NASA gives the luminosity of the Sun as 382.8 \times 10^{24} \text{ J/s}. I assume this is radiant power since it is given in watts. Georgia State University says I can convert radiant power to luminous flux by an efficiency factor:
In terms of radiant power (also called radiant flux) it can be expressed as:
Luminous flux in lumens = Radiant power (watts) x 683 lumens/watt x luminous efficacy
The luminous flux is the part of the power which is perceived as light by the human eye, and the figure 683 lumens/watt is based upon the sensitivity of the eye at 555 nm, the peak efficiency of the photopic (daylight) vision curve. The luminous efficacy is 1 at that frequency.
I want to find the luminous flux of the Sun itself, but I’m not sure what wavelength to use in the efficiency table. 2.614524 \times 10^{29} \text{ lm} is an upper limit, but how do I find something more accurate?
I suspect NASA’s figure covers a whole range of wavelengths (maybe the range for visible light?). I also remember learning that stars approximate black body radiation, and therefore radiate energy on all wavelengths according to a curve. What do I need to do here? I have the vague idea that I need to calculate the volume of a three-dimensional object, measured by wavelength, radiant power, and luminous efficiency.
In fact, the entire range of wavelengths, from radio to gamma. If it’s ever ambiguous, this sort of luminosity measure, over the full range of the entire spectrum, is called “bolometric luminosity”, and it is properly measured in watts.
Lumens are rather a bastard unit, and ought not to be considered part of the SI standard for base units, as they depend on a frequency-response curve designed to approximate the human eye. But you can probably look up the luminous efficacy of a 6000-K blackbody. Blackbodies are typical enough sources that this information should be out there somewhere, already computed (though I suppose that if you really like numerical calculus, you could find it yourself from the monochromatic tables you have).
Radiometery and phootmetry is a fascinating and confusing topic. There is much math, a lot of definitions, and a mind-boggling assortment of units with names that, if you say them rapidly in succession, might cause you to say obscene words.
Radiometry is concerned with the measurement of radiant energy of all types emitted by a source. Photometry is the measurement of VISIBLE light emitted from a source. Visible light extends from about 400 to 650 nanometers. Consequently, you can’t simply multiply radiant flux by some factor and get the photopic flux. You have to multiply the light at each wavelength by a conversion factor that has the eye’s response at that wavelength built into it, then integrate over all wavelengths. There are sites on the internet that can walk you through this kind of thing, or you can get a book on radiometry and photometry.Here’s one that explains this and works out the luminous flux from the sun:
SI units are there to be the best units for the job.
It is rather important to measure the actual utility of a light source for human use…
6000 K black body, 100 lumens per Watt, this means that for each watt in the entire spectrum only 0.14 watts of visible light, or 100 lumens, is found in the visible spectrum.
Its important not to use earth readings of the luminous efficacy, as that measuring the spectrum as altered by earths atmosphere and earths surface (reflections…)