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Notes on surface brightness

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#1 GlennLeDrew

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Posted 27 March 2013 - 06:32 PM

I tried this little experiment several years ago, with my then brand new Sky Quality Meter. Under the fairly dark sky at Starfest, I laid a sheet of white paper on the ground. I crouched so as to block the smallest amount of sky, and pointed the meter down at the paper. I got a reading that was pretty close to the average of a number of sky readings.

This told me that a flat surface will have a surface brightness equal to that of the mean surface brightness of the hemisphere it faces, multiplied by that surface's albedo.

If we take the paper's reflectance as 85%, the surface brightness should be 0.85 times the averaged surface brightness of the sky and any objects protruding higher than the horizon (including me). I forget the numbers, but indeed the paper reading was a bit dimmer than the sky readings by about the expected degree.

By extension, if the light source does not occupy a full hemisphere, the surface brightness on a surface facing it scales as the ratio of the solid angle of the source to the area of a hemisphere. An example.

The Sun has a diameter of 1/2 degree and a surface brightness of -10.6 magnitudes per square arcsecond. What is the surface brightness of a tropical beach having an albedo of 0.45, under the midday sun?

Because of the small angle, we can calculate the Sun's angular area in the usual way, obtaining 0.196 deg^2. And we can use the area of a hemisphere in degrees, which is ~20,000 deg^2. We see that the Sun occupies 0.196 / 20,000 = 0.0000098 of the sky. The beach has a surface brightness fainter than that of the Sun by a factor of 0.0000098 * 0.45 = 0.0000044. In magnitudes, this is 2.5 * LOG (0.0000044) = 13.39. And so the beach has a surface brightness of -10.6 + 13.4 = 2.8 magnitudes per square arcsecond.


Another problem. What is the contribution to surface brightness at the ground when a -8m Iridium flare occurs?

We can integrate the light over a hemisphere. 20,000 deg^2 is 20,000 * 12,960,000 = 2.592 * 10^11 arcsec^2. With the light of a -8m point source spraed over the whole sky, one square arcsecond would be 2.592 * 10^11 times fainter, or 2.5 LOG ( 2.592 * 10^11) = 28.5 magnitudes fainter. The surface brightness contribution from a -8m Iridium flare would be -8 + 28.5 = 20.5 magnitudes per square arcsecond.

If we are concerned with the *relative* contribution to ground illumination, albedo need not be considered. Suppose the sky's average brightness was 20.5 MPSAS. The contribution from it and the -8m flare would then be equal. This means the ground will become twice as bright, and shadows will be seen.

But suppose the sky's mean brightness was 19 MPSAS. The flare's light is 1.5 magnitudes, or 4X fainter. Compared to the 19 MPSAS shadow, the ground outside is the combination of the 19 sky and the 20.5 flare, which is 25% brighter. Shadows should still be seen, but not easily.


A final problem, involving an angularly large source. This is illustrate the concept of solid angle.

What is the surface brightness of a surface having an albedo of 0.45 (same as our beach earlier), but when the Sun is only 0.01 AU distant?

Naively, we would say it's 10,000X brighter because we're 100X nearer to the Sun. But we shall encounter a divergence...

From a distance of 0.01 AU the Sun subtends an angular diameter of 55 degrees (not the 50 degrees perhaps at first assumed, due to the viewing geometry). Because of the large angular size, we must calculate the solid angle, for the source behaves as a section of a sphere occupying a not considerable fraction of a hemisphere. The solid angle of the 55 degree wide (27.5 degree radius) Sun in steradians (sr) is

2 Pi SIN^2 (27.5) = 1.34 sr

A hemisphere occupies 2 Pi, or 6.28 sr. We see the Sun occupies 1.34 / 6.28 = 0.21 of a hemisphere, and so the surface brightness of our surface is 0.21 * 0.45 = 0.095 times as bright as the Sun's surface. This is 2.5 * LOG (0.095) = 2.6 magnitudes fainter, making our surface -10.6 + 2.6 = -8 MPSAS.

How much brighter is our surface when 100X closer to the Sun? This scales directly as the solid angle of the solar surface. As we've seen, at 0.01 AU, the solid angle is 1.34 sr. From 1 AU, the 1/2 degree disk occupies 0.00012 sr, an area ratio of 12,000 (not the 10,000 based on the 100X distance ratio.) The surface has this got 12,000 times brighter when placed 100X nearer to the Sun.


Can you think of other problems to solve with these tools?

#2 azure1961p

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Posted 27 March 2013 - 08:11 PM

Ok . Ill confess. I glazed. I see where you are going but itd be wrong for me to say "I've got it". I'm interested to see what others bring here.

Pete

#3 buddyjesus

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Posted 27 March 2013 - 10:24 PM

I knew it!

#4 azure1961p

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Posted 28 March 2013 - 06:08 AM

Hey Im honest!!!!

Pete

#5 Tony Flanders

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Posted 28 March 2013 - 06:12 AM

I tried this little experiment several years ago, with my then brand new Sky Quality Meter. Under the fairly dark sky at Starfest, I laid a sheet of white paper on the ground. I crouched so as to block the smallest amount of sky, and pointed the meter down at the paper. I got a reading that was pretty close to the average of a number of sky readings.


Interesting! That makes perfect sense, though I suspect it depends a good deal on the paper. Better with matte than glossy. This seems like an inexpensive way to convert the narrow-field SQM-L to a wide-field averaging SQM.

#6 careysub

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Posted 28 March 2013 - 03:13 PM

I had wanted to research the sources of light in the night sky, and saw references to this expensive little book:
"The Light of the Night Sky" by F. E. Roach, Janet L. Gordon (1973); but also to an earlier paper one of the authors wrote on the same subject: "The Light of the Night Sky: Astronomical, Interplanetary and Geophysical" by F. E. Roach, Space Science Reviews, Volume 3, Issue 4, pp.512-540:

http://adsabs.harvar...SSRv....3..512R

One problem - it uses units called "S10" not MPSAS (there is an notational convention for MPSAS by the way: mag/(")^2).

How to convert?

Perusing a whole list of on-line references fails to find many that have this notion of converting these astronomical brightness units, and most units conversion resources seem to disdain photometry.

An S10 also called an S10vis or S10(vis) is defined as 10th Stellar Magnitudes/square degree, it is however a linear, not a logarithmic scale.

Let us recall the relationships between magnitude (an inverse logarithmic scale) and luminous intensity (a normal non-logarithmic physical scale):

(m1 - mref) = -2.5*log(I1/Iref)
[magnitude differences are equal to the logarithm of the intensity ratio multiplied by the conversion factor -2.5; used so that that an intensity ratio of 100 equals a 5 magnitude difference]

(I1/Iref) = 10^(-0.4*(m1 - mref))
[simply the converse of the above; converting from the magnitude difference to the intensity ratio]

Since we have
1 S10 = Mag. 10/deg^2

The S10 unit covers an solid arc (3600)^2 times larger than the MPSAS, also the scale is with reference to a dim magnitude 10 source, while the mag/(")^2 scale implicitly references a 0th magnitude which is 100*100=10000 times brighter. So the S10 is 1/10000 as much light spread over an area (3600)^2 larger so the intensity is 1/(100000*3600^2) = 7.716*10^-12 times dimmer.

That is:
10^(-0.4*(10 - 0))/(3600^2) = 7.71604938272*10^-12

Now:
m1 - mref = -2.5*log(10^(-0.4*(10 - 0))/(3600^2))

= -2.5*log(1/(10000*3600^2)) = 27.7815125038

And we can now write:
S10 = 10^(0.4*(27.78 - MPSAS))
MPSAS = -(log(S10)*2.5 - 27.78)

See also:
http://trs-new.jpl.n...5/1/97-1547.pdf

#7 GlennLeDrew

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Posted 28 March 2013 - 06:39 PM

Good info! I've long wanted that book, but it's pricey.

How about working the conversions and supply a short list of the contributors to sky glow in mag/arcsec\^2? Thanks!

#8 careysub

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Posted 28 March 2013 - 11:24 PM

Good info! I've long wanted that book, but it's pricey.

How about working the conversions and supply a short list of the contributors to sky glow in mag/arcsec\^2? Thanks!


I am working on that very thing.

I am also inspired to do a write-up on photometry units - introducing them from first principles and providing a grand view of how they all inter-relate and how to convert them. There are a passel of units new, old (and very olde) and the literature is confusing. Bigger project though.

#9 BillFerris

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Posted 29 March 2013 - 09:52 AM

Brian Skiff of Lowell Observatory has written a nice, concise summary of contributors to night sky brightness. It's titled, "How Dark Can the Night Sky Get," and is available, here: link.

Skiff writes, "The main contributions to the natural skyglow are: the zodiacal light, the night-airglow ("permanent aurora"), and scattered starlight in the atmosphere. There are also contributions from such things as scattered extragalactic "cosmic" light, but these are so small that even now their actual values are known only as upper limits." The darkest professional observatory sites on the planet have natural night sky brightness of about 22 magnitudes per square arcsecond, which amateur observers can experience by visiting star parties hosted at true dark sky sites in the western U.S. and other locations around the world. To experience a significantly darker "night" sky would require billions of dollars and, literally, a team of rocket scientists. As Skiff writes, "To escape these major effects completely you'd have to go into space well out of the plane of the solar system. Just going into Earth orbit doesn't gain you much as far as dark sky since the zodiacal light is still there."

As observers, we are also interested in how much light from celestial objects is lost or absorbed before it reaches our eyes, binoculars and telescopes. This is called extinction and the primary contributors are Rayleigh scattering (scattering light by the atmosphere) and absorption of light by the ozone layer. The baseline absorption for Rayleigh scattering plus ozone extinction varies by the elevation of your site. At sea level, the figure is about 0.25 magnitude per airmass. At 2000 meters, baseline extinction is reduced to about 0.11 magnitude per airmass. Add to this baseline the extinction caused by aerosols (dust, volcanic ash, chemicals and other human pollutants), which can vary seasonally and is, of course, sporadically impacted by natural and human-caused events.

In addition to the Roach and Gordon book, Skiff cites several relevant publications in the scientific literature. Those articles include calculations anyone can use to accurately predict sky brightness for any site based on natural and artificial factors.

Bill in Flag

#10 GlennLeDrew

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Posted 29 March 2013 - 01:20 PM

Near sea level, those factors which cause extinction operate equally for both the upper atmosphere glows and celestial sources. In other words atmospheric extinction, to first order, does not affect contrast.

Scattered starlight must necessarily take a back seat to scattered airglow and scattered zodiacal light. Whatever source which most contributes to sky glow must be the primary source of atmospheric scatter. Scattered starlight is a minor contributor.






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