# How to calculate magnitudes in photometry?

I have a passband filter $$b$$ and a spectrum (SED) $$f_{\lambda}(\lambda)$$. Often the formula I find to get the flux in that filter $$F_b$$ is $$F_{b}=\dfrac{\int f_{\lambda}(\lambda) T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)d\lambda}\,,$$ where $$T_b$$ is the filter response in band $$b$$, which is similar to the filters shown in figure 2 of The PLAsTiCC team et al (2018).

However, when I try to compute a apparent magnitude, I get the formula $$m_b=-2.5\log_{10}\left(\dfrac{\int f_{\lambda}(\lambda)\,\lambda\,T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)\,\lambda\,d\lambda}\right)\,.$$

I don't understand from where does the $$\lambda$$ in the second formula arises, since I thought that $$m_b=-2.5\log_{10}\left(\dfrac{F_b}{F_{\mathrm{reference}}}\right)\,.$$

For example, Casagrande & VandenBerg (2014) show both expressions in equations 1 and 3. However Bessell & Murphy (2012) (eq. 2) and Hogg, Baldry, Blanton, & Eisenstein (2002) (eq. 4) only show the second approach. Additionally, Hogg (2000) shows the K-correction without the added $$\lambda$$'s: $$K=-2.5\log_{10}\left(\dfrac{1}{(1+z)}\dfrac{L_{\lambda/(1+z)}}{L_{\lambda}}\right)\,.$$

I don't know how to reconcile these different views. Can you help me?

• @usernumber, I have fixed the link. When I feed equation (1) into equation (3), the $\lambda$ factor does not appear. That is why I am confused. – Catarina Alves Feb 25 at 16:40
• @usernumber there's an additional lambda in the denominator's integral – planetmaker Feb 25 at 17:05