PC3692: Physics of Stellar Structure (and Evolution)

The goal of this course is to apply fundamental physics laws to understand the internalstructure and evolution of stars. In this course, you will employ virtually all the physics you havelearned so far, in particular thermal physics, quantum mechanics and nuclear physics. You willunderstand how energy is generated inside stars and transferred outwards. In short, you willunderstand how stars shine. The lectures will be divided into six chapters.

1. Observed properties of stars

2. Equations of stellar structure

3. Equation of state

4. Energy transport in stars

5. Nuclear fusion in stars

6. Applications of stellar equations

1. Observed properties of stars

1.1. Why do we study stars?

• They are beautiful and seemingly eternal

Twinkle, twinkle, little stars,how I wonder what you are,Up above the world so high,like a diamond in the sky– Jane Taylor (1806)

A sense of wonder is universal in all civilisations. We wonder what the stars are, as in thepoem. Although we have observed stars for millions of years, they are really physicallyunderstood only in the 20th century, because, as we will see in the course, it requiresknowledges of nuclear physics. Such information has only become available in the 20thcentury.

However, the long history in astronomy means that some of the terminologies in astronomyare peculiar, such as apparent magnitudes and metallicities.

• Stars, in particularly the Sun, plays a crucial role in our lives

Stars → nuclear reaction → Energy+weather (seasons) → life

Stars → synthesise elements (C, O, N) → found in air and our human bodies

• Stars are also interesting astrophysically for at least three reasons:

1. Stars → star clusters → galaxies → universe. Stars to the cosmology is like as atoms inphysics. To understand the whole universe, it is crucial to understand stars.

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2. Stars provide extreme conditions for studying physics. The high density, high temperatureand high pressure are usually not found on Earth. White dwarfs, neutron stars are extremelydense and are a good testing ground for theories such as general relativity.

3. Stars are fascinating for yet another reason: while the general principle of stars iswell-understood, there are still many puzzles in stars, such as the solar neutrino problem,star formation and convection; the latter two are long-standing unsolved problems inastrophysics.

1.2. Stellar parameters

How do we describe stars? In fact, the internal properties of stars can be primarily describedby just a few parameters, M,R, T and chemical composition (the fractions of different elements,such as H, He, inside a star), and a related concept luminosity, L.

The stars also have several important external parameters, such as distance and their motionsin space. The distance measurement is essential for determining the internal properties of thestars, so we discuss it first.

1.2.1. Distances

Th distance of stars can be measured with a number of different methods. The mostfundamental one, parallax, is also the simplest; it is a purely geometric method (see Fig. 1.2.1).

The viewing angle of the stars from Earth changes due to the Earth motion around the Sun.The parallax angle is usually quite small. The nearest star has a parallax of about 1 arcsecond.The distance is then simply

d = 1AU/π, AU = 1.5× 1011m (1)

When the parallax is one arcsecond, the corresponding distance is called parsec (pc). Since 1radian=180/π × 60× 60 ≈ 2× 105 arcsecond. This means

1 pc = 2× 105 AU ≈ 3× 1016m (2)

Fig. 1.— Schematic diagram of parallax. The parallax angle is usually denoted as π (labelled as ρ

in the figure)

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1 pc ≈ 3.26 ly (3)

Example 1.1 A star at 500pc will have a parallax of 1/500=0.002 arcsecond.

The recent satellite Hipparcos can measure parallax as accurate as 0.002 arcsecond, so we canroughly the star measure distances out to few hundred pc. Future satellite missions, such as GAIAwill be able to measure parallax to even higher accuracies, hence they can measure distances toall bright stars in our Galaxy.

Astronomers also frequently use kpc=103pc, Mpc=106pc, Gpc=109pc.

1.2.2. Luminosity of stars and related concepts

Stars radiate and we see the photons from them. The luminosity of a star is defined as theenergy radiated per second,

Luminosity =energysecond

(4)

Suppose we have a solar panel with perfect efficiency receiving light from stars like the Sun, thenwe can measure how much energy is received per square meters per second, which is usually calledflux:

flux =energy

second m2(5)

The luminosity is related to the flux by

L = 4πd2f, (6)

where d is the distance to the star. And so if we know the distance to a star, then from theobserved flux, its luminosity can be determined.

For our sun, L� = 3.8 × 1026W , including photons in all wavelengths. The solar energyradiated in the “visual” part of the electromagnetic radiation, L�,V , (i.e., at a wavelength of about5500A) is about 4.64× 1025W , while that for the blue (‘B’) part of the electromagnetic radiation(i.e., λ ∼ 4000A) is about 4.67× 1025W .

In astronomy, for historical reasons, luminosity is frequently measured in so-called absolutemagnitudes. For example, the B-band and V-band magnitudes of the Sun are ‘defined’ asM�,B = 5.48 and M�,V = 4.83, respectively. For other stars, we have

MB = M�,B − 2.5 log(LB/L�,B) = 5.48− 2.5 log(LB/L�,B) (7)

MV = M�,V − 2.5 log(LV /L�,V ) = 4.83− 2.5 log(LV /L�,V ) (8)

Example 1.2 If LB = 100L�,B, then we have

MB = 5.48− 2.5 log(100) = 0.48 (9)

Note that the brighter a star is, the smaller its absolute magnitude!

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The apparent brightness of a star obviously depends on its distance to us, the closer a staris to us, the brighter it appears. Astronomers therefore use the so-called apparent magnitude todescribe the brightness of a star.

The apparent magnitude of a star is related to the absolute magnitude by

mB = MB + 5 log(d/10 pc), (10)

mV = MV + 5 log(d/10 pc), (11)

mB and mV are also commonly written as B and V . One can show that the apparent magnitudesare directly related to flux, and therefore it can be measured without knowing the source distance.The absolute magnitude (luminosity) of a star, in most cases, can only be determined if we knowthe distance to the source.

Example 1.3 The distance to the Sun is d = 1AU, so

B = 5.48 + 5 log(1 AU/10 pc) = −26.08 (12)

Again, the brighter a star is, the smaller its B magnitude is.

1.2.3. Mass and Radius of stars

Masses and radii of stars are difficult to measure. Most measurements are from studies ofbinary stars; most stars (∼> 80%) are in binaries. In this subsection, we discuss the procedure fora special class of binaries, the so-called eclipsing binaries.

In eclipsing binaries, the brightness declines periodically as one stars passes in front of theother. Of course, for an eclipse to occur, the orbit of the binary must be nearly edge-on. Fig. 3shows one sample geometrical configuration. The principle of mass determination can be seen theclearest if the orbits are circular, which we will assume in the following discussion.

From the observations, we can determine the orbital velocities from Doppler shifts, v1 and v2.From these two velocities, we can measure the mass ratio.

Fig. 2.— Schematic orbits of a binary in circular orbits. Various quantities are defined. In generalthe orbits will be elliptic.

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Fig. 3.— Schematic diagram showing the orbits and corresponding light curves of eclipsing binaries.

M1

M2=

v2

v1→ M1

M2(13)

Exercise 1.1 Treat the solar system as a binary with only the Sun and Jupiter. Estimate thevelocity of the Sun around the centre of mass. Assuming the measurement accuracy in velocity is5 m s−1, can the solar motion be detected? From this, discuss what kind of extrasolar planets arethe easiest to be detected in terms of their distances to the parent star and their masses. Compareyour answer with the observed ones at http://exoplanets.org/

Now from the observed period, P , and the two velocities, we can infer the two orbital radii,a1 and a2, and the separation, using

v1 =2πa1

P, v2 =

2πa2

P, a = a1 + a2, (14)

so we can infer the separation, a.

From Kepler’s third law, we have

P 2 =4π2a3

G(M1 + M2)→ M1 + M2 (15)

So eqs. (13) and (15) can be combined to determine the two masses.

Clearly the time duration of the eclipses is related to the stellar radii. Only a detailed analysisof the eclipse shape and duration can give the stellar radii. In fact, the exercise is often quiteinvolved. Notice the depth of the eclipse is related to the relative brightness of the two stars.

Let us suppose a low-mass, tiny star rotates around a very massive giant star, as depicted inFig. 3. The separation between the stars is assumed to be much larger than both stellar radii.The massive star is essentially stationary. Let us denote the relative rotation velocity of the less

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massive star around the giant companion as v; in our case, v = v1 + v2. For this special case, thetotal duration (t) of the eclipse will be related to the period by

t

P=

2R1 + 2R2

2πa, (16)

while the duration of the descent from the maximum light to the minimum light is related to thediameter of the small star by

t1 =2R1

v(17)

We can determine t1 and t2 from the observed light curve, and combined with the measuredvelocity v, we can find the stellar radii for both stars from the previous two equations.

1.2.4. Effective temperature and Chemical Composition

The energy distributions of a star as a function of wavelength (or frequency) is called itsspectrum. The spectrum can be regarded as the finger prints of the star. The spectra of stars canbe crudely approximated as a blackbody. In fact, in astronomy it is common to define an effectivetemperature of a star. This is the temperature that a star would have if it were a perfect blackbody and had identical luminosity and radius as the real star.

Now the energy per unit time per unit area radiated by a blackbody is given by

f = σ T 4eff (18)

Therefore, we haveL = 4πR2 × σ T 4

eff (19)

So the radius of a star, the effective temperature and luminosity are related, and if we can measuretwo of them, then we can determine the remaining third parameter.

Fig. 4 illustrates the spectrum of the Sun. The top panel shows a low-resolution spectrumwhile the right panel shows a high-resolution spectrum between 4660Aand 4990A. As one can see,the solar spectrum is approximately a blackbody of temperature 5770K. The absorption lines inthe spectra are caused by photons being absorbed by atoms in the stellar atmosphere when theyemerge from the solar interior. The number and depths of the absorption lines depend on theelement abundances, i.e., the chemical composition. Detailed analysis of stellar spectra can yieldthe effective temperature, the chemical composition of stars and ‘surface gravity’ (= GM/R2).

The effective temperature of a star can also be crudely measured using the so-called colourindex. A colour index is simply the magnitude difference in two colours (or filters), e.g., B − V .Note that the colour index is independent of the source distance. Very hot stars are blue, whichmay have B − V = −0.3, while cooler stars are red and may have B − V = 2.

Example 1.4 For the Sun, the B − V colour index is B − V = MB −MV = 5.48− 4.83 = 0.65

Stars can be divided into various classes according to their spectra. The sequence is historicallytermed as O, B, A, F, G, K, M, R, N1. Each type corresponds to a certain range of effective

1To be remembered as the initials of Oh, Be A Fine Girl (Guy), Kiss Me, Right Now

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Fig. 4.— The solar spectrum in low-resolution (top) and in high-resolution (bottom). In the toppanel, the smooth dashed curve is a blackbody with a temperature of about 5770K. There aremany dips in the spectra because photons are absorbed by atoms in the stellar atmosphere. Thecontinuum in the high-resolution spectrum has been renormalised to unity. It resolves many moreabsorption lines. A careful analysis can yield many important stellar parameters, such as the Sun’schemical composition, the effective temperature and the ‘surface gravity’ (= GM/R2).

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temperature and colour index. From O stars to M stars, they become cooler (temperature from40,000K drops to 2,800K) and redder (with B− V changes from −0.35 to 1.6). In comparison, thehottest white dwarfs have Teff ≈ 3× 105K, while the hottest neutron stars have Teff ≈ 3× 107K.

1.3. Range of stellar parameters

Now we know how to measure the stellar parameters. What are the typical values of thestellar parameter values?

Parameter Sun StarsRadius R� = 7× 108m 10−2 − 102R�Mass M� = 2× 1030 kg 10−1 − 102M�Teff Teff,� = 5770K 103 − 105 KLuminosity L� = 3.8× 1026W 10−5 − 106 L�Chemical composition Z� = 0.02 ∼ 10−3 − 5 Z�

Comments:

• Sun is a typical star, with intermediate parameter values.

• Mass has a lower cutoff and upper cutoff. This is because lower mass objects can not initiatenuclear burning while higher mass stars are unstable.

• The luminosity range varies by about 11 orders of magnitude, much larger than any otherparameters. Some stars are extremely luminous, while others are very faint. We will learnwhy this is the case.

1.4. Properties of Stellar Populations

So far, we have discussed how to measure the parameters for individual stars. In the following,we will discuss the statistical properties of stellar populations. We can often gain insight into theastrophysics by examining a large sample at the same time, rather than individually. We shall alsodiscuss what makes stars to have different properties, such as their temperatures and luminosities.

Spatial distribution of stars

Stars in the universe are not uniformly distributed in space. They are highly clustered. It isstill not completely understood how this highly clustered pattern arises.

Milky Way is a typical disk galaxy (see Fig. 5). It consists a bulge and a thin disk. Thedisk is thin because it’s rotationally supported, like an audio CD. You also see black lanes in thegalaxy, these are caused by dust obscuration.

The diameter of our luminous Milky Way is about 30 kpc. The Sun is about 8 kpc off center.In the picture you can also see some star clusters, called globular clusters. They are spherical

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distributed around the Galactic center. They consist off 105 − 106 stars, and they play animportant role in our understanding of stellar structure and evolution.

1.4.1. Mass-Luminosity relation

The first relation we examine is that between the mass and luminosity (obtained for nearbystars). This is shown in Fig. 6. In this figure, both the luminosity and mass are plotted onlogarithmic scales.

Fig. 5.— An image of the Milkyway in the optical (top panel). The bottom panel shows theapproximate linear scales (in light years) of the Milky Way. Notice that the globular clusters areroughly spherically distributed around the Galactic centre.

Fig. 6.— The Mass-luminosity relation for nearby stars. The scales on both axes are logarithmic.

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The data points lie approximately on a straight line. This line is approximated by 2

L ∝ M3.5 (20)

Since the mass of stars vary from 0.1 to 100 solar masses, we immediately see that the luminosityof main sequence stars will vary by about 10 or 11 orders of magnitude.

This also has profound consequences for the life times of stars. The total energy available fora star is, from Einstein’s famous formula, proportional to E = MT 2. We therefore have

t ∝ Mc2

L∝ M

M3.5∝ M−2.5 (21)

A more massive star has a larger energy reservoir, however, it also consumes its energy at a muchfaster rate. As a result, the life time of a massive star is shorter.

More precise modelling gives

t ≈ 1010 yr(

MM�

)−2.5

(22)

For the Sun,t� ≈ 1010 yr (23)

For a star with mass below the Sun, the life is then comparable to the age of the universe,tuniverse ≈ 1.3 × 1010 years. In other words, very low mass stars evolve very slowly, and they donot evolve beyong hydrogen burning.

However, for a star with M = 10M�, we have

t ≈ 1010 × 10−2.5 = 107.5 = 3× 107 yr � tuniverse (24)

Since the life time is much shorter than the age of the universe, so these stars could not haveformed at the beginning of the universe. Instead, they must have formed very recently, about tensof millions years ago. So stars are not eternal! they are forming and dying today. In fact, they arejust like human beings, they have births, they grow, evolve, and eventually die.

Unfortunately, how a star is borne is not yet well understood. Star formation is currently oneof the most active research areas in astrophysics.

1.4.2. Hertzsprung-Russell diagrams

Hertzsprung-Russell diagram for the solar center

In Fig. 7, we show the Hertzsprung-Russell diagram for 41453 nearby stars with accuratedistances measured by the Hipparcos satellite. The Hertzsprung-Russell diagram is also oftencalled HR digram or colour-magnitude diagram.

In the figure, the horizontal axis plots V − I, a colour index similar to B−V that we discussedin §1.2.4 The stars become hotter and bluer as you move from the right to the left. The verticalaxis plots the absolute magnitude in the Hipparcos filter. The stars become more luminous as youmove from the bottom to the top.

2Rigorously, this relation only applies to massive stars; for low-mass stars, this relation has to be modified.

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Fig. 7.— HR diagram for 41453 nearby stars with accurate distance measured by the Hipparcossatellite. The horizontal axis is the V − I colour index, while the vertical axis is the absolutemagnitude in the Hipparcos passband. The I-band is a filter centred around 8000A.

One striking feature is there is a sequence of stars running from the top left to the bottomright. This sequence is called main sequence.

You also see a clump of to the right of the main sequence, these stars are called red clumpstars, and the stars further to the right, red giants.

You can also vaguely see some stars in the bottom left; these are white dwarf stars, they arehot and very faint.

90% of stars belong to the sequence, and it appears that one single parameter, M , determinesthe position of a star on the HR diagram. There are about 10% of stars that do not belong tothe main sequence. One important goal of this course is to understand why most stars are on themain sequence, while other stars do not, and what are the physical differences between differenttypes of stars.

HR diagram for globular clusters

We already discussed that stars are not uniformly distributed in space, they are highlyclustered. Globular clusters are very dense star clusters with 105 − 106 stars with a size of a fewpc. In Fig. 8 we show one example, M5. The stars in such clusters move with typical speeds of∼ 5 km s−1. It is yet not well understood how globular clusters form.

The right panel shows the HR diagram for M5. The x-axis now shows the colour index B− V

while the vertical axis shows the apparent V magnitude. Again the temperature increases fromthe right to the left while the luminosity increases from the bottom to the top.

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Fig. 8.— Image (left) and HR diagram (right) for globular cluster M5.

There are a few key features, these are main sequence (labelled as A), red giant branch(labelled as B), and then the horizontal branch stars (labelled as D). The horizontal branch starsare so named because they are nearly horizontal in the HR digram. Notice in a globular cluster,we do not have very hot and luminous stars.

The stars in (most) globular clusters are thought to have formed at the same time and a longtime ago, in fact, about 10 billion years ago, about the age of the universe. They are also thoughtto have the same chemical compositions.

The fact that they are old offers a crucial clue to understand their HR diagram. As we haveseen, the high mass stars are luminous and blue, so they have short life times. These stars wouldhave disappeared (more precisely, evolved away) from the main sequence. In fact, as we willdiscuss, these stars evolve to form the red giant and horizontal branch stars in the HR diagram.

A key goal in stellar astronomy is to understand the fine features in the HR diagrams ofglobular clusters. Different globular clusters have different HR diagrams, and these differences arethought to arise mainly from the difference in ages. So a detailed study of the HR diagram inglobular clusters can also be used to determine the age of these clusters.

HR diagram for open clusters

There are also another kind of cluster of stars in galaxies, these are called open clusters. Theyhave only hundreds of stars. They are a few pc across, similar to the size of globular clusters. Fig.9 shows an example, the Pleiades open cluster. As you can see from this figure, there are bluishstars in the figure.

The right panel shows the HR diagram for this cluster. This is very different from the globular

cluster M5, nearly all stars are on the main sequence, including some hot and blue giant stars atthe top left.

Fig. 9.— Image (left) and HR diagram (right) for the open cluster Pleiades.

Open clusters are, like globular clusters, thought to have formed essentially at the same timeand with the same chemical composition.

Why are stars in open clusters lie essentially on the main sequence? The key of understandingis very simple: they are thought to be young. So they still do not have time to evolve off the mainsequence. That’s why all the stars essentially still lie on the main sequence.

1.4.3. Summary

We have studied the mass-luminosity relation and HR diagrams for the solar neighbourhood,open and globular clusters. HR diagrams play an essential role in the study of stellar structureand evolution, and in fact also for cosmologies. So a clear understanding of the basic features inHR diagrams is very important.

Problem set 1

1. The binding energy per nucleon for 56Fe is 8.8 MeV per nucleon. Estimate the total energyreleased per kilogram of matter by the sequence of reactions which fuse hydrogen to iron.

2. The main sequence of the Pleiades cluster of stars consists of stars with mass less than 6M�;the more massive stars have already evolved off the main sequence. Estimate the age of thePleiades cluster.

3. Given that the luminosity of the sun is 3.8 × 1026W and that the absolute bolometricmagnitude of the sun is M ≈ 4.7, estimate the distance at which the sun could just be seenby the naked eye (The naked eye can detect a star of apparent magnitude 6). The apparentmagnitude, m, is related to the absolute magnitude, M , by

m = M + 5 logd

10pc. (25)

Estimate the number of photons incident on the eye per second in this situation.

4. Sketch the HR diagram for an old globular cluster, label the main sequence, the red giantand horizontal branch. (2001 final exam, 3 marks)

Write down the approximate observed range of stellar masses. What imposes the lower andupper limits on the mass of stars? [2001 final exam, 5 marks]

Using the observed mass-luminosity relation for nearby stars, argue stars must be formingat present day. [2002 final exam, 5 marks]