<p>In that formula, <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> are not the flux of two stars in a binary system. They are the flux from two different stars.  In the typical use of that formula <span class="math-container">$f_2$</span> is the amount of light coming from a reference star with known magnitude <span class="math-container">$m_2$</span>, and <span class="math-container">$f_1$</span> is the amount of light measured from your star that you are studying. Knowing <span class="math-container">$m_2$</span>, <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> you can calculate <span class="math-container">$m_1$</span>.</p>
<p>If you are observing a binary system, and you don't know either <span class="math-container">$m_1$</span> or <span class="math-container">$m_2$</span>, you can't use the formula.</p>
<p>However, if you know that the two stars in the binary system have the same brightness,  and you know the brightness of the system you can work out the brightness of each.  Suppose the system has a magnitude of 5.0 Then this means that relative to a magnitude 0 star the total flux from the system (f) is given by</p>
<p><span class="math-container">$$ 5.0 - 0 = 2.5 \log(f/F)$$</span></p>
<p>where F is the flux from a magnitude 0 star, ie <span class="math-container">$f=F/10^{5.0/2.5}  = 0.01F$</span></p>
<p>As the flux from each star is half the flux from the system, the flux of each star in the binary is <span class="math-container">$f_* = 0.005F$</span></p>
<p>And the magnitude of each star is <span class="math-container">$m_* - 0 = 2.5 \log (f_*/F)$</span></p>
<p>That is <span class="math-container">$m_* = 2.5\log(0.005) = 5.75$</span></p>
<p>That is a close binary in which each star has magnitude 5.75 would appear as a single magnitude 5.0 star.</p>
<p>More generally if the binary has magnitude <span class="math-container">$m$</span>, and the two stars are assumed to be equally bright with magnitude <span class="math-container">$m_*$</span>, then:</p>
<p><span class="math-container">$$m_*= 2.5\log (10^{m/2.5}/2) = m + 2.5 \log(2) = m+0.75$$</span></p>

CC BY-SA 4.0
2023-01-27T06:14:03.313Z
2023-01-27T17:09:56.763Z
2023-01-27T17:09:56.763Z
James K
<p>I'm used to use this formula to find the apparent magnitude of a star</p>
<p><span class="math-container">$$m_1 - m_2 = -2.5 log(\frac{f_1}{f_2})$$</span></p>
<p>If both star are at the same distance from us and their luminosity is the same, does it means that <span class="math-container">$f_1/f_2 = 1 ?$</span></p>
<p>Knowing the magnitudes of the system, is there a way to find the magnitude of each star?</p>

<star><binary-star><apparent-magnitude><absolute-magnitude>
calculate the absolute and apparent magnitude for a binary star system
InvisibleParticle
# calculate the absolute and apparent magnitude for a binary star system

(Asked by: username_0 on 2023-01-27)

I'm used to use this formula to find the apparent magnitude of a star

$$m\_1 - m\_2 = -2.5 log(\frac{f\_1}{f\_2})$$

If both star are at the same distance from us and their luminosity is the same, does it means that $f\_1/f\_2 = 1 ?$

Knowing the magnitudes of the system, is there a way to find the magnitude of each star?

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

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## 1 answers

### Answer by username_1 on 2023-01-27. Score: 2 | Accepted answer

In that formula, $f\_1$ and $f\_2$ are not the flux of two stars in a binary system. They are the flux from two different stars. In the typical use of that formula $f\_2$ is the amount of light coming from a reference star with known magnitude $m\_2$, and $f\_1$ is the amount of light measured from your star that you are studying. Knowing $m\_2$, $f\_1$ and $f\_2$ you can calculate $m\_1$.

If you are observing a binary system, and you don't know either $m\_1$ or $m\_2$, you can't use the formula.

However, if you know that the two stars in the binary system have the same brightness, and you know the brightness of the system you can work out the brightness of each. Suppose the system has a magnitude of 5.0 Then this means that relative to a magnitude 0 star the total flux from the system (f) is given by

$$ 5.0 - 0 = 2.5 \log(f/F)$$

where F is the flux from a magnitude 0 star, ie $f=F/10^{5.0/2.5} = 0.01F$

As the flux from each star is half the flux from the system, the flux of each star in the binary is $f\_\* = 0.005F$

And the magnitude of each star is $m\_\* - 0 = 2.5 \log (f\_\*/F)$

That is $m\_\* = 2.5\log(0.005) = 5.75$

That is a close binary in which each star has magnitude 5.75 would appear as a single magnitude 5.0 star.

More generally if the binary has magnitude $m$, and the two stars are assumed to be equally bright with magnitude $m\_\*$, then:

$$m\_\*= 2.5\log (10^{m/2.5}/2) = m + 2.5 \log(2) = m+0.75$$


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