In V band, extinction is about 0.1 mag/airmass.
The Sun’s apparent brightness is -26.74.
0.1 mag of extinction means that about 10% of the light is taken from the sun and spread over the sky, about 21,000 square degrees. Half of this is scattered out of the atmosphere, half down to make the blue sky.
The integrated magnitude of the sky would then be of order -26.74 + 2.5*log10(10*2) = -23.5.. i.e. 3.2 magnitudes dimmer than the Sun.
Spreading this over half the sky (21000 square degrees or 2.7E11 square arcseconds) gives a reduction in surface brightness by 2.5*log10(2.7E11) = 28.6 magnitudes.
This yields a daytime sky brightness in V band of -24.24+28.6 = 4 mag square arcsecond.
If we have any daytime exposures with the ES telescope — we could check this and get numbers in other bands as well…
I agree with you on comment 2: combination of Rayleigh and something else. The PSF measurements of the Sun show that there is a power law falloff, typically, of the PSF — just as we find in Hawaii — but this must eventually merge with the sky at a level of a few magnitudes / square arcsec – we never got that quite sorted out where it happens in angular distance from the Sun. It will depend very much on local conditions though — e.g. in Antarctica there is almost no halo around the Sun – you can block it out with your thumb and see blue sky (I’ve been told by people down there), whereas haze almost anywhere else means that the sky near the Sun is quite white.
Good point on the scattering — I think this doesn’t change the result because it would be the same for stars as well as the sun, so that stellar extinction value of 0.1 is OK. Well, I am not sure about this : the computation might be off by a factor of 2, or 0.75 magnitudes — but it is very rough any way. I can ask some people who measure the sky brightness in Antarctica about this…
ACtually, I was looking at this idea of ‘collectinga ll the scattered light’ in images of the Moon – or Jupiter. I got as far as realizing that all the light seems to be collected once you are a small distance – perhaps 10-20 arc minutes – from the source: after that point all is lost in the noise. This does not really match the expectation that ‘the scatteredlight is spread all over the sky’. I suppose the effect coul dbe there but be confounded by noise and other sources? Hmm. Why doesn’t the curve of growth keep changing as your radius from source increases? That it doesn’t change seems to imply that most of the scattered light is near the source and any other scattered light is indeed spread across the sky so thinly that for manageable anulus-radii you see no ‘pickup’ or slope in the curve? That in turns implies that scattering is not all due to Rayleigh – because Rayleigh scattering is proportional to 1+cos^2(theta) . Yep – this confirms our picture of what extinction is about – it is Raeyligh + ‘something else’; and the something else is probably Mie scattering which is strongly forward concentrated.
Great stuff this! We do have some dawn and dusk observations where stars are vissible – but no real daytime observations. This could be obtained however …
One note: I am figuring that Rayleigh scattering is symmetric so that 50% scatters forward and 50% backwards – if Rayleigh scattering is mainly single-scattering then it is all right so say that of the light removed from the beam half is spread across the sky, the rest goes into space.