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The magnitudes of stars theory

How bright a star looks is given by its apparent magnitude. This is different from its absolute magnitude. The absolute magnitude of a star is defined as the apparent magnitude that it would have if placed at a distance of 10 parsecs from the Earth.



Consider two stars A and B. Star A appears to be brighter than star B. In other words the intensity of the light reaching the observer from star A is greater than that from star B.


Let the apparent magnitude of star A = mA and the apparent magnitude of star B be mB.

Referring back to the magnitude difference of 5 being a difference in intensity by a factor of 100 we can write:

IA/IB = 100(mB mA)/5

Since if (mB mA) = 5 then IA/IB = 100(5)/5 = 100

Therefore taking logs of both side : lg(IA/IB) = 2/5(mB mA)

Therefore: mB mA = 5/2[lg(IA/IB)]

Now let the magnitude of A (mA) be that at 10 parsecs, in other words the absolute magnitude of the star (M) and let mB be the magnitude (m) at some other distance d (also measured in parsecs).

Therefore : m M = 5/2[lg(IA/IB)]

But from the inverse square law: (IA/IB) = (dB/dA)2 because the intensity is inversely proportional to the square of the distance of the star.

Therefore:

m M = 5/2lg(IA/IB) = 5/2[lg(dA/dA)2] = 5lg(dA/dA) = 5lg(d/10) = 5lgd 5
Therefore :

Apparent magnitude (m) Absolute magnitude (M) = 5lg(d/10) = 5 - 5lg(d)



In our example if A appears to the observer to be brighter than B, and if we use mA to be the absolute magnitude (M) then its apparent magnitude (mB) is less and so its distance must be more than 10 pc.


The apparent and absolute magnitudes of a number of stars are given in the following table.


Object Apparent
magnitude
Absolute
magnitude
Distance
(light years)
Sun -26.7 - -
Venus -4.4 - -
Jupiter -2.2 - -
Sirius -1.46 +1.4 8.7
Rigel 0.1 -7.0 880
Arcturus -0.1 -0.2 35.9
Proxima Centauri 10.7 +15.1 4.2
Vega 0.0 +0.5 26.4
Betelguese -0.4 -5.9 586
Deneb (a Cygni) 1.3 -7.2 1630
Andromeda galaxy 5 -17.9 2 200 000
Our Galaxy - -18.0 -


Example problems
1. Calculate the absolute magnitude of a star of apparent magnitude +2.5 which is at a distance of 25 pc from the Earth.
M = m + 5 5lg(d) = 2.5 + 5 5lg(25) = 7.5 6.99 = +0.51

2. Calculate the distance of a star with an apparent magnitude of +6.0 and an absolute magnitude of +4.0.
6 4 + 5 = 5lg(d) Therefore: lg(d) = 0.6 d = 3.98 pc

3. Calculate the apparent magnitude of a star whose absolute magnitude is -1.8 if the star is at a distance of 35.5 parsecs from the Earth.
m = M 5 + 5lg(d) = -1.8 5 + 5x1.55 = + 0.95

4. Calculate the apparent magnitude of a star that has an absolute magnitude of +2.0 if the star is 25 parsecs from the Earth.
m = 5lg(d/10) + M = 5lg(25/10) + 2.0 = 1.98 + 2.0 = 3.989 = 4
 

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© Keith Gibbs 2020