# Computation of magnitude of the secondary

How to approximately compute the magnitude of the secondary in a binary when I know the magnitude of the primary, masses, radii, and approximate luminosities for both of them, please? Thank you very much.

• What particular problem did you encounter? What is your own research on these two(!) questions? How did the readily available sources not help you? Jul 4, 2022 at 9:02
• I do not have the magnitude of the secondary - that is the problem. What are the available sources, I have not found anything that is clear for me. Jul 4, 2022 at 10:25
• Are you aware of the definition of luminosity and magnitude? What did you search for? What ressources did you check? Why did they not help? With all respect, but your real issue seems to be the google-foo here. Jul 4, 2022 at 11:30

So the trick here will be to use the magnitude of the primary ($$m_1$$) and the luminosities of the two ($$L_1$$ and $$L_2$$) to get to $$m_2$$. When first googling for the relationship between luminosity and magnitude you could easily get confused because apparent magnitude isn't a measurement of luminosity, its a measurement of flux! The difference between two magnitudes is related to the ratio of the two fluxs like this: $$$$\label{eq:1} m_1-m_2=-2.5\log\frac{f_1}{f_2}$$$$ This is done because the definition of the magnitude scale was so that a difference of 5 magnitudes is equivilent to a flux ratio of 100:1 (try the math out to test this!).
But, you don't have fluxs, you have luminosities! Since your stars are in a binary we can use a little trick. To go from flux to luminosity you use the following relationship: $$f = \frac{L}{4\pi d^2}$$ where $$d$$ is the distance to the object in question. Substituting this into the first equation we get $$m_1-m_2=-2.5\log\frac{L_1/(4\pi d_1^2)}{L_2/(4\pi d_2^2)} = -2.5\log(\frac{L_1}{L_2}\frac{d_2^2}{d_1^2})$$ Since you are looking at a binary $$d_1=d_2$$! So this simplifies down to the relation you want, $$m_2$$ as a function of $$m_1$$, $$L_1$$, and $$L_2$$ $$\boxed{m_2=m_1+2.5\log\frac{L_1}{L_2}}$$ Hopefully this helps! the magnitude system is the bane of my existence so I have to look this up about once a week even though I'm getting a PhD in astronomy D: