Topics: Post-main-sequence stellar evolution, degeneracy pressure, and white dwarfs Summary of reading: Review section 2 of Ch. 17. Read the beginning and first section of Ch. 18 (up through the middle of p. 416) Optionally, you can read ahead into section 2 of Ch. 18, which will be assigned for our next class Summary of work to submit: For Tuesday s class, please answer the three questions on the first page of the worksheet posted on the website as part of the assignment for this class. Scan/photo your answers to these three questions and email them to me by midnight on Monday. Here are the things you should be thinking about: Well, you should look back at this things to think about section for the last couple of assignments. You ll note that degeneracy pressure plays an important role in post-main sequence stellar evolution basically any time nuclear energy production is in a shell of material surrounding an inert core, the core is held up against its own gravity by degeneracy pressure. This phenomenon a type of pressure that traces its physical origins to fundamental quantum mechanics: the Heisenberg uncertainty principle and the Pauli exclusion principle is not described in detail in 17.2, but is in the first section of Ch. 18, as it is the key to the existence of white dwarfs and neutron stars. This is the planetary nebula Abell 39. The white dwarf that ionizes, excites, and heats the nebula can be seen at its center. All the other stars (and galaxy) that appear to be in the nebula are really just chance Page 1 of 5
projections of objects behind or in front of the nebula. More information: http://apod.nasa.gov/apod/ap121008.html A planetary nebula is a late-stage of the evolution of solar type stars but it also is physically very much like an H II region, with the main difference being that the central star is a newly formed white dwarf. With its spherical shape and single ionizing star, Abell 39 is the best example I know of a simple Strömgren sphere. (3) Post-main sequence evolution once again! Starting on the next page, we ve got new stuff. Note how the increase in µ as the Sun converts H to He in its core leads to the perfect gas law providing less and less pressure. But HSEQ demands that pressure so the core sill very slowly contract and heat up and this will increase the nuclear reaction rates and hence the Sun s luminosity. Here are key points/physical processes that you should be paying attention to as you read 17.2: There is such a thing as degeneracy pressure that can hold up the core of a star even if no nuclear reactions are happening there. Each given nuclear energy source (e.g. hydrogen) will provide energy first in the core, and then when the supply runs out, in a shell right above the core. Make sure you re clear on that geometry. Such shell burning tends to happen at high temperatures, produce a lot of luminosity, and make the envelope of the star (its outer layers) expand. Note where the helium flash is in the HR diagram. What happens right at that moment? And do you see how the horizontal branch is very much like the main sequence, only with the triple-alpha process generating energy instead of the p-p chain? You should be able to describe, physically, what happens at the end of a solar-type star s life; why it doesn t fuse carbon to yet heavier elements and why it forms a planetary nebula. We will learn about end-states of stars what s left behind after all the nuclear fuel is exhausted white dwarfs, neutron stars, black holes, next. Page 2 of 5
Read the intro to Ch. 18 Stellar Remnants, after reviewing the key stellar evolution points (previous page). On p. 409 there s a short review of the fusion life history of stars having different masses. Read this carefully and relate the facts in each bullet point to your knowledge of stellar interiors and evolution from the previous chapter(s). See how the physics involves the interplay between HSEQ ( how much pressure will this object need? ) and nuclear reaction physics ( is it hot enough to fuse hydrogen to helium? ). Note that the second bullet point describes stars whose main sequence lifetimes are longer than the current age of the universe! We probably understand well what will happen to these M stars after they leave the main sequence, but no actual M stars have evolved off the main sequence yet, anywhere in the universe. As you look down the list, more massive stars fuse heavier and heavier elements. physical cause of this? Can you explain the (1) Big picture Once sources of nuclear energy are used up (either because there are no more light elements from which nuclear energy can be extracted via fusion or because the star isn t massive enough to generate sufficient heat and pressure to initiate fusion of the light elements that do remain) it (or its core) will start to contract until the core of the star is dense enough to provide pressure to resist further gravitational contraction via degeneracy pressure. After the thermal pressure associated with gravitational energy and nuclear energy, degeneracy pressure is the last option for a star that wants to keep from totally collapsing (and forming a black hole). For class on Tuesday, when discussing the Ch. 18 material, I d like to focus on the physics of degeneracy pressure and the resulting equation of state and the consequences for the physics and properties of white dwarfs and also on the degenerate cores of red giants. We can explain the helium flash via the properties of degenerate gases, too. (2) Degeneracy pressure A fundamental tenet of quantum mechanics is the Pauli Exclusion Principle which says that no two identical particles (e.g. two electrons) can occupy the same part of phase space six-dimensional space with three dimensions of true space, or location (x, y, z) and three dimensions of momentum (mv x, mv y, mv z ). What is the criterion for how close together two particles can get and still not count as being in the same location in phase space? It is the Heisenberg Uncertainty Principle, which states that the fundamental limit is on the product of complementary variables, like position and momentum (or time and energy), given by x p h, where h is the reduced Planck constant, h = h/2π. Remember, these weird quantum effects are descriptions of physical reality, not statements about the limits of human knowledge. So if an electron is packed in very tightly with its neighbors and it can t move very far at all, then its positional uncertainty, x is small. Thus its momentum uncertainty will be big. And since this uncertainty really means that the electron takes on all values within that range if it makes you feel more comfortable, you can imagine measuring its momentum over and over again and finding a wide range of values and so some average value (of the modulus of the momentum vector). In other words, being constrained in position really makes the momentum of an electron get very big. It will make it on average p p h/ x. (1) Page 3 of 5
How much is the position of an electron constrained by its neighbors? Imagine a square lattice of electrons like three-dimensional graph paper, with an electron at every point where lines cross. Please think about it hard enough to convince yourself that the volume associated with each electron is x 3 = n 1 e, where n e is the number density of electrons (electrons per cubic meter). See how a number density is the inverse of a volume per particle? So, electrons in a densely packed environment will have a Heisenberg speed as described on p. 411 associated with this momentum, of v v hn1/3 e. (2) m e Does this equation make sense? The denser the material, the more electrons per cubic meter, the less volume each electron has. The less space it has the bigger its momentum. And finally, the bigger its mass, for a given amount of momentum, the lower its velocity is. So, light particles (like electrons, rather than protons or neutrons) have higher Heisenberg speeds for a given density that s why we re focusing on electrons at first. They have a bigger effect than heavier particles like neutrons do, at any given density. Page 4 of 5
(2) Degeneracy pressure continued The next conceptual step is to realize/remember that pressure is related to momentum. It is the momentum flux. To compute the flux of any quantity that has a velocity and density associated with it, you simply multiply that quantity by the product of the density and velocity. Try it! Write down the units for number density and velocity: m 3 times m s 1 gives s 1 m 2, and just as energy flux (like in the inverse square law) is energy per time per area, momentum flux (or pressure can you manipulate the units of momentum flux into pressure units, maybe by using the fact that pressure is force per unit area?) is momentum per time per area. OK! So, eqn. 18.10 shows that. Do you see how nmv 2 is density times momentum times velocity, given that momentum is mv? So, we can compute the pressure, P, from P e = n e m e v 2 e. (3) Note as an aside that you can actually derive the perfect gas law from this expression if you make it an integral over a Maxwell-Boltzmann velocity distribution. Substitute eqn. 2 (in these notes, above) for v e in eqn. 3 and we have an equation of state for electron degeneracy pressure: P e degen h 2 n 5/3 e. (4) m e An equation of state (EOS) is a pressure-density-temperature relationship. The perfect gas law is one. Solids tend to have complicated EOSs that can t readily be computed analytically. Note something interesting about the equation of state for degeneracy pressure the pressure depends only on the density and not at all on the temperature. More mundanely but still quite importantly the pressure increases with increasing density. If you squeeze a degenerate ball of electrons its pressure will increase. Of course gravity can do that squeezing, especially if the ball of degenerate electrons is very massive. Finally, an important but simple point there are protons and neutrons, in the form of normal nuclei mixed in with the electrons in a white dwarf or in the degenerate core of a red giant star. That s why we don t have to worry too much about the electromagnetic repulsion of the electrons for each other. We aren t talking about the protons and other nuclei right now, though, because they just don t produce much degenerate pressure of their own compared to the much less massive electrons. Note the inverse dependence of the pressure on the particle mass in equation 4. Page 5 of 5