Tuesday November 13
We have now discussed the global observable properties of stars (their temperatures, luminosities, radii, masses and surface composition). The next step is to study the mechanism of stellar energy generation. The first serious physical study of this problem dates to work by Lord Kelvin and von Helmholtz in the 1800s. They both examined the possibility that the Sun generates its luminosity from gravitational contraction.
Consider a spherical cloud of gas of some constant density rho, and radius R One can estimate the gravitational potential energy of such a sphere as follows:
Now consider the mass and volume increments of a thin spherical shell at a radius r < R. From the above, one can show:
Recall the general definition of the gravitational potential energy between any two masses (quoted below). Note that a mass within a gas cloud only feels a net inward force from gravity due to the material that is closer to the center of mass than it is, AND that the force felt by the test mass is equal to the force felt if all the enclosed mass were concentrated at a point at the center of mass. Given these points, we can procede as follows:
Note that the numerical factors derive from the assumption of a constant density sphere. Other geometries will give different numerical factors, but will not change the result in any physically significant way.
Now, given the above, how long can the Sun radiate if it is powered by gravitational contraction?
The time an object takes to radiate away its stored energy is just the ratio of the total stored energy to the luminosity of the object. If we take the total stored energy to be the gravitational potential energy derived above, the result for the Sun is as follows:
Given that the age of the Earth is more than two orders of magnitude larger than this, it is clear that gravitational compression cannot be the energy source of normal stars.
Combustion energy comes from the breaking of molecular bonds. Such bonds have a characteristic energy of around 1 eV. Thus one can make an estimate of the total energy from combustion one can extract from a Solar mass of material, and the resulting timescale for powering the Sun as follows:
Well that's even worse! This presented late 19th century astrophysics with a thorny problem. The physics of the day could manage no way to power something like the Sun for the apparent age of the Earth. The resolution of this problem had to await a digestion of the implications of Einstein's theory of special relativity.
We define Z, the Atomic Number of a nucleus, to be the number of protons in that nucleus. Thus a nucleus will have a charge of +Ze. Nuclei are composed of protons and neutrons. Neutrons are slightly more massive than protons, and have no electric charge.
We define A, the Atomic Mass of a nucleus, to be the sum of the number of protons and neutrons in the nucleus. Thus a nucleus will have a Mass Number of A = Z + N. It is possible to have nuclei with the same atomic number, but differing atomic masses. Such nuclei are different Isotopes of the same Element. As an example below, I show the nomenclature for the isotopes of hydrogen:
Note that hydrogen is the only element for which we have special names for the isotopes. Deuterium is (p,n), and Tritium is (p,2n).
Elements with "small" Z usually have Z ~ N. For larger Z, Z < N as a rule.
Our study of the nature of interactions between objects in the Universe has led to the realization that physics is governed by four forces. In order of strength these are
Neutrons help stabilize nuclei because they only contribute to the Strong attraction, not to the E&M repulsion. They also make the nuclei larger, thus spreading out the distribution of positive charges.
One can think of nuclear binding energy in a way analogous to atomic binding energy. If one takes an atom in its electronic ground state, one must add energy to remove a valence electron. The typical energy required is about 1eV. Similarly, if one takes an atomic nucleus in its nuclear ground state, one must add energy to remove a nucleon. The typical energy required is around 1MeV. One can define the nuclear binding energy of a nucleus as follows:
Now consider the rest-energy of a proton, and compare it to the binding energy of a deuterium nucleus:
In other words, the binding energy is about a part in a thousand of the rest-mass energy. Thus the mass-difference between a nucleus and the sum of the masses of its individual protons and neutrons is a measurable amount.
While this will be a fairly general discussion, one should keep in mind that our agenda is understanding the energy generation in stars, thus we are primarily interested in exothermic reactions.
There are three principle sorts of nuclear decay:
Gamma emission: Gamma "particles" are high-energy photons. They are emitted by nuclei in excited states in a way analogous to the emission of optical photons from atoms in excited atomic states. A given isotope has a characteristic set of spectral features it will emit, and the study of such features tells us about nuclear structure.
Beta emission: Beta particles are electrons. Beta decay of nuclei is driven by the Weak nuclear force, and involves the transformation of a neutron into a proton plus an electron. The result of such a decay is to transform (Z,A) into (Z+1,A)
Study of weak decays in the 1930s produced something of an embarressment. Neither total energy, nor total angular momentum appeared to be conserved in such reactions. The energy deficit was variable, but the angular momentum deficit was fixed (at one unit of h-bar). This led theorists to invent a particle they called a neutrino. Neutrinos were discovered experimentally some 20-30 years after they were predicted. This is because they, in fact, very difficult to detect. Neutrinos interact with other matter only by the weak nuclear force and gravity.
The basic beta-decay reaction is the disintegration of a free neutron:
For a free neutron, the timescale for this is about 11 minutes. Neutrons in nuclei are generally much more stable. However, if a nucleus has many more neutrons than protons, it can be unstable to spontaneous beta decay.
The binding energy per nucleon reaches a broad peak around Iron. Thus medium-mass nuclei are the most stable. Very massive nuclei are much less stable. Certain isotopes are sufficiently unstable that they spontaneously disintegrate into smaller chunks. This process is called Fission.
Fusion is the process of building up more massive from less massive nuclei. Recall the peak in the binding energy curve at Iron. This means that one can extract energy from fusion of lighter nuclei up to iron. In order to do this, one needs very high temperature and pressure. To understand why this is so, consider the case of two protons at some initial seperation r. These two protons have an electrostatic potential energy (repulsive!) that goes as 1/r. If we take the electrostatic potential energy as equal to zero when r goes to infinity, we can then evaluate the required initial kinetic energy in order for the protons to approach each other to within about a proton radius:
That's a VERY large temperature. But that's the temperature for an average collision to have enough energy for a fision reaction. Recall the general form of the Maxwell-Boltzmann temperature distribution:
There is a tail at large energies. The atoms in that high-energy tail are the ones that can undergo fusion reactions.
The above is a classical argument that the temperature for fusion should be much less than 10e10 Kelvin. There is an additional factor that arises from quantum mechanical considerations. If one considers the energy between two protons as a function of radius, one sees that it can be expressed in terms a a potential barrier (due to Coulomb repulsion) with a central (bound) potential well (due to Strong interaction). Classically, an incoming proton MUST have a kinetic energy larger than the potential barrier in order for a reaction to occur. Quantum mechanics allows for particles with less energy than the potential barrier to react due to Quantum tunnelling. This means that an even larger fraction of the protons, at any given temperature, can fuse to heavier nuclei than one would expect based on pure classical calculations. The upshot of this, is that one needs temperatures of around 10e7 Kelvin for hydrogen fusion to procede.
I continue with bit more detail on tunnelling. The typical tunnelling distance for any particle will be on the order of its deBroglie wavelength. The probability of tunnelling to a given depth obeys an exponential distribution:
Where a in the expression above is a constant. Recall that the classical definition of the reaction point follows from equating the incoming particle's kinetic energy with the potential energy from Coulomb repulsion. One can use this argument to express the tunnelling probability as follows:
Where b is a constant that subsumes all the individual constants. The only variable in the exponential is v, which is equivalent to E^1/2.
Now recall that the distribution of particle velocities in a gas is given by the Maxwell-Boltzmann distribution. This means we can express the reaction probability for the ensemble of particles in the gas as a convolution of the Maxwell-Boltzmann distribution and the tunnelling probability expression just arrived at:
Note that this means that a gas of a given temperature (and composition) will have a "sweet spot" in the energy distribution where most of the fusion reactions are occuring. Note also that the reaction-rate is an exponential function of gas temperature, thus small changes in the temperature translate into large changes in the reaction rate. This is the fundamental reason why there is such a steep relationship between mass and luminosity for main sequence stars (recall the discussion in week 9). Large mass implies larger central pressure, and therefore higher central temperature.
Thursday November 15
The net reaction that powers Main-Sequence stars is
In stars with masses up to about 2 Solar masses, the reaction mechanism that dominates in called the pp-chain:
More massive stars have hotter cores, and will thus have collisions energetic enough to overcome the Coulomb barrier of heavier nuclei. One consequence of this is that in stars more massive than ~2 Solar masses, fusion procedes by the CNO cycle:
This is a catalytic cycle. It takes a 12C nucleus as input, and returns a 12C nucleus at the end. Recall (again from week 9) that the shape of the Mass-Luminosity relation changes at 2-3 Solar masses. One reason for that change is the cross-over from pp-chain dominated fusion to cno-cycle dominated fusion.
A collapsing gas cloud becomes a Main-Sequence star when hydrogen fusion in the core ignites, and produces sufficient thermal pressure to support the gas cloud against further collapse.
Now, consider the result of those core fusion reactions. They cause a net decrease in the number of nuclei in the core *AND* the cause a net increase in the mean molecular weight of the gas. The consequence of this is that the core pressure is constantly (and VERY slowly) decreasing. Thus, while we can treat the star as an instantaneous equilibrium system, it really is changing in a secular way. The result of this slow change in the core is that the total luminosity of the star increases slightly over the course of its main sequence lifetime. This can manifest as a slight increase in radius or in temperature or both.
One observational consequence of this slow, secular evolution is that the Main Sequence isn't really a line, it is a band. There is a line that one can draw on an H-R diagram that corresponds to where a star of a given mass will lie when it first reaches stable hydrogen core burning. This line is called the Zero-Age Main Sequence or ZAMS.
Recall the discussion of logical models presented in the context of the Sun. We use Solar models as the basis of the general study of stellar structure. That is, one must first be able to construct a self-consistent Solar model. Then one can take that Solar model, and see what happens when you change the mass. These results can then be compared to the (integrated, and generally MUCH poorer) observational data on other stars.
In order to start with the simplest useful situation, we shall begin with the assumption of spherical symmetry. Note that this is incorrect in detail, as stars rotate. And, given that stars are made of gas, their rotation makes them oblate. This is a complication we shall ignore for the moment.
The assumption of spherical symmetry means that we can express mass continuity in stars in a fairly simple way:
Consider a slab of material at some distance r from the center of the star. The statement that the star is an equilibrium system means that the total force on that slab of material must be zero. The forces on the slab are an inward force from gravity due to the mass at radii less than r, and an outward force due to thermal pressure from the hot material below the slab.
The geometry of the slab had BETTER be irrelevant to the discussion, so we can take it to be anything we want. Following the text, we take it to be a cylinder, oriented with its length in the radial direction. The cylinder has some length dr and some cross-sectional area dA. One can thus write its mass as
The inward force of gravity on the cylinder can be expressed as follows:
The thermal pressure on the cylinder results in a buoyant force that is evaluated as follows:
As noted above, equilibrium requires the sum of the forces to be zero. From this the result follows:
The last equation expresses the terms (GM/r^2) as the local acceleration due to gravity.
The Equation of State of a material is the relation amongst pressure, density and temperature for that material. For the study of main sequence stars, we can typically get by using the Ideal Gas Law. Note that there are situations at very high temperatures (relativistic gases) and densities (degenerate gases) for which the Ideal Gas Law is grossly incorrect. But discussion of such situations can be deferred until ASTR 225.
If we take the ideal gas law, and the equation of hydrostatic equilibrium together, we reach the following result:
Thus the equilibrium equation is actually a first-order coupled differential equation.
Recall that energy can be transported by Conduction, Radiation, or Convection. Conduction is unimportant in stars. Radiation is generally the dominant mode of energy transport. I will thus defer a consideration of convection for the moment.
If one considers a spherical shell of material in a stellar interior, and makes the assumption that the shell will behave like a blackbody, one can analyse the radiative flux differential as follows:
Now recall our discussion of opacity from week 10. Realizing that "intensity" is the same as "flux", one can make the following argument:
In week 10, we defined the absorption coefficient. It is also conventional to define the absorption coefficient per unit density. This changes the notation in the above to the following:
One can combine the two expressions for df given above, and evaluate the luminosity from the shell as follows:
Note that dT/dr is negative (heat flows from hot to cold)! Thus L(r) is positive. Note further that increases in the density or the absorption coefficient decrease the luminosity (due to radiation). In either case, the opacity increases, and radiation becomes less efficient. Convection occurs when the radiative opacity becomes too high. It also occurs when the radial temperature derivative becomes too large. We shall return to this point.
If we ignore all the ugly details of the nuclear physics, we can simply define a quantity epsilon that represents the energy generated per unit time per unit mass in some shell of radius r. Given the volume of a spherical shell, we can thus write the following:
Why don't stars just blow themselves apart?
There are many ways to look at this, but the simplest of them is to view it as a heat capacity problem. Consider the total energy of the system, and recall that a star is gravitationally bound and will thus obey the virial theorem (recall week 2):
Now recall that the macroscopic manifestation of the "kinetic energy" of a gas is the temperature of the gas. This means that if we add energy to a stellar envelope, the gas expands and cools. This *decreases* the thermal pressure of the layer, causing it to contract and re-heat. This negative heat capacity means that a star with a steady-state energy source at its core will reach a stable configuration.
Boundary Conditions:
Such calculations were done (with hand-crank adding machines) in the 1940s. This technological limitation forced workers at the time to make the incorrect approximation that one could treat stars with static models. As I have already noted, as fusion proceedes in a stellar core, the pressure drops and the mean molecular weight increases (because 4 particles get turned into 1 particle, and that 1 particle has a larger mass than any of the 4 original particles). This causes the core to contract and heat up while the star is on the main sequence. As a result, one must build a time-variable stellar model.
One begins with the static stellar model for a zero-age main sequence star. Once this is in hand (or in memory), one then adjusts the core properties by an amount corresponding to a "small" time step, and then recomputes the entire radial model. This continues for each time step. As in the case of the radial steps, one must make the time-steps just small enough to properly treat all the important physics.
Tuesday November 20
Hot gas is buoyant. Thus if you have a gas that is being heated from below, there will be an upward buoyant force on that gas. An idealized picture of convection has hot blobs of gas moving upward into cooler environments. As the blob rises, it is thus surrounded by lower-pressure gas, and it will expand and cool. But the blob does not exchange energy with its environment as it rises. This is called an Adiabatic Process. Gas of a given composition, initial temperature, and initial density will have a particular radial temperature gradiant (called the Adiabatic Lapse Rate that obeys these constraints. Convection will occur if the Adiabatic Lapse Rate is smaller than the Radiative Temperature Gradiant:
In what follows, I will treat convection with what is called the Plane-Parallel Approximation. That is, I will ignore the fact that stars are spherical. This works because the size-scale of a convection cell is much smaller than the radius of curvature of a star. I will also explicitly state that I adopt the approximation that the density in a convecting blob is roughly constant during convection. We begin by considering the implications of the ideal gas law:
Recall that we are interested in the radial temperature gradiant. Thus we take the derivative of the above, and work through the algebra. In the steps below I use (without proof) the result from thermodynamics that the radial pressure gradiant is equal to -g rho.
The above can be reformulated by once again appealing to the ideal gas law:
Note that this is only a slowly changing function of radius once we get to the outer layers of a star. The mass is increasing slowly (most of the mass is concentrated toward the center), and the local gravitational acceleration is decreasing (more rapidly, as it falls almost as r^-2). If the opacity increases rapidly (driven by the onset of H- formation), this causes the absolute value of the radiative temperature gradient to increase strongly, and drives the outer layers of the star to be convectively unstable.
A fundamental problem with Solar models is that the boundary conditions do not give us a direct probe of the current properties of the Solar core. This is because the photons generated by the nuclear fusion reactions in the core have a VERY short mean-free path compared to the Solar radius. Thus any photon only travels a small fraction of the Solar radius before it is absorbed and re-emitted. The re-emission is isotropic. Thus the photons undergo a random walk through the Solar interior. The reason they make their way out at all is that there is a density gradient in the Solar interior. Thus a photon that is scattered "up" will travel slightly farther on average than one that is scattered "down". A consideration of the details of the Solar interior leads us to a timescale for radiative diffusion of around 10 000 000 years.
Now, recall the pp-chain:
One of the things that motivated physicists in the 1930s to propose the existence of neutrinos has to do with reactions like this. Two normal matter particles came together and produced a heavier nucleus of normal matter, and an antimatter particle (the positron). Most reactions that people had seen preserved the number of "anti-matter" + "matter" particles of a given type. Also, the total mass+energy on the right side should equal the total mass+energy on the left side. Likewise, the total angular momentum in the reaction should be conserved. The observed products of nuclear reactions appeared to be violating all these constraints. So physicists at the time "invented" the neutrino to take care of this problems. In order to be consistent with the experiments of the day, physicists realized that neutrinos had to have the following properties:
The advantage and difficulty of neutrinos is that they do not interact electromagnetically. This means that the neutrinos generated by weak interactions in the Solar core will travel through the Solar interior at essentially the speed of light, and will reach us 8 minutes after they are generated, instead of 10 000 000 yrs after. The problem is that they are difficult to detect.
The principle reaction chain shown above yields a neutrino at step 1, but the neutrino energy from this process is relatively low. The lower the energy, the more difficult it is to detect neutrinos. The variant in the reaction chain for step 3) also produces neutrinos, and they are at somewhat higher energies. Yet another branch of the reaction chain is the following:
The neutrinos from this reaction are the most energetic neutrinos produced in the Sun.
Neutrinos were experimentally detected in the late 1950s, some 20-30 years after they were predicted based on weak-decay studies. As soon as they were detected, people began thinking of how to use them as probes of the Solar core. The neutrino-capture process that has the highest known cross-section is
So one is left with designing an experiment that takes advantage of this reaction. Such an experiment is the Homestake Mine Neutrino Detector that ran for about 35 years from the mid 1960s until the early 2000s. The person responsible for designing and running this experiment since its inception, Dr. Raymond Davis, was awarded the Nobel Prize in Physics in 2002. In this experiment, a chamber about a mile down in the mine is filled with cleaning fluid. You fill up the tank, let the stuff sit for awhile, and then count the Argon. That gives you a total number of reactions per unit time, and from that you can work out what the neutrino flux is.
The results of this experiment have been stubbornly consistent: The Solar neutrino flux at the energies measured by the experiment is only about 1/3 of the value predicted by Solar models.
BUT! This experiment is only sensitive to the highest energy neutrinos. The neutrinos released by the typical p-p chain reactions aren't energetic enough to drive the reaction. Other experiments were needed to see if this was a minor problem with the reaction rate of one of the branches of the p-p chain, or a fundamental problem for our understanding of either the Sun, or of basic particle physics.
A great deal of effort was expended for three decades on trying to sort this out. And *all* available avenues ended up being pursued. Was the Solar model wrong in some fundamental way? Was the treatment of the details the problem (an opacity problem, a problem with the relative frequency of the side-chain reaction that yields the high-energy neutrinos)? Was it a numerical problem with the computer models? Was it a systematic problem with the Homestake experiment? Was it a problem with the Cl/Ar cross-section? And so forth.
The resolution now appears to be in hand, and it turns out that the problem was that we did not understand neutrino physics well enough. So a little more background on particle physics.
The neutrino that they proposed was a particle that was associated with the electron. But over the next few decades, two other particles were discovered that are similar to electrons. They differ in that they are much more massive than electrons, and they have very short lifetimes. Each of these particles also has a neutrino associated with it:
But the Solar neutrino experiments that were run from the 1960s until the late 1990s were only sensitive to electron neutrinos. When neutrino detectors that could also detect the muon and tau neutrinos were developed, and put into service, they saw the missing Solar neutrinos. This means that the electon neutrinos are able to convert into muon or tau neutrinos (a feat that particle physicists call "flavor mixing" or "neutrino oscillation"). This is only possible if neutrinos have a non-zero mass.
The current best-estimate of the electron neutrino mass from the Sudbury Neutrino Observatory data is about 0.01 eV (recall the electron has a mass of 511 keV!).
Tuesday November 27
1) Consider a sphere of some radius R with a density distribution as follows:
Derive an expression for M(r).
We begin by considering that we are given a density law, and asked to determine a mass-profile. Further, the problem has spherical geometry. From here, the calculation procedes:
2) Consider a gas with some constant density and absorption coefficient. a) give an expression for the mean free path, b) solve this numerically for conditions typical of the Solar core, c) determine the time-scale for photons to diffuse from the Solar core to the photosphere assuming a constant density Solar interior.
The answer to the first part can be expressed in different ways, depending on the use or not of the density-weighted form for opacity. For the second part, we take numbers typical for the Solar core:
For the third part, we make use of analysis from the text that shows on geometric grounds that it takes some total number of steps N, each of length equal to the mean-free path, to go some distance d. Here, we take d to be R, the radius of the star.
Note that the assumption of constant density means this is clearly an overestimate of the true diffusion timescale. This follows because the mean-free path will increase as the density decreases.
3) Use the virial theorem to derive an estimate for the average temperature in the Solar interior.
The algebraic part of this is straightforward if you trust your physics:
The meat of the question is really in figuring out what to put in this expression, now that it is in hand. I assert that a typical atom in the Solar interior is about half-way out from the core (that is an overestimate), and has about half the Solar mass interior to it. The factors of two cancel, and we have that M is the Solar mass, and r is the Solar radius.
If we consider that we are treating the Sun as a bound, relaxed gravitating system of point masses, the obvious choice for m is thus the mass of a hydrogen atom. Taking these as input, we arrive at the numerical result:
This is a shockingly good estimate for the average temperature of the Solar interior, considering how crude the calculation is.
4) Consider two stars with the properties given below. From these properties, and assuming a Main-Sequence lifetime for the Sun of 10 Gyr, estimate the expected Main-Sequence lifetimes of the two stars.
The key to the problem is to realize that the luminosity of the star is equivalent to its fuel consumption rate, and that its mass scales with its fuel supply. For star 1, the calculation can procede from here:
For star 2, there is an additional subtlety. The Sun and star 1 both have largely radiative envelopes. Thus there is no deep mixing during the Main Sequence, and both stars have about 10 percent of their total mass available for hydrogen fusion. Star 2 is fully convective, thus all of its mass is available for hydrogen fusion. One must therefore include the order of magnitude increase in the estimate for star 2:
5) Consider a sphere with an inverse-square law density profile, as given below. Determine the total graviational potential energy of such a sphere with mass M and radius R.
The analysis begins with the consideration of the spherical geometry, and continues as follows:
Note that this means we can make the following statements:
Now that the pieces are all prepared, we can assemble the dish:
