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.
B
A
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(dB/dA)2] = 5lg(dB/dA) = 5lg(d/10) = 5lgd – 5
Therefore :
m – M = 5lg(d/10) = 5 - 5lg(d)
Star A
10 pc
Star B
d
1
In our example if A appears to the observer to be brighter than B, and if we use m A 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
Sun
Apparent
magnitude
-26.7
Absolute
magnitude
Distance (light
years)
Venus
-4.4
Jupiter
-2.2
Sirius
-1.46
+1.4
8.7
Rigel
0.1
-7.0
880.0
Arcturus
-0.1
-0.2
35.86
Proxima Centuari
10.7
15.1
4.2
Vega
0.0
+0.5
26.4
Betelguese
0.4
-5.9
586
Deneb ( 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
2