<blockquote>
<p>Does this mean that a white pixel on a typical display actually emits the inverse of this?</p>
</blockquote>
<p>You're right to think that the luminance formula is related to the brightness of display elements, but it is a directly proportional relationship, not inverse. Take a close look at some white pixels, like the ones I'm using to write this answer right now:</p>
<p><a href="https://i.sstatic.net/jt6thYuF.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/jt6thYuF.jpg" alt="Pixel grid from LCD monitor" /></a></p>
<p>If you have typical color vision, the brightest color in this image will be green, followed by red, and blue will be nearly invisible. Their perceived brightness will (approximately) match the coefficients in that formula for luminance. But if you look at it from a sufficient distance, this will look like a shade of grey — that is, “white” but less than 100% bright. (Note that if you try to get this effect by resizing the image smaller, or zooming out, you're making an assumption that the resizing algorithm is accurate!)</p>
<hr />
<p>To understand why this relationship exists and why it is direct, not inverse, consider how the formula is used in RGB image processing. We use it to map R, G, and B components to luminance, or gray levels. But what are we measuring those R, G, and B components in? We're measuring them in <em>fractions of an arbitrary white level.</em> We have already built into the system that (some particular choice of) <em>white</em> is what we measure everything relative to. Therefore, if we plug white <span class="math-container">$(R, G, B) = (1, 1, 1)$</span> into the luminance formula, we should get 1 out, and we do:</p>
<p><span class="math-container">$$ (0.2126 \cdot 1 + 0.7152 \cdot 1 + 0.0722 \cdot 1) = 1 $$</span></p>
<p>When the formula tells you to multiply <span class="math-container">$B$</span> by 0.0722, what this means is that the blue component makes up 0.0722 of the relative luminance of the white; if you take the white and discard everything but the blue component, then it is 0.0722 times as bright as the whole white. The blue component of white is dim, and the green part of white is bright.</p>
<hr />
<p>Finally, note that none of this has <em>anything</em> to do with “the sensitivity of the human eye”, its relationship of perceived brightness to optical power (watts of photons). <strong>Both sides</strong> of the luminance formula are <a href="https://en.wikipedia.org/wiki/Photometry_(optics)" rel="nofollow noreferrer">photometric</a>, not <a href="https://en.wikipedia.org/wiki/Radiometry" rel="nofollow noreferrer">radiometric</a>. The candela is a photometric unit — it is <em>already</em> weighted by the (typical) sensitivity of the human eye.</p>
<p>“The human eye is most sensitive to green” and “white is mostly made of green” probably have a common <em>cause</em>, but when you're using RGB display devices that already have been calibrated to <span class="math-container">$(1, 1, 1)$</span> = some-shade-of-white, not working with spectra, you don't care about that.</p>

CC BY-SA 4.0
2024-05-20T20:53:11.293Z
This is a really good explanation.
pmw1234
CC BY-SA 4.0
2024-05-19T15:56:59.503Z
2024-05-19T15:56:59.503Z
Kevin Reid
<p>The human eye is considerably more sensitive to green light than red or blue; this is expressed in formulae which calculate an approximate overall &quot;luminance&quot; value for a RGB color (<a href="https://stackoverflow.com/a/596243/2439932">example</a>).</p>
<p>Does this mean that a white pixel on a typical display actually emits the inverse of this?  For example, would a common 100cd/m² SDR display output approximately 24cd/m² of red, 6cd/m² of green and 70cd/m² of blue (physically forming a blue-ish purple, but perceived by us as white)?</p>

<color><light>
Is the light emitted by a typical white pixel actually a blue-ish purple to compensate for our greater sensitivity to the color green?
jameswilddev
# Is the light emitted by a typical white pixel actually a blue-ish purple to compensate for our greater sensitivity to the color green?

(Asked by: username_0 on 2024-05-19)

The human eye is considerably more sensitive to green light than red or blue; this is expressed in formulae which calculate an approximate overall "luminance" value for a RGB color ([example](https://stackoverflow.com/a/596243/2439932)).

Does this mean that a white pixel on a typical display actually emits the inverse of this? For example, would a common 100cd/m² SDR display output approximately 24cd/m² of red, 6cd/m² of green and 70cd/m² of blue (physically forming a blue-ish purple, but perceived by us as white)?

**Tags:** color, light


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## 1 answers

### Answer by username_1 on 2024-05-19. Score: 3

> Does this mean that a white pixel on a typical display actually emits the inverse of this?

You're right to think that the luminance formula is related to the brightness of display elements, but it is a directly proportional relationship, not inverse. Take a close look at some white pixels, like the ones I'm using to write this answer right now:

[![Pixel grid from LCD monitor](https://i.sstatic.net/jt6thYuF.jpg)](https://i.sstatic.net/jt6thYuF.jpg)

If you have typical color vision, the brightest color in this image will be green, followed by red, and blue will be nearly invisible. Their perceived brightness will (approximately) match the coefficients in that formula for luminance. But if you look at it from a sufficient distance, this will look like a shade of grey — that is, “white” but less than 100% bright. (Note that if you try to get this effect by resizing the image smaller, or zooming out, you're making an assumption that the resizing algorithm is accurate!)

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To understand why this relationship exists and why it is direct, not inverse, consider how the formula is used in RGB image processing. We use it to map R, G, and B components to luminance, or gray levels. But what are we measuring those R, G, and B components in? We're measuring them in *fractions of an arbitrary white level.* We have already built into the system that (some particular choice of) *white* is what we measure everything relative to. Therefore, if we plug white $(R, G, B) = (1, 1, 1)$ into the luminance formula, we should get 1 out, and we do:

$$ (0.2126 \cdot 1 + 0.7152 \cdot 1 + 0.0722 \cdot 1) = 1 $$

When the formula tells you to multiply $B$ by 0.0722, what this means is that the blue component makes up 0.0722 of the relative luminance of the white; if you take the white and discard everything but the blue component, then it is 0.0722 times as bright as the whole white. The blue component of white is dim, and the green part of white is bright.

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Finally, note that none of this has *anything* to do with “the sensitivity of the human eye”, its relationship of perceived brightness to optical power (watts of photons). **Both sides** of the luminance formula are [photometric](https://en.wikipedia.org/wiki/Photometry_(optics)), not [radiometric](https://en.wikipedia.org/wiki/Radiometry). The candela is a photometric unit — it is *already* weighted by the (typical) sensitivity of the human eye.

“The human eye is most sensitive to green” and “white is mostly made of green” probably have a common *cause*, but when you're using RGB display devices that already have been calibrated to $(1, 1, 1)$ = some-shade-of-white, not working with spectra, you don't care about that.

> **Comments:**
> **username_2**: This is a really good explanation.  

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