# I Lumens and the Luminosity Function

1. Mar 24, 2017

### JohnnyGui

Hello,

I have a small confusion regarding the formula that calculates the luminous flux of a light source, which is given by:

(See this Wiki)

The $\bar y (λ)$ is the so-called luminosity function that corrects for the sensitivity of the human eye for a particular chosen wavelength.
The $Φ_{e,\lambda}$ is the "objective" radiant flux (in watts) per nanometer which is the total power per nanometer, regardless of the sensitivity of the human eye (hence "objective"). You can see from the formula that if a light source emits more than 1 wavelength, an integration has to be done with the limits being the endpoints of the emitting wavelength range, so that it gives the total luminous flux of that light source.

Now, I understand that this formula gives the luminous flux $Φ_V$ in lumen because of the multiplication with $683.002 lm/W$. However, if we remove this constant, then this means that:

Will give a luminous flux in Watts, right? This means that you can calculate the subjective power according to the human eye's sensitivity in Watts. According to the definition of 1 candela, if we now plug in a wavelength of 555nm of 1 Watt in the formula, we'd get 1/683 Watts out of this formula. This means that the subjective power of 1 candle according to the human eye is 1/683 Watts.

If that's the case, then I don't get why this Wiki says the following:

"The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that direction of  1⁄683 watt per steradian"

Notice that it says that 1 candela is the radiant intensity of 1/683 watt, which is the overall objective power (per steradian in this case) and not the luminous intensity (per steradian) as I concluded, which is the subjective power according to the human eye.

What am I concluding wrong here?

2. Mar 30, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Mar 31, 2017

### JohnnyGui

Ok, so I’ve figured out what my misunderstanding was. The Luminous Flux (the formula above) merely shows how strong a light source is relative to a candle that emits light of 555nm at 1/683 Watts, all based on the human eye’s sensitivity.

I have one other question. I understand that if a light source emits several wavelengths, that you’d have to integrate the formula above over dλ as discussed. For each dλ you’d have to give the specified amount of Watts that the light source is emitting for that specified dλ.

What I don’t understand is that this is given by a constant $Φ_{e,λ}$ that gives the amount of Watts per nanometer in the luminosity function. This constant says that the larger the wavelength, the larger the amount of Watts it emits and that it's proportional. But a light source doesn’t necessarily emit larger wavelengths at larger Watts and it surely doesn't emit wavelengths that is proportional to the amount of Watts. So why is this constant in the luminosity function?

4. Mar 31, 2017

### Eric Bretschneider

You are assuming a spectrum in which power increases with wavelength. That is almost never the case. When you restrict the range of interest to the visible spectrum (360nm to 830nm) then that is case for a Planckian or black body emitter for temperatures below about 3500°K. According to Wein's Displacement Law, the peak wavelength of a Planckian emitter scales as 1/absolute temperature (http://hyperphysics.phy-astr.gsu.edu/hbase/wien.html).

Even allowing for that, the luminosity function has a maximum at 555nm and drops rapidly as you move to longer and shorter wavelengths. So at 555nm it is 1.0000, at 600nm it is 0.6292, at 650nm 0.1069, at 700nm 0.0041, at 750nm 0.0001 and it keeps dropping.

For your statement to be true, the power at each wavelength would have to increase faster than the luminosity function is decreasing. Anything that does that would appear as a very pure red color.

5. Mar 31, 2017

### JohnnyGui

But I agree on that power shouldn't increase with wavelength. My problem is that the formula of the luminosity function shows a constant of Watts/nm which implies that the amount of Watts increases with wavelength My post was directed at the fact that this shouldn't be the case.

This Wiki gives the following formula for the luminosity function:

In which $Φ_{eλ}$ is the spectral radiant flux, in watts per nanometer. How can this be while the wavelength is never proportional to the amount of watts in the spectrum of a lightsource (which is a peak curve)?

6. Mar 31, 2017

### Eric Bretschneider

Watts/nm isn't a constant - it is the units!

Φ is a function of the wavelength. Don't confuse units with numbers.

7. Mar 31, 2017

### JohnnyGui

Ah, I got confused there! That's why it has a $(λ)$ after it since it differs with the wavelength.

So is $Φ_{eλ}$ the derivative of a spectrum curve of a light source so that integrating it over a specified spectrum range would give the total energy in that spectrum range in Watts?

8. Apr 1, 2017

### JohnnyGui

I am kind of surprised that the luminosity function needs to be an integral if a light source is emitting a spectrum of wavelengths.

From what I understand, each specific wavelength which is emitted, has a certain amount of watts and a certian sensitivity of $\overline{y}(λ)$. So it's actually not the amount of watts and sensitivity in a given range of $dλ$, which is what the integral does, but for each specific wavelength.
So instead, I'd expect that it has to be a cumulative sum that adds every amount of watts and sensitivity of each specific wavelength over the range of the spectrum of a light source.

Shouldn't therefore the luminosity function be a summation formula that adds $Watts(λ) \cdot \overline{y}(λ)$ for each wavelength that the light source emits?

9. Apr 3, 2017

### Eric Bretschneider

The sensitivity function is also known as V(λ) and has units of lumens/W. So the spectral power distribution (Watts/nm) x V(λ) gives lumens (when you multiple by a wavelength interval).

Integration is a summation process. You can calculate an analytical integral (which would be an equation) or a numerical integral which would be the process you are describing. They are one and the same.

10. Apr 3, 2017

### JohnnyGui

But doesn't multiplying the spectral power distribution in Watts/nm with a wavelength interval $dλ$ merely give the difference in Watts between 2 wavelengths with a difference of $dλ$? So that basically integration is merely giving the total difference of Watts between the 2 ends of a spectrum range? Basically like this (which is in this case a total difference of 680nm):

What I meant with summation is more like a function with $∑$ so that the formula adds the amount of Watts of each specific wavelength in the spectrum range together. Like this:

11. Apr 3, 2017

### Eric Bretschneider

No it doesn't. Integration gives the area under the curve. The spectral power distribution has units of W/nm. If you integrate you multiple W/nm by nm and get W. When you take the product of the spectral power distribution and the V(λ) function you get W/nm x lumens/W = lumens/nm. Integrating that gives you lumens.

In numerical integration you would take the value at every wavelength and multiple by the spacing and add it all up. The simplest form of numerical integration is adding the area of a bunch of rectangles that just touch the curve you are integrating (start, mid point or end of the wavelength interval). You could also do trapezoidal integration or something like Simpson's Rule (fits a parabola to 3 data points). Integration is the same thing as summation!

Here is a simplified tutorial https://www.mathsisfun.com/calculus/integration-introduction.html

12. Apr 4, 2017

### JohnnyGui

I think my whole misunderstanding here is that I think that $Φ_{eλ}(λ)$ is a derivative of that spectrum graph I posted in my post #10 above.
Integrating a derivative will give the area beneath its own graph line like you said. However, that area beneath that derivative graph line is equal to the total difference between your 2 chosen limits in the original function (the one you took the derivative of).

Since I thought that $Φ_{eλ}(λ)$ is the derivative function of that spectrum graph, I expected it to merely give the difference in amount of watts between 2 chosen wavelengths in the spectrum, when you integrate it.

13. Apr 5, 2017

### Eric Bretschneider

Definitely not. It isn't even an analytical function. Although you can go to some extremes to create a piece-wise analytical version, in practice it is much easier to use a lookup table.

14. Apr 6, 2017

### JohnnyGui

Apologies for my stubborness but I've found a simpler way to explain what I don't understand.

Suppose a light source is emitting light only between 2 and 5 nm, and the spectral power (Wtts/nm) being constant at 4 Wtts/nm over that range of wavelengths (I know it's not supposed to be constant, but just for the sake of argument). The spectral power graph would then look like this:

The luminosity function would integrate this graph. Integration will give an area of 12 Watts.

However, this graph does not represent the absolute power in Watts that each wavelength is emitted at. Spectral power is more like the power concentration of each emitted wavelength. And from what I understand, to know the amount of lumen a wavelength has, one has to look at its absolute power that it is emitted at.

The spectral power (power concentration) of 4 Wtts/nm multiplied by the corresponding wavelength will give the absolute power ($P$) for that wavelength. Multiplying that absolute power (in Watts) of each wavelength with the sensitivity function (Lumens/Wtts) will give the total amount of lumens for that specific wavelength. I'd have to do this for each wavelength and add them all together to get the total amounts of lumens over the whole spectrumrange (2nm -5nm)
So the formula would look something like this:

$P(λ=2) \cdot \overline{y}(λ=2) + P(λ=3) \cdot \overline{y}(λ=3) + P(λ=4) \cdot \overline{y}(λ=4) + P(λ=5) \cdot \overline{y}(λ=5)$

This formula however will give way more total amounts of Watts than the 12 Watts calculated in the spectral power graph.

So my confusion lies in the fact that the luminosity formula uses the spectral power of a wavelength x the corresponding sensitivity function to get the amount of lumens at that wavelength while I think that it should take the absolute power of that wavelength and multiply it with the wavelength to get the amount of lumens at that wavelength. This is what I can't seem to grasp.

Last edited: Apr 6, 2017
15. Apr 6, 2017

### Eric Bretschneider

You have it correct, but remember that when you multiply by V(λ) the units of the product are no longer Watts. V(λ) has units of lumens/Watt so the number that you get (larger than 12 Watts) is the number of lumens. You should be using the absolute power at a wavelength x V(λ) at that wavelength.

Lumens are a perceptual unit and that makes things peculiar at times. 100 lumens should appear as the same "brightness" regardless of the spectral power distribution.

16. Apr 6, 2017

### JohnnyGui

My bad, I indeed meant that I'd get way more lumens than when I integrate the spectral power graph according to: $Φ_{eλ}(λ) \cdot V(λ)\cdot dλ$.
If I have it correct, why would integration of that luminosity formula then give way less lumens than my conclusion?

17. Apr 12, 2017

### JohnnyGui

Could someone please help me with this question? Still not really understand why my conclusion is different from how it is actually calculated while it should be the absolute power for each wavelength.

18. Apr 25, 2017

### JohnnyGui

Bump

19. Apr 25, 2017

### Andy Resnick

So far, so good- the formula above is the power ('light watts') that gives rise to a response by the human eye. The prefactor 683 lm/W at 540 THz is a normalization factor for human vision (hence the Wiki entry). There are other normalization factors: 'Photosynthetic watts', for example, or 'sunburn watts'. 1 sunburn watt = 1 finsen.

These normalization factors (and others- the sidewinder missile uses 'lead sulfide watts') are used to avoid the need for spectral radiometric measurements.

Don't forget- 'intensity' is W/sr.

Does that help?

20. Apr 25, 2017

### JohnnyGui

Thanks for the reply. I indeed was able to understand this eventually. However, a new confusion came regarding calculating the spectral sensitivity. I can't formulate my question better than I have in my post #14. Hope you're able to help me with that