RSS feed for White dwarfs and neutron stars

Introduction

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

In this course, you will explore the exotic end-points in the lives of most stars – namely white dwarfs and neutron stars. To do this we consider the properties of so-called degenerate matter and how these properties govern the later stages of stellar evolution. To set the scene, we begin by outlining the life cycle of stars in general, before moving onto consider their end-points. We next look at planetary nebulae and white dwarfs, which are the end points of low- and medium-mass stars, and then see how high-mass stars end their lives as supernovae, and may produce neutron star remnants.

This course shall refer to masses, radii and luminosities of stars in terms of the mass, radius and luminosity of the Sun (represented by M_☉ = 1.99 × 10³⁰ kg, R_☉ = 6.96 × 10⁵ km and L_☉ = 3.83 × 10²⁶ W respectively) and will sometimes refer to energies in the alternative units of electronvolts (where 1 eV = 1.60 × 10^-19 J, with 1 keV = 10³ eV and 1 MeV = 10⁶ eV for convenience).

This OpenLearn course is an adapted extract from the Open University course S384 Astrophysics of stars and exoplanets.

Learning outcomes

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

After studying this course, you should be able to:

appreciate how stars end their lives
understand how planetary nebulae and supernova remnants form
calculate the physical properties of white dwarfs and neutron stars
understand the properties of electron-degenerate and neutron-degenerate matter
calculate the energetics of supernovae explosions.

1 The life cycle of stars

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

Stars form from clouds of gas that collapse under the influence of gravity until their central temperatures and pressures become high enough for nuclear fusion to be initiated in their cores. At this point, the clouds stop contracting further and are maintained in hydrostatic equilibrium with a balance between the inward force of gravity and the outward force due to their internal pressure. Stars are typically initially composed of about 75% hydrogen and 25% helium, with small traces of other elements.

Stars are observed with a wide range of masses, ranging from around one-tenth that of the Sun to over a hundred times the Sun’s mass. Other characteristics of stars include their radii, surface temperature and luminosity. Usually it is only stars’ surface temperatures (characterised by their colour) and luminosities (characterised by their brightness) that are directly observable. When these two quantities are plotted against each other for a population of stars, on a so-called Hertzsprung–Russell diagram (see Figure 1), certain patterns emerge in the stars’ locations.

Figure 1 A schematic H–R diagram indicating the locations of four regions where most stars are found. The vertical axis shows stellar luminosity in solar units or absolute visual magnitude; the horizontal axis shows surface temperature in kelvins, spectral type, or colour index defined as the difference between B and V magnitudes. (Notice that temperature increases to the left.) Diagonal lines of constant stellar radius are also shown and some notable named stars are indicated.

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The figure illustrates a schematic H–R diagram indicating the locations of four regions where most stars are found. The horizontal axis has three ranges of information: (1) T in kelvin, (2) spectral type, and (3) colour index B minus V. The following points are marked on the horizontal axis, from left to right in logarithmic spacing at unequal distances. For T in kelvin: 40000, 30000, 20000, 10000, 5000, 3000, and 2000. For spectral type: O (about 30000), B (between 20000 and 10000), A (below 10000), F (between 7000 and 6000), G (about 5000), K (below 5000), M (above 3000), and L (above 2000). Colour index: negative 0.4 (between 40000 and 30000), 0 (at about 10000), and 1.5 (above 3000). The vertical axis is labelled M subscript V and ranges from 15 (at the bottom) to negative 10 (at the top) in decrements of 5 units, and L divided by L subscript solar and ranges from 10 raised to the negative fourth power (at the bottom) to 10 raised to the sixth power (at the top) in increments of 100 units. Diagonal parallel lines of constant stellar radius labelled 0.001 R subscript solar, 0.01 R subscript solar, 0.01 R subscript solar, 1 R subscript solar, and 10 R subscript solar from the vertical axis move toward the bottom right until the horizontal axis or the right vertical axis. The main-sequence region starts from the top left end and moves toward the bottom right end. The following stars are marked on the main sequence from the top to bottom: Zeta Puppis, Achernar, Regulus, Sirius A, Procyon A, Sun, 61 Cygni A, 61 Cygni B, Barnard’s Star, Proxima Centauri, and Wolf 359. The white dwarfs region is in the bottom left corner of the graph. Some of the white dwarfs are Sirius B (top left), Procyan B (bottom right), and Van Maanen 2 (bottom right of Procyan B). The red giants are on the top right. Some of the red giants (from bottom to top) are Pollux, Arcturus, and Aldebaran A. The supergiants region is at the top of the graph. Some of the supergiants are as follows: Rigel A (top left), Betelgeuse (top right), and Canopus (below and between Rigel A and Betelgeuse). The stars in the region between 10000 K and 7000 K are bluish-white. Moving to the right, the colours become white (around 5000 K), yellowish-white (between 5000 K and 3000 K), and orangey-white (between 3000 K and 2000 K).

Figure 1 A schematic H–R diagram indicating the locations of four regions where most stars are found. The vertical axis shows stellar ...

Most stars are found in a band running from the top left to bottom right of the H–R diagram, known as the main sequence. This is where all stars spend most of their lives as they undergo nuclear fusion in their cores, converting hydrogen into helium, and releasing energy. On the main sequence, the hottest, most luminous stars sit at the top left; these stars are the most massive and the largest. The coolest, least luminous stars sit at the bottom right; these stars are the least massive and the smallest. The most massive stars have relatively short main-sequence lifetimes of only a few tens of millions of years, whilst the least massive ones have main-sequence lifetimes of hundreds of billions of years (longer than the current age of the Universe).

When most of the hydrogen in the core of a star is depleted, its outer layers will expand and the star will move towards the right of the main sequence, such that its surface appears cooler. This is the region known as the red giant branch. To begin with, hydrogen fusion occurs in a shell around the inert helium core, but as the central temperature and pressure rise further, helium fusion is initiated in the centre of the star. The star then moves back down to somewhat lower luminosities and hotter temperatures, at the lower end of the red giant branch. This new fusion process converts helium into carbon and oxygen in the star’s core, again releasing energy. Low mass stars (< 3 M_☉) and intermediate mass stars (in the range 3–8 M_☉) ultimately begin a second ascent of the red giant branch (known as the asymptotic giant branch, or AGB) when core helium is exhausted and helium fusion only continues in a shell surrounding an inert carbon/oxygen core.

Massive stars (> 8 M_☉) leave the upper main sequence when their core hydrogen is exhausted and then track back and forth along the supergiant branch as new sources of fusion in turn become active and are then depleted. Unlike lower mass stars whose fusion ends with carbon and oxygen, they can initiate further fusion reactions, converting carbon and oxygen into elements such as neon, magnesium, silicon and ultimately iron. They will end up with an ‘onion-like’ structure with shells of less massive elements surrounding shells of heavier elements towards the core. Once the core is composed of iron, no further fusion reactions are possible as reactions to produce nuclei heavier than iron require more energy to be put in than is released.

The subsequent evolution of stars once all nuclear reactions cease is the main topic of this course, which will be explored in the following sections.

2 Planetary nebulae

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

As a low-mass or intermediate-mass star ascends the AGB, mass loss from the star becomes more and more significant. A strong stellar wind may develop in which a substantial fraction of the mass is carried away from the star. This usually corresponds to most of the hydrogen envelope! One consequence of extreme mass loss can be the formation of a planetary nebula, described below. Another effect is that even intermediate-mass stars, which had main-sequence masses in the range 3–8 M_☉, end up with a mass < 1.4 M_☉. The significance of this final mass will become clear when white dwarfs are discussed later.

The mass lost from stars at the end of their AGB phase is not immediately visible in optical light, because dust grains in the cool ejecta obscure the star (although it may be seen in the infrared). The underlying helium-rich star – with a carbon and oxygen core – gets smaller and hotter, and a fast, radiation-driven stellar wind develops. Once the central star’s surface temperature reaches ~ 10 000 K it ionises the ejected envelope, which is then seen as a planetary nebula, expanding at ~ 30–60 km s^-1. (Note: the term planetary nebula is a misnomer originating from when these objects were first observed as fuzzy planet-like objects through 18th century telescopes.)

Figure 2(a) Hubble Space Telescope (HST) image of the bipolar planetary nebula NGC 6302. (b) A composite of data from the Chandra X-ray Observatory and HST showing the multipolar planetary nebula NGC 6543. (c) A composite view of the AGB star L₂ Puppis in visible light (blue) and radio (orange) revealing a giant planet shaping the outflow from the star. (d) ALMA radio observations reveal a spiral structure in the material around the AGB star R Sculptoris.

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The figure consists of four parts labelled (a) to (d) that illustrate images of different astronomical sources. Part (a): Image of the nebula NGC 6302 with left and right lobes of a central star. Part (b): The multipolar planetary nebula NGC 6543 with two axes of symmetry and a blue cloud at the centre of a reddish brown cloud at the intersection of the axes. Part (c): A composite view of the AGB star L2 Puppis in the visible light (blue). A red giant is seen as a yellow dot at the centre between top and bottom light regions labelled ‘plume’. On the left of the red giant is a small reddish patch. On the right of the red giant is a larger bright orange patch labelled ‘ALMA candidate planet’. Part (d): A reddish brown spiral in the material around the AGB star R Sculptoris, as viewed tunnelling through the spiral.

Figure 2(a) Hubble Space Telescope (HST) image of the bipolar planetary nebula NGC 6302. (b) A composite of data from the Chandra X-ray ...

The majority of planetary nebulae are asymmetric, with a small fraction being largely spherical or elliptical. Most are bipolar (with lobes either side of a central star) or multipolar (with at least two axes of symmetry), as shown in Figures 2a and 2b. The origin of these asymmetries has been the subject of considerable debate over recent decades. However, recent observations with the ALMA radio telescope and the SPHERE coronagraphic adaptive optics system on the Very Large Telescope (VLT) show that many of the asymmetries in planetary nebulae derive from structures that first formed while the star was on the AGB. They appear to be the result of orbiting companions, either a binary companion star or a massive planet (see Figure 2c). These can either transfer angular momentum to the ejecta and form a constraining torus, leading to a bipolar nebula, or create an orbital motion of the AGB star around the system’s centre of mass, giving rise to a spiral pattern in the ejecta (see Figure 2d).

The physical diameter of a planetary nebula can be derived from its angular extent and distance. Because the expansion velocity of the nebula gas can be measured by its Doppler shift, the duration of the expansion – and hence of the post-AGB phase – can be calculated. By extrapolating the observed expansion of the nebula backwards, it can be shown that the whole evolution of a planetary nebula lasts only about 20,000 years.

The star at the centre of the planetary nebula continues to get smaller and hotter. Its hydrogen-burning shell is extinguished when it gets too close to the surface and hence too cool, and the same happens eventually to the helium-burning shell. What remains is a hot carbon-and-oxygen core, surrounded by thin shells of helium and hydrogen, with a surface temperature of ~ 100,000 K. All nuclear burning has been extinguished by this stage, and there is no prospect of thermonuclear reactions being re-ignited. Observations show that, although the nebula continues to expand, the central star now begins a long slide to lower luminosity and temperature at almost-constant radius. The star has become a cooling white dwarf.

3 Degeneracy

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

Before considering the physics of white dwarfs, let’s pause to look at the concept of degeneracy. In this context, degeneracy means that there is more than one quantum state with the same energy.

In quantum mechanics, all particles display wave-like behaviour. In particular, any particle has a characteristic wavelength, known as the de Broglie wavelength, defined by

Equation label: (1)

where h is Planck’s constant ( or ) and p is the particle’s momentum.

Now, inside a star, a particle’s kinetic energy can be written as , where T is the temperature and k_B is Boltzmann’s constant (). Another expression for the kinetic energy of a particle (with mass m and speed v) is simply . So equating these two we obtain or . (Strictly, particles exist with a range of speeds, but is close to the average.) The momentum of a particle is given by , so . Hence the de Broglie wavelength of a particle may be written as

Equation label: (2)

Degeneracy becomes relevant when particles are packed so closely together that their separation is similar to their de Broglie wavelength. It turns out to be important at various stages in the lives of stars. Firstly, it determines the lower mass limit for stars: below a certain mass (about 0.075 times that of the Sun) matter in the star’s core becomes degenerate as the star collapses and hydrogen fusion cannot begin. Such a star will instead become a brown dwarf. Secondly, degeneracy also determines how helium fusion begins: for stars with a mass below about 2.25 times that of the Sun, after hydrogen fusion finishes in the core, matter in the star’s core becomes degenerate and helium fusion begins explosively in an event called a helium flash. Both these phenomena are therefore due to degeneracy. This concept will now be explored in detail as we consider the compact remnants left behind at the end of a star’s life.

3.1 The de Broglie wavelength of electrons and nucleons

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

Let’s begin by thinking in terms of the quantum mechanical properties of particles. A volume of space containing a gas may be considered to be occupied by particles whose wave properties give rise to standing waves that fit neatly within that volume. They therefore have quantised values of wavelength, and also of energy and momentum. This quantisation means that there is a discrete (rather than continuous) distribution of possible particle energies.

The number of possible quantum states available to particles below some energy is finite; often it is very large, but nevertheless it is finite. In the very dense interiors of some astronomical objects, such a finite number of available quantum states, though large, can be insufficient for the huge number of particles squeezed into the confined volume. Under these circumstances, physics then takes new twists, giving rise to some remarkable properties.

The critical point of having too few available quantum states for the number of particles present is reached when the separations of the gas particles become smaller than their de Broglie wavelength. At that point, the wave properties of the gas dominate its classical properties, and a quantum gas is said to exist. The following activity explores how the de Broglie wavelengths for protons and electrons compare in the core of the Sun.

Activity 1

a.The de Broglie wavelength of non-relativistic particles is given by Equation (2) as where m is the particle’s mass, k_B is Boltzmann’s constant () and T is the temperature. Calculate the de Broglie wavelength, in the core of the Sun at a temperature of , of:
- i. a proton (with mass kg)
- ii.an electron (with mass kg)
b.Which has the larger de Broglie wavelength, and by what factor?
c.How does this ratio vary from star to star?

Discussion

a.(i) For a proton in the core of the Sun:
(ii) For an electron in the core of the Sun:
b.The ratio of their de Broglie wavelengths is therefore:
i.e. the de Broglie wavelength of the electron is greater than that of the proton, by a factor of about 40.
c.From part (b), this ratio depends only on the mass of the particles, and hence is independent of the environment and hence of the temperature. The electron’s wavelength is about 40 times longer than the proton’s wavelength in all stars.

Note that neutrons have a similar de Broglie wavelength to protons because both types of particle have similar mass. Since the de Broglie wavelength of electrons is roughly 40 times longer than that of nucleons (i.e. protons or neutrons), then as the stellar core contracts and its density increases, electrons run out of space and hence reach the quantum limit corresponding to degeneracy sooner than nucleons.

3.2 Conditions for degeneracy

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

Degeneracy can be described in two equivalent ways:

1. The separation between particles is less than their de Broglie wavelength,

Equation label: (3)

2. The number of particles per unit volume, n, is greater than the number of available quantum states n_Q known as the quantum concentration,

Equation label: (4)

Before going further, complete the following activity to confirm to yourself that these two conditions are equivalent.

Activity 2

Beginning with the second statement of the condition for degeneracy, , use the definition of the quantum concentration to derive an expression for the mean separation of the particles, in terms of the de Broglie wavelength. Show that this leads to the first statement for the condition of degeneracy.

Hint: if the mean separation of particles is , then the number of particles per unit volume is .

Discussion

The second condition is , i.e. .

Because , substituting gives .

Taking the (1/3)-power and rearranging the result gives

That is, from the second degeneracy condition, , we obtain the first degeneracy condition, .

You have now shown that the two conditions for degeneracy are actually equivalent; they are merely alternative expressions of the same physics. In the following activity, you will see how a third such expression is derived, which is also equivalent.

Activity 3

a.Rewrite the second description of degeneracy, , to find a limit for the thermal energy . State a third, equivalent condition for degeneracy as a limit on the temperature.
b.Calculate values of the quantity in the core of the Sun for (i) protons and (ii) electrons. Assume that the mass fractions of hydrogen and helium are and that the core density of the Sun is . Note that in the solar core, all atoms are fully ionised and that the number density is for each type of particle.
The mass of an electron is kg, the mass of a hydrogen nucleus is kg and the mass of a helium nucleus is where u is the atomic mass unit given by kg. As usual Planck’s constant is and Boltzmann’s constant is .
c.Consider the results to part (b), and state whether (i) protons and (ii) electrons are degenerate in the Sun’s core. Assume that the core temperature of the Sun is .

Answer

a.Since , so the degeneracy condition implies that .
Taking the (2/3)-power and multiplying both sides by gives , i.e. .
The third, equivalent condition for degeneracy, in relation to temperature, is: the gas is degenerate if its temperature .
b.(i) For protons, (i.e. hydrogen nuclei per m³).
Therefore, for protons, the temperature-related degeneracy condition is
(ii) In the solar core, all atoms are ionised. The electrons are provided by the hydrogen (1 electron per atom) and helium (2 electrons per atom), which each account for 0.5 of the composition by mass. So
Therefore, for electrons, the temperature-related degeneracy condition is
c.The temperature condition for degeneracy is . Since , so the temperature in the core of the Sun is (i) much too high for proton degeneracy to have set in, and (ii) marginally too high for electron degeneracy to have set in.

You have seen in the previous activity that a third equivalent expression for the condition of degeneracy can be written as follows:

Equation label: (5)

Sometimes, this third condition is used to describe a degenerate gas as cold, because its temperature falls below some limit. However, this can be misleading because electrons may become degenerate at temperatures of millions of kelvins, which is obviously not cold in the common usage of the word.

3.3 Fermions

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

Electrons, along with protons and neutrons, are fermions, which means they have particle spin . The occupation of quantum states by fermions is restricted by the Pauli exclusion principle, which says that no more than one identical fermion can occupy a given quantum state. The only way to distinguish one fermion from another of the same particle is its spin, and because this has only two possible values, +1/2 or –1/2, at most two fermions of opposite spin can occupy a given quantum state, so that the number of fermions per quantum state is g_s = 2.

At low temperatures, all quantum states with energy less than their chemical potential are filled, and all quantum states with energy greater than their chemical potential are empty. In a cold electron gas, the energy of the most energetic degenerate electron is called the Fermi energy, E_F, and the momentum of particles with this energy is called the Fermi momentum, p_F. The Fermi energy is the sum of a particle’s kinetic energy and rest-mass energy. At low speeds, in the so-called non-relativistic limit, the Fermi kinetic energy and Fermi momentum for electrons are related by

Equation label: (6)

where m_e is the electron mass. The total number of degenerate electrons in the gas is evaluated as

Equation label: (7)

Because the number density of degenerate electrons is simply the total number of electrons per unit volume, , and there are two spin states for electrons (g_s = 2), this equation can be rearranged to express the magnitude of the Fermi momentum as

Equation label: (8)

Combine Equations (6) and (8) to express the Fermi kinetic energy in terms of the electron number density.
For electrons we have

Fermi energies are conveniently expressed in units of eV, keV or MeV. For comparison, the gap between atomic energy levels in a hydrogen atom is a few eV, the rest mass energy of an electron is 511 keV and that of a proton or neutron is about 940 MeV.

3.4 The pressure of a non-relativistic degenerate gas

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

In the non-relativistic limit (i.e. when v << c), the total energy of the gas may be expressed as

Equation label: (9)

The second term in the brackets in Equation (9) is simply the rest-mass energy per particle, and the first term is the kinetic energy per particle. Furthermore, the kinetic energy per unit volume is the kinetic energy per particle multiplied by the number of particles per unit volume. So the kinetic energy per unit volume is .

Now, the pressure provided by non-relativistic particles is 2/3 of the kinetic energy per unit volume, so

Then, using Equation (8), we have

The equation of state for non-relativistic degenerate electrons may therefore be written as follows:

Equation label: (10)

3.5 The pressure of an ultra-relativistic degenerate gas

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

We can also develop a similar expression for the equation of state for ultra-relativistic particles (i.e. when v ~ c). In this case, the total energy of the gas may be expressed as

Equation label: (11)

Because, in this case, the rest-mass energy is negligible, the kinetic energy per particle is . Furthermore, the kinetic energy per unit volume is (once again) the kinetic energy per particle multiplied by the number of particles per unit volume. So the kinetic energy per unit volume is .

Now, it turns out that the pressure provided by ultra-relativistic particles is 1/3 of the kinetic energy per unit volume, so

Then using Equation (8) we have

The equation of state for ultra-relativistic degenerate electrons may therefore be written as follows:

Equation label: (12)

Notice that the equations of state for degenerate electrons, whether non-relativistic or ultra-relativistic (or, indeed, somewhere in between), show that the pressure is independent of temperature and depends in each case only on the number density of electrons. In both the non-relativistic and ultra-relativistic cases, it is the independence of the pressure from the temperature that helps give a degenerate gas its interesting properties.

What is the key difference between the equation of state for a non-relativistic electron-degenerate gas and that of an ultra-relativistic electron-degenerate gas?
The pressure of an ultra-relativistic electron-degenerate gas has a weaker dependence on density than that of a non-relativistic gas (i.e. compared to ). So as the density increases, the pressure in an ultra-relativistic gas increases less markedly.

Whether the gas of electrons is considered to be non-relativistic or ultra-relativistic, it is clear that white dwarfs are supported against further collapse by electron degeneracy pressure. Notice that the equations of state for a degenerate gas (whether non-relativistic or ultra-relativistic) show that the electron degeneracy pressure does not depend on temperature. This is quite different from a so-called ideal gas (which is a good approximation for most normal gases) where the equation of state is , where n is the particle number density, k_B is Boltzmann’s constant and T is the temperature. This decoupling of pressure and temperature in a degenerate gas is what is responsible for many of the unusual properties of white dwarfs.

4 White dwarfs

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

We now have all the tools in place to consider the physics of white dwarfs – the end points in the evolution of low- and intermediate-mass stars.

4.1 The electron density

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

Because we will be considering electron degeneracy, we begin by deriving a new way of expressing the electron density. The general equation for the number density n of some type of particle is , where ρ is the gas density, X is the mass fraction and m is the particle mass.

We can write this for electrons just as easily as for nucleons: . However, it is more convenient to write where Y_e is the number of electrons per nucleon. This leads to the expression

Equation label: (13)

What is the value of Y_e in: (a) pure ionised hydrogen; (b) pure ionised helium?
(a) Hydrogen has one electron and one nucleon (its proton), so Y_e = 1; (b) helium has two electrons and four nucleons (two protons and two neutrons), so Y_e = 0.5.

You can readily convince yourself that Y_e = 0.5 for the dominant isotopes of carbon-12 and oxygen-16 also (since carbon-12 has 12 nucleons and 6 electrons, while oxygen-16 has 16 nucleons and 8 electrons). For white dwarfs, the amount of hydrogen is negligible – it has all been burnt to helium or on to carbon and/or oxygen – so Y_e = 0.5 for white dwarfs also.

For an electron-degenerate gas, you know that the pressure is independent of temperature, and is given by a constant multiplied by some power of the electron density. Using Equation 13, we can therefore write:

for non-relativistic gas, .
for ultra-relativistic gas,

The so-called Clayton model for the internal structure of a star gives the core pressure required to support a star in terms of its core density as:

Equation label: (14)

Here G is Newton’s gravitational constant (). By saying that degenerate electrons provide the required internal pressure to support the star, that is, by equating this to the degenerate-electron pressure, it is possible to write these equations purely in terms of the mass and core density, which therefore allows us to write one variable in terms of the other, as the following activity demonstrates.

Activity 4

a.By equating the core pressure in a star to the degenerate pressure of non-relativistic electrons, derive an expression for the core density in terms of its mass M and the number of electrons per nucleon, Y_e. Leave physical constants unevaluated.
b.Using the relation between electron number density n_e and gas density ρ_c in the core of the star, express the core electron density as a function of stellar mass.

Answer

a.Equating the core pressure to the pressure of non-relativistic degenerate electrons , where , we have
Collecting terms in ρ on the left-hand side, and all others on the right, we get
Cubing this gives
and substituting for K_NR gives
Consolidating the two numerical factors, we obtain
b.The electron number density in the core of the star is , so substituting in the core density from part (a) gives

Activity 4 shows that in the non-relativistic case, the density in the core of a star may be written as

Equation label: (15)

or, in terms of the degenerate-electron number density, as

Equation label: (16)

4.2 The Chandrasekhar limit

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

We have now derived an expression for the electron number density inside a star in the non-relativistic case, but is the material inside a white dwarf really non-relativistic? In the non-relativistic limit, the Fermi kinetic energy of degenerate electrons is given by (Equation 6), where p_F is the Fermi momentum given by (Equation 8). Therefore, the Fermi kinetic energy of the degenerate, non-relativistic electrons in the core of a white dwarf is

So, using Equation 16, this becomes

Equation label: (17)

The expression above relates the Fermi kinetic energy to a set of physical constants and the mass of the white dwarf, M. Substituting for these values therefore we have keV. Evaluating this expression for a white dwarf with mass 0.4 M_☉ reveals that the Fermi kinetic energy in this case is about 107 keV which is already more than 20% of the electron rest-mass energy ( keV). This makes it doubtful that the non-relativistic treatment is reliable for such a white dwarf. In more massive white dwarfs, the Fermi kinetic energy is even higher (because ) and the non-relativistic treatment will be progressively less reliable. Equation 17 implies that when M = 1.3 M_☉, indicating that the ultra-relativistic treatment is certainly required instead when masses become this large.

Activity 5

a.By equating the core pressure in a star to the degenerate pressure of ultra-relativistic electrons, derive an expression for the mass of a star supported by ultra-relativistic electrons.
b.Evaluate the mass, assuming Y_e = 0.5, in SI units and solar masses. Note that , , and

Discussion

a.For the ultra-relativistic case, equating P_c to P_UR gives where
Note that, in contrast to the non-relativistic case, the density term is the same on both sides, so cancels out, leaving
Collecting M on one side and swapping left and right sides gives
Taking the (3/2)-power and substituting for K_UR gives
b.Putting in the numbers, we have
So Converting to solar units, this is .

The mass of a star when the ultra-relativistic limit is reached may be viewed as the maximum stellar mass that can be supported by degenerate electron pressure. The approximate calculation in the previous activity shows that this maximum mass is of the order of the mass of the Sun.

The maximum mass for a white dwarf is called the Chandrasekhar limit; the most realistic computations estimate it as .

If the mass is increased further, then the pressure required to support the star also increases. But the ultra-relativistic degenerate electrons will not be able to increase their pressure to support it. Something has to give.... That something is the material supporting the star. There is a stable configuration of material at higher mass, but it does not consist of degenerate electrons. Rather, the electrons and protons of higher-mass objects are forced to combine as neutrons, and the stable object above the maximum white-dwarf mass is called a neutron star. Such objects will be studied later in this course.

4.3 The white dwarf mass-radius relationship

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

The mean density of a star is equal to its mass divided by its volume, . It can be related to the core density if the density profile of the star is known. So-called polytropic stellar models are ones in which the pressure at some radius is proportional to the density at that radius to some power. For a star described by a polytropic model with (a reasonable approximation), the mean density turns out to be one-sixth of the core density: . It is therefore possible to write a relation between the radius, mass and core density of a star.

Where the core density is dictated by non-relativistic degenerate electrons, we have already seen how the core density and mass are related (Equation 15). We can eliminate the density from this equation and find an expression for the radius of the object as a function of its mass, as shown by the following activity.

Activity 6

a.Rearrange the definition of the mean density to get an expression for the radius in terms of the mass and mean density.
b.Use the result for a polytropic model with (i.e. ) to re-express the radius in terms of mass and core density.
c.Substitute in the expression for the core density of non-relativistic degenerate electrons (Equation 15) to derive an expression for the radius of a white dwarf as a function of mass. Give your final answer in units normalised to the solar mass and solar radius. Note that , , and
d.Re-express the final result in solar masses and Earth radii, where the Earth’s radius is m.

Discussion

a.The radius as a function of mass and average density is
b.If the average density is 1/6 of the core density, then
c.Substituting in the core density from Equation 15 gives
Rearranging the terms with negative powers:
Now consolidate the mass terms and the numerical factors:
Now, for convenience, express the mass in solar units:
Then putting in the numbers:
And so
Because , the white dwarf radius is
d.Because the radius of the Earth is ,

The result of the previous activity

Equation label: (18)

is remarkable for two reasons. First, it shows that the radius of a white dwarf with the same mass as the Sun is two orders of magnitude smaller, comparable in radius to the Earth, as illustrated in Figure 3.

Figure 3 An artist’s impression showing the relative sizes of the white dwarf Sirius B and the Earth.

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An artist’s impression of the white dwarf Sirius B and the Earth shows them to be of the same size.

Figure 3 An artist’s impression showing the relative sizes of the white dwarf Sirius B and the Earth.

Second, notice the mass-dependence of the radius: . Unlike low-mass main-sequence stars fusing hydrogen in helium, where , note the important result that the sign of the exponent is negative for white dwarfs!

While the more massive a main-sequence star is, the larger its radius, the more massive a white dwarf is, the smaller its radius!

The derivation above of the white dwarf mass-radius relationship was based on the assumption that the electrons are non-relativistic. We showed this to be incorrect as the mass increases and it would imply infinite density (zero radius) at the Chandrasekhar limit, 1.4 M_☉. There is another formula which gives a more accurate picture of the radius of white dwarfs, but it is an empirical formula, simply fitted to observed data. This formula, derived by Michael Nauenberg in 1972, is

Equation label: (19)

The formula derived in Equation 18, while unable to correctly describe white dwarfs near the Chandrasekhar limit, does encapsulate real physical ideas that are valid for low-mass stars. There is a place in science for both formulae!

4.4 White dwarf composition and cooling

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

The simplest type of white dwarf, resulting from the collapse of a star of very low main-sequence mass () in which no helium burning has occurred, would be composed entirely of helium. Such objects are referred to as He white dwarfs, with masses . However we are unlikely to find isolated He white dwarfs in the Galaxy for the following reason. The He white dwarfs form from stars with very low mass. Such stars have main-sequence lifetimes of ~ 10¹¹ years, or around 10 times the age of the Universe, so have not had time to evolve into white dwarfs yet. Any He white dwarfs that are seen must have formed as part of a binary star system that has undergone a phase of mass transfer which has altered the mass and composition of one or both of its components.

Low-mass stars, with , undergo helium burning, producing carbon and oxygen in their cores. They lose some mass while on the AGB and so typically end their lives with masses of . Even intermediate-mass stars, with , lose so much mass as giants that only remains to form the white dwarf. These carbon/oxygen cores are supported against further collapse by degenerate electrons, and hence do not attain the temperatures required to initiate carbon burning. The remnants of low- and intermediate-mass stars are therefore CO white dwarfs, with masses .

There is a narrow range of massive stars, with , whose remaining cores are massive enough to collapse and ignite carbon (forming oxygen, neon and magnesium), but which ultimately fall below the Chandrasekhar limit owing to continued mass loss. When this carbon burning terminates, these stars will be supported by degenerate electrons again and will contract no further, leaving ONeMg white dwarfs, with masses typically in the range .

Stars with retain enough of their mass during mass loss that their cores continue to exceed the Chandrasekhar limit; at the end of each burning phase, their cores contract and heat up more. These stars complete all of the advanced burning phases, including silicon burning to form iron. Their endpoints will be explored in later sections of this course.

As there are no nuclear energy sources active in a white dwarf, its luminosity is due solely to the slow leakage of thermal energy into space, as radiation. It takes ~ 10⁹ y for white dwarfs to fade to . As white dwarfs fade, they also cool. The luminosity of white dwarfs is provided by thermal leakage rather than slow gravitational contraction because white dwarfs are degenerate, so their temperature and pressure are decoupled. As they cool, the pressure remains constant so the radius is unchanged. Hence there is no release of gravitational potential energy. As white dwarfs cool and fade in the H–R diagram, they do so along lines of constant radius (see Figure 1).

Figure 1 (repeated) A schematic H–R diagram indicating the locations of four regions where most stars are found. The vertical axis shows stellar luminosity in solar units or absolute visual magnitude; the horizontal axis shows surface temperature in kelvins, spectral type, or colour index defined as the difference between B and V magnitudes. (Notice that temperature increases to the left.) Diagonal lines of constant stellar radius are also shown and some notable named stars are indicated.

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Figure 1 (repeated) A schematic H–R diagram indicating the locations of four regions where most stars are found. The vertical axis shows ...

5 Supernovae

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

In this section, you’ll see what happens to a star whose final core mass exceeds the Chandrasekhar limit, and subsequently, discover what remnant it may leave behind.

A star with a main-sequence mass will complete all the stages of nuclear fusion that are available. Silicon burning will result in a core composed mainly of iron-56, surrounded by concentric shells of silicon, oxygen, neon, carbon, helium and hydrogen. No energy can be released by the thermonuclear fusion of iron, so the core collapses and the degenerate electrons within it become more and more relativistic. When the mass of the core exceeds the Chandrasekhar limit (about ), the degenerate electrons are no longer able to support the core, and a catastrophic collapse follows. The core will essentially collapse on a free-fall timescale, liberating gravitational potential energy.

The free-fall timescale is given by . Calculate the free-fall timescale for a stellar core with a density of .
Putting in the numbers to the equation above,

The gravitational potential energy released in the collapse of a stellar core of mass M, from an initial radius R₁ to a final radius R₂, is . Calculate the gravitational potential energy released when a stellar core of mass collapses to a radius of about 10 km.
In this case, the final radius is very small, so , and we can neglect the initial gravitational potential energy as it is so small. Therefore the gravitational potential energy released is . So the energy released by the collapse of a stellar core is

As the previous two questions show, the collapse of the core is very rapid and it liberates a vast amount of energy.

The initiation of exothermic (energy-liberating) fusion reactions provides pressure support for stars during their long-lasting burning phases. However, the initiation of endothermic (energy-absorbing) reactions draws kinetic energy out of the material and hence eliminates the pressure support. There are two processes that can absorb energy in the collapsing core: photodisintegration of nuclei by high-energy gamma rays and electron capture processes. In the first, the energy is used to unbind the nuclei, while in the second, energy is converted into the kinetic energy of neutrinos, which stream out of the star largely unhindered. Let’s consider each of these two processes in more detail.

5.1 Nuclear photodisintegration

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

There are many ways in which high-energy gamma rays can disintegrate nuclei of iron-56, but as an illustration of the physics involved, and the amount of energy that may be absorbed by this photodisintegration process, consider the situation in which a nucleus of iron-56 is broken down into helium nuclei:

The amount of energy absorbed by this process in the forward direction is . At the temperature and density of a stellar core, the result is that about three-quarters of the iron nuclei are dissociated.

Activity 7

If each iron nucleus absorbs 124.4 MeV of energy by photodisintegration, and three-quarters of a core of mass 1.4 M_☉ is dissociated in this way, calculate the total energy absorbed by this process. (The mass of an iron-56 nucleus is 56u, where kg is the atomic mass unit.)

Answer

The number of particles N in a sample of material is given by the total mass of the sample M divided by the individual masses of the particles m, i.e. N=M/m. Three-quarters of the core mass is , so the number of iron-56 nuclei is

Each nucleus absorbs 124.4 MeV of energy, which is equivalent to . So the core absorbs via the photodisintegration of iron-56 nuclei.

At still higher temperatures, helium nuclei will also undergo photodisintegration as follows:

The energy absorbed by this process in the forward direction is . At the temperature and density of a stellar core, the result is that about half of the helium-4 nuclei are dissociated.

Activity 8

If each helium-4 nucleus absorbs 28.3 MeV of energy by photodisintegration, and half of the core of mass 1.4 M_☉ is dissociated in this way, calculate the total energy absorbed by this process. (The mass of a helium-4 nucleus is 4u, where kg is the atomic mass unit.)

Discussion

The number of particles is N = M/m. Half the core mass is , so the number of helium-4 nuclei is

Each nucleus absorbs 28.3 MeV of energy, which is equivalent to So the core absorbs via the photodisintegration of helium-4 nuclei.

Therefore, the total amount of energy absorbed by nuclear photodisintegration is from the dissociation of iron-56 plus from the dissociation of helium-4, or about in total.

5.2 Electron capture

The Open University — Wed, 30 Oct 2024 00:00:00 GMT

The second mechanism by which energy is absorbed in the iron core of a star is that of electron capture, allowing energy to be carried away by neutrinos. The conversion of protons (in nuclei) to neutrons by electron capture is possible if the gas is sufficiently dense for degenerate electrons to have an energy greater than the 1.29 MeV mass-energy excess of neutrons relative to protons (, while ). Actually, the energy excess required for electron capture by bound nuclear protons, rather than free protons, is usually somewhat higher and depends on the nucleus, but is nevertheless usually just a few MeV.

The general reaction may be written as

and is sometimes referred to as neutronisation.

At densities above about 10¹² kg m^-3, iron-56 nuclei will capture electrons in the following reaction which produces a neutron-rich manganese nucleus:

where is a neutrino. Following this, further electron-capture reactions occur in successive nuclei, converting more and more protons into neutrons.

Because electrons are rapidly used up by these reactions, the pressure support provided by degenerate electrons quickly disappears, and the core collapses rapidly, as noted earlier. The neutrinos produced by the electron-capture reactions carry away most of the energy. Although the passage of the neutrinos is hindered by the high densities in the core, which raises the likelihood of interactions, all of the neutrinos are still able to escape within a few seconds.

Activity 9

If each neutrino produced in the reactions described above carries away 10 MeV of energy, how much energy (in joules) is removed by neutrinos if the whole core (comprising 1.4 M_☉ of iron-56) undergoes neutronisation? (The mass of an iron-56 nucleus is 56u, where kg is the atomic mass unit.)

Discussion

The number of nucleons contained in the stellar core is where u is the atomic mass unit. Because an iron-56 nucleus contains 26 protons (and 30 neutrons), the number of protons in the core is . This is evaluated as

Assuming charge neutrality, there will also be electrons. If every proton and electron undergoes neutronisation, this will produce neutrinos.

If each neutrino carries away 10 MeV of energy, then the amount of energy removed by each neutrino in joules is . So the total amount of energy removed is .

As the result of the previous activity shows, neutronisation, like nuclear photodisintegration, can also remove ~ 10⁴⁵ J of energy from the collapsing stellar core within a few seconds!