Where Does Planck’s Formula Come From?

— Following the derivation in lecture notes (8) (for second- and third-year undergraduates)

The simulator computes the spectrum of light from the temperature as $U (ν) = \frac{8 π h}{c^{3}} \cdot \frac{ν^{3}}{e^{h ν / k T} - 1}$ — a formula made of two parts: “how much energy one mode carries on average” and “how many modes there are.” The notes calculate the former, and here we assemble everything through to the end, including the latter.

The conclusion first In classical theory, the average energy of a single vibrational mode was always

k T

(the equipartition theorem). However, if energy can be taken only in steps of $h ν$ , then a high-frequency mode with

h ν ≫ k T

“can’t afford even a single quantum,” and so it is effectively frozen out. This one point alone eliminates the ultraviolet catastrophe and brings the theory into agreement with experiment.

0. What Do We Want to Find?

Consider the electromagnetic waves in a cavity kept at temperature T. What we want to find is the energy per unit volume and per unit frequency $U (ν)$ , and multiplying this by $d ν$ gives “the energy per unit volume at frequencies between $ν$ and $ν + d ν$ .” Our approach has two stages.

\underset{what we want}{\underset{⏟}{U (ν)}} = \underset{number of modes (§3)}{\underset{⏟}{g (ν)}} \times \underset{average energy per mode (§1–2)}{\underset{⏟}{⟨ E ⟩}}

1. What Happens in Classical Theory, and Why It Breaks Down

This is the equation at the start of the notes. Regard a single vibrational mode as a harmonic oscillator with energy $E = p^{2} / 2 m + \frac{1}{2} m ω^{2} q^{2}$ and take the average with the Boltzmann distribution.

(1)

⟨ E ⟩ = \frac{\iint E e^{- E / k T} d p d q}{\iint e^{- E / k T} d p d q} = k T

The integrals are Gaussian, so they can be done by hand. The $p^{2}$ term and the $q^{2}$ term each yield $\frac{1}{2} k T$ , so together they give $k T$ . Regardless of frequency, every mode has the same $k T$ — this is the equipartition theorem.

⚠ It breaks down here As we will see in §3, the number of modes is

g (ν) \propto ν^{2}

and hence the classical prediction is

U = 8 π ν^{2} k T / c^{3}

(the Rayleigh–Jeans formula). It grows without bound as the frequency is raised, so the total energy

\int U d ν

diverges. A body at any temperature would end up with infinite energy — this is the ultraviolet catastrophe.

2. Planck’s Quantum Hypothesis — Turning the Integral into a Sum

The heart of the notes is this single line. We assume that energy can take only integer multiples of $h ν$ ( $E = n h ν$ ). Then the integral in Eq. (1) turns into a sum.

(2)

⟨ E ⟩ = \frac{\sum_{n = 0}^{\infty} n h ν e^{- n h ν / k T}}{\sum_{n = 0}^{\infty} e^{- n h ν / k T}}

2.1 The denominator — just a geometric series

With $y \equiv h ν / k T$ the sum is a geometric series with common ratio $e^{- y} < 1$ .

(3)

Z \equiv \sum_{n = 0}^{\infty} e^{- n y} = \frac{1}{1 - e^{- y}}

2.2 The numerator — obtained by differentiation

The numerator carries an extra factor $n h ν$ . Summing it straightforwardly is laborious, but with the clever trick from the notes it takes an instant. That is because, setting $x \equiv 1 / k T$ , the operation of multiplying by $n h ν$ can be produced by $- \partial / \partial x$ .

(4)

\sum_{n = 0}^{\infty} n h ν e^{- n h ν x} = - \frac{\partial}{\partial x} \sum_{n = 0}^{\infty} e^{- n h ν x} = - \frac{\partial}{\partial x} (\frac{1}{1 - e^{- h ν x}}) = \frac{h ν e^{- y}}{{(1 - e^{- y})}^{2}}

2.3 Dividing to finish

(5)

⟨ E ⟩ = \frac{h ν e^{- y}}{1 - e^{- y}} = \frac{h ν}{e^{h ν / k T} - 1}

This equation determines everything

$h ν ≪ k T$ (low frequency): $e^{y} - 1 ≃ y$ , so $⟨ E ⟩ \to k T$ . We recover classical theory.
$h ν ≫ k T$ (high frequency): $⟨ E ⟩ ≃ h ν e^{- h ν / k T}$ and thus it vanishes exponentially. If the “minimum unit” of energy is larger than the thermal energy, that mode cannot be excited even by a single quantum — this intuition is what kills the ultraviolet catastrophe.

Incidentally, Eq. (5) is nothing other than the Bose–Einstein distribution

⟨ n ⟩ = 1 / (e^{h ν / k T} - 1)

that bosons (photons) obey, multiplied by

h ν

(as noted in the margin of the notes).

3. How Many Modes Are There? — The Part the Notes Pass Over with “From the Above”

The notes take the prefactor in $U (ν) = (8 π ν^{2} / c^{3}) ⟨ E ⟩$ as known. Only that part is supplied here.

Standing waves in a cube of side $ℓ$ must have nodes at the walls, so they are restricted to $k = (π / ℓ) (n_{1}, n_{2}, n_{3})$ , with $n_{j} = 1, 2, 3, \dots$ (we write the volume of the box as $V = ℓ^{3}$ ). In k-space, the volume occupied by one mode is ${(π / ℓ)}^{3}$ . Since $n_{j} > 0$ , only $1 / 8$ of the sphere can be used, and in addition there are two polarizations.

(6)

N (k) = 2 \times \frac{1}{8} \times \frac{\frac{4}{3} π k^{3}}{{(π / ℓ)}^{3}} = \frac{k^{3} V}{3 π^{2}} ⟹ g (ν) = \frac{1}{V} \frac{d N}{d ν} = \frac{8 π ν^{2}}{c^{3}}

The last step is just substituting $k = 2 π ν / c$ . The important point is that the box volume $V$ cancels out: for a sufficiently large cavity, the result does not depend on its shape or size, so the formula is universal.

You can also check this

g (ν)

by actually counting the modes. In a cube of side 100 µm, if

(n_{1}, n_{2}, n_{3})

runs through all its values and the modes are counted, the count comes to 96–98 % of Eq. (6). The remainder is a surface effect due to the finite size of the box, and the larger the box, the closer the ratio approaches 100 %.

4. Assembling Planck’s Formula

(7)

U (ν) = \frac{8 π ν^{2}}{c^{3}} \cdot \frac{h ν}{e^{h ν / k T} - 1} = \frac{8 π h}{c^{3}} \cdot \frac{ν^{3}}{e^{h ν / k T} - 1}

This is the equation about which the notes say, “This formula agrees astonishingly well with experiment!!” Let us look at the limits at both ends.

Limit	Formula	Name	Behavior
$h ν ≪ k T$	$U \to 8 π ν^{2} k T / c^{3}$	Rayleigh–Jeans	Agrees with classical theory (which was correct on the low-frequency side)
$h ν ≫ k T$	$U \to (8 π h / c^{3}) ν^{3} e^{- h ν / k T}$	Wien	Falls off exponentially, so the integral converges

5. The Peak Moves When Viewed in Wavelength — Wien’s Displacement Law

5.1 The conversion requires a Jacobian

$U (ν)$ is a quantity “per unit frequency.” When we want to view it per unit wavelength, simply replacing $ν$ with $c / λ$ is not allowed. Instead, we start from $U (λ) d λ = U (ν) d ν$ so that no light is left out of the count.

(8)

U (λ) = U (ν) | \frac{d ν}{d λ} | = U (ν) \frac{c}{λ^{2}} = \frac{8 π h c}{λ^{5}} \cdot \frac{1}{e^{h c / λ k T} - 1}

5.2 So the two peaks are different things

Differentiating $U (λ)$ with respect to $λ$ and setting the result equal to 0 gives, for $x = h c / λ k T$ , $x = 5 (1 - e^{- x})$ . For $U (ν)$ , it is $y = 3 (1 - e^{- y})$ . The difference between 5 and 3 comes from multiplying by the Jacobian $c / λ^{2}$ in Eq. (8) (that is, dividing by $λ^{2}$ ).

(9)

x = 4.965114 ⟹ λ_{max} T = 2.8978 \times 10^{- 3} m·K, y = 2.821439 ⟹ ν_{max} / T = 5.8789 \times 10^{10} Hz/K

⚠

c / ν_{max} \neq λ_{max}

Taking the ratio gives

c / ν_{max} = 1.76 λ_{max}

— they are off by as much as 76 %. This is not because one of them is wrong, but because “the most abundant color when counted per unit wavelength” and “the most abundant color when counted per unit frequency” are simply different questions. Whenever you speak of “the peak wavelength,” always state which way you counted. The same pitfall appears whenever a distribution is rewritten in terms of a different variable. In single-slit diffraction, too, depending on whether the intensity distribution is measured by the position x on the screen or by the dimensionless

α x

, the normalization factor is

α / π

1 / π

, respectively (

α

is a constant with the dimension of inverse length).

The same light at the same temperature, counted by wavelength in the left panel and by frequency in the right panel. The solid vertical line is the peak of that graph itself; the dotted line is the position of the other graph’s peak, converted and carried over. The two lines do not coincide.

Everyday examples Temperature T

6. Adding Everything Up Gives Stefan–Boltzmann

Integrate Eq. (7) over all frequencies. Substituting $y = h ν / k T$ , the factor $T^{4}$ comes out in front, and only a number remains.

(10)

u = \int_{0}^{\infty} U (ν) d ν = \frac{8 π h}{c^{3}} {(\frac{k T}{h})}^{4} \underset{= π⁴/15}{\underset{⏟}{\int_{0}^{\infty} \frac{y^{3}}{e^{y} - 1} d y}} = \frac{8 π^{5} k^{4}}{15 c^{3} h^{3}} T^{4}

The radiation going out (the radiant exitance) is $M = (c / 4) u$ , so

(11)

M = σ T^{4}, σ = \frac{2 π^{5} k^{4}}{15 c^{2} h^{3}} = 5.6704 \times 10^{- 8} W/(m²·K⁴)

The Stefan–Boltzmann constant can be written with nothing but $h, k, c$ — very strong evidence that Planck’s formula is correct. Its exact agreement with the measured $σ$ was what proved decisive at the time (in the current SI, $h, k, c$ are defined values, so $σ$ is determined exactly by this formula).

7. What Does the Simulator Calculate?

The spectrum is just Eq. (7) evaluated as it stands. The laborious part is the color, which is less a matter of physics than of “how the human eye perceives it.”

Apply the eye’s sensitivity. Multiply the spectrum by the CIE 1931 color-matching functions $\overline{x}, \overline{y}, \overline{z}$ and integrate over 360–830 nm to obtain the three numbers of the XYZ colorimetric system.
Convert to screen color. XYZ is converted to sRGB. Colors outside the gamut are desaturated by adding white.
Brightness is handled separately. The large color swatch shows only the “hue,” with the brightness equalized. The total radiation increases as $T^{4}$ according to Eq. (11), but the brightness of the visible light reaching the eye lies in the Wien tail and so changes far more steeply; taking the Sun’s surface as 1, iron at 900 K has only about $10^{- 10}$ . The band below shows this ratio of visible-light brightness compressed onto a logarithmic scale.

⚠ No “color” is shown at low temperatures When the visible range lies far out in the Wien tail, the chromaticity is determined by nothing but the approximation error of the color-matching functions, and so has no meaning as a number. At body temperature (310 K) the visible-light brightness is only

1.6 \times 10^{- 30}

times that of the Sun’s surface, which is as much as 15 orders of magnitude dimmer than the detection limit of a dark-adapted eye. Below roughly 600 K the color swatch is made black and a message to that effect is displayed.

Reading temperature from color Heated iron appears to change from red → orange → yellow → white because, as the temperature rises, the peak moves toward shorter wavelengths (§5) and the mixture of light falling in the visible range changes. Stars are read the same way: in observations, the surface temperature is estimated from the color (the color index). However, since neither iron nor stars are perfect blackbodies, what is obtained this way is the temperature of the blackbody that looks the same color — a quantity called the color temperature, which gives a rough guide to the effective temperature.

8. Checking with Numbers

What was checked	Agreement
Eq. (3), geometric series: direct sum vs. closed form	relative difference < 2×10⁻¹⁶
Eq. (4), numerator: direct sum vs. result of differentiation	relative difference < 5×10⁻¹⁶
Eq. (5): $⟨ E ⟩$ from the direct sum	relative difference < 6×10⁻¹⁶
Eq. (6): counting the modes (100 µm cube)	96–98 % (finite-box effect)
Eq. (9): $λ_{max}$ found directly by numerical search	relative difference 2×10⁻⁷
Eq. (10): $\int y^{3} / (e^{y} - 1) d y = π^{4} / 15$	relative difference 2×10⁻¹²
Eq. (11): value of $σ$	5.670374419×10⁻⁸
Eq. (8): $U (λ) = U (ν) c / λ^{2}$	relative difference < 1×10⁻¹⁵

At 1300 K, the total energy density of 2.161×10⁻³ J/m³ and the radiant exitance of 1.620×10⁵ W/m² both agree with the values read from the simulator.

9. Summary

What we want is “the number of modes × the average energy per mode.”
In classical theory the average is $k T$ regardless of frequency, and $\int U d ν$ diverges (the ultraviolet catastrophe) — Eq. (1).
Quantizing the energy as $E = n h ν$ turns the integral into a sum, and the average follows from a geometric series and a single differentiation — Eqs. (2)–(5).
At high frequencies, $⟨ E ⟩$ freezes out exponentially. This is what removes the catastrophe.
Counting standing waves, the number of modes is $8 π ν^{2} / c^{3}$ . The size of the box drops out — Eq. (6).
Converting to a wavelength representation requires a Jacobian. Hence $c / ν_{max} \neq λ_{max}$ — Eqs. (8), (9).
Adding everything up gives $σ T^{4}$ . $σ$ can be expressed with $h, k, c$ alone — Eqs. (10), (11).

Mochizuki Research Group, Tokyo University of Science / From lecture notes (8), “Planck’s Formula” (in Japanese). The mode density in §3 and the Stefan–Boltzmann law in §6 supplement the parts that the notes use as known. The numerical checks use direct summation of series, explicit counting of modes, and Simpson’s rule. The equations are typeset in the browser’s native MathML (no external libraries or network communication are used at all).