Apparent magnitude in binary systems

[Geordie mc]

If two stars have the same apparent magnitude in a close binary system and you know the mag og the system, do you just half it to find the mag of each star. ie if mag is 2.2 then each star is 4.4?

Thanks for any pointers on this one.

[Stephen]

Google:

combined magnitude

The total brightest of two or more celestial objects, such as the stars of a binary system that appear as a single point to the naked eye. Unfortunately, this can't be found simply by adding the separate magnitudes, m1 and m2, because the magnitude scale is logarithmic. Instead the combined magnitude, m, has to be calculated from the formula:

m = m1 - 2.5 log {1 + antilog [-0.4 (m2 - m1)]}.

[Stephen]

Then work backwards, so substitute m1 = m2, m = 2.2 (in your same magnitude question) and solve for m2?

[Stephen]

m = m1 - 2.5 log {1 + antilog [-0.4 (m2 - m1)]}

2.2 = m2 - 2.5 log {1 + antilog [-0.4 (m2 - m2)]}

2.2 = m2 - 2.5 log {1 + antilog [0]}

2.2 = m2 - 2.5 log 2

2.2 = m2 - 2.5 * 0.3 (approx)

m2 = 2.2 + (2.5 * 0.3)

m2 = 2.2 + 0.75

m2 = 2.95

Fixed it

[Geordie mc]

Many thanks. Most helpful.

[acey]

If two stars have the same apparent magnitude in a close binary system and you know the mag og the system, do you just half it to find the mag of each star. ie if mag is 2.2 then each star is 4.4?

Thanks for any pointers on this one.

You add 0.75, as has been shown. We can do it more quickly, though, starting from the definition

magnitude difference = 2.5log(brightness ratio).

In this case the ratio is 2 (one star is half as bright as the combined pair) hence the magnitude difference is 2.5log(2) = 0.75. For the particular example we get 2.2 + 0.75 = 2.95, but you would add the same amount whatever the combined magnitude, e.g. an equal pair of combined magnitude 3 would have individual stars of magnitude 3.75.

We can easily generalise it further: if there were three close stars of equal magnitude then their individual magnitudes would differ by 2.5log(3) = 1.19 from the combined figure.