# Finding individual band magnitudes: Why and how is the integration of individual flux values for each wavelength used?

What differentiates the distance modulus equation to find the difference between two apparent magnitudes? For example,

$$m_1 = -2.5\cdot log_{10}\left(\frac{F_1}{F_2}\right) + m_2$$ leads to the following, in the U band: $$\implies m_U = -2.5\cdot log_{10}\left(F_U\right) + C_U$$

and the relationship involving taking the sum of the fraction of the flux that a star outputs at each wavelength rather than the ratio between the fluxes of two stars, here in the UV band for example:

$$m_U = -2.5\cdot log_{10} \left(\displaystyle\int_0^\infty F_{\lambda}S_U \,\, d\lambda\right) + C_U$$

*$S(\lambda)$ = the fraction of the star's flux detected at wavelength lambda.

Succinctly, what is the difference between the above two equations?

More specifically, how would this be used in a real-world situation to find the apparent magnitude of an object? My deduction is that apparent magnitude isn't as simple as taking the flux in a certain filter to find the apparent magnitude, and that each individual wavelength must be accounted for. However, with that notion, why can't we just remove the sensitivity function and change the bounds to those wavelengths within the U band?

The other option would be that this is just more accurate than taking the flux within the U band (referring to the above example), because you would take smaller intervals, possibly over 1 $\buildrel _\circ \over {\mathrm{A}}$ intervals since infinitesimal ones would be impossible.

• Your question isn't clear, but I suspect what you're getting at is that one should only use the U-band when you already know the star-type and can compare it with other stars of the same type. Otherwise, as you suggest, one must integrate over the visible spectrum. BTW the added accuracy from 1-Angstrom bands vs, say, 1-nm bands is not generally worth the trouble. – Carl Witthoft Dec 18 '17 at 14:35
• When you are looking only for the U band magnitude, why would you need to integrate through all wavelengths? That was part of what I was trying to get at. – Rithwik Sudharsan Dec 19 '17 at 8:25
• I guess more succinctly, what's the difference between the second and third equations I've given. – Rithwik Sudharsan Dec 20 '17 at 20:25