<p>So the trick here will be to use the magnitude of the primary (<span class="math-container">$m_1$</span>) and the luminosities of the two (<span class="math-container">$L_1$</span> and <span class="math-container">$L_2$</span>) to get to <span class="math-container">$m_2$</span>. 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:
<span class="math-container">\begin{equation}
\label{eq:1}
m_1-m_2=-2.5\log\frac{f_1}{f_2}
\end{equation}</span>
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!).</p>
<p>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:
<span class="math-container">$$f = \frac{L}{4\pi d^2}$$</span>
where <span class="math-container">$d$</span> is the distance to the object in question. Substituting this into the first equation we get
<span class="math-container">$$m_1-m_2=-2.5\log\frac{L_1/(4\pi d_1^2)}{L_2/(4\pi d_2^2)} = -2.5\log(\frac{L_1}{L_2}\frac{d_2^2}{d_1^2})$$</span>
Since you are looking at a binary <span class="math-container">$d_1=d_2$</span>! So this simplifies down to the relation you want, <span class="math-container">$m_2$</span> as a function of <span class="math-container">$m_1$</span>, <span class="math-container">$L_1$</span>, and <span class="math-container">$L_2$</span>
<span class="math-container">$$\boxed{m_2=m_1+2.5\log\frac{L_1}{L_2}}$$</span>
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:</p>

CC BY-SA 4.0
2022-07-07T06:14:05.353Z
Thank you so much
Anna-Kat
CC BY-SA 4.0
2022-07-06T23:11:07.663Z
2022-07-06T23:11:07.663Z
Keyan Gootkin
<p>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.</p>

CC BY-SA 4.0
2022-07-04T09:02:01.080Z
What particular problem did you encounter? What is your own research on these two(!) questions? How did the readily available sources  not help you?
planetmaker
CC BY-SA 4.0
2022-07-04T10:25:43.003Z
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.
Anna-Kat
CC BY-SA 4.0
2022-07-04T11:30:01.863Z
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.
planetmaker
<binary-star><apparent-magnitude>
Computation of magnitude of the secondary
Anna-Kat
# Computation of magnitude of the secondary

(Asked by: username_0 on 2022-07-04)

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.

**Tags:** binary-star, apparent-magnitude

> **Comments:**
> **username_1**: What particular problem did you encounter? What is your own research on these two(!) questions? How did the readily available sources  not help you?  
> **username_0**: 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.  
> **username_1**: 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.  

---

## 1 answers

### Answer by username_2 on 2022-07-06. Score: 2 | Accepted answer

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:

> **Comments:**
> **username_0**: Thank you so much  

---

