Calculating the luminance of the sun.

[Andrew_M]

Hello everyone,

 

I have been reading about photometric optics and photometric units recently. I keep seeing several documents saying that the luminance of the sun is 1.6×10⁹ cd/m² at noon, eg Wikipedia.

 

I tried to use the definition of luminance and some facts about the sun to see if I could calculate what the luminance should appear to be to someone on Earth, just to see if I got the same number as in Wikipedia. Well not only did I not get the same number, but my answer was off by a factor of 10³⁰ too big!

 

I am puzzled because I think I am using the right formulae and all the correct numbers, but still it does not match, so I am wondering where I have gone wrong.

 

So generally the formula is Lum = Flux / SolidAngle / Area

 

Wikipedia includes a cosine theta term there too, but in this case we are looking at the sun and the sun is looking at us, we're all in the same orbital plane.

 

The Flux will be half of the total flux of the sun, because we can only see one half at a time. From WP, this is 3.75×10²⁸ / 2 lumens.

 

According to the setup for a luminance photometer, the Solid Angle is the angle subtended by the receiving area as seen by a point on the source. Since the receiving area is going to be the pupil of an observer's eye (hopefully he is wearing powerful sunglasses) then the solid angle can be found by formula for solid angle of the base of a cone with full angle theta:

sr = 2*pi*(1 - cos(theta))

 

I'll assume the pupil radius is 1mm. The distance of 1AU is 1.496×10⁸ km or 1.496×10¹¹ m

sr = 2*pi*(1 - cos( 2*arctan(pupil radius / 1 AU) ))

= (2π×(1−cos(2× tan⁻¹(0.001÷(1.496×10¹¹)))))

 

This is going to be rather a small solid angle, obviously.

 

The Area is the area of the source being measured, which here is half the total area of the sun, so 3×10¹² km² which is 3×10¹⁸ m²

 

So putting it altogether...

 

Lum_sun = (1.8×10²⁸) ÷ (2π×(1−cos(2× tan⁻¹(0.001÷(1.496×10¹¹)) ))) ÷ (3×10¹⁸)

= 1.068573927×10³⁷

 

Hah, quite a bit different to 1.6x10⁹, so where did I go wrong?

[Schrödinger's hat]

Not quite sure exactly where you went wrong, but there's a much shorter way, at least to get as far as power per area (watts per square metre rather than candela).

If you want candela you have to take into account the sensitivity of the human eye.

The logic for getting flux over one square metre from the luminosity of the sun is

[math] \frac{L_{sun}}{A_{earth\, orbit}} \times 1 m^2[/math]

Or for power per square metre [math]\frac{L}{A}[/math]

Also the figure found for luminosity disagrees with yours.

[math]\frac{3.85\times10^{26}}{4\pi \times1.496^2\times10^22}[/math]

This comes to about 1300Watts

To get luminous intensity from this you'll have to do an integral, or look up the ratio between power per square metre and luminous intensity for a black body at around the sun's temperature. I can explain this step in more detail if you need it.

[Andrew_M]

Well it would seem nobody here knows the answer to this problem.

 

I will try to find other explanatory texts on luminance measurement and calculation.

I will also check if the difference in magnitude between my answer and the Wikipedia figure is the same as the difference in magnitude between the area subtended by the pupil and the area subtended by the sun. Perhaps someone has used the wrong solid angle in the calculation, or the wrong area.

[Alper Ülkü]

Here is the solution:

 

Sun subtends about 0.5 degree on earth surface solid angle Omega becomes = 2pi (1- cos (0.5/2)) = 5,98e-5 steradians

 

At bright day light and at 90 degree normal incident, illuminance of sun at earth's surface is around 10.000 lux = lm/m2

 

Luminance = illuminance / Omega = 10.000/5,89e-5 =1.67e9 cd/m2

 

QED

Edited February 6, 2015 by Alper Ülkü