being too many low-frequency modes."

False. The mode count

g (ν) = 8 π ν^{2} / c^{3}

grows without bound at high

ν

; it is the pile-up of high-frequency modes each holding

k_{B} T

that makes the classical integral diverge.

TF2. "Planck's quantization changes how many modes exist in the cavity."

False. Quantization only changes the average energy per mode

⟨ ϵ ⟩

. The density of states

g (ν) = 8 π ν^{2} / c^{3}

is purely geometric and is identical in the classical and quantum treatments.

TF3. "At very low frequency the Planck curve and the Rayleigh–Jeans curve agree."

True. For

h ν ≪ k_{B} T

(i.e.

x \to 0

) we expand

e^{x} - 1 \approx x

, giving

⟨ ϵ ⟩ \to k_{B} T

, which is exactly the classical (Rayleigh–Jeans) result — see how the curves hug on the left of figure s01.

TF4. "Because more modes exist at high

ν

, a hot body radiates most strongly at the highest frequencies."

False. The

ν^{2}

mode growth is beaten by the exponential factor

e^{- h ν / k_{B} T}

⟨ ϵ ⟩

, so the spectrum turns over at a finite peak

ν_{m a x} \approx 2.82 k_{B} T / h

and falls off afterward.

TF5. "The Bose–Einstein occupation number

\overset{n}{ˉ} = 1/ (e^{h ν / k_{B} T} - 1)

can exceed 1."

True. Photons are bosons with no exclusion; for

h ν ≪ k_{B} T

(cheap, low-frequency modes)

\overset{n}{ˉ}

becomes very large, meaning many photons pile into the same mode.

TF6. "Doubling the temperature doubles the total radiated energy density."

False. Stefan–Boltzmann gives

U \propto T^{4}

, so doubling

T

multiplies

U

2^{4} = 16

, not 2.

TF7. "The peak of

u (ν)

and the peak of

u (λ)

point to the same frequency/wavelength pair."

False. Writing the dimensionless ratios

x_{ν} = h ν / k_{B} T

(frequency form) and

x_{λ} = h c / λ k_{B} T

(wavelength form), the Jacobian

∣ d ν / d λ ∣ = c / λ^{2}

shifts the maximum:

x_{ν} \approx 2.82

but

x_{λ} \approx 4.97

. Same physics, different peak location.

TF8. "A perfect blackbody is black because it never emits any light."

False. "Black" means it absorbs all incident radiation; in thermal equilibrium it also emits the full Planck spectrum. A hot blackbody glows brightly.

TF9. "The zero-point energy

\frac{1}{2} h ν

matters for the shape of the thermal radiation spectrum."

False. The equilibrium photon picture uses

⟨ ϵ ⟩ = h ν \overset{n}{ˉ}

with no zero-point term; the constant

\frac{1}{2} h ν

per mode is

T

-independent and drops out of the observable radiated spectrum.

Spot the error

Each statement below hides one flaw. Name it and correct it. (Figure s02 shows the $n$ -space octant for SE1.)

SE1. "We count modes as a full sphere of radius

n

n

-space, volume

\frac{4}{3} π n^{3}

The octant error. Since each

n_{i} = 1, 2, 3 \dots

is positive, only

1/8

of the sphere is physical:

N = \frac{1}{8} \cdot \frac{4}{3} π n^{3} = \frac{π n ^{3}}{6}

SE2. "Each wavevector

k

contributes one mode, so

g (ν) = 4 π ν^{2} / c^{3}

The polarization factor is missing. Every

k

carries two independent transverse polarizations, so multiply by 2:

g (ν) = 8 π ν^{2} / c^{3}

SE3. "By equipartition each oscillator carries

\frac{1}{2} k_{B} T

, so

u = \frac{4 π ν ^{2}}{c ^{3}} k_{B} T

Each 1-D oscillator has two quadratic degrees of freedom (kinetic + potential), giving

2 \times \frac{1}{2} k_{B} T = k_{B} T

, not

\frac{1}{2} k_{B} T

. And equipartition itself is invalid once levels are discrete: the theorem is derived by integrating over a continuous energy that appears quadratically; with allowed energies stuck at

0, h ν, 2 h ν, \dots

that integral is replaced by a sum, and when the gap

h ν ≫ k_{B} T

the mode cannot even reach its first excited level, so it carries far less than

k_{B} T

SE4. "The partition function is

Z = \sum_{n} e^{- nh ν β} = \frac{1}{1 - e ^{- x}}

(with

β = 1/ k_{B} T

and

x = β h ν

), so

⟨ ϵ ⟩ = - \frac{\partial ln Z}{\partial x}

The energy average is

⟨ ϵ ⟩ = - \partial ln Z / \partial β

, differentiating with respect to

β

, not

x = β h ν

. Getting the variable wrong drops the factor of

h ν

(since

\partial x / \partial β = h ν

SE5. "Since

⟨ ϵ ⟩ = h ν / (e^{h ν / k_{B} T} - 1)

, at high

ν

each mode holds about

h ν

At high

ν

e^{h ν / k_{B} T} ≫ 1

⟨ ϵ ⟩ \approx h ν e^{- h ν / k_{B} T} \to 0

. High-frequency modes hold almost nothing, not

h ν

SE6. "In the Stefan–Boltzmann integral, substituting

x = h ν / k_{B} T

leaves a

T

-dependent integral."

The whole point of that substitution is to make the integral

\int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x = π^{4} /15

a pure number with no

T

; all

T

dependence is pulled out front as

T^{4}

SE7. "Wien's law comes from setting

⟨ ϵ ⟩

maximum."

You maximize the full spectral density

u (ν) \propto ν^{3} / (e^{x} - 1)

(density of states × energy), not

⟨ ϵ ⟩

alone. This gives

3 (1 - e^{- x}) = x

, i.e.

x \approx 2.821

Why questions

WQ1. Why does the classical result diverge but the quantum result stay finite?

Classically every mode gets

k_{B} T

, so

\int ν^{2} k_{B} T d ν

diverges. Quantization makes

⟨ ϵ ⟩

decay like

e^{- h ν / k_{B} T}

at large

ν

, and exponential decay overwhelms the

ν^{2}

growth, so the integral converges.

WQ2. Why is energy quantized in lumps of

h ν

(proportional to frequency) the specific fix, rather than just any discreteness?

The lump size

h ν

scaling with frequency is exactly what makes high-frequency modes expensive: a fixed thermal budget

k_{B} T

can excite cheap low-

ν

lumps but not costly high-

ν

ones, freezing them out. Frequency-independent lumps would not selectively suppress the UV modes.

WQ3. Why can we treat each standing-wave mode as an independent harmonic oscillator?

In a linear cavity the EM field decomposes into normal modes that do not exchange energy directly; each mode's amplitude oscillates sinusoidally in time exactly like a 1-D harmonic oscillator, so its energy levels are those of a quantum oscillator.

WQ4. Why does a hotter star look bluer?

Wien displacement gives

ν_{m a x} \propto T

, so raising

T

pushes the spectral peak to higher frequency (shorter wavelength), shifting the perceived color from red toward blue-white.

WQ5. Why does the density of states depend on

ν^{2}

and not

ν

ν^{3}

Modes fill a shell in 3-D

k

-space; the number in a shell of radius

\propto ν

scales with its surface area

\propto ν^{2}

. In 2-D it would be

\propto ν

, in 1-D constant — the exponent tracks spatial dimension minus one.

WQ6. Why is the factor a geometric series in the partition function, and why must the ratio be less than 1?

The allowed energies

nh ν

are equally spaced, so

Z = \sum_{n} (e^{- x})^{n}

is geometric with ratio

e^{- x}

. Since

x = h ν / k_{B} T > 0

we have

e^{- x} < 1

, so the sum converges to

1/ (1 - e^{- x})

; a ratio

\geq 1

would mean an unphysical infinite

Z

. (The geometric decay of these Boltzmann weights is drawn in figure s03.)

WQ7. Why does the photon picture have a chemical potential

μ = 0

(unlike a gas of atoms)?

Recall

μ

is the energy cost of adding one particle. For atoms particle number is fixed, so

μ

adjusts to enforce it and can be positive or negative. But photon number is not conserved — the walls freely create and destroy photons — so the system minimizes free energy with respect to number itself, which forces

μ = 0

. That is exactly why Bose–Einstein gives

\overset{n}{ˉ} = 1/ (e^{h ν / k_{B} T} - 1)

with no

e^{- β μ}

factor.

Edge cases

EC1. What is

⟨ ϵ ⟩

in the strict limit

T \to 0

for a fixed

ν > 0

x = h ν / k_{B} T \to \infty

, so

\overset{n}{ˉ} \to 0

and

⟨ ϵ ⟩ \to 0

. Every mode falls into its ground state; the cold cavity radiates nothing.

EC2. What happens to

⟨ ϵ ⟩

ν \to 0

at fixed

T

x \to 0

gives

⟨ ϵ ⟩ \to k_{B} T

, the classical equipartition value — the very low-frequency corner is where quantum and classical predictions merge.

EC3. Does the total energy density

U

stay finite even though

g (ν)

diverges at large

ν

Yes. Although

g (ν) \to \infty

, the product

g ⟨ ϵ ⟩

decays exponentially at large

ν

, so

U = \int_{0}^{\infty} g ⟨ ϵ ⟩ d ν

converges to

\frac{8 π ^{5} k _{B}^{4}}{15 c ^{3} h ^{3}} T^{4}

EC4. Is there a mode with exactly

ν = 0

No physical standing wave has zero frequency in the cavity (that would need

n_{x} = n_{y} = n_{z} = 0

, i.e. no wave).

ν \to 0

is a limit, not an actual mode, and it contributes negligibly because

g (ν) \to 0

there.

EC5. In the limit

h \to 0

(imagining Planck's constant shrinking), what spectrum do we recover?

With

h \to 0

x = h ν / k_{B} T \to 0

for all

ν

, so

⟨ ϵ ⟩ \to k_{B} T

everywhere and we recover the divergent Rayleigh–Jeans law — the classical catastrophe returns, confirming

h

is what tames it.

EC6. What is the occupation of a mode when

h ν = k_{B} T

exactly (

x = 1

\overset{n}{ˉ} = 1/ (e^{1} - 1) \approx 0.58

, i.e. on average about half a photon per mode — the crossover region between the "cheap, populated" and "expensive, frozen" regimes sits near

x \sim 1

Recall One-line summary of every trap

The count $g \sim ν^{2}$ is classical geometry; the energy $⟨ ϵ ⟩$ is where quantum mechanics enters and where the catastrophe dies. Watch the factor of 2 (polarization), the octant (1/8), the $β$ -vs- $x$ derivative, the $T^{4}$ (not $T$ ) scaling, and the $ν$ -peak vs $λ$ -peak Jacobian.