# Aperture photometry uncertainties

Suppose I have a 2D data array with number of counts in each pixel (i.e., this is the image array). Suppose I have another 1-1 mapped same-shape data array that gives the 1-sigma Gaussian standard deviation in the number of counts in each pixel in the image (i.e., this is the error array).

If I do circular aperture photometry, I just calculate -2.5*log10(summed counts within aperture) + magnitudeZeropoint.

On the other hand, for the magnitude uncertainty on my aperture magnitude, I read that I'm supposed to do a quadrature sum of the errors within my aperture, then compute the fractional flux uncertainty as the ratio of my quadrature-summed uncertainty divided by my measured counts from the image, and then magnitude error = 2.5*log10(1 + fractional flux uncertainty).

How come the uncertainties have to be quadrature summed ($\sqrt{\Sigma \sigma_{x,y}^2}$), instead of just added up like I do with the image pixel values? The quadrature sum results in a smaller errorbar but is that realistic? Also, I emphasize that my data arrays are in units of counts (i.e., ADU's), not electrons.

Perhaps a good way of seeing this is that suppose you took 100 independent measurements of the same thing, each with its own, roughly equal uncertainty. If I asked what the uncertainty in the average was, you wouldn't just add up all the errors and divide by 100 because that would give an uncertainty that was identical to that of an individual measurement. Instead you would do the quadrature sum of the uncertainties and divide by 100, which reduces the uncertainty in the average by a factor of $\sqrt{100}$.
• @quantumflash You deal with it using error propagation formulae in a similar way. The number of background counts in the aperture (assuming that the background has been predicted as $b \pm \delta b$) is $b \pm ((\delta b)^2 + b)^{0.5}$. – Rob Jeffries Jul 8 '17 at 6:16