Stellar Evolution and the HR Diagram

SPEC 3.9.2.5 & 3.9.2.6 - CLASSIFICATION OF STARS

The Hertzsprung-Russell diagram, stellar life cycles, supernovae, neutron stars, pulsars, and black holes

The Hertzsprung-Russell Diagram

Definition

The Hertzsprung-Russell (HR) diagram shows the variation of absolute magnitude (or luminosity) of a star against its spectral type (or surface temperature).

The HR diagram reveals patterns in stellar properties and allows us to understand how stars evolve. Hot, bright stars appear at the top-left; cool, dim stars appear at the bottom-right.

Key Concept

The HR diagram is not a conventional graph - both axes are inverted from normal orientation, and the temperature scale is logarithmic (it halves at every interval).

HR Diagram Axes and Scales

Vertical Axis (y-axis)

Absolute Magnitude M

Scale: -15 (top) to +15 (bottom)

Inverted: smaller/negative = brighter

Can also show luminosity

L / L_{Sun}

on log scale

Horizontal Axis (x-axis)

Temperature / Spectral Class

Temperature: 40,000 K → 2,500 K

Logarithmic scale - halves each interval

Spectral class: O B A F G K M

Common Mistake

Both axes are inverted. Temperature decreases left to right. Magnitude scale means brighter stars have more negative values at the top.

Structure of the HR Diagram

Regions and Star Populations

RegionPosition% of StarsExamples

Main SequenceDiagonal band (top-left to bottom-right)~90%Sun, Sirius A, Vega

White DwarfsBottom-left (hot but dim)~10%Sirius B, Procyon B

Red GiantsUpper-right (cool but luminous)~1%Arcturus, Aldebaran

SupergiantsTop (red or blue)Betelgeuse, Rigel

Why are Red Giants Luminous?

Despite being cool, red giants have enormous surface area. From Stefan's law

L = σ A T^{4}

, the large area compensates for low temperature.

Position of the Sun

Students must know the Sun's position on the HR diagram:

Sun's Properties

Spectral Class:G (yellow-white)Surface Temperature:~5,800 KAbsolute Magnitude:M = +4.83Luminosity:

L_{Sun} = 3.83 \times 1 0^{26}

WMass:

M_{Sun} = 2 \times 1 0^{30}

kg

The Sun is a typical main sequence star - neither particularly hot nor cool. It has been fusing hydrogen for ~4.6 billion years and will continue for another ~5 billion years.

01Worked Example

Star X has absolute magnitude -5 and temperature 3,500 K. Star Y has absolute magnitude +10 and temperature 15,000 K.

(a) Identify which region of the HR diagram each star occupies.

(b) Which star has the larger radius? Explain using Stefan's law.

Solution

•

Part (a):

Star X: M = -5 (very bright), T = 3,500 K (cool, class K/M) → upper-right = Red Giant

Star Y: M = +10 (very dim), T = 15,000 K (hot, class B) → lower-left = White Dwarf

•

Part (b): From

L = 4 π r^{2} σ T^{4}

Magnitude difference = 15 → brightness ratio =

2.5 1^{15} \approx 1 0^{6}

Star X is

1 0^{6}

times more luminous but cooler. Since

L \propto r^{2} T^{4}

, for X to be brighter despite lower T, it must have vastly larger radius.

Star X (the red giant) has the much larger radius

Stellar Evolution: Overview

Stars move across the HR diagram as they evolve. The path depends critically on the star's initial mass.

Specification Requirement

Students must trace the evolutionary path of a Sun-like star (mass < 3 solar masses) on the HR diagram from formation to white dwarf.

Low Mass (< 3 solar masses)

Nebula → Protostar → Main Sequence → Red Giant → Planetary Nebula → White Dwarf → Black Dwarf

High Mass (> 3 solar masses)

Nebula → Protostar → Main Sequence → Red Supergiant → Supernova → Neutron Star or Black Hole

Complete Stellar Evolution Paths

Stages 1-2: Protostar to Main Sequence

1. Protostar

Clouds of gas and dust (nebulae) clump together under gravity

Irregular clumps rotate and spin inwards forming a denser centre

Surrounded by a circumstellar disc

Gravitational PE → thermal energy

Not yet visible on HR diagram

2. Main Sequence

Core reaches ~15 million K

Hydrogen fusion begins

Inward gravity = outward radiation pressure → equilibrium

Star is now stable

Greater mass = shorter main sequence period (uses fuel faster)

Key Concept

The star's position on the main sequence is determined by its mass. High-mass stars are at the top-left (hot, bright); low-mass stars are at the bottom-right (cool, dim).

Stage 3: Red Giant (for star < 3 solar masses)

Once the hydrogen in the core runs out:

Process

The temperature of the core increases as it contracts

Core begins fusing helium nuclei into heavier elements (C, O, Be)

The outer layers expand and cool

Star moves up and right on the HR diagram

A red giant is brighter and cooler than a main sequence star

The Sun will expand to ~100× its current radius in this phase, possibly engulfing Mercury and Venus.

Movement on HR Diagram

Main sequence → Red giant: moves UP (more luminous due to larger area) and RIGHT (cooler surface temperature)

Stage 4-5: White Dwarf (for star < 1.4 solar masses)

Process

When the red giant uses up all its fuel, fusion stops

The core contracts as gravity becomes greater than outward force

The outer layers are thrown off, forming a planetary nebula

The remaining core becomes very dense (~10⁸ - 10⁹ kg m⁻³)

This is a white dwarf - hot but dim (small surface area)

Black Dwarf

A white dwarf will eventually cool to become a black dwarf - a cold, dark stellar remnant. This takes longer than the current age of the Universe, so none exist yet.

Movement on HR diagram: Red giant → White dwarf: moves DOWN (dimmer) and LEFT (hot exposed core)

Sun-like Star Path on HR Diagram

Worked Example02

Describe how the position of a Sun-like star changes on the HR diagram as it evolves from the main sequence to a white dwarf. Explain the physical processes causing each change.

Solution

•

Move 1: Main Sequence → Red Giant (UP and RIGHT)

Core hydrogen exhausted → core contracts and heats → shell hydrogen burning → outer layers expand → surface cools (RIGHT) → larger area increases luminosity (UP). Red giant is brighter and cooler.

•

Move 2: Red Giant → White Dwarf (DOWN and LEFT)

Helium exhausted → outer layers ejected as planetary nebula → hot core exposed (LEFT) → tiny surface area despite high temperature → much lower luminosity (DOWN). White dwarf is hotter and dimmer.

•

Final cooling: White dwarf gradually cools (moves RIGHT) over billions of years, eventually becoming a black dwarf.

MS → Red giant (↗ up-right) → White dwarf (↙ down-left) → Cooling (→)

Stage 5: Red Supergiant (for star > 3 solar masses)

When a high-mass star runs out of hydrogen, the same process occurs but on a larger scale:

Red Supergiant Properties

Much larger and more luminous than red giants

Can fuse elements up to iron in successive shells

The collapse of red supergiants causes gamma ray bursts

Core eventually collapses catastrophically → supernova

Why Does Fusion Stop at Iron?

Fusion of iron is endothermic - it absorbs energy rather than releasing it. When the core becomes iron, there is no energy source to prevent gravitational collapse.

Stage 6: Supernova (for star > 1.4 solar masses)

Definition

A supernova is the explosive, cataclysmic end to the life of a star. A defining characteristic is its rapidly increasing absolute magnitude.

When all fuel runs out, fusion stops and the core collapses inwards very suddenly, becoming rigid as matter can no longer be forced closer together.

The outer layers fall inwards and rebound off the core

This launches them into space as a shockwave

Elements heavier than iron are fused and scattered

Energy release: ~ 10⁴⁴ J (same as Sun's entire 10 billion year output)

Luminosity: up to 10¹⁰ times the Sun - can outshine entire galaxies

Type 1a Supernovae: Standard Candles

There are two main types of supernovae:

Type I (Type 1a)

A star accumulates matter from a companion in a binary system and explodes after reaching critical mass (1.4 solar masses - the Chandrasekhar limit).

Type II

The death of a high-mass star after it runs out of fuel. Outer layers rebound off the collapsing core.

Standard Candles

All Type 1a supernovae occur at the same critical mass, so they all have a very similar peak absolute magnitude (about -19.3) and produce very consistent light curves. This allows astronomers to use them as standard candles to calculate distances to far-off galaxies using

m - M = 5 lo g (d /10)

.

Type 1a Supernova Light Curve

Students should be familiar with the characteristic light curve:

03Worked Example

A supernova releases $2.0 \times 1 0^{44}$ J of energy over 10 seconds. The Sun's luminosity is $3.83 \times 1 0^{26}$ W.

(a) Calculate the average power output of the supernova.

(b) For how many years would the Sun need to shine to emit the same energy?

Solution

•

Part (a):

P = \frac{E}{t} = \frac{2.0 \times 1 0 ^{44}}{10} = 2.0 \times 1 0^{43}

W

•

Part (b):

Time =

\frac{E}{L _{Sun}} = \frac{2.0 \times 1 0 ^{44}}{3.83 \times 1 0 ^{26}} = 5.2 \times 1 0^{17}

s

Convert to years:

\frac{5.2 \times 1 0 ^{17}}{3.15 \times 1 0 ^{7}} = 1.7 \times 1 0^{10}

years

(a) 2.0 × 10⁴³ W (b) 17 billion years - longer than the Sun's entire lifetime!

Stage 7: Neutron Star (1.4 - 3 solar masses remnant)

Formation

When the core of a large star collapses, gravity is so strong that it forces protons and electrons together to form neutrons:

p + e^{-} \to n + ν_{e}

PropertyValue

CompositionAlmost entirely neutrons

Remnant mass1.4 - 3 solar masses

Radius~10 km (city-sized)

Density~10¹⁷ kg m⁻³ (density of nuclear matter)

Pulsars

Pulsars are spinning neutron stars that emit beams of radiation from their magnetic poles as they spin (up to 600 times per second). We detect regular pulses as the beam sweeps past Earth.

Stage 8: Black Hole (for remnant > 3 solar masses)

Definition

When the core of a giant star collapses, the neutrons are unable to withstand gravity, forcing them together. The gravitational pull is so strong that not even light can escape.

Forms from stars with original mass > ~10 solar masses (remnant > 3 solar masses)

The event horizon is the point at which the escape velocity becomes greater than the speed of light

Nothing can escape from within the event horizon

Schwarzschild Radius

The Schwarzschild radius is the radius of the event horizon:

R_{s} = \frac{2 GM}{c ^{2}}

where G is the gravitational constant, M is the mass, and c is the speed of light.

Deriving the Schwarzschild Radius

At the Schwarzschild radius, the escape velocity equals the speed of light:

Derivation

Escape velocity:

v_{E} = c = \frac{2 GM}{R _{s}}

Square both sides:

c^{2} = \frac{2 GM}{R _{s}}

Rearrange for

R_{s}

:

R_{s} = \frac{2 GM}{c ^{2}}

At the event horizon, the escape velocity equals c. Inside

R_{s}

, escape is impossible.

ObjectMassSchwarzschild Radius

Sun (if compressed)

2 \times 1 0^{30}

kg~3 km

Earth (if compressed)

6 \times 1 0^{24}

kg~9 mm

04Worked Example

Calculate the radius of the event horizon for a black hole with mass 50 times that of our Sun.

Data:

G = 6.67 \times 1 0^{- 11}

N m² kg⁻², mass of Sun

= 2 \times 1 0^{30}

kg, speed of light

= 3 \times 1 0^{8}

m s⁻¹

Solution

•

Step 1: Calculate mass

M = 50 \times 2 \times 1 0^{30} = 1 \times 1 0^{32}

kg

•

Step 2: Apply Schwarzschild formula

R_{s} = \frac{2 GM}{c ^{2}} = \frac{2 \times 6.67 \times 1 0 ^{- 11} \times 1 \times 1 0 ^{32}}{( 3 \times 1 0 ^{8} ) ^{2}}

•

Step 3: Calculate

R_{s} = \frac{1.33 \times 1 0 ^{22}}{9 \times 1 0 ^{16}} = 1.5 \times 1 0^{5}

m = 150 km

Rₛ = 150 km (or 1.5 × 10⁵ m)

P

Practice Problem

A black hole has a Schwarzschild radius of 30 km.

(a) Calculate the mass of the black hole in kg.

(b) Express this mass in solar masses.

(c) What type of stellar remnant is this?

Data:

G = 6.67 \times 1 0^{- 11}

N m² kg⁻²,

c = 3.00 \times 1 0^{8}

m s⁻¹,

M_{Sun} = 2 \times 1 0^{30}

kg

Solution

•

Part (a): Rearrange

R_{s} = \frac{2 GM}{c ^{2}}

for M:

M = \frac{R _{s} c ^{2}}{2 G} = \frac{30000 \times ( 3 \times 1 0 ^{8} ) ^{2}}{2 \times 6.67 \times 1 0 ^{- 11}} = \frac{2.7 \times 1 0 ^{21}}{1.33 \times 1 0 ^{- 10}} = 2.0 \times 1 0^{31}

kg

•

Part (b):

\frac{M}{M _{Sun}} = \frac{2.0 \times 1 0 ^{31}}{2 \times 1 0 ^{30}} = 10

solar masses

•

Part (c): A 10 solar mass black hole is a stellar black hole, formed from core-collapse supernova of a massive star.

(a) 2.0 × 10³¹ kg (b) 10 solar masses (c) Stellar black hole

Supermassive Black Holes

Definition

Scientists believe there are supermassive black holes at the centre of every galaxy. Stars and gas near galactic centres appear to be orbiting very quickly, indicating a very strong gravitational field.

Supermassive black holes can form from:

The collapse of massive gas clouds while the galaxy was forming

A normal black hole that accumulated huge amounts of matter over millions of years

Several normal black holes merging together

Milky Way's SMBH: Sagittarius A*

Mass: ~4 × 10⁶ solar masses

Located 26,000 light-years from Earth

Dark Energy and the Accelerating Universe

Hubble's law shows the Universe is expanding. If expansion was slowing down, more distant objects would appear to be receding more quickly (since light takes longer to reach us from the past).

The Discovery

Type 1a supernovae have been seen to be dimmer than expected, meaning they are more distant than Hubble's law predicted. This suggests the expansion of the Universe is accelerating.

Dark Energy

Dark energy is thought to be the reason behind the Universe accelerating. It has an overall repulsive effect throughout the Universe. Since gravity follows the inverse square law (decreases with distance), but dark energy remains constant everywhere, dark energy has a greater effect than gravity and causes the expansion speed to increase.

Dark energy is controversial because there is evidence for its existence but no one knows what it is or what is causing it.

Summary: Key Points

HR Diagram (3.9.2.5)

Axes: M (-15 to +15) vs T (log scale)

Both axes inverted

90% main sequence, 10% white dwarfs, 1% giants

Sun: G, M = +4.83, T = 5,800 K

Low mass: MS → Giant → White dwarf → Black dwarf

Stellar Remnants (3.9.2.6)

Type 1a SNe: M ≈ -19.3, standard candles

Neutron stars: ρ ~ 10¹⁷ kg m⁻³, pulsars

Black holes:

R_{s} = \frac{2 GM}{c ^{2}}

SMBHs at galactic centres

Dark energy: accelerating expansion, controversial

Key Equations

R_{s} = \frac{2 GM}{c ^{2}}

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

L = σ A T^{4}