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25.2Stellar properties and the HertzsprungâRussell diagram
The HertzsprungâRussell diagram
For this section you may need to review the material on luminosity and apparent brightness in SectionÂ 7.3. In the early part of the twentieth century, the Danish astronomer Ejnar Hertzsprung and the American Henry Norris Russell independently pioneered plots of stellar luminosities. They plotted luminosity on the vertical axis and surface temperature on the horizontal axis (increasing to the left). Such plots are now called HertzsprungâRussell (HR) diagrams. In the HR diagram of Figure 25.1 luminosity is given in terms of the sunâs luminosity (i.e. 1 on the vertical axis corresponds to the solar luminosity Lâ = 3.9 Ã 1026 W; astronomers use the symbol â to denote quantities pertaining to the sun). The surface temperature is given in thousands of kelvin.
The luminosity in this diagram varies from 10â4 to 104, a full 8 orders of magnitude, whereas the temperature varies from 3000 K to 30 000 K. For this reason, the scale on both axes is not linear so that all data can fit.
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EXAM TIP
It is important that you understand the HR diagram very well.
The slanted dotted lines represent stars with the same radius. So our sun and Procyon A have about the same radius. The symbol Râ stands for the radius of our sun.

Figure 25.1: The HertzsprungâRussell diagram. The (surface) temperature increases to the left. Note that the scales are not linear. The dotted lines represent stars with the same radius. Râ is the radius of the sun.

CHECK YOURSELF 2
Sirius A and Sirius B occupy very different places on the HR diagram. Can we conclude that these stars are physically far from each other in space?
As more and more stars were placed on the HR diagram, three clear features emerged:
Most stars fall on a strip extending diagonally across the diagram from top left to bottom right. This is called the main sequence.
Some large stars, reddish in colour, occupy the top rightâthese are the red giants (large and cool). Above these are the super red giants (also known as red supergiants) and are even larger and brighter than red giants.
The bottom left is a region of small stars known as white dwarfs (small and hot).
As we move along the main sequence toward hotter stars, the mass of the stars increases. Thus, the right end of the main sequence is occupied by red dwarfs and the left by blue giants.
An almost vertical strip above the main sequence is known as the instability region. We will discuss this region a bit later in the section on stellar evolution. It contains stars that have a variable luminosity.
Note that, once we know the temperature of a star (for example, through its spectrum), the HR diagram can tell us the luminosity of the star with an acceptable degree of accuracy, provided we know what type of star it is.
Main sequence stars
Our sun is a typical member of the main sequence. It has a mass of 2 Ã 1030 kg, a radius of 7 Ã 108 m, an average density of 1.4 Ã 103 kg mâ3 and radiates at a rate of 3.9 Ã 1026 W. Main sequence stars produce energy in their core from nuclear fusion of hydrogen into helium. The outflow of energy from the core creates an outward radiation pressure that exactly counterbalances the tendency of the star to collapse under its own weight. The common characteristic of all main sequence stars is the fusion of hydrogen into helium.
KEY POINT
Stable stars are in equilibrium due to a balance of the inward gravitational pressure and the outward radiation pressure.
Red giants and super red giants
Red giants are very large, cool stars with a reddish colour. The luminosity of red giants is considerably greater than the luminosity of main sequence stars of the same temperature. Treating them as black bodies radiating according to the StefanâBoltzmann law means that a luminosity of 106 times larger corresponds to an area of 106 times larger, which means a radius of 103 times larger, Figure 25.2. The mass of a red giant can be as much as 1000 times the mass of our sun, but their huge size also implies small densities. A red giant will have a central hot core surrounded by an enormous envelope of extremely tenuous gas.
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Figure 25.2: Artistâs impression of the comparison between our sun and the red supergiant VY Canis Majoris.

White dwarfs
These are common stars, but their faintness makes them hard to detect. A well-known white dwarf is Sirius B, the second star in a binary star system (double star system), the other member of which, Sirius A, is the brightest star in the evening sky, Figure 25.3.

Figure 25.3: Sirius A is the bright star in the middle of the photograph. The white dwarf companion Sirius B is the tiny speck of light lower left. The rings and spikes are artifacts of the telescopeâs imaging systems. The photograph has been overexposed so that the faint Sirius B could be seen.

Sirius A and B have about the same surface temperature (about 10 000 K), but the luminosity of Sirius B is about 10 000 times smaller. This means that Sirius B has a radius that is 100 times smaller than that of Sirius A. Here is a star of mass roughly that of the sun with a size similar to that of the earth. This means that its density is more than about 106 times the density of the earth!
Type of star
Main characteristic
Main sequence star
Fuses hydrogen into helium
Red giant
Bright, large, cool, reddish, tenuous
Super red giant
Even larger and brighter than red giants
White dwarf
Dim, small, hot, whitish, dense
Neutron star
Dimmer and smaller than a white dwarf with a density similar to that of nuclei. Too dim to show on an HR diagram
Black hole
The end point of the very massive stars.
Table 25.1
CHECK YOURSELF 3
One star appears reddish and another appears bluish. Explain which is the hotter of the two.
Worked example 25.3 shows how we can get information on stellar radii. The key idea is the StefanâBoltmann law that states that L = ÏAT4. Since the area is given by A = 4ÏR2 we find
R=L4ÏÏT4âââââââ
WORKED EXAMPLE 25.3
The HR diagram (Figure 25.4) shows three stars, A, B and C.

Figure 25.4: For worked example 25.3.

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aCalculate the ratio of radii RARC and RBRA.
bEstimate the wavelength at which star C emits most of its light.
Answer
aLALC=108=Ï(4ÏR2A)T4Ï(4ÏR2C)T4=R2AR2C
Hence:
RARC=104
Similarly:
LBLA=1=Ï(4ÏR2B)T4BÏ(4ÏR2A)T4A=(R2BR2A)(300030000)4
Hence:
RBRA=(300003000)2=100.
bFrom Wienâs law: Î» Ã 30000 = 2.9 Ã 10â3 K m and so Î» = 9.7 Ã 10â8 m.
To measure the radius of a star through its temperature and lumininosity assumes we know the temperature and luminosity! Temperature can be measured using Wienâs law. But what about luminosity? We can find it, if we know its apparent brightness (which is easily measured) and the distance. So what about distance? There are various methods for doing that. In this course we will learn about a method that works for nearby stars. It is called the parallax method. It is based on the fact that when you look at an object from two different positions, the object appears to shift relative to the very distant background. You can easily see this: hold up a finger in front of you and look at it with the left eye closed. Now look at it with the right eye closed. The finger looks shifted relative to distant objects.
Figure 25.5 shows two observations of the star from two positions six months apart.
The parallax angle is the angle at which the star subtends a distance equal to the earth orbit radius R.

Figure 25.5: The earth (blue circles) in orbit around the sun. The star is observed from two positions six months apart.

Figures 25.6a and 25.6b show the two images of the star taken six months apart. In Figure 25.6c the two images are put together.

Figure 25.6: The white stars represent the distant background. a The star observed from one position. b The star observed from another position six months later. c The two images are put together to determine the parallax angle. The shift shown here is greatly exaggerated.

Parallax angles are very small. It is convenient to express them in arc seconds (arcsec): one degree contains 60 arc minutes and one arc minute contains 60 arc seconds. So 1Â° = 3600 arcsec. Now suppose we define a new unit of distance, the parsec, pc. One parsec is the distance at which the parallax is 1 arcsec. So if we are told that a star has parallax p in arcsec, the distance in pc is
d(inÂ pc)=1p(in arcsec)
How many metres are there in a pc? From Figure 25.5, we see that
tanp=Rd
The distance d is 1 pc when p = 1 arcsec; R = 1.496 Ã 1011 m and so
1pc=1.496Ã1011tan(13600)â=3.086Ã1016â3.09Ã1016Â m
This is a convenient unit for distance in astronomy. Another is the light year, ly, which is the distance travelled by light in one year: 1 ly = 3.0 Ã 108 Ã 365 Ã 24 Ã 60 Ã 60 = 9.461 Ã 1015 m. It follows that: 1pc=3.086Ã10169.461Ã1015=3.26Â ly
So, a star with a parallax angle of 0.125 arcsec is at a distance of 8.0 pc or 26 ly or 2.5 Ã 1017 m.
The method is restricted to nearby stars. Stars far away have parallax angles that are too small to measure accurately. The limit for earth-bound telescopes is about 100 pc. However, satellite-based systems such as the Hubble space telescope, the JWST (James Webb Space Telescope) and ESAâs Gaia can measure parallaxes to distances up to 3000 pc.
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The protonâproton chain
The main series of nuclear fusion reactions taking place in the cores of low mass main sequence stars is the protonâproton chain:
H11+H11âH12+e+10+Î½00
H11+H12âHe23+Î³00
He23+He23âHe24=2H11
Notice that the deuterium and helium-3 produced in the first and second stages are used up again. In this cycle, the net effect is to turn four hydrogen nuclei into one nucleus of helium. We can see this by rewriting the reaction equations above after having multiplied the first two by 2. Nuclei in the same colour âcancel outâ, i.e. what is produced is used up again:
2Ã11H+2Ã11Hâ2Ã21H+2Ã01e++2Ã00Î½2Ã11H+2Ã21Hâ2Ã32He+2Ã00Î³32He+32Heâ42He+2Ã11H4Ã11Hâ42He+2Ã01e++2Ã00Î½+2Ã00Î³
WORKED EXAMPLE 25.4
Our sun emits energy at a rate (luminosity) of about 3.9 Ã 1026 W.
The mass of the sun = 2.0 Ã 1030 kg.
When the sun was created, 75% of its mass was hydrogen.
Estimate the mass of hydrogen that undergoes fusion in a year.
Assuming that the energy loss is maintained at this rate, find the time required for the sun to convert 12% of its original hydrogen into helium.
(Assume that energy is produced in the protonâproton chain with an energy released per reaction of about 3.98 Ã 10â12 J.
Answer
Since the luminosity of the sun is 3.9 Ã 1026 W, it follows that the number of fusion reactions required per second is
3.9Ã10263.98Ã10â12=9.8Ã1037
For every such reaction, four hydrogen nuclei turn into helium, and thus the mass of the fused hydrogen is:
9.8Ã1037Ã4Ã1.67Ã10â27Â kgÂ sâ1 = 6.5Ã1011Â kgÂ sâ1Â orÂ 2Ã1019Â kgÂ per year.
At the time of its creation, the sun consisted of 75% hydrogen, corresponding to a mass of
0.75Ã2.0Ã1030Â kg=1.5Ã1030Â kg
The limit of 12% results in a hydrogen mass to be fused of 1.8 Ã 1029 kg. The time for this mass to fuse is thus
1.8Ã102â¿¶6.5Ã1011Â s=2.8Ã1017Â s=9Ã109yr
Since the sun has existed for about 5 billion years, it still has about 4 billion years left in its lifetime as a main sequence star.
In stars more massive than our sun the CNO cycle dominates. Its net effect is to convert four hydrogen nuclei into one of helium just like the proton-proton chain. See extension material at the end of the chapter.
Stellar evolution
When about 12% of the hydrogen in the star is used up in nuclear fusion, a series of instabilities develop in the star. These have the result that the delicate balance between the radiation and gravitational pressures is upset. The star will then begin to move away from the main sequence. What happens next is determined mainly by the mass of the star. Other types of nuclear fusion reactions will take place and the star changes in size and surface temperature (and hence colour).
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KEY POINT
The mass of the star is the main factor that determines its evolution off the main sequence.
Figure 25.7 shows a schematic summary of the life history of a star.

Figure 25.7: The birth and death of a star. The star begins as a protostar, evolves to the main sequence and then becomes a red giant or supergiant. After a planetary nebula or supernova explosion, the core of the star develops into one of the three final stages of stellar evolution, a white dwarf, a neutron star or a black hole.

The star is called a protostar when it is first formed. It then settles on the main sequence according to its temperature and luminosity.
A low mass star (this means of mass less than about 12 solar masses) will follow the top path leading to red giant, an explosion called a planetary nebula and the final remnant of a white dwarf. A heavier star will follow the lower path leading to a red supergiant and a more violent explosion called a supernova leaving behind a neutron star or a black hole.
The paths shown schematically in Figure 25.8 can also be shown on HR diagrams.

Figure 25.8: a Evolutionary path of a low mass star. This is the path of a star of one solar mass that ends up as a white dwarf, which continues to cool down, moving the star ever more to the right on the HR diagram. b This is the path of a star of 15 solar masses. It becomes a red supergiant that explodes in a supernova. After the supernova, the star becomes a neutron star, whose luminosity is too small to be plotted on the HR diagram.

As seen in Figure 25.8a, a star will leave the main sequence following an almost horizontal path on the HR diagram. A star more massive than the sun may go back and forth on the HR diagram several times as shown in Figure 25.9; the star is now in the instability region of the HR diagram.
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Figure 25.9: The instability region of the HR diagram.

The delicate balance between radiation and gravitational pressures has been disturbed; the star tries to find a new equilibrium state but instead the star pulsates, i.e. it grows and then shrinks in size. So the star cools and heats up during these pulsations and that affects the rate and type of nuclear fusion reactions taking place. This makes its luminosity variable, and this is a general characteristic of all stars in the instability region.
The Chandrasekhar limit
Low mass stars after ejecting mass into space in the planetary nebula explosion leave behind a core, Figure 25.10. The conditions in the core mean that the electrons behave as a gas, and the pressure they generate is what keeps the core from collapsing further under its weight. This pressure (electron degeneracy pressure) is a quantum effect and beyond the level of this course.
The core has now become a white dwarf star. Now exposed, and with no further energy source, the star is doomed to cool down and will then become what is called a black dwarf.
This pressure is able to stop the further collapse of the core only if the mass of the core is less than a certain limit.
KEY POINT
This is known as the Chandrasekhar limit and is the largest mass a white dwarf star can have. It equals 1.4Mâ.
The limit is named after the astrophysicist S. Chandrasekhar, Figure 25.12, who discovered it in the 1930s.

Figure 25.10: The Helix: a planetary nebula. The star that produced this nebula can be seen at the exact centre.

CHECK YOURSELF 4
What is the minimum mass that must be ejected by a 3 solar mass star so that it ends up as a white dwarf?
The OppenheimerâVolkoff limit
If the Chandasekhar limit is exceeded, electrons will be forced into protons, turning them into neutrons. The star is now made almost entirely of neutrons, and it is still contracting rapidly. The neutrons get too close to each other, and a pressure develops to prevent them from getting any closer. But the neutrons have overshot. They have become too close together, and the entire core will now rebound to a larger size. The rebounding of the core is catastrophic for the star. It creates an enormous shockwave travelling outward that tears apart the outer layers of the star. The explosion that takes place is much more violent than a planetary nebula and is called a supernova, Figure 25.11. The core that is left behind is called a neutron star.
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The core is kept stable by quantum effects similar to those in the white dwarf case provided the mass of the core is not more than the OppenheimerâVolkoff limit. The OppenheimerâVolkoff limit is the largest mass a neutron star can have and equals about 3Mâ.

Figure 25.11: a The Crab nebula, the remnant of a supernova observed by the Chinese in 1054 and perhaps also by Native Americans. It was visible for weeks even during daytime. b This majestically serene bubble of gas is the remnant of a supernova that happened 400 years ago.

If the original mass of the star is very high (M > 40Mâ), nothing can stop further collapse and the core will become a black hole. Black holes are predicted by Einsteinâs theory of general relativity.
CHECK YOURSELF 5
A main sequence star increases its luminosity when it becomes a red supergiant even though its temperature decreases. How is this explained?
