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Magnitude of a component of a triple star

  1. Dec 7, 2016 #1
    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?
     
  2. jcsd
  3. Dec 7, 2016 #2

    haruspex

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    Leaving magnitudes aside for the moment, what do you think the relatioship would be between the individual flux densities and the combined flux density?
     
  4. Dec 7, 2016 #3
    Well, picking up my second edition of Understanding Physics by Mansfield and O'Sullivan, flux density is defined as follows: [tex]F:= \lim_{\Delta A \rightarrow 0} \frac{\Delta \Phi}{\Delta A}\hat{t},[/tex] 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: Dec 7, 2016
  5. Dec 7, 2016 #4
    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
    [tex]
    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,
    [/tex]
    which was the desired answer.

    Thanks. :smile:
     
    Last edited: Dec 7, 2016
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