I know that in order to get the Flux of a star (or something else) in a particular filter from its SED (luminosity per unit wavelength), I need to convolve the spectrum (SED) with the filter response. Most of the formulas I see to do this are $$F_{b}=\dfrac{\int f(\lambda) T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)d\lambda}\,,$$ where $f(\lambda)$ is the SED, and $T_b$ is the filter response in band $b$.

This formula seems very different from a mathematical convolution that I would write as $$f*g(\lambda)=\int_{-\infty}^{\infty} f(x) g(\lambda-x)dx\,.$$

Are these two convolutions the same thing? Or is the "astronomy convolution" a different thing (e.g. SED Fitter python package)?


To get the flux of an SED through a particular filter, you actually multiply the the SED by the filter's response. Talking about convolution in this context is a bit of a misnomer.

Basically, for each wavelength, you look at what fraction of the light will go through the filter, and you sum up the values you get at those wavelengths. The division by $\int T_{b}(\lambda)d\lambda$ is just a normalization term.

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    $\begingroup$ It is a bit of a misnomer, but I suppose the justification is that the convolution theorem says that convolution in the time domain equals point-wise multiplication in the frequency domain. And of course spectra & filter responses are in the frequency (or wavelength) domain. $\endgroup$ – PM 2Ring Feb 24 at 9:34
  • $\begingroup$ @PM2Ring I never thought of it that way, but it would make sense ! $\endgroup$ – usernumber Feb 24 at 9:39
  • $\begingroup$ @PM2Ring, does this means I should be able to find a $T_b$ defined in time domain? Additionally, I always need to integrate over frequency/ wavelength in the Fourier space or over time in the time domain (i.e. $\int_{t^*=-\infty}^{t^*=\infty}\int_{t^=-\infty}^{t=\infty}f(t)T_b(t^*-t)dtdt^*$), right? $\endgroup$ – Catarina Alves Feb 24 at 11:33
  • $\begingroup$ I don't know about the physical relevance of switching to time domain for an SED... $\endgroup$ – usernumber Feb 24 at 13:05
  • $\begingroup$ Thank you. I was just trying to reason further why is this operation called "convolution" in astronomy. $\endgroup$ – Catarina Alves Feb 24 at 15:12

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