Why normalize the spectrum on the graph?

B_λ changes by many orders of magnitude with temperature. Normalizing to the peak in the displayed λ range keeps the curve on screen while preserving the shape and the marked Wien peak.

Why is ∫πB_λ dλ on the page not exactly equal to σT⁴?

σT⁴ equals the integral of πB_λ over all wavelengths from zero to infinity. The simulator integrates only from 50 nm to 3500 nm, so the ratio is below 100% until almost all power lies in that band.

Is the colored bar what a blackbody “looks like”?

It is a rough perceptual mix: each visible wavelength is weighted by B_λ at your temperature. Human color vision and display gamuts are more complex, but the trend (cooler → redder, hotter → whiter/bluer) is right.

Black Body: Planck Spectrum

This page plots Planck’s spectral radiance B_λ(T, λ) for an ideal black body. Wien’s displacement law locates the wavelength of peak emission (λ_max T ≈ 2.898×10⁻³ m·K). Stefan–Boltzmann’s law states that the total hemispherical exitance is M = σT⁴ with σ ≈ 5.67×10⁻⁸ W·m⁻²·K⁻⁴. The chart normalizes B_λ to unit peak on a fixed wavelength window so the shape is easy to compare across temperatures; a vertical segment marks the theoretical λ_max. A trapezoidal integral of πB_λ over the same window is shown next to σT⁴ so you can see how much of the total power lies inside the visible band’s context window.

Who it's for: Introductory thermal radiation, astrophysics (stellar colors), and engineering heat transfer students.

Key terms

Planck law
Wien displacement
Stefan–Boltzmann law
spectral radiance
black body

Live graphs

How it works

Spectral radiance of a black body: Planck’s law, Wien’s displacement (λ_max ∝ 1/T), and Stefan–Boltzmann total emitted power ∝ T⁴.

Key equations

B_λ = (2hc²/λ⁵) / (e^{hc/(λkT)} − 1)

λ_max T ≈ 2.898×10⁻³ m·K · M = σT⁴

Frequently asked questions

Why normalize the spectrum on the graph?: B_λ changes by many orders of magnitude with temperature. Normalizing to the peak in the displayed λ range keeps the curve on screen while preserving the shape and the marked Wien peak.
Why is ∫πB_λ dλ on the page not exactly equal to σT⁴?: σT⁴ equals the integral of πB_λ over all wavelengths from zero to infinity. The simulator integrates only from 50 nm to 3500 nm, so the ratio is below 100% until almost all power lies in that band.
Is the colored bar what a blackbody “looks like”?: It is a rough perceptual mix: each visible wavelength is weighted by B_λ at your temperature. Human color vision and display gamuts are more complex, but the trend (cooler → redder, hotter → whiter/bluer) is right.