Can you please explain how the IRAC zmags were derived?
Posted by Seppo Laine on 29 January 2009 02:23 PM


To understand where the IRAC zmag comes from, you can start with the fundamental equation between magnitudes and flux densities. In one incarnation, it becomes m  M_{0} = 2.5*log(F/F_{0}) Here m is the magnitude of the source you want to measure, M_{0} is the zero magnitude (= 0), F is the flux density in Jy of the source you want to measure and F_{0} is the flux density of a zero magnitude source. For IRAC channel 1, F_{0} = 280.9 Jy. Expanded out, this becomes thus m = 2.5*log(F) + 2.5*log(F_{0}) Here 2.5*log(F_{0}) is the same as zmag. Now, since the IRAC images are in units of MJy/sr, we have to do some manipulation to get the equation to this form. Specifically, the measurable F that we have in IRAC images is the surface brightness, not the flux density. So therefore the equation becomes m = 2.5*log(SB*C) + 2.5*log(F_{0}) where SB is the measured surface brightness in the image in MJy/sr and C is a conversion factor from MJy/sr to Jy/pixel. For IRAC channel 1 mosaics with 0.6 arcsec x 0.6 arcsec pixels it equals C = 8.461595E06 Jy/pixel/(MJy/sr). Therefore the equation becomes m = 2.5*log(SB) + 2.5*log(F_{0}/C) where zmag now corresponds to the latter term, 2.5*log(F_{0}/C). Inserting the values of F_{0} and C mentioned above, we get zmag = 2.5*log(280.9/8.461595E06) = 18.80 mag These are zmags in the Vega system using a Kurucz model spectrum of an A0 V star matched to the visible photometry of Vega. Note that zero magnitude fluxes are not necessarily what you would measure for Vega as Vega does have circumstellar dust. Please remember that this is true only for the 0.6 arcsec x 0.6 arcsec pixel scale mosaics. For other pixel scales you will get a different value. Also, please remember the required corrections (e.g., aperture correction) that are needed for high accuracy photometry.  
