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ScatterinaRayleigh Scattering / Why the Sky Is BlueSticker

Scatterina – Rayleigh Scattering / Why the Sky Is Blue Physics Sticker

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Size
3" × 3"
Surface
White
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Why is the sky blue? Scatterina is a nitrogen molecule in the air. Molecules much smaller than a wavelength scatter short-wavelength light far more strongly than long, in proportion to 1/λ⁴, so blue sunlight is sent across the sky while most red carries straight on. Lord Rayleigh explained this in 1871. This is a 3" pixel-art sticker from Pixelated Physics.

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Free US shipping$10 flat shipping outside the USMade to order, ships in 2–5 business days

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Size3″ × 3″

3″ × 3″ kiss-cut sticker.

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Free US shipping · $10 flat shipping outside the US, whatever's in your order.

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Returns & replacements21 days

Every item is printed to order, so we can’t accept returns or exchanges for change of mind, or if you ordered the wrong size or colour.

Print defect, misprint, damage or the wrong item? We’ll send a free replacement, or a refund if you prefer. Report it within 21 days of delivery with a photo; no need to send it back.

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The physics

Scatterina is a nitrogen molecule, two atoms and a triple bond, and white sunlight keeps walking into her.

A molecule is far smaller than a wavelength of light. Light shakes its electrons, and they re-radiate it in new directions. The power scattered this way goes as 1/λ⁴, so short wavelengths are scattered far more than long ones.

Blue at 450 nm is scattered (700/450)⁴ ≈ 5.9 times more than red at 700 nm. In the art she tosses six blue photons for every red one. Red is scattered too, just much less, so most of it carries straight on.

Lord Rayleigh worked this out in 1871. It is why the daytime sky is blue, and why the low sun, seen through a long path of air, looks red.

Blue gets tossed. Red goes through.

Equations

\[I_s \propto \frac{1}{\lambda^4}\]
\[\frac{I_s(450\,\mathrm{nm})}{I_s(700\,\mathrm{nm})} = \left(\frac{700}{450}\right)^4 \approx 5.9\]
Symbols
SymbolMeaningUnit
\(I_s\) power scattered by a molecule much smaller than the wavelength \(\mathrm{W}\)
\(\lambda\) wavelength of the light \(\mathrm{m}\)
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