Computation of magnitude of the secondary

Question

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.

Answer 1

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: \begin{equation} \label{eq:1} m_1-m_2=-2.5\log\frac{f_1}{f_2} \end{equation} 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: