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| Neutron Stars and Black Holes |
1. A white dwarf is about 100 times smaller than the Sun and a neutron star is about 1000 times smaller than a white dwarf. Assuming all contain about a solar mass, the ratio of densities goes as the cube of the diameters, so the Sun has a density of about 1 g/cm3, a white dwarf about 106 g/cm3, and a neutron star about 1015 g/cm3. These estimates turn out to be approximately correct.
2. Primarily because their surface areas are so small. Also, they often emit at a range of wavelengths, which makes their emission less concentrated at particular wavelengths.
3. The energy is primarily gravitational, from the enormous contraction down to neutron star size. This huge energy is slowly being radiated away by neutrino and photon emission.
4. The total luminosity of the neutron star should be only about 15% as much as the Sun, and it would mostly be X-ray.
5. No, normally the fastest pulsars are the youngest, and globular clusters are old with few recent supernova events that could have produced young pulsars. The millisecond pulsars thus must be old pulsars that have been spun up since their births. Some form of mass transfer in binary systems is suspected as the culprit.
6. First, a young, rapidly rotating, neutron star only produces a pulsar if the "lighthouse beams" sweep over the Earth. Second, over time the rotation slows and the magnetic field weakens and the pulsar mechanism is expected to fade out.
7. If the answer depends only on the gravitational field, the answer is no because the external gravitational field of a black hole is exactly the same as the external gravitational field of a star with the same mass. However, it might be that the environment near a black hole (accretion disks and possible jet outflows) alters the probability (either up or down, depending on details) of accreting matter from the surrounding region.
8. There are 365.25 x 24 = 8766 hours in a year. The period of the binary pulsar is 7.75 hours, so it makes (8766 hr) / (7.75 hr/revolution) = 1169 revolutions per year.
9. The present size of the orbit is about one million kilometers. The orbit shrinks by about 3 mm = 0.000003 km per revolution. There are 1169 revolutions per year because the period is 7.75 hours per revolution (see Exercise 8), so the orbit shrinks by about 1169 x 0.0000003 = 0.0035 km/yr. At that rate the time to shrink to zero is
10. You can't. Since the only three distinguishing characteristics for a black hole are charge, mass, and angular momentum, all nonrotating, uncharged black holes of the same mass are identical.
11. No. Any object has a Schwarzschild radius. It is just the radius that you would have to squeeze the object down to in order for its gravitational field to become strong enough to form a black hole. Of course the Schwarzschild radius of a watermelon is incredibly small because it has small mass, but it is not zero.
12. Type II Cepheids are Pop II and type I are Pop I. but Pop I and Pop II stars have very different metal abundances, and metals are crucial to the surface opacities. Thus, it is plausible that the period-luminosity relation is different for a low-metal Pop II Cepheid relative to a higher metal Pop I Cepheid.
13. From the electromagnetic spectrum:
| Spectrum of Electromagnetic Radiation | ||||
|---|---|---|---|---|
| Radio | > 109 | > 10 | < 3 x 109 | < 10-5 |
| Microwave | 109 - 106 | 10 - 0.01 | 3 x 109 - 3 x 1012 | 10-5 - 0.01 |
| Infrared | 106 - 7000 | 0.01 - 7 x 10-5 | 3 x 1012 - 4.3 x 1014 | 0.01 - 2 |
| Visible | 7000 - 4000 | 7 x 10-5 - 4 x 10-5 | 4.3 x 1014 - 7.5 x 1014 | 2 - 3 |
| Ultraviolet | 4000 - 10 | 4 x 10-5 - 10-7 | 7.5 x 1014 - 3 x 1017 | 3 - 103 |
| X-Rays | 10 - 0.1 | 10-7 - 10-9 | 3 x 1017 - 3 x 1019 | 103 - 105 |
| Gamma Rays | < 0.1 | < 10-9 | > 3 x 1019 | > 105 |
X-rays have wavelengths of approximately 10 to 0.1 Angstroms. From the Wien Law this corresponds to blackbody temperatures of approximately 3 million to 300 million K (see also this applet).
14. The radius is R = 2MG/c2 . The product 2G/c2 is
| 2G/c2 | = | 2 x 6.67 x 10-8 cm3 g-1 s-2 / (3 x 1010 cm/s x 3 x 1010 cm/s) |
| = | 1.482 x 10 -28 cm/g |
so that R = ( 1.482 x 10-28 cm/g ) x M. Plugging in the mass M in grams gives the radius in centimeters. The results are summarized in the following table.
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15. The Crab Nebula is about 10 light years in diameter and the Crab Pulsar about 10 km in diameter; a hydrogen atom is of order 10-8 cm in diameter. Thus, both ratios are of order 1013.