Stellar Luminosity and Magnitude

SPEC 3.9.2.1 & 3.9.2.2 - ASTROPHYSICS

Understanding how we measure the brightness and distance of stars

Luminosity and Intensity

Definition

Luminosity (L) is the total power output of a star - the rate of light energy released per second. It is measured in watts (W).

What Determines Luminosity?

A star's luminosity depends on:

The size (radius) of the star - larger stars radiate from a greater surface area

The temperature of the star - hotter stars radiate more energy per unit area

Definition

Intensity (I) is the power received from a star per unit area at a given distance. It is measured in W m⁻². This is the effective, measurable brightness.

Key Distinction

Luminosity is an intrinsic property of the star. Brightness is subjective (varies by observer). Intensity is the measurable quantity that depends on both luminosity and distance.

The Inverse Square Law

As light travels outward from a star, it spreads over an ever-increasing spherical surface. At distance d, the light is spread over a sphere of surface area 4πd².

Formula

I = \frac{L}{4 π d ^{2}}

where L = luminosity (W), d = distance (m), I = intensity (W m⁻²)

Important

Intensity is inversely proportional to d². If distance doubles, intensity falls to ¼. If distance triples, intensity falls to ⅑.

Worked Example01

The Sun has luminosity 3.83 × 10²⁶ W. Calculate the intensity of sunlight at Earth, given Earth is 1.50 × 10¹¹ m from the Sun.

1

Write the inverse square law:

I = \frac{L}{4 π d ^{2}}

2

Substitute values:

I = \frac{3.83 \times 1 0 ^{26}}{4 π \times ( 1.50 \times 1 0 ^{11} ) ^{2}}

3

Calculate denominator:

4 π \times (1.50 \times 1 0^{11})^{2} = 4 π \times 2.25 \times 1 0^{22} = 2.83 \times 1 0^{23}

4

Final calculation:

I = \frac{3.83 \times 1 0 ^{26}}{2.83 \times 1 0 ^{23}} = 1350 W m^{- 2}

Answer: I ≈ 1350 W m⁻² (the solar constant)

Worked Example02

Star A and Star B have the same luminosity. Star A is 20 light-years away and Star B is 80 light-years away. Calculate the ratio of their observed intensities I_A : I_B.

1

Since luminosity is the same:

I \propto \frac{1}{d ^{2}}

2

Find the distance ratio:

\frac{d _{B}}{d _{A}} = \frac{80}{20} = 4

3

Apply inverse square relationship:

\frac{I _{A}}{I _{B}} = (\frac{d _{B}}{d _{A}})^{2} = 4^{2} = 16

Answer: I_A : I_B = 16 : 1

Star A appears 16 times brighter than Star B, even though they have identical luminosities.

Astronomical Distance Units

Stars are so far away that we need special units to describe their distances:

Astronomical Unit (AU)

The mean distance between the centre of the Earth and the centre of the Sun.1 AU = 1.50 × 10¹¹ m

Light Year (ly)

The distance that light (or any EM wave) travels in a vacuum in one year.1 ly = 9.46 × 10¹⁵ m

Parsec (pc)

The distance at which an object subtends an angle of 1 arcsecond using the radius of Earth's orbit (1 AU) as the baseline.1 pc = 3.086 × 10¹⁶ m = 3.26 ly = 2.06 × 10⁵ AU

Useful Conversions

1 parsec = 3.26 light years | 1 Mpc (megaparsec) = 10⁶ pc = 3.26 × 10⁶ ly

Stellar Parallax

Definition

Parallax is the apparent change of position of a nearer star in comparison to distant stars in the background, as a result of the orbit of the Earth around the Sun.

Parallax Formula

d = \frac{1}{θ}

d in parsecs, θ in arcseconds

Key Point

A parsec is defined as the distance at which a star has a parallax angle of exactly 1 arcsecond.

The greater the parallax angle, the closer the star. Distant stars have very small parallax angles.

Worked Example03

A star has a parallax angle of 0.25 arcseconds. Calculate its distance from Earth in (a) parsecs, (b) light years, and (c) metres.

1

(a) Apply parallax formula:

d = \frac{1}{θ} = \frac{1}{0.25} = 4 pc

2

(b) Convert to light years:

d = 4 \times 3.26 = 13.0 ly

3

(c) Convert to metres:

d = 4 \times 3.086 \times 1 0^{16} = 1.23 \times 1 0^{17} m

Answer: (a) 4 pc, (b) 13.0 ly, (c) 1.23 × 10¹⁷ m

Worked Example04

Proxima Centauri is 4.24 light years from Earth. Calculate its parallax angle in arcseconds.

1

Convert to parsecs:

d = \frac{4.24}{3.26} = 1.30 pc

2

Rearrange parallax formula:

θ = \frac{1}{d}

3

Calculate:

θ = \frac{1}{1.30} = 0.77 arcseconds

Answer: θ = 0.77 arcseconds

This is a relatively large parallax because Proxima Centauri is the nearest star to the Sun.

Apparent Magnitude

Definition

Apparent magnitude (m) is a measure of how bright an object appears in the sky. It depends on both the star's luminosity and its distance from Earth.

The Hipparcos scale was devised by the Greek astronomer Hipparchus around 150 BCE. He classified the brightest stars as magnitude 1 and the faintest visible stars as magnitude 6.

Key Features of the Magnitude Scale

Logarithmic: Each step of 1 magnitude = intensity ratio of 2.51

Inverted: Lower (more negative) numbers = brighter objects

5 magnitudes = factor of 100 in intensity (2.51⁵ = 100)

ObjectApparent Magnitude (m)

Sun−26.7

Full Moon−12.6

Sirius (brightest star)−1.46

Naked eye limit+6

Large telescope limit+15

Pogson's Equation

In 1856, Norman Pogson formalised the magnitude scale mathematically. He defined that a difference of 5 magnitudes corresponds to an intensity ratio of exactly 100.

Pogson's Ratio

\frac{I _{1}}{I _{2}} = 10 0^{\frac{m _{2} - m _{1}}{5}} = 2.5 1^{(m_{2} - m_{1})}

where 2.51 ≈ ⁵√100 = 100^0.2

Understanding the Ratio

Magnitude DifferenceIntensity Ratio1 magnitude2.51 : 12 magnitudes6.31 : 14 magnitudes40 : 15 magnitudes100 : 110 magnitudes10,000 : 1

Exam Tip

A smaller (more negative) magnitude means a brighter object. The brighter star goes in the numerator of the intensity ratio.

Worked Example05

If the apparent magnitudes of two stars differ by 4, what is the difference in their brightness (intensity ratio)?

1

Apply Pogson's equation:

\frac{I _{1}}{I _{2}} = 2.5 1^{Δ m}

2

Substitute Δm = 4:

\frac{I _{1}}{I _{2}} = 2.5 1^{4} = 40

Answer: The brighter star is 40 times brighter than the fainter star.

Worked Example06

Sirius has apparent magnitude m = −1.46 and Polaris has apparent magnitude m = +1.98. Calculate how many times brighter Sirius appears than Polaris.

1

Find magnitude difference: Δm = m₂ − m₁ = 1.98 − (−1.46) = 3.44

2

Apply Pogson's equation:

\frac{I _{S i r i u s}}{I _{P o l a r i s}} = 2.5 1^{Δ m} = 2.5 1^{3.44}

3

Calculate:

2.5 1^{3.44} = 23.7 \approx 24

Answer: Sirius appears approximately 24 times brighter than Polaris

Absolute Magnitude

Apparent magnitude doesn't tell us how intrinsically bright a star is - a dim nearby star might appear brighter than a brilliant distant one. We need a fair comparison.

Definition

Absolute magnitude (M) is defined as the apparent magnitude a star would have if it were placed at a standard distance of 10 parsecs from Earth.

Interpreting M vs m

M > m: Star is closer than 10 pc (appears brighter than its intrinsic brightness)

M = m: Star is exactly at 10 pc

M < m: Star is further than 10 pc (appears dimmer than its intrinsic brightness)

The Distance Modulus

Distance Modulus Formula

m - M = 5 lo g (\frac{d}{10})

where d is the distance in parsecs

Rearranged Form

M = m - 5 lo g (\frac{d}{10})

StarDistance (pc)Apparent mag (m)Absolute mag (M)

Sirius2.7−1.46+1.4

Vega8.1+0.03+0.6

Arcturus11−0.04−0.2

Aldebaran21+0.85−0.3

Rigel*260+0.12−7.1

*Rigel is a variable star so its magnitude changes slightly with time.

Important

The distance d must be in parsecs for this formula. Convert from light-years if needed (1 pc = 3.26 ly).

Worked Example07

Calculate the absolute magnitude of a star of apparent magnitude +2.5 that is at a distance of 25 pc from the Earth.

1

Use:

M = m - 5 lo g (\frac{d}{10})

2

Substitute:

M = 2.5 - 5 lo g (\frac{25}{10})

3

Calculate:

M = 2.5 - 5 lo g (2.5) = 2.5 - 5 \times 0.398 = 2.5 - 1.99

Answer: M = +0.51

Worked Example08

Calculate the distance of a star with an apparent magnitude of +6.0 and an absolute magnitude of +4.0.

1

Use distance modulus:

m - M = 5 lo g (\frac{d}{10})

2

Substitute:

6 - 4 = 5 lo g (\frac{d}{10})

3

Simplify:

2 = 5 lo g (\frac{d}{10})

, so

\frac{2}{5} = lo g (\frac{d}{10})

, giving

0.4 = lo g (\frac{d}{10})

4

Take antilog:

\frac{d}{10} = 1 0^{0.4} = 2.51

5

Solve:

d = 2.51 \times 10 = 25 pc

Answer: d = 25 pc

Worked Example09

Two stars have the same absolute magnitude. Star X has apparent magnitude +4.0 and Star Y has apparent magnitude +9.0. Calculate the ratio of their distances from Earth.

1

For Star X:

m_{X} - M = 5 lo g (\frac{d _{X}}{10})

2

For Star Y:

m_{Y} - M = 5 lo g (\frac{d _{Y}}{10})

3

Subtract equations (M cancels):

m_{Y} - m_{X} = 5 lo g (\frac{d _{Y}}{d _{X}})

4

Substitute:

9.0 - 4.0 = 5 lo g (\frac{d _{Y}}{d _{X}})

5

Solve:

1.0 = lo g (\frac{d _{Y}}{d _{X}})

, so

\frac{d _{Y}}{d _{X}} = 10

Answer: d_Y : d_X = 10 : 1

Star Y is 10 times further away (5 magnitudes = factor of 100 in intensity = factor of 10 in distance)

Practice

Question 1: A star has a parallax angle of 0.04 arcseconds. Calculate its distance from Earth in (a) parsecs and (b) light years.

1

(a) Apply parallax formula:

d = \frac{1}{θ} = \frac{1}{0.04} = 25 pc

2

(b) Convert to light years:

d = 25 \times 3.26 = 81.5 ly

Answer: (a) 25 pc, (b) 81.5 ly

Practice

Question 2: A star has luminosity 2.5 × 10²⁷ W. An observer is 4.0 × 10¹⁷ m from the star. Calculate the intensity of radiation received by the observer.

1

Apply inverse square law:

I = \frac{L}{4 π d ^{2}}

2

Substitute:

I = \frac{2.5 \times 1 0 ^{27}}{4 π \times ( 4.0 \times 1 0 ^{17} ) ^{2}}

3

Calculate:

I = \frac{2.5 \times 1 0 ^{27}}{2.01 \times 1 0 ^{36}} = 1.2 \times 1 0^{- 9} W m^{- 2}

Answer: I = 1.2 × 10⁻⁹ W m⁻²

Practice

Question 3: A Cepheid variable star has absolute magnitude M = −4.5 and is observed with apparent magnitude m = +12.5. Calculate the distance to this star in parsecs and light-years.

1

Calculate distance modulus:

m - M = 12.5 - (- 4.5) = 17.0

2

Apply formula:

17.0 = 5 lo g (\frac{d}{10})

3

Divide by 5:

3.4 = lo g (\frac{d}{10})

4

Antilog:

\frac{d}{10} = 1 0^{3.4} = 2512

, so

d = 25, 120 pc \approx 25, 000 pc

5

Convert:

d = 25, 120 \times 3.26 = 82, 000 ly

Answer: d ≈ 25,000 pc ≈ 82,000 ly

Summary

Key Formulas

QuantityFormulaUnits

Intensity (inverse square)

I = \frac{L}{4 π d ^{2}}

W m⁻²

Parallax distance

d = \frac{1}{θ}

d in pc, θ in arcsec

Pogson's ratio

\frac{I _{1}}{I _{2}} = 2.5 1^{(m_{2} - m_{1})}

ratio

Distance modulus

m - M = 5 lo g (\frac{d}{10})

d in parsecs

Key Concepts

Luminosity (L) = total power output (depends on size and temperature)

Apparent magnitude (m) = how bright a star appears (depends on L and distance)

Absolute magnitude (M) = apparent magnitude at 10 pc (intrinsic brightness)

Parallax = apparent shift due to Earth's orbit; d = 1/θ

5 magnitudes = factor of 100 in intensity

Distance Unit Conversions

1 AU = 1.50 × 10¹¹ m | 1 ly = 9.46 × 10¹⁵ m | 1 pc = 3.086 × 10¹⁶ m = 3.26 ly