Talk:Luminosity function

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[edit] Older comments

The integral sucks. What is it supposed to show / equal?? Can anyone correct this, please? Ian Cairns 23:50, 12 Nov 2004 (UTC)

Hes right, if you go to Planckian locus and multiply all constants units together, you will have W*m^2 (and this matches value given in ISO/CIE 10526:1991(E) sec5.1, so it must be right); divide by lambda^5 and you have W/m^3. Now, 683 lm/W is given in ISO/CIE 10527:1991(E) sec7.1, so it is correct, too. Plugging it into integral, we have W/m^3 * lm/W * m = lm/m^2, *NOT* just lm. WTF ?? 91.124.43.185

The J(lambda) specified is watts per wavelength, not W/m^3, so it's correct as stated. If you'd rather use W/m^3, then the answer would indeed be lm/m^2 as you suggest (assuming the dlambda is also in meters in that case). By the way, the comment you responded to is over two years old; I checked and the integral and units were substantially the same back then. Dicklyon 00:21, 14 January 2007 (UTC)

Could you please refer to formula for J(λ) ? 82.207.27.24

It's not from a formula; think of it as the independent variable in this equation, the spectrum of the light flux in watts per unit wavelength, as a function of wavelength. Dicklyon 00:03, 16 January 2007 (UTC)

What I mean it is, well, it should be, flux x some surface area, so I would like to know what area it is, 195.137.203.137

The area is the surface of a large sphere around the source, since J(λ) is based on the total power emitted. You may be being confused by the word "flux". There are two incompatible definitions of "flux" used in physics. One is a vector quantity, with dimensions of something per unit area. The other is the integral of that over some surface, i.e. it is a scalar that represents the total amount of that "something" that flows through the surface, per unit time. Here we're talking about the second type of flux—it's a scalar with dimensions of power. J(λ) is the spectrum of the total radiant flux , in all directions. Note that the description of it as an "intensity" was wrong, and I have removed that from the article.--Srleffler 13:15, 16 January 2007 (UTC)

The area can be any surface, not necessarily spherical, not necessarily related to any source. It can be the area of the detector in a light meter, for example, which can then be positioned in any way you want relative to any set of sources. Or, instead of any area, you can apply the equation to the total power of a source; or other ways; The equation doesn't have much to say about where you can apply it. Dicklyon 16:38, 16 January 2007 (UTC)

Thanks I seem to getting to it. Could you (or anyone) point me to some page online where actual numbers are calculated, say for A source, or T=1234K, or anything, so that I could make sure I use the formula correctly? Whole this thing is because I am getting pretty big lux values, I think. 195.137.203.137

I see where the confusion came from. The article mistakenly described J(λ) as a "spectral intensity". It is not. It is a power spectral density. I(λ) is the spectral intensity.--Srleffler 06:19, 14 January 2007 (UTC)

but you know what lm/m^2 = Lux and if you integrate by that formula you will get reasonable values 195.137.203.137 (aka 91.124.43.185)

Errr... I was too fast in my conclusions, they are not reasonable. 82.207.27.24

Not sure what you're getting at, but if you have questions perhaps someone here can answer them.

I assume all the anon posts above are "you". You don't have to sign with your IP address. You could sign with your name or a pseudonym so others can tell that all these messages are from one person, even though the IP keeps changing. You can register your choice of username so no one else uses it. It only takes a few minutes and doesn't even require you to give your real name. This makes it easier for others to communicate with you, and gives you a few benefits that are only available to signed-in users.--Srleffler 22:36, 15 January 2007 (UTC)

This page could really use a graph of the function.--Srleffler 06:37, 12 November 2005 (UTC)

The graph on this page is utterly unreadable to anyone with even slight colorblindness. can someone add "on the left" and "on the right" to the descriptions in the caption?--Deglr6328 04:26, 14 November 2006 (UTC)

Srleffler fixed it. Wouldn't it have been easier for you to just do that than ask? Dicklyon 06:32, 14 November 2006 (UTC)

I presume he is colourblind, and could not do this for himself.--Srleffler 07:48, 14 November 2006 (UTC)

Probably is. But he said exactly what to do, and had it right, so I figured he had worked it out. Dicklyon 15:12, 14 November 2006 (UTC)

Is blue supposed to be shown in the graph? I assumed there would be three peaks, for red/green/blue, but there are only two.—The preceding unsigned comment was added by Atamido (talk • contribs) 10:32, November 14, 2006.

You have misunderstood. The peaks on the graph have nothing whatsoever to do with RGB; the choice of red and green for the two curves was completely arbitrary. Note also that (despite statements to the contrary that one sometimes sees) human vision is not an RGB system. There are three types of color receptor in the eye, but their peak responses are to yellowish-green, bluish-green, and blue-violet respectively. The brain determines color by comparing the responses of the three types of receptor. The red curve in this graph shows the combined response of all three.--Srleffler 15:42, 14 November 2006 (UTC)

I suppose it would have been smarter of me to choose two colors that would not lead to such a false impression. Maybe I'll fix it. Dicklyon 15:54, 14 November 2006 (UTC)

OK, I changed to green and black curves. Should be no need for color-blindness disambiguation any more, and no confusion with RGB. You might need to refresh if you still see green and red. Dicklyon 18:18, 14 November 2006 (UTC)

Of course, you've used black for the color vision curve and green for the monochrome vision curve. :) --Srleffler 01:45, 15 November 2006 (UTC)

Well, that's something I hadn't thought of. I used black for set with solid, dashed, and dotted so that they'd show up more clearly. But feel free to swap colors if you think it will help. Dicklyon 02:46, 15 November 2006 (UTC)

Ahhh now I can read it. Thank you!--Deglr6328 06:17, 15 November 2006 (UTC)

[edit] Astronomical definition

This needs some kind of disambiguation page -- "luminosity function" in astronomy means the number of objects of a given luminosity per unit volume of space. It's used in the context of stars, galaxies, and even quasars, so it's very broadly used. I'm not in a position to write such an article at the moment, but maybe someone could write a stub with just that info and create the disambiguation page? -- JThorstensen —The preceding unsigned comment was added by 67.142.130.36 (talk) 03:06, 3 May 2007 (UTC).

[edit] Human cone response

Quick plot of the mentioned Stockman, MacLeod and Johnson (1993) 2-deg cone fundamentals (colors) and corresponding photopic luminosity function (black)

Please provide the complete un-normalized response curves of the three types of humans color receptors, and show how this adds up to (or differs from) the Luminosity function.-69.87.203.133 02:23, 25 May 2007 (UTC)

The cone responses are physical; I'm not sure where to find good data on them, but I'll see. The luminosity function, on the other hand, is derived from psychovisual measurements. So there's not necessarily any simple relationship, is there? Presumably, though, the luminosity function is a linear combination of cone curves, since the XYZ color space is supposed to span the same 3D subspace of spectrum space that human color vision does. What are you hoping to get out of this exercise? Dicklyon 02:44, 25 May 2007 (UTC)

I found a bunch of cone datasets at the CVRL. Most seem to been constructed from color-matching functions, and based on some measurements that you'll have to chase down in the refs. They seem to be normalized. It's not clear what you can get from "un-normalized", but maybe there's something there. Some indicate the weights needed to get the (modified) photopic luminosity function, like this one, which indicates no weight at all for the short-wavelength cones. Dicklyon 04:40, 25 May 2007 (UTC)

[edit] Integral should not be presented as an inner product?

I don't think the approximation to the integral should be described as an "inner product". I understand that considering the functions inside the integral to be infinite dimensional vectors and the integral to be an inner product is the way to go when considering coordinate system changes in that vector space (e.g. Fourier transform and so forth) and that approximating these functions as finite-dimensional vectors is appropriate in estimating the integral. However, the behavior of these functions under such transformations is not really relevant here, is it? So I think that introducing these functions as elements of an irrelevant space just serves to complicate matters. In other words there is no point in thinking of them as vectors unless their vector nature (invariance under transformation) is relevant, which it is not, I think - correct me if I am wrong. PAR (talk) 16:07, 24 January 2008 (UTC)

I don't understand your point. It is conventional to treat spectra as vectors in a Hilbert space. The operation is a inner product, whether on the discretized vector or the Hilbert space vector. Dicklyon (talk) 18:08, 24 January 2008 (UTC)

I think he's just saying that readers might find it confusing? I think describing it as an inner product is fine. --jacobolus (t) 18:54, 24 January 2008 (UTC)

Perhaps so. And the Poynton ref doesn't really support what's said there any more. But if it said that making it discrete makes it an inner product, which is about what Poynton says, that would be incorrect and confusing, wouldn't it? Maybe we need to find another good source to explain this, or just ditch all the vector talk. Dicklyon (talk) 19:20, 24 January 2008 (UTC)

Well, I guess my question would be: When, in color science or vision science, is it necessary to appeal to the vector nature of a spectrum, rather than just treating it as a simple function of wavelength, or frequency? I mean, a spectrum is also an element of a complex Hilbert space too, so why don't we say that the luminous flux is the inner product on this space:

where * denotes the complex conjugate. The answer is, because we never need to use the properties of the functions as elements of a complex Hilbert space in any application of color and vision science. I don't think we need to use any properties of the functions as elements of a vector space either, so, for the same reason, lets not do it. PAR (talk) 21:41, 24 January 2008 (UTC)

[edit] Uncited speculation removed.

I removed the following from the article, since it is clearly identified as speculation/original research, and is uncited.

An argument can be made that the difference between rod and cone response is an evolutionary adaptation to survival under star-lit skies. The apparent-magnitude-weighted average spectrum of stars in the Earth's sky is bluer than the spectrum of the Sun (this statement is based on conversations with other professional astronomers and needs verification and a reference), hence the optimal vision for predation - and avoiding predation - at night is bluer than the optimal for daylight. P.r.newman (talk) 09:36, 3 April 2008 (UTC)

--Srleffler (talk) 05:04, 28 February 2008 (UTC)

[edit] Lower limit

Quick comment: The last section says that the lower spectral limit for eye responsivity is 507 nm. I assume that this should be 407 nm, to be consistent with the plot. 128.111.74.4 (talk) 21:53, 14 March 2008 (UTC)Scott

You seem to have misread it. The last section (on scotopic vision) says that the peak of the eye's sensitivity in dim light is 507 nm. This is the green curve on the graph.--Srleffler (talk) 02:23, 15 March 2008 (UTC)