### How Bright is the Daytime Sky?

A friend of mine has a popular astronomer’s toy—a gizmo called a Sky Quality Meter. It measures light pollution. Point it at the nighttime sky and it tells you how bright the sky is—in magnitudes per square arcsecond. Where he lives, the skies are pretty dark at night (at least for New Jersey). His gizmo...

A friend of mine has a popular astronomer’s toy—a gizmo called a Sky Quality Meter. It measures light pollution. Point it at the nighttime sky and it tells you how bright the sky is—in magnitudes per square arcsecond. Where he lives, the skies are pretty dark at night (at least for New Jersey). His gizmo tells him his skies are mag 19.5 per square arcsecond. He took his toy to a “dark-sky” site, also here in NJ, and it read 20.6. Not bad.

For those not familiar with the mag per square arcsecond unit of measure, here’s a brief explanation.

Stars are rated in brightness using something called apparent magnitude, usually referred to simply as mag. Higher numbers represent fainter stars. Traditionally, mag 1 was meant to refer to the very brightest stars in the night sky and mag 6 was used for the faintest stars visible to the naked eye under dark skies. In modern times, we are more precise. It turns out that mag 1 stars are actually about 100 times brighter than mag 6 stars and hence this ratio was taken as a definition. It therefore follows that mag 2 stars are about 2.5 times fainter than mag 1 stars, mag 3 stars are about 2.5 times fainter still, etc. (2.51*2.51*2.51*2.51*2.51 = 99.6, which is about 100).

Stars are point sources of light. For them a simple brightness concept, such as apparent magnitude, suffices. But, the night sky is an extended object. For extended objects we need to give the brightness per unit area. The “unit area” of choice is the square arcsecond. The moon is about 2000 arcseconds in diameter. Therefore, the full moon is about 3.14*1000*1000 = 3,140,000 square arcseconds in area (recall that area is pi times the square of the radius). Clearly, a square arcsecond is a very small area. It is, however, a useful measure because stars in most telescopes appear to be about 1 arcsecond in diameter.

Out of curiosity, I asked my friend: “How bright is the daytime sky?” He said he’d check. When he got back to me, he said that the instrument was overloaded by the daylight and simply reported that it is brighter than magnitude zero per square arcsecond (i.e., a negative number). This made me wonder: is that right? That seems too bright to me. It turns out that the answer is somewhere between 1.5 and 3. Here’s how I get that…

I’ve viewed Venus in the daytime through binoculars and small telescopes. Venus is brighter than the daytime sky. How bright is Venus? Well digging through old observing logs, I viewed (and photographed) Venus at noon on Oct 13, 2007 (the picture shown below). At that time, Venus was magnitude -4.5, it was 29 arcseconds in diameter and its face (the face facing us) was 41.8 percent illuminated. Doing the math, Venus’s illuminated face was 276 square arcseconds in apparent area. Hence, a one square arcsecond piece of Venus is 276 times fainter than Venus as a whole. As we explained earlier something 100 times fainter is 5 magnitudes fainter. Similarly, a 250 times fainter object is 6 magnitudes fainter. Therefore, the “surface brightness” of Venus was about magnitude -4.5 + 6 = 1.5. And, on the day the picture was taken, Venus was clearly brighter than the daytime sky.

Using the picture shown at the top and consulting JL Hilton for the brightness of Venus in its crescent phase, I get that Venus had apparent magnitude of -0.75 and illuminated surface area of 65.2 square arcseconds. That works out to a surface brightness of 3.8 (i.e., -0.75 + 2.5*log(65.2)). On this day, I would say that the daytime sky was brighter than Venus (the picture shown here is enhanced to make Venus appear brighter than it actually was relative to the sky). So, daytime skies are somewhere between 1.6 and 3.8 magnitudes per square arcsecond. I’m guessing about magnitude 3 but I don’t know exactly.

Details on the images and their acquisition can be found here.