# Homework Help: Magnitude of a component of a triple star

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1. Dec 7, 2016

### 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?

2. Relevant equations
Since according to Norman Pogson (1856), the ratio of two subsequent classes of brightnesses (flux densities)

\frac{F_m}{F_{m+1}} = 100^{1/5} = 10^{2/5} = 10^{0,4},

the difference between two apparent magnitudes:

m_x - m_y = -\frac{5}{2}\lg{ \left( \frac{F_x}{F_y} \right)}

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

m = -\frac{5}{2}\lg{ \left( \frac{F}{F_0} \right)}

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:

m_3 = -\frac{5}{2}\lg{ \left( \frac{F_3}{F_0} \right)},

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

\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)}

Solving these for $F_3$:

F_3 = 10^{-0,4(m_3 - m_1)}F_1 = 10^{-0,4(m_3 - m_2)}F_2

Plugging into $(4)$:

m_3 = -\frac{5}{2}\lg{ \left( 10^{-0,4(m_3 - m_1)}\frac{F_1}{F_0} \right)},

Now

\frac{F_1}{F_0} = 10^{-0,4m_1},

which when plugged into $(7)$ yields $m_3 = m_3$, so I am obviously missing something.

But what exactly?

2. Dec 7, 2016

### haruspex

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

3. Dec 7, 2016

### TheSodesa

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

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,

where $n$ is the number of elements producing flux. At least that's what my intuition says.

Last edited: Dec 7, 2016
4. Dec 7, 2016

### TheSodesa

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

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)

Again, we can solve for the ratios $\frac{F_1}{F_0}$ and $\frac{F_2}{F_0}$ as follows using $(3)$:

\frac{F_1}{F_0} = 10^{-0,4m_1} \text{ and } \frac{F_2}{F_0} = 10^{-0,4m_2}

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