Synthetic Photometry - Calculating a colour index I have a theoretical black body spectrum as described by plancks law. I also have the bandpass sensitivity function for various filters. I would like to calculate a colour index from this information, so I can compare it to an experimental result.
My proposed method is to take the black body spectrum and convolve it with the passband. I would then bin the resulting spectrum and convert to photons using the bin's average wavelength. Summing up the photons should give me counts that can be used to calculate a colour index. This method is described at the bottom of the page here http://spiff.rit.edu/classes/phys440/lectures/filters/filters.html
My question is - is this the correct method? Why is the passband convolved with the blackbody spectrum, rather than multiplied together? Which is the correct method?

 A: It isn't a convolution, you are just integrating the product. I.e. if your (normalised) filter bandpass is $b(\lambda)$ and the spectral flux from the star is $f(\lambda)$, then the thing you are trying to calculate is a magnitude, which will be given by
$$ m_b = -2.5 \log_{10}\left[ \int b(\lambda) f(\lambda)\ d\lambda \right] + 2.5\log_{10} f_0 , $$
where the $-2.5\log_{10}$ puts the integrated flux onto the logarithmic magnitude system, and $f_0$ defines the flux zero point of that system
A: In the link, the spectrum is very noisy (possibly by the way of absorption peaks). Then, convolving it by a band-pass filter is simulating the integration - or average - over the band, resulting in a smooth curve, that you can then sample (i.e. pick up discrete values) without suffering aliasing artifact (as you would if the sampling rate of your bining is less than twice the highest frequency of these peaks in the curve).
In your problem (that is not 100% clear to me), you start with a spectrum curve that is already smooth, so you don't need to filter it. (Or you might want to integrate if you want ultra acurate results in non-affine areas, or if you BB is narrow peaked, or your bins are very sparse. But your figure doesn't seem to correspond to any of these cases :-) ).
