- #1

TheSodesa

- 224

- 7

## Homework Statement

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

## Homework 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}

## 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?