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 :
or:
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.
