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## What calculation make the pieces of this puzzle go together?

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PlutonianEmpire Posts: 1359
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### What calculation make the pieces of this puzzle go together?

(Title limit made me take the "s" out of the word "calculations.")

Okay, back to Delta Tri (again ).

Right now, I'm going by Wikipedia and other sources, where the Primary has an individual AppMag of +4.865, at 35.1735.... LY from Earth, giving it a luminosity of 1.12609 suns. Distance calculated from SIMBAD data.

The secondary, however, according to sources other than Wikipedia, has an AppMag precisely two magnitudes dimmer than the primary, giving it an AppMag of +6.865, with the distance resulting in a luminosity of 0.178473 suns.

However, at least one source that I know of seems to give the whole system an overall AbsMag of +4.69, resulting in a luminosity of 1.13763 suns. If you subtract the Primary's luminosity from the AbsMag luminosity, you get 0.01154 suns, which conflicts with the known luminosity of the secondary.

If I add the Secondary's luminosity of 0.178473 to the Primary's, I get 1.304563 suns, giving an overall AbsMag of +4.54134, with the distance giving it an overall AppMag of +4.70527.

My question is, is there some sort of mechanism or calculation that reduces a dimmer secondary's contribution to the overall AbsMag or AppMag? Or, is the Secondary simply not taken into account when calculating an AbsMag, thus making said AbsMag apply only to the primary?

If the answer is the former, then, how do I adjust my AbsMag or AppMag calculations when having a dimmer Secondary contribute to the overall AbsMag or AppMag of a binary system?
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### Re: What calculation make the pieces of this puzzle go toget

Hi,

Maybe my answer comes a bit too late but...

The contribution of a second star to the total apparent magnitude decrease very quickly when it becomes dimmer than the bightest one.

This is a simple matter of calculation.

By definition the apparent magnitude m = -2.5 log10(F) where F is the flux observed relative to a reference flux.
Reversely, knowing m gives F = 10^(-m / 2.5)

- calculate the flux observed from each star,
- and calculate the resulting magnitude.

Let m1 and F1 the magnitude and the flux of the primary star and m2 and F2 the corresponding values for the secondary star.
F1 = 10^(-m1/2.5) = 0.01132400
F2 = 10^(-m2/2.5) = 0.00179473
F1 + F1 = 0.01311873
and the resulting magnitude
m = -2.5 log10(0.01311873) = 4.70527

This is exactly what you had ! This formula is very useful because you don't need to know the distance nor the absolute luminosity of the star. Now when you know the distance D (in parsecs) you can set the absolute magnitude (M) by applying the formula :
M = m - 5 log10(D) + 5

If it is too late for these stars, I hope it will be useful for your future work. @+
Last edited by jogad on 09.09.2012, 15:50, edited 1 time in total.

Topic author
PlutonianEmpire Posts: 1359
Joined: 09.09.2004
Age: 35
With us: 15 years 4 months
Location: MinneSNOWta

### Re: What calculation make the pieces of this puzzle go toget

Maybe my answer comes a bit too late but...

The contribution of a second star to the total apparent magnitude decrease very quickly when it becomes dimmer than the bightest one.

This is a simple matter of calculation.

By definition the apparent magnitude m = -2.5 log10(F) where F is the flux observed relative to a reference flux.
Reversely, knowing m gives F = 10^(-m / 2.5)

- calculate the flux observed from each star,
- and calculate the resulting magnitude.

Let m1 and F1 the magnitude and the flux of the primary star and m2 and F2 the corresponding values for the secondary star.
F1 = 10^(-m1/2.5) = 0.01132400
F2 = 10^(-m2/2.5) = 0.00179473
F1 + F1 = 0.01311873
and the resulting magnitude
m = -2.5 log10(0.01311873) = 4.70527

This is exactly what you had ! This formula is very useful because you don't need to know the distance nor the absolute luminosity of the star. Now when you know the distance D (in parsecs) you can set the absolute magnitude (M) by applying the formula :
M = m + 5 log10(D) - 5

If it is too late for these stars, I hope it will be useful for your future work. @+
Umm.... What?

The AppMag you gave me, as you said, was identical to mine.

However, at Delta Tri's distance, that AppMag results in an AbsMag of 4.54134, which contradicts the AbsMag of 4.69 I got from the source.
Terraformed Pluto: Now with New Horizons maps! :D

Posts: 458
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### Re: What calculation make the pieces of this puzzle go toget

Hi,

@ bdfd
J'appr?cie ton intervention et je te remercie particuli?rement pour le sous-titrage.
Si tu pouvais aussi traduire mes posts ?a me permettrait enfin de comprendre de quoi je parle ! PlutonianEmpire wrote:Okay, back to Delta Tri (again ).
PlutonianEmpire wrote:The AppMag you gave me, as you said, was identical to mine.
However, at Delta Tri's distance, that AppMag results in an AbsMag of 4.54134, which contradicts the AbsMag of 4.69 I got from the source.

First I apologize. I was mistaken in the signs of the last formula. I edited my previous post to correct it.
Anyway you did not use because your calculation is right.

with apparent mags of 4.865 and 6.865 as source data, this actually yields to an overall absolute mag of 4.54.
These are roughly the values used in Celestia.

The problem is that you're trying to do calculations with data that are not compatible with each other.
Different sources of data lead to different results.

The data in Celestia are pretty well controlled. Values that deviate too much are probably not trustworthy.
In my opinion the value of 4.69 is suspect and you can discard it. There are several methods of measurements or in different wavelengths that lead to different results.

Finally, we must be careful with accuracy. We do calculations with all decimals available on our calculators but you can not expect much more than 3 accurate digits for a magnitude.
The apparent magnitude is easier to measure than the absolute magnitude which also suffers from inaccuracies in the distance or the absorption of light by the interstellar medium. 