# Magnitude of a component of a triple star

• TheSodesa
In summary, the apparent total magnitude of a triple star is 0,0. The apparent magnitudes of two of its components are 1,0 and 2,0. The third component has an apparent magnitude of 0,9. If the flux densities of the individual stars are unknown, the relatioship between the individual flux densities and the combined flux density can be calculated using flux density definition and the attempt at a solution.
TheSodesa

## 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?

## Homework 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)}

## 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?

TheSodesa said:
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
haruspex said:
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

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:
haruspex said:
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

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

Thanks.

Last edited:

## What is the magnitude of a component of a triple star?

The magnitude of a component of a triple star refers to the brightness of that particular star when observed from Earth. It is measured on a logarithmic scale, with lower numbers indicating brighter stars and higher numbers indicating dimmer stars.

## How is the magnitude of a component of a triple star determined?

The magnitude of a component of a triple star is determined using a photometer, which measures the amount of light received from the star. The brightness of a star is compared to that of a standard star, and the difference in magnitude is calculated.

## Can the magnitude of a component of a triple star change?

Yes, the magnitude of a component of a triple star can change over time. This can be due to changes in the star's luminosity, distance from Earth, or any factors that may affect the amount of light received from the star.

## Why is the magnitude of a component of a triple star important?

The magnitude of a component of a triple star is important because it gives us information about the star's brightness and can help us understand its properties, such as temperature and size. It is also used to classify stars and determine their position in the Hertzsprung-Russell diagram.

## What is the difference between apparent and absolute magnitude of a component of a triple star?

Apparent magnitude refers to the brightness of a star as seen from Earth, while absolute magnitude refers to the brightness of a star if it were placed at a standard distance of 10 parsecs from Earth. The absolute magnitude is a more accurate measure of a star's true brightness.

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