# Magnitude of a component of a triple star

#### TheSodesa

1. The problem statement, all variables and given/known data
The apparent total magnitude of a triple star is $m_0 = 0,0$. The apparent magnitudes of two of its components are $m_1 = 1,0$ and $m_2 = 2,0$.

What is the apparent magnitude of the third component?

Answer: 0,9

2. Relevant equations
Since according to Norman Pogson (1856), the ratio of two subsequent classes of brightnesses (flux densities)
\begin{equation}
\frac{F_m}{F_{m+1}} = 100^{1/5} = 10^{2/5} = 10^{0,4},
\end{equation}
the difference between two apparent magnitudes:
\begin{equation}
m_x - m_y = -\frac{5}{2}\lg{ \left( \frac{F_x}{F_y} \right)}
\end{equation}

If the magnitude $0$ is chosen to represent a certain flux density $F_0$, generally corresponding to the flux density $F$ there is a corresponding magnitude
\begin{equation}
m = -\frac{5}{2}\lg{ \left( \frac{F}{F_0} \right)}
\end{equation}
3. The attempt at a solution

From the assignment, the total magnitude of the system is chosen to be zero. Therefore we can get the apparent magnitude of the third component from $(3)$ as follows:
\begin{equation}
m_3 = -\frac{5}{2}\lg{ \left( \frac{F_3}{F_0} \right)},
\end{equation}
where $F_0$ is the flux density of the system.

Since I have no idea, what the flux densities are, I think I might be better off trying to solve for the ratio $\frac{F_3}{F_0}$. Now
\begin{equation}
\frac{F_3}{F_1} = 10^{-0,4(m_3 - m_1)} \text{ and } \frac{F_3}{F_2} = 10^{-0,4(m_3 - m_2)}
\end{equation}
Solving these for $F_3$:
\begin{equation}
F_3 = 10^{-0,4(m_3 - m_1)}F_1 = 10^{-0,4(m_3 - m_2)}F_2
\end{equation}

Plugging into $(4)$:
\begin{equation}
m_3 = -\frac{5}{2}\lg{ \left( 10^{-0,4(m_3 - m_1)}\frac{F_1}{F_0} \right)},
\end{equation}
Now
\begin{equation}
\frac{F_1}{F_0} = 10^{-0,4m_1},
\end{equation}
which when plugged into $(7)$ yields $m_3 = m_3$, so I am obviously missing something.

But what exactly?

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#### haruspex

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I am obviously missing something
Leaving magnitudes aside for the moment, what do you think the relatioship would be between the individual flux densities and the combined flux density?

• TheSodesa

#### TheSodesa

Leaving magnitudes aside for the moment, what do you think the relatioship would be between the individual flux densities and the combined flux density?
Well, picking up my second edition of Understanding Physics by Mansfield and O'Sullivan, flux density is defined as follows: $$F:= \lim_{\Delta A \rightarrow 0} \frac{\Delta \Phi}{\Delta A}\hat{t},$$ where $\Phi$ is the flux through area $A$ on the enclosing surface, and $\hat{t}$ a unit vector pointing in the direction of the flux. If we have 3 sources producing flux inside of a surface, at each point on the surface their flux should be summed, meaning the total flux density
\begin{equation}
F_{tot} = \lim_{\Delta A \rightarrow 0} \frac{\sum_{i=1}^{n}\Delta \Phi_i \hat{t_i}}{\Delta A} = \sum_{i=1}^{n}F_i,
\end{equation}
where $n$ is the number of elements producing flux. At least that's what my intuition says.

Last edited:

#### TheSodesa

Leaving magnitudes aside for the moment, what do you think the relatioship would be between the individual flux densities and the combined flux density?
If my assumption in post #3 is correct, the total flux detected at the observation point $F_0 = F_1 + F_2 + F_3 \iff F_3 = F_0 - (F_1 + F_2)$. Then $(4)$ becomes
\begin{equation}
m_3 = -\frac{5}{2}\lg \left( \frac{F_3}{F_0} \right) = -\frac{5}{2}\lg \left( 1 - \frac{F_1 + F_2}{F_0} \right) =
-\frac{5}{2}\lg \left( 1 - \frac{F_1}{F_0} - \frac{F_2}{F_0} \right)
\end{equation}

Again, we can solve for the ratios $\frac{F_1}{F_0}$ and $\frac{F_2}{F_0}$ as follows using $(3)$:
\begin{equation}
\frac{F_1}{F_0} = 10^{-0,4m_1} \text{ and } \frac{F_2}{F_0} = 10^{-0,4m_2}
\end{equation}

Therefore
$$m_3 = -\frac{5}{2}\lg \left( 1 - \frac{F_1}{F_0} - \frac{F_2}{F_0} \right) = -\frac{5}{2}\lg \left( 1 - 10^{-0,4m_1} - 10^{-0,4m_2} \right) = 0,883002 \approx 0,9,$$
which was the desired answer.

Thanks. Last edited:

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