# Thread: Alpha Centauri in lumens

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## Alpha Centauri in lumens

So I just calculated the lumens the human retina should measure by Alpha Centauri and I get:

(1.5×10^8 km)^2 ÷ (4.1×10^13 km)^2 × 150 lumens = 0.000000002 lumens

That's way too low for the human eye to detect. How come we can see stars if they are that far away?

2. I'm not sure what formulas you are using to calculate this.
Our Sun puts out about 36 Octillion Lumens as we see it from here. Where did your 150 lumens come from?

3. I don't know where your calculation comes from, but you're using the wrong units. Lumens are a measure of luminous flux. The SI equivalent of apparent visual magnitude is illuminance, which is measured in lumens per square metre (lux). A star with an apparent visual magnitude of zero has an illuminance of 2.54x10-6 lx. You can tweak that a little to get the overall illuminance of Alpha Centauri. Then you need to calculate the luminance of the image of the star formed by the eye, which is diffraction-limited to a disc about a minute of arc across. The units of luminance are lx.sr-1.
The luminance of a zero-magnitude star is easily brighter than the retina's detection threshold, and easily brighter than the luminance of the night sky. But it's much lower than the luminance of the daytime sky -- which is why we can see stars at night but not during the day.

Grant Hutchison

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From here:
https://kogalla.com/blogs/tech-trail...ens-is-the-sun

"So at earth's orbit, for each square meter, our sun puts out 127,000 lumens."

So for each square centimeter, it's 12.7. So yeah sorry it's 12.7 not 150 but this won't affect the amplitude of the finding.

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The inverse square law is right here:
https://en.wikipedia.org/wiki/Invers...re_law#Formula

6. Related to this, take a 6.5 star (or whatever the minimum is) that you can just see with averted vision in a Bortle Class 1 sky.
How many photons per second are entering your pupil?

7. Related to this, take a 6.5 star (or whatever the minimum is) that you can just see with averted vision in a Bortle Class 1 sky.
How many photons per second are entering your pupil?

8. Originally Posted by philippeb8
From here:
https://kogalla.com/blogs/tech-trail...ens-is-the-sun

"So at earth's orbit, for each square meter, our sun puts out 127,000 lumens."

So for each square centimeter, it's 12.7. So yeah sorry it's 12.7 not 150 but this won't affect the amplitude of the finding.
Originally Posted by philippeb8
The inverse square law is right here:
https://en.wikipedia.org/wiki/Invers...re_law#Formula
Yes, I understood you were applying the inverse square law, but for obvious reasons I had no idea what you were applying it to.

So you were trying to estimate the illuminance we'd received from the sun if it were placed at the distance of Alpha Centauri. Which is sufficiently low that we wouldn't be able to see by its light. But being able to see other things illuminated by a light source (illuminance) is not the same thing as being able to see the light source itself (luminance).
Then, for some reason, you calculated the luminous flux passing through a square centimetre. Did that have something to do with the area of a human retina? (I'm guessing.)

But a single star does not illuminate the whole retina, because its light is focussed to form a diffraction-limited image on the retina. To calculate the apparent luminance of that spot of light, we need to divide the illuminance by the angular area.

So the sun would actually deliver an illuminance of 1.7x10-6 lx if it were as far away as Alpha Centauri, from a spot in the sky with an apparent angular area of 6.6x10-8 sr (a disc one minute across). That's a luminance of about 26 lx.sr-1 (the units are equivalent to the usual cd.m-2, but I've chosen them for clarity). That corresponds to the level of luminance you get from a sheet of white paper under moderate indoor lighting. You can't directly convert the physiological effect of large diffusers to point sources, but basically, if you can see to read black letters on white paper under indoor lighting, you could see the sun at 4.3 light years against a black sky.

Grant Hutchison

9. Originally Posted by Tom Mazanec
Related to this, take a 6.5 star (or whatever the minimum is) that you can just see with averted vision in a Bortle Class 1 sky.
How many photons per second are entering your pupil?
I'm guessing you want visual photons.
Several spherical cows are required.

Sun's visual brightness (mv=-26.75) relative to limiting visual magnitude (mv=6.5) = 10^(0.4*[6.5+26.75]) = 1.995x1013

Solar constant = 1366 W.m-2

Proportion of black body radiation in visible spectrum at 5800K = 0.37 (You need to integrate the black body curve to derive this. My visual-band limits are 400-700nm.)

Therefore, approximate visual-band irradiance from a sunlike mv 6.5 star = 1366*0.37/1.995x1013 = 2.533x10-11 W.m-2

Approximate the energy per photon in visual band from that of a 550nm photon = h*c/lambda = 3.632x10-19 J

Therefore, visual photon irradiance from sunlike mv 6.5 star = 7.013x107 count.s-1.m-2

Area of seven-millimetre pupil = 3.848x10-5 m2

Therefore, visual photon flux at seven-millimetre pupil = ~2700 s-1

Grant Hutchison

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Originally Posted by grant hutchison
Yes, I understood you were applying the inverse square law, but for obvious reasons I had no idea what you were applying it to.

So you were trying to estimate the illuminance we'd received from the sun if it were placed at the distance of Alpha Centauri. Which is sufficiently low that we wouldn't be able to see by its light. But being able to see other things illuminated by a light source (illuminance) is not the same thing as being able to see the light source itself (luminance).
Then, for some reason, you calculated the luminous flux passing through a square centimetre. Did that have something to do with the area of a human retina? (I'm guessing.)
Yes exactly.

But a single star does not illuminate the whole retina, because its light is focussed to form a diffraction-limited image on the retina. To calculate the apparent luminance of that spot of light, we need to divide the illuminance by the angular area.

So the sun would actually deliver an illuminance of 1.7x10-6 lx if it were as far away as Alpha Centauri, from a spot in the sky with an apparent angular area of 6.6x10-8 sr (a disc one minute across). That's a luminance of about 26 lx.sr-1 (the units are equivalent to the usual cd.m-2, but I've chosen them for clarity). That corresponds to the level of luminance you get from a sheet of white paper under moderate indoor lighting. You can't directly convert the physiological effect of large diffusers to point sources, but basically, if you can see to read black letters on white paper under indoor lighting, you could see the sun at 4.3 light years against a black sky.
Maybe that mixture of equations was not the right choice and thank you for the explanation. But another way to see it is using the perspective effect which uses the inverse law only this time:
https://en.wikipedia.org/wiki/3D_pro...ive_projection

(1.5e8 km) / (4.1e13 km) * 5 cm = 1.8e-5 cm

So if we see a 5 cm wide Sun from the Earth then we should (or not) see a 1.8e-5 cm wide Alpha Centauri.

11. Originally Posted by philippeb8
https://en.wikipedia.org/wiki/3D_pro...ive_projection

(1.5e8 km) / (4.1e13 km) * 5 cm = 1.8e-5 cm

So if we see a 5 cm wide Sun from the Earth then we should (or not) see a 1.8e-5 cm wide Alpha Centauri.
But length is the wrong unit in this setting, since the sun's apparent diameter is angular, rather than linear.
It's about half a degree across in Earth's sky, which would make it about two millionths of a degree across if it were placed at the distance of Alpha Centauri. That's well below the diffraction limit of the human eye, though--we cannot resolve any star as a visible disc. So the sun's light would be blurred by our eyes into a disc about a sixtieth of a degree in diameter, and its surface brightness diluted in proportion.

Grant Hutchison

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Originally Posted by grant hutchison
But length is the wrong unit in this setting, since the sun's apparent diameter is angular, rather than linear.
I am simply projecting a width at different distances in order to have an estimate.

It's about half a degree across in Earth's sky, which would make it about two millionths of a degree across if it were placed at the distance of Alpha Centauri. That's well below the diffraction limit of the human eye, though--we cannot resolve any star as a visible disc. So the sun's light would be blurred by our eyes into a disc about a sixtieth of a degree in diameter, and its surface brightness diluted in proportion.
Which means what exactly in relation with Alpha Centauri?

13. Originally Posted by philippeb8
Which means what exactly in relation with Alpha Centauri?
It means what I already said, in some detail. That the Sun at 4.3 lightyears (as a proxy for Alpha Centauri), would appear as a diffraction-limited disc about a minute of arc across with a luminance of about 26 lx.sr-1, which is easily bright enough to see against the dark background of the night sky, which has a luminance that is generally well below 1 lx.sr-1.

Grant Hutchison

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Originally Posted by grant hutchison
It means what I already said, in some detail. That the Sun at 4.3 lightyears (as a proxy for Alpha Centauri), would appear as a diffraction-limited disc about a minute of arc across with a luminance of about 26 lx.sr-1, which is easily bright enough to see against the dark background of the night sky, which has a luminance that is generally well below 1 lx.sr-1.
Ok thanks. The projection is dim but the luminance is high enough.

15. Originally Posted by Tom Mazanec
Related to this, take a 6.5 star (or whatever the minimum is) that you can just see with averted vision in a Bortle Class 1 sky.
How many photons per second are entering your pupil?
Ignoring things like spectral slope across the visual band, a starting point is that a star of V=0 means about 1000 photons/(cm^2 S Angstrom) at the top of the atmosphere. (IIRC closer to 1084, but still.) At V=6.5, that makes 1084 x 10^(-0.4x6.5) = 2.7 photons/(cm^2 s A). A young person's full dark-adapted pupil has diameter ~7mm or area 0.38 cm^2, and we can assign the effective bandwidth of the eye's response as (somewhat generously) 3800-6500 A or 2700 A. So that puts 2770 photons/second into the pupil in a useful wavelength range. Lab experiments (and a couple of observers documented with very special vision) take that ultimate limit to about magnitude 8, dropping the rate by a factor 4.0. That's still tens of photons per time interval usually associated with visual response (0.1 second or maybe a bit shorter).

16. Originally Posted by grant hutchison
Therefore, visual photon flux at seven-millimetre pupil = ~2700 s-1
Originally Posted by ngc3314
So that puts 2770 photons/second into the pupil in a useful wavelength range.
Heh. I love it when stuff like that happens. Two different routes to the same answer.

Grant Hutchison

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