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 = m_A and the apparent magnitude of star B be m_B.

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


Since if (m_B – m_A) = 5 then I_A/I_B = 100(5)/5 = 100

Therefore taking logs of both side : lg(I_A/I_B) = 2/5(m_B – m_A)

Therefore: m_B – m_A = 5/2[lg(I_A/I_B)]

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

Therefore : m – M = 5/2[lg(I_A/I_B)]

But from the inverse square law: (I_A/I_B) = (d_B/d_A)² because the intensity is inversely proportional to the square of the distance of the star.

Therefore:

m – M = 5/2lg(I_A/I_B) = 5/2[lg(d_A/d_A)²] = 5lg(d_A/d_A) = 5lg(d/10) = 5lgd – 5
Therefore :




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 (m_B) 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	-