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Physics Lesson 22.4.2 - The Illumination of Stars
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Welcome to our Physics lesson on The Illumination of Stars, this is the second lesson of our suite of physics lessons covering the topic of Stars, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
The Illumination of Stars
Like all the other light sources, the quantity used to evaluate the illumination of stars is the light flux, Φ, which represents the total light energy emitted by a light source in the unit time. Light flux (measured in lumen) has been explained in Section 12.
Another relevant parameter used when dealing with light illumination is illumination, b, otherwise known as illuminance (or brightness). It represents the light flux Φ incident on a given surface area, A. If the distance star-Earth (observer) is denoted by d (it represents the radius of light sphere formed when light is emitted from a star located at its centre), we obtain for illumination b of the given star:
The figure below clarifies this point.
The illumination of stars is measured in lux (1 lux = 1 lumen/m2 where 1 lumen = 1 candela/steradian. Hence, 1 lux = 1 cd/str·m2). Illuminaion is a quantitative indicator of which star is brighter and which is dimmer. However, scientists prefer to use the logarithm of illumination known as apparent magnitude, m, to estimate how bright a star is. Apparent magnitude m is a dimensionless quantity. The choice to use logarithm of illumination instead of its actual value is related to the human perception of light. It is a proven fact that if a number of light sources emit light of true illumination having a ratio 10:100:1000:10000, etc., to each other, the human eye perceives them as varying according the ratio 1:2:3:4, etc. This is similar to when we have used (in 11.5) the level of sound (which is a logarithmic quantity) instead of intensity of sound to describe how strong a sound is.
Since antiquity, visible stars have been classified in 6 categories depending on the illumination they brought on Earth. Thus based on this classification, the brightest stars were considered as first class stars and the dimmer ones as sixth class stars. The stars belonging to the last category were barely visible at naked eye. Astronomers of the 19th century realized that two stars having a five-class difference have a ratio of 1:100 in the light flux they bring on Earth. Therefore, they created the logarithmic system discussed above to measure the illumination of stars. Thus, according to this system, the apparent magnitudes m1 and m2 of two different stars relate to each other according the equation
Example 1
A star S2 looks 39.81 times brighter than another star s1 when observed from Earth. How many classes of difference do these stars have with each other?
Solution 1
Clues:
b2 = 39.81 x
m1 - m2 = ?
We have:
= -2.5 log x/39.81 x
= -2.5 log 1/39.81
= -2.5 log 39.81-1
= (-1) ∙ (-2.5) ∙ log 39.81
= 2.5 ∙ 1.6
= 4
This means the star s1 may be of class 5 and s2 of class 1 or the star s1 may be of class 6 and s2 of class 2.
From the definition of apparent magnitude, it results that stars with the smallest apparent magnitude are the brightest. This is because we have a negative sign in the equation, which makes the result big when the expression on the right side is small. Indeed, if we take b = 10 lux in the equation
the value of apparent magnitude m is
= -2.5 ∙ 1
= -2.5
But is we take b = 100 lux, we obtain for the apparent magnitude m
= -2.5 ∙ 2
= -5
For example, m(Sun) = -26.8; m(full Moon) = -12.5; m(Venus) = -4; m(Mars) = -2, m(Sirius) = -1.4, etc. (Sirius is the second brightest star after the Sun).
Example 2
Calculate how many times the Sun is brighter than the full Moon.
Solution 2
Clues:
mMoon = -12.5
bSun/bMoon = ?
To find the given ration, we use the formula
Substituting the known values and then doing the operations, we obtain
-14.3 = -2.5 log bSun/bMoon
5.72 = log bSun/bMoon
bSun/bMoon = 105.72
bSun/bMoon = 524 807 times brighter
All the other bright objects in the Northern hemisphere of the sky besides those mentioned above have a positive apparent magnitude. We can see, using the naked eye, in a clear sky night stars of apparent magnitude up to 6.5, and when using binoculars we can see stars of apparent magnitude up to 10.
The brightness of a star may give us the wrong impression of its radiating ability. This is because of the different distances that stars are from Earth. One star may look dimmer than another star but this may occur because it farther from Earth and not because it radiates less. Therefore, it is not sufficient to know the light flux Φ of a star in isolation, we also need the illumination b as dictated by the formula given at the beginning of this tutorial - a value that depends on the distance d from the star.
In order to create an analogy with sound topic, we use a quantity known as absolute illumination b0 as a reference unit. It corresponds to the illumination produced by a star that is 10 parsecs away from Earth. A parsec (pc) is a unit of length used for astronomical distances outside the Solar System; 1 parsec = 3.26 light years = 3.086 × 1017 m. Thus, the absolute illumination produced by a star represents the light detected by an observer who is 32.6 light years away from this star. Likewise, we can introduce a quantity known as absolute magnitude M (which does not depend on distance) obtained when a logarithmic scale (as we did for apparent magnitude) is used. Based on this approach, it is clear that the absolute magnitudes M1 and M2 of two stars 1 and 2 relate to their light fluxes Φ1 and Φ2 through the following formula:
Obviously, absolute magnitude is a dimensionless unit as it is obtained by dividing two same quantities before taking their logarithm.
In simpler words, absolute magnitude M of an luminous object represents the apparent magnitude if this object was at a distance of 10 pc.
Example 3
We can write the formula
in another form, as
where A is a constant, because one of illuminations can be taken as the reference value discussed above.
Giving that apparent magnitude of Sun is mSun = -26.8, its flux is ΦSun = 4 × 1026 lumen and the distance Sun-Earth is 149.6 million km, calculate:
- The value of constant A
- The absolute magnitude of the Sun
Solution 3
- First, we have to calculate the illumination b produced by the Sun. We are given the following clues for this task: Φ = 4 × 1026 lumenThus,
d = 149.6 million km = 1.496 × 1011 mb = Φ/4π ∙ d2Now, we find the value of constant A. Giving that m = -26.8, we have
= )4 × 1026 lumen/4 ∙ 3.14 ∙ (1.496 × 1011 m)2
= 1.423 × 103 lux
= 1423 luxm = -2.5 log b + A
-26.8 = -2.5 log 1423 + A
-26.8 = -7.8 + A
A = -19 - Absolute magnitude M represents the apparent magnitude if Sun was at a distance of 10 pc = 10 · 3.086 × 1017 m = 3.086 × 1018 m. We must divide this value by 2 because we are only in one of directions of the Sun. First, we find the illumination b at this distance. We haveb = Φ/4π ∙ d2Thus, giving that
= 4 × 1026 lumen/4 ∙ 3.14 ∙ (3.086 × 1018 m)2
= 3.34 × 10-12 luxm = -2.5 log b + Awe obtain after substitutionsm = -2.5 log b + AAs we said earlier, we must divide this value by 2 to obtain the absolute magnitude M of the Sun. Thus, we have
= -2.5 ∙ log (3.34 × 10-12 ) - 19
= -2.5 ∙ (log 3.34 + log 10-12 ) - 19
= -2.5 ∙ (log 3.34 - 12 ∙ log 10 ) - 19
= -2.5 ∙ (0.524 - 12) - 19
= -2.5 ∙ (-11.476) - 19
= 28.69 - 19
= 9.69M = b/2
= 9.69/2
= 4.845
You have reached the end of Physics lesson 22.4.2 The Illumination of Stars. There are 5 lessons in this physics tutorial covering Stars, you can access all the lessons from this tutorial below.
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