# ls10 - class 10 tues march 18 chapter 15 and 19 last things...

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class 10. tues march 18 chapter 15 and 19. last things about chapter 15: - example of what to do with "per steradian". sky brightness. The sky is not dark - it has light. some of that's terrestrial -- the sky is glowing, really! because of the Moon, or because of 'airglow' molecular processes in the atmos, or because of light pollution. But also there's a 'real' source to the night-sky brightness: 'faint fuzzies', faint distant galaxies that are unresolved. their light combines together to show us a smear of light in the background. It's not uniformly bright but it's there. And it's important to know how to understand what the sky brightness is like. Often sky brightness is measured in "mag per sq. arcsec.". That means within a solid angle of 1 sq. arcsec., the light we see is equivalent to a magnitude of such-and-such. This is a very odd unit because mag is not a linear unit. E.g. suppose the night sky is given msky = 21.7 mags per sq. arcsec. in V band. What kind of flux density is this? what is the equivalent per sq. degree.? You have to do this in flux units and switch betw mags and flux. In 1 sq. arcsec. you have equiv of 21.7 mag of flux density. what is that? F0 * 10^(-0.4*21.7). handout #17 would give you F0, but let's not worry about that now. F = F0 * 10^(-0.4*msky). 1 sq.deg. = 1.296e7 sq.arcsec., so the flux density in 1 sq.deg. is going to be F0 * 10^(-0.4*msky) * 1.296e7. so what magnitude is that? = -2.5*alog10( F0*10^(-0.4*msky)*1.296e7 ) = -2.5*(alog10( F0*10^(-0.4*msky) ) + alog10(1.296e7)) = msky - 2.5*alog10(1.296e7) = msky - 17.78 = 21.7 - 17.78 = 3.92. So a sky brightness of 21.7 mag per sq. arcsec. is the same as 3.92 mag per sq. deg. Notice that multiplying by 1.296e7 sq.deg/sq.arcsec. would definitely not have worked. Another example: say you want to look at a star that's magnitude mV = 20.0. Is it going to be a lot brighter than the night sky of 21.7 mag/sq.arcsec? How much brighter or fainter? Well it depends on how much solid angle that star covers -- in reality it's small but the atmosphere blurs it out -- let's say that it's magnitude is 17.0 and that's all encapsulated within a circle of diameter 3 arcsec. how much brighter is that compared to the sky? Well the sky is 21.7 mag / sq.arcsec. And the area is !pi * 1.5^2 = 7.07 sq. arcsec. So how many magnitudes is within that area? again, don't multiply by 7.07. do it in flux units. flux density w/i 1 sq. arcsec. = F0 * 10^(-0.4 * 21.7) but agan, don't worry about sticking in F0. flux density within circle = F0 * 10^(-0.4* 21.7) * 7.07 now go back to magnitudes: = -2.5*alog10( F0*10^(-0.4*21.7)*7.07 )

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= -2.5*(alog10(F0*10^(-0.4*21.7)) + alog10(7.07)) = msky - 2.5*alog10(7.07) = 21.7 - 2.12 = 19.58. So! the sky will actually contribute more light within
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ls10 - class 10 tues march 18 chapter 15 and 19 last things...

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