3. MORE MECHANICS : NEUTRON STARS AND CLUSTERS OF GALAXIES


  3.1 Gravity on a neutron star
      density of nuclear matter, centrifugal force

Purpose: A stretch of the imagination.

The problem: Estimate the density of a neutron star assuming that it consists of idealized neutrons that are hard balls with radius of about 0.7x10-15m. If the mass of this star is 2Mo, compute its radius (compare to the size of the main city of your country), the gravity on the surface (compare to that on Earth), and the escape velocity (compare to the speed of light, c, and ignore relativistic effects).

   How high a "mountain" on the surface of the neutron star would you have to climb to gain the same gravitational energy as climbing one of the high mountains on Earth, say Kilimanjaro (Africa), Aconcagua (South America), or Mount Everest (Asia)? Identify some object on Earth with about as much mass as resides in a hand full of neutron star matter.

   Finally, compute the centrifugal force acting on gas at the equator of the neutron star for an arbitrary rotational period P. Find that value of P for which the centrifugal force equals the surface gravity (evaluated assuming a spherical star). Explain why one would not expect to observe periods shorter than this P and compare it to observed pulsar periods P > 0.0016 s.

The setting: The lives of stars are the subject of section V. Here we summarize the life of stars only to introduce neutron stars.

   Stars are generally static objects. Each layer is supported against gravity by a gradient in the pressure. In a normal star like the Sun, the pressure is that of an ideal gas (p = nkT). The pressure at the center is maintained sufficiently high because nuclear reactions maintain a high temperature. Once nuclear fuel is exhausted, the star contracts, either until some new kind of pressure supports the star or until it becomes a black hole. For a star like the Sun, the new pressure is provided by (degenerate) electrons, and the Sun becomes a white dwarf. For stars with mass between 1.4Mo and about 4Mo, electrons provide insufficient pressure, the star shrinks until electrons are squeezed into protons, the new pressure is provided by the resulting neutrons, and the star becomes a neutron star. For stars with mass above about 4Mo, even neutrons provide insufficient pressure, no new pressure is available and the star contracts forever, becoming a black hole.

   Why can neutrons support a star against its own gravity? The simplest model says that neutrons act like extremely hard balls. That is the model of the present problem. The values of density, surface gravity and escape speed truly stretch everyone's imagination. Neutron stars were theoretically predicted, but the astronomical community remained extremely skeptical that they would actually exist.

   Now neutron stars have actually been observed, though only indirectly. Within about 107years of the formation of a neutron star, a powerful beacon of radiation rotates with the neutron star and is observable as a pulsar (see problem 3.4). Also, when a neutron star is a member of a binary, gases falling from the normal star toward the neutron star begin to revolve around the neutron star. As they gradually spiral toward the neutron star they become very hot and become observable because of their x-ray (and other) emission (see problem 3.2). In 1997, the Hubble Space Telescope discovered directly the first non-pulsating "lone" neutron star (see problem 3.3).

The solution: The volume per neutron is 4pi/3 r3 = 1.5x10-45m3. If we ignore the small amount of space between the neutrons, their number density is 0.7x1045m-3 and the mass density is 1018kg m-3. The mass in a "handful", say ten cubic centimeters, has the mass of rock (density 4 x water) on Earth with a volume of 2.5 km3, a small mountain.

   Given the density and the mass, the volume of the neutron star is 4x1012m3, and the radius is 10 km. This is smaller than the sizes of many capitol cities of the Earth's countries. The surface gravity is GM/R2 = 2.7x1012m s-2, huge compared to 10 m s-2 on Earth. A climb of 8000 meters on Mt. Everest is equivalent to climbing 8x103x 10 / 3x1012 = 3x10-8m on the neutron star, a height less than the wavelength of visible light. The escape velocity is (2GM/R) 1/2 = 2.3x108m/s, about 3/4 c .

   When the centrifugal force equals local gravity, v2/R = GM/R2. With 2piR = Pv, P2 = 4pi2R3/GM = 3pi/Grho, and P = 4x10-4s. If a neutron star ever had such a period, the star must have lost matter from the equator so rapidly that we would not observe it. The fastest observed pulsars are "safe" from rotational disruption, but only by a factor of 4.

Didactics:

  1. The parameters of neutron stars are truly astounding. No wonder that most astronomers doubted their existence until pulsars required it. The escape velocity is near the speed of light. Indeed, the neutron star's radius is only slightly larger than the Schwarzschild radius of 6 km (see problem 1.5).

  2. The neglect of the space between the spherical neutrons is a simplification no worse than assuming that neutrons are hard spheres.

  3. Mathematically, the critical rotational period P can be expressed either in terms of M and R or in terms of rho. In this case, rho is actually the primary quantity known physically and, therefore, writing the critical P in terms of rho is much to be preferred. It not only minimizes the carrying of errors that arose in computing R and M, but shows physically much more comprehensibly what the critical P depends on.

Interpretation: Neutrons are not really simply hard balls. They are made of more fundamental particles. The appropriate equation of state is still uncertain in detail, but one expects most neutron stars to consist of a liquid interior, a crystallized crust, and a gaseous (non-degenerate) atmosphere a few centimeters thick. If the star contains more than about 4 Mo (uncertainty 3 Mo to 5 Mo), then the pressure due to neutrons is inadequate and the star collapses to become a black hole.


3.2 Accretion disks around neutron stars
      activity: the orbital period of Her X-1

Educational purpose: An example of measurement uncertainties and simple error analysis.

Astrophysics purpose: Introduction to the notion of an accretion disk, a common phenomenon around young planet-forming stars, white dwarfs, neutron stars, and black holes.

The observational problem: The graph shows the x-ray flux versus time for an x-ray source called Her X-1. The measured data points are shown by the small crosses. The vertical lines show the uncertainty of the individual data points, based on the Poisson statistical error from photon counting. The horizontal lines represent the widths of the detector energy channels and are not strictly error bars. A computer program has connected the data points.

 


 

Data for Her X-1 measured by the Earth satellite Rossi X-ray Timing Explorer (NASA)

 


 

   These data, looked at from afar, clearly show an oscillation with a period of roughly two days. Measure the period as accurately as you can and estimate the uncertainty in the deduced period: the correct value probably lies between and days. Explain how you arrive at this conclusion.

The solution: Much judgment is involved in determining the period, and that is the main point of this exercise. The most straightforward attempt is to count the number of periods over the entire interval. But obviously there are some intervals such as at t = 121 days when no maximum is observed. Somehow one has to "correct" for such problems by mentally adding maxima to be counted, based on the rest of the plot. That way students may well determine there are 27 to 30 peaks in 40 days, for a period of 1.3 to 1.5 days. To avoid the "correction", perhaps it is better to select a part of the plot where maxima are very clear, such as between days 129 to 136. But then the time interval for a whole number of periods is difficult to measure accurately. For instance, the high data points with counts of 6 to 8 per second are probably not exactly at the midpoint of the entire peak when one takes into account the data points just before or afterwards. Similarly it is difficult to find the midpoint of a steep slope. Different individuals or groups tend to find values of the period in the range 1.4 to 1.8 days. Which method is better? Should the resulting P simply be averaged? It is not obvious.

Didactics: It is useful to give one copy of the graph (and a tool to measure lengths) to each group of three to five students. Let them work out the period before they receive any description of the radiation or the object that emits the radiation. The groups should talk quietly enough so that neighboring groups cannot hear the answers discussed. When (almost) all groups are finished, ask for the periods from one group, then ask whether any groups have different periods. Let the students suggest reasons for the differences in period derived by different groups. The end result should be the realization that the derived period has an inherent uncertainty, and an estimate of the magnitude of the uncertainty.

   Once the discussion is finished, students may be told: If one Fourier-analyzes a much longer data set, one obtains a "signal" in a small range of periods centered on a period of 1.7 days.

The astrophysical setting: (not really needed for this practical problem) : The object Her X-1 is so named because it was the first x-ray source observed in the part of the sky that includes the star constellation of Hercules. It was first carefully observed in the early1970's by the X-ray satellite with the Swahili name Uhuru. The satellite was so named because it was launched off the coast of Kenya on December 12, 1970, on the anniversary of the country's independence.

   Her X-1 is a binary star: a normal star and a neutron star orbit around each other. The neutron star is an x-ray pulsar with a rotational period of 1.24 sec. (Problem 3.4 includes a diagram for a pulsar.) The x-ray data in the graph are taken with low time resolution, so that the pulsar signal is averaged out. Clearly, there is a period of roughly two days. This is the period at which the two stars orbit about each other.

   Why the x-ray emission? A binary pair of stars necessarily has angular momentum. If gas from a normal star escapes toward a neutron star, it has angular momentum relative to that neutron star, and cannot fall directly toward it. Instead, the gas swirls about the star, gradually spiraling inward. The gas becomes part of an "accretion disk", centered on the neutron star. In this disk, each part of the gas moves very nearly at the Keplerian velocity around the star, but there is also a very small component of velocity inward.

   The small inward velocity has a major physical implication: the gas must lose angular momentum. Viscosity is far too small. Probably turbulence within the gaseous disk acts like a viscosity, gradually transporting angular momentum outward, thus allowing the gas to migrate inward. (The turbulence, in turn, may be caused by magnetic fields.) Inevitably, as the gas sinks deeper into the gravitational well of the neutron star, some of the orbital kinetic energy is turned into heat. The detailed process by which this occurs is not known, though some kind of viscous heating is probably involved. The gas in the disk is heated to tens of millions of degrees and so radiates largely in x-rays (hv of the order of kT, see problem 4.4).

   We observe the orbital period in the x-rays because we view different parts of the accretion disk as it revolves, with the neutron star, around the other star. Sometimes we see the "top" of the disk, sometimes the edge, sometimes the "bottom" of the disk.

   Finally, when the gas has swirled very close to the star, the star "grabs" the gas (probably by magnetic forces) so that the gas falls the remaining distance onto the neutron star. The orbital kinetic energy which the gas had just before infall is released in the form of heat on or near the surface of the neutron star, and this heat is radiated away as x-rays. Probably the gas is funneled onto the neutron star near its magnetic poles. Then we see a "pulsating" x-ray source because the rotation of the neutron star repeatedly carries the x-ray emitting region into and out of our view. An x-ray pulsar is thus analogous to a radio pulsar.


3.3 Radiation from neutron star, accretion disk
      energy in Keplerian orbits

The problem: Gas spirals slowly toward a (non-magnetic) neutron star, forming an accretion disk, and then falls a short distance (by much less than a neutron star radius) onto the neutron star. What is the ratio of the energy radiated away by the gas while in the accretion disk to the energy released by the gas upon falling onto the neutron star? (Assume that gas in the accretion disk at any one time is in a practically circular orbit satisfying Kepler's third law.)

The theoretical answer: One.

Interpretation: Why "one"? Every Keplerian orbit involves a kinetic energy equal to half the gravitational energy released while the gas approached from far away, starting approximately at rest. Therefore, each element of gas in a Keplarian orbit has converted half the gravitational energy to heat and radiation. Specifically, this is valid for those elements of gas just ready to fall onto the star. Once they have fallen onto the star, the entire gravitational energy must have been converted to heat. Since half of that heat was accounted for in the accretion disk, the second half must be heat released on the neutron star.

Interpretation: This situation may be approximately true in the class of x-ray sources called Low Mass X-ray Binary Sources. However, in many other accretion disks, including objects like Her X-1, magnetic fields are important. The physics of these accretion disks is still a very active subject of research.


3.4 Crab Nebula and pulsar slow-down
      rotational energy, moment of inertia

The setting: The Crab Nebula is the gas ejected by a supernova explosion seen on Earth in the year 1054 A.D. Even now, centuries after the explosion, the nebula is still filled with highly relativistic electrons and with electrical currents and their magnetic fields. The energies inherent in these phenomena cannot be left over from the explosion. Apparently, energy is supplied continuously, at the rate of about 1.2x105Lo (Lo = solar luminosity). What is the source of this enormous energy?

 


 

A sequence of images showing the flashes at visible wavelengths from the Crab pulsar, located at the center of the Crab Nebula. Credit: National Optical Astronomy Observatories

 


 

   F. Pacini (of Italy) suggested that the energy needed for energizing the Crab Nebula might come from the slowing-down of a rotating neutron star at the center of the nebula. But how would one detect a neutron star? One year later, in 1967, pulsars were discovered. Pulsars are beacons of radiation that sweep past us. Presumably the beacons are attached to some rotating compact object, such as a neutron star. Initially the pulsars were detected in the radio range, and one pulsar was promptly observed at the center of the Crab Nebula, sweeping past us 30 times per second.

   Pulsars have a very precisely defined period, but over many years the period increases, indicating that the rotation of the central object gradually slows down. The energy made available by the slow-down of the pulsar in the Crab Nebula, if the central object is a neutron star, neatly accounts for the energy needed in the Crab nebula. This agreement is one of the main reasons that pulsars were promptly identified with neutron stars, thus providing good evidence that neutron stars exist.

The problem: Write down the kinetic energy of rotation, E, of a neutron star in terms of M, R, and the rotational period P. Assume uniform density. Then write an equation for the rate of change of rotational kinetic energy, dE/dt, in terms of M, R., P, and dP/dt, where dP/dt measures the slow-down of the rotation (with dE/dt << E/P, R = constant). Evaluate dE/dt for the pulsar in the Crab Nebula using observed values P = 0.0333 s and dP/dt = 4.21x10-13 s/s, and theoretical values for a neutron star R = 10 km and M = 1.4Mo. Compare dE/dt to the power needed to energize the Crab Nebula.

The solution: The moment of inertia of a homogeneous spherical object is I = 2/5 MR2. The rotational energy is E = 1/2 I omega2 = 2pi2I/P2, and dE/dt = -8/5 pi2MR2 (dP/dt) / P3 = 1.2x105Lo.

Interpretation: The decrease in rotational energy matches the energy needed by the Crab Nebula remarkably precisely! Of course, we assumed a mass of 1.4Mo for the neutron star, and there is no independent measurement of M. But this value is in the correct range expected from theory.


3.5 Clusters of galaxies and cosmology
      integral conditions

Physics purpose: Theoretical integral conditions provide useful information when detailed observational information is lacking. Integral conditions suggest that clusters of galaxies consist mostly of invisible "dark matter".

Didactic purpose: When we deal with clusters of galaxies, we are truly dealing with "astronomical numbers" : huge distances, sizes, and time scales. And yet the human mind can deal with such phenomena!

Physics Introduction to the integral condition called the Virial theorem. Consider any number of planets circling around the Sun. For each, the kinetic energy is -1/2 x their gravitational energy. Add them up, and you find that the total kinetic energy in the system K = -1/2 OMEGA, where OMEGA is the total gravitational energy. More generally, consider a star circling around the center of a cluster of stars, experiencing gravity g(r) = GM(r)/r2. For each star, the kinetic energy is -1/2 times its gravitational energy. For the total system of the stars, K = -1/2 OMEGA.

   In fact, K = -1/2 OMEGA is true for any system of stars or galaxies, held together by their own gravity, and kept from collapsing by orbital motions. It also holds for a star, held together by its own gravity, and kept from collapsing by thermal motions (gas pressure). (See didactics of problem 5.3). The relation K =-1/2 OMEGA is called the Virial Theorem. Its general proof is given in the Appendix, but it is not very useful for most students.

   Often in astrophysics, we have insufficient information to make detailed models of observed objects. But the Virial Theorem provides important constraints. Here it is used for the establishing the "missing mass" in clusters of galaxies.

The problem: A spherically symmetric cluster of galaxies, diameter 3x106 light years, contains 102 galaxies, distributed uniformly through the cluster, traveling at a (root mean square) speed of 103km/s, each containing stars with a total mass 1011Mo.

  1. Calculate the gravitational and the kinetic energies residing in the cluster of galaxies. (For the gravitational energy, assume uniform density, so that OMEGA = -3/5 GM2/R.)

  2. In view of the Virial theorem, are the galaxies likely to be kept within the cluster by the gravity due to the stars in the galaxies?

  3. Compare the time needed for a galaxy to cross the cluster to the probable age of the cluster, roughly the age of the Universe, 1.3x1010years. Might this cluster of galaxies simply be moving apart if it was created early in the age of the Universe?

  4. If "dark matter" is distributed uniformly in the cluster, with the same random velocities, how much matter would be needed to satisfy the Virial theorem? Compare to the amount of matter observed in galaxies.

The solution: The kinetic energy per galaxy (on average) is 1/2 2x1041 (106)2 =1053j . For the cluster K = 1055 j. The gravitational energy of the cluster is 3/5 6.7x10-11 (2x1043)2/1.5x1022 = 1054j. This gravitational energy cannot constrain the random motions. The crossing time is 3x1022/106 = 3x1016s = 109years. Galaxies could easily have escaped by now. Since they have not, the Virial Theorem must be satisfied. The total mass in the cluster must be about twenty times the mass of stars observed in galaxies.

Interpretation: Typical clusters of galaxies may be 107light years in diameter and contain roughly 103galaxies with a size of 105 light years, separated by 106light years. Typical motions among the galaxies are 103km/s. These clusters have presumably lasted for over 1010years, and so the galaxies cannot merely be flying apart. Yet, if we measure the mass of each galaxy according to the amount of starlight we receive, the total mass of the galaxies does not yield enough gravity. For a while there was the possibility that much gas resides in the vast space between the galaxies. The gas is now observed by its x-ray emission. Its mass is not enough.

   There must be much "dark matter" in clusters of galaxies. What is this matter? We know nothing more than we know regarding the "dark matter" in our Galaxy (Problem 1.6.) However, the problem is more significant for clusters of galaxies: It is estimated that the dark matter in clusters of galaxies amounts to roughly 0.2 of the density that closes the Universe (that is, 0.2 of the density for which the Universe will expand forever with forever decreasing velocity of expansion). For comparison, one can compute the amount of baryonic mass (protons, neutrons) in the Universe that was needed to create, during the Big Bang, the now observed deuterium, 3He and 7Li. One finds an upper limit of baryonic mass of 0.2 times the density that closes the Universe. Therefore, it is possible that the dark matter resides in galaxies in the form of white dwarfs, neutron stars, or black holes left over from earlier stars, or large planets or very cold gas. However, many astronomers take this upper limit on baryonic mass as an additional argument that the universe contains much non-baryonic matter, such as neutrinos or axions. For instance, neutrinos might have a finite mass such that they are gravitationally confined to clusters of galaxies.

Didactics:

  1. Students may have heard that we look back in time as we look at ever further objects. They may ask: Perhaps the clusters of galaxies are so far away that they are young and have not had time to expand? No, these clusters of galaxies are within 109light years from us.

  2. The gas between galaxies has been observed to emit primarily x-rays, indicating a temperature of the gas about 108oK. Why is the gas so hot? At this temperature, the thermal speed of protons and the sound speed are about 103km/s. Perhaps, when the gas was less hot, the galaxies moved through the gas supersonically and created powerful shock waves. They churned up and heated the gas until the galaxies were no longer supersonic relative to the gas.

  3. The Virial theorem must hold true for equilibrium, but it is not sufficient to guarantee equilibrium. Indeed, galaxies of a uniform density throughout the cluster would not be a distribution in equilibrium. Therefore, the use of the Virial Theorem in problems such as this one only yields estimates, surely no better than a factor of two. But that is quite satisfactory for exploratory problems such as this one.

  4. The gravitational energy of a self-gravitating spherically symmetric object can be derived if one considers the object having been built up to a radius r with mass M(r) and then adding mass m: the gravitational energy released is -GM(r)m/r . [See problem 1.6 and the Appendix for discussion of g(r).] The whole object has the gravitational energy OMEGA = -integral[GM(r)/r]dM(r). With dM(r) = 4pirhor2dr and uniform density rho, the result is OMEGA = -3/5 GM2/R.