Astronomy - Stars

(View Cosmos DVD 5, episode 9, on solar activity; mov of solar surface (source) and Hinode mpg of same (source); also Hinode mpg and magnetic field images from 12/13/06 flare (source) and Hinode mov of solar X-ray jets (source).)

(source).

Sunspot activity defines the solar cycle; note the Maunder Minimum between 1645 and 1715:

(source)

The "solar wind", particles streaming from the surface of the Sun, gives us auroras and comet tails. It is a ubiquitous phenomenon, especially in hot young stars like LL Orionis:

(source)

Even radiation pressure sculpts the interstellar medium.

(source, source, source, source, source, source, source)

r = 1000 / 5.31 = 188.3 parsecs, times 3.084 * 10¹⁶ m/parsec

= 5.81 * 10¹⁸ m

L = 4 p * (5.81 * 10¹⁸ m)² * 9.9775 * 10^-11 W/m²

= 4.23 * 10²⁸ W, divided by 3.85 * 10²⁶ W for our Sun

= 110 L_solar

We can find the effective temperature of a star as follows. Use the "Black Body" button and adjust the temperature scroll bar so that the black body curve is as close to the spectrum as possible. Since a star is not a perfect black body, we have some choices here: we can match the peak wavelength (in this example, around 3970 Angstroms, giving us a temperature of 7300 K) or we can match the leading or trailing slope (in this example, closer to 9300 K). In general, matching the trailing slope tends to be more accurate than the other options.

A given atom, ion or molecule only exists in a star's atmosphere for a range of temperatures. For lower temperatures, an ion may become neutral, or for higher temperatures, an atom may become ionized, in either case changing the wavelengths of its absorption lines. At sufficiently high temperatures, molecules will break up and their lines (or bands) disappear altogether. We can use the following table to look for some characteristic lines as a check on our black body temperature. Choose the element from the drop-down menu ("Reference Spectra Element") and look for consistent matches for the lines displayed, indicating the presence of that element in the star's atmosphere. If the lines in the table are present, the temperature should be in or close to the range given, and if the lines are extremely strong (dark), the temperature should be close to the peak temperature. The wavelengths for the CH and TiO bands are the wavelengths at the left-hand edge of the bands.

specie	lines (Angstroms)	T Range (K)	Peak T (K)
H	3966, 4097, 4336, 4856, 6555	5000-40000	9000
He	4471, 4542	10000-50000	29000
Ca	4227	2000-5500	3000
Ca+	3934, 3968	3000-7000	5000
Na	5890, 5896	2000-5500	3000
Mg+	2796, 2802, 4481, 5173	8000-30000	9200
Si+	4128, 4131	8000-20000	9200
Si++	4552	20000-40000	25000
Si+++	1394, 1403, 4089, 4116	30000-50000	40000
Fe	4045, 4143, 4299, 4325, 4384, 5270	2000-7000	4500
Fe+	4173, 5316	4000-8000	5700
CH band	4300-4315	5000-6000	5500
TiO bands	4762, 4955, 5167, 5448, 5862, 6159, 6384	2000-4000	3000

(source)

The spectrum for 108 Virginis is a low resolution spectrum, so all we can see are the broad Hydrogen lines, which do not help much in verifying our black body temperature. This check is much more useful with high resolution spectra, especially spectra in the visible and near ultraviolet.

With the temperature and the luminosity, we can compute the star's radius using

luminosity = 4 p s radius² temperature⁴

r = (4.23 * 10²⁸ / 4 p s 9300⁴)^1/2

= 2.82 * 10⁹ m, divided by the radius of our Sun (7 * 10⁸ m)

= 4 r_solar

Note that the presence of metallic absorption lines indicates that the star is a later-generation star: the metals were inherited from earlier-generation stars.

Portfolio Exercise 1: You will be given a list of 9 stars which you are to analyze for each of the first 4 portfolio exercises. In this exercise, compute each star's distance in parsecs and in meters; compute each star's luminosity in both Watts and solar units; find the black body temperature for each star; and compute each star's radius, both in meters and in solar units.

The broadening of most narrow absorption lines is primarily due to the motion of the atoms involved, although pressure and other factors can be equally important. For instance, Hydrogen and Helium lines are broadened by the Stark Effect: since their nuclear charge is small, they are particularly susceptible to electric fields created by ions in the stellar atmosphere. These fields perturb their energy levels and are primarily responsible for their line broadening.

Atomic speeds in the photosphere can be estimated using the width of some absorption lines:

line width = 2 * speed * central wavelength / c

The spectrum for 108 Virginis has too low a resolution to be able to compute Doppler Shifts; only the broad Hydrogen lines are present, and these are unsatisfactory for computing atomic velocities, because of the Stark Effect. But using the prominent Sodium line of Epsilon Indi (at 5890 Angstroms), we can compute an approximate speed for these atoms in the star's atmosphere as follows. Click on the graph at 5890 Angstroms, and "Zoom In"; if the central wavelength is not 5890, click again or use the "Central Lambda" scroll bar until it is. Then click the "Gaussian Fit" button; a small curve appears around the line. Change the Standard Deviation using the scroll bar until the curve is a close fit to the line on the graph. In this example, it is a close fit between 1.5 and 2.4. Taking the average (1.95), we compute the approximate speed of these atoms as

v = 1.95 * 3 * 10⁸ / 5890 = 99000 m/s

Portfolio Exercise 2: Using both the Calcium lines near 3900 Angstroms, and the Sodium lines near 5900 Angstroms, compute the approximate atomic speeds from the line width, for each star. Are they consistent for each star? What does the range tell you about this computation?

Spectral Classes and Effective Temperatures

Spectral Class	T (Main Sequence)	T (Giants)	T (Supergiants)
O3	48000
O5	44000
O6	43000
O8	37000
B0	31000		30000
B1	24100
B2	21080
B3	18000
B4	15870
B5	14720
B8	11950
A0	9572		9550
A2	8985		9000
A5	8306		8500
A7	7935		8300
F0	7178	7178	8030
F2	6909	6909	7780
F5	6528	6528	7020
F8	6160	6160	6080
G0	5943	5943	5450
G2	5811	5811	5080
G5	5657	5657	4850
G8	5486	5486	4700
K0	5282	5282	4500
K2	5055	5055	4400
K3	4973	4973	4230
K5	4623	4623	3900
K7	4380	4380	3870
M0	4212	4212	3850
M2	4076	4076	3800
M5	3923	3923
M8	2400

Portfolio Exercise 3: Using the information you have collected so far, assign a spectral class to each of your stars.

For each star, compute the absolute and apparent magnitudes. Assign a luminosity class to each star.

Look each star up on SIMBAD. Comment on your success rate.

The volume in the immediate neighborhood of the Earth has a mean density of about 5 atoms per cubic centimeter (mostly Hydrogen (source); 1 cm³ = 10^-6 m³). This falls to about 0.3 in the Solar neighborhood and to less than 0.001 in the "Local Bubble". In interstellar space the average density is around 1 atom per cm³, and it is even less in intergalactic space. Larger dust particles are about a million times rarer, except in nebulae and clouds, where the density runs between 10 to a million atoms, molecules or dust particles per cm³. But since space is so vast, some of the electromagnetic radiation traveling through the interstellar medium is absorbed, and because shorter wavelengths are scattered more than longer ones, there is additional attenuation in the red and infrared wavelengths. The overall loss is called extinction, and the preferential scattering causes reddening. Molecular Cloud Barnard 68 illustrates both phenomena.

While the spectra in the Spectrum Viewer have been corrected for extinction and reddening, it is important to know how astronomers can compute the necessary corrections. Armed with an independent measure of distance (for instance, from the parallax), we can compare the observed apparent magnitude with the expected apparent magnitude. The result is

extinction = (m_observed - m_expected) / distance

Spectral classification is particularly useful here. Along the main sequence of a Hertzsprung-Russell diagram, there is essentially a 1:1 correspondence between spectral class and absolute magnitude. Because of this, stars of a given spectral class but at various distances can be used to compute extinction. For instance, the star BD +31 640 is located in the open galactic cluster IC 348. It is an A3V star located 236 parsecs away, with an absolute magnitude of 1.5. This means the expected apparent magnitude (m = M - 5 + 5 log D) is

1.5 - 5 + 5 log 236 = 8.36

(11.4 - 8.36) / 0.236 = 12.86 magnitudes / kpc

and it is more reasonable to quote the extinction simply as 3.04 magnitudes.

The Hertzsprung-Russell Diagram

(source)

The position of a star on the main sequence (the line from upper left to lower right) is determined by its mass. Main sequence stars of spectral type G0 are about 1 M_solar, while those of type B0 are about 15 M_solar (stars begin their main sequence lives with masses varying from 0.04 to about 200 M_solar (source)). Almost half of the stars in the neighborhood of our Sun have a mass less than about a third of the Sun's; about one in 300 of the neighborhood stars have a mass in excess of 8 M_solar.

Straight diagonal lines parallel to the tangent at the center of the main sequence correspond roughly to lines of equal radius. Our B0 main sequence star has a radius around 10 R_solar. The spacing of these lines of equal radius is roughly logarithmic.

Another thing to notice is that the white dwarf region of the diagram is distributed over a relatively wide range of luminosities and temperatures. This means that knowledge of temperature does not translate as easily into knowledge about luminosity as it does with main sequence stars. At first glance one might be tempted to say the same thing about the giant and supergiant regions, but because they represent definite paths in the life cycles of those stars, and the stars themselves are much larger and brighter, they do not represent so much of a problem. Because they are so small and relatively dim, white dwarfs are very hard to measure.

These graphs illustrate the lifetimes of stars of 0.5, 1, 1.5, 3, 5 and 9 solar masses, from various perspectives:

• relative to the end of the main sequence phase of a 0.5 solar mass star
• relative to the lifetime of a 1 solar mass star
• relative to the lifetime of a 1.5 solar mass star
• relative to the lifetime of a 3 solar mass star
• relative to the lifetime of a 5 solar mass star
• relative to the lifetime of a 9 solar mass star
• showing the birth of a 9 solar mass star

• The brown phase represents the Hyashi Track, during which the protostar shrinks significantly, dropping in luminosity but increasing in temperature due to the increased internal pressure. At the beginning of the Hyashi Track, the core is convective. For stars less than 1.3 solar masses, the core is radiative by the end of the track, and the pp chain dominates the fusion reactions that begin then. For stars greater than 3 solar masses, the core is convective throughout and the CNO cycle dominates.
• The green phase represents the approach to equilibrium that precedes the main sequence. The time of the beginning of the main sequence is called the Zero Age Main Sequence; a star arrives on the main sequence then, and stays essentially at that one place on the H-R diagram until the end of its main sequence life.
• During the main sequence (shown in yellow), the outward pressure from the fusion reactions taking place in the core balances the inward pull of gravity.
• What happens next is a tug of war between those opposing forces: as fusion products build up in the core, the core shrinks, the pressure increases, the temperature increases, and fusion reactions move outward into shells. As the fusion reactions take place in regions closer to the surface of the star, the outward pressure from those reactions enlarge the star. Depending on the size of the star, this process can repeat, sometimes with multiple shells supporting different reactions simultaneously.

  The period of Hydrogen shell burning is shown in orange, and the red giant branch is shown in red. During this portion of stellar evolution, the star's luminosity and radius increase enormously, but the expansion is cooling and results in a drop in surface temperature.

• At the end of the red giant branch, if the star is less than 0.5 solar masses, no Helium fusion occurs. If the star is less than 2.25 solar masses, there is a Helium flash which begins Helium fusion; larger stars begin burning Helium in a smooth transition. Stars larger than about 9 solar masses do not pass through the red giant branch. Here is a graph showing the change in stellar radii from the beginning of the main sequence to the end of the red giant branch. The black vertical line indicates 1 A.U.:
  
• The horizontal branch is shown in purple. This is another equilibrium state, during which the star's luminosity is almost constant. Its surface temperature and radius may change depending on the nuclear reactions underway. For stars greater than 4 solar masses, Carbon fusion also occurs during the horizontal branch (and a Carbon flash can occur). For stars greater than 8 solar masses, Oxygen fusion also occurs, and for stars larger than 10 solar masses, Silicon burning occurs in the terminal stages before supernova. These stars may also begin burning Helium before completing the red giant branch.

Nuclear Reactions

• ¹H + ¹H -> ²H + e⁺ + n_e
• ²H + ¹H -> ³He + g
• ³He + ³He -> ⁴He + ¹H + ¹H

(The first two of these reactions occurs twice for each occurrence of the third reaction.)

energy = mass * c²

Most people think of this as "converting matter into energy". Since energy is a quality of matter, and not really a separate entity, it is better to think of this process in terms of the constituent quarks of the protons and neutrons involved. Then we can see that each reactant has an energy content associated with the fact that they are bound states of particles which are electrically as well as strongly interacting. The product has a different binding energy, and the fusion process liberates the difference as gamma rays (with electromagnetic energy) and thermal energy of motion. The energy of any neutrinos which are produced is usually included as part of the energy released, because the neutrino has a very small and not very accurately measured mass in comparison with the other particles.

E = (4 * 1.673 * 10^-27 - 6.647 * 10^-27 - 2 * 9.109 * 10^-31) * (3 * 10⁸)²

= 3.881 * 10^-12 J

(= 24.225 MeV)

9.92 * 10³⁷ * 4 * 1.673 * 10^-27

= 6.64 * 10¹¹ kg / s.

• Hydrogen fusion ignites at 10⁷ K (pp chain)

  • ¹H + ¹H -> ²H + e⁺ + n_e (0.38)
  • ²H + ¹H -> ³He + g (5.5)
  • ³He + ³He -> ⁴He + ¹H + ¹H (12.9)

  This process takes place not only in the interior of main sequence stars, but on the surfaces of white dwarfs that have accreted Hydrogen from a binary partner. This is called a nova, and the process usually repeats over time.

  or (CNO Cycle, dominant beyond 16 MK)

  • ¹²C + ¹H -> ¹³N + g (2)
  • ¹³N -> ¹³C + e⁺ + n_e (1.1)*
  • ¹³C + ¹H -> ¹⁴N + g (11.2)
  • ¹⁴N + ¹H -> ¹⁵O + g (3.6)
  • ¹⁵O -> ¹⁵N + e⁺ + n_e (1.8)*
  • ¹⁵N + ¹H -> ¹²C + ⁴He (4.9)

• Helium fusion ignites at 10⁸ K (Triple Alpha Process)

  • ⁴He + ⁴He -> ⁸Be + g (-0.09)
  • ⁴He + ⁸Be ->¹²C + g (7.4)

• Carbon fusion ignites at 2-6 * 10⁸ K

  • ¹²C + ⁴He -> ¹⁶O + g (7.2)
  • ¹²C + ¹²C -> ²⁰Ne + ⁴He (4.7)
  • ¹²C + ¹²C -> ²³Na + ¹H (2.2)
  • ¹²C + ¹²C -> ²³Mg + n + e⁺ + n_e (-3.1)
  • ¹³C + ⁴He -> ¹⁶O + n + e⁺ + n_e (1.7)

  Again, this process does not just occur during the horizontal branch of a massive star. If a Carbon white dwarf accretes enough material from a binary companion to raise its internal temperature sufficiently, the white dwarf will begin Carbon fusion throughout its mass and explode in a Type Ia Supernova (Type I Supernovae spectra lack Hydrogen lines).

• Oxygen fusion ignites at 1.4-2 * 10⁹ K

  • ¹⁶O + ⁴He -> ²⁰Ne + g (4.8)
  • ¹⁶O + ¹⁶O -> ²⁸Si + ⁴He (0.47)

• Silicon fusion ignites at 3.5 * 10⁹ K

  • ²⁸Si + ⁴He -> ³²S + g (16)
  • ³²S + ⁴He -> ³⁶Ar + g (6.7)
  • ³⁶Ar + ⁴He -> ⁴⁰Ca + g (7)
  • ⁴⁰Ca + ⁴He -> ⁴⁴Ti + g (5.1)
  • ⁴⁴Ti + ⁴He -> ⁴⁷V + ¹H (-0.37)
  • ⁴⁰Ca + ⁴He -> ⁴³Ti + n + e⁺ + n_e (-11.7)
  • ⁴⁴Ti + ⁴He -> ⁴⁸Cr + g (7.7)
  • ⁴⁸Cr + ⁴He -> ⁵²Fe + g (7.9)
  • ⁵²Fe + ⁴He -> ⁵⁶Ni + g (8)
  • ²⁸Si + ²⁸Si -> ⁵⁶Ni + g (29.2)
  • ⁵⁶Ni -> ⁵⁶Co + e⁺ + n_e (1.1)*
  • ⁵⁶Co -> ⁵⁶Fe + e⁺ + n_e (3.5)*

The following masses (in "atomic mass units"; 1 amu = 1.66055 * 10^-27 kg) will enable you to compute more accurate energy yields from these reactions.

n	1.0087	15N	14.9963	40Ca	39.9516
1H	1.00725	15O	14.9987	43Ti	42.9564
2H	2.01355	16O	15.9905	44Ti	43.9476
3He	3.0149	20Ne	19.9869	47V	46.9423
4He	4.0015	23Na	22.9838	48Cr	47.9408
8Be	8.00311	23Mg	22.9875	52Fe	51.9338
12C	11.9967	28Si	27.979	56Fe	55.9206
13C	13.0001	32S	31.9633	56Co	55.925
13N	13.0019	36Ar	35.9576	56Ni	55.9267
14N	13.9954

You can find the atomic numbers of these atoms from the Periodic Table of the Elements.

It is often said that ⁵⁶Fe is the most stable nucleus. A nucleus is said to be stable if its energy per baryon is a minimum for its atomic number. Unstable nuclei undergo radioactive decay, emitting Helium nuclei (alpha decay, which heavier nuclei often do because they lose energy quicker), electrons or positrons (beta decay, accompanied by neutrinos), or photons (gamma decay, if nothing else is allowed). The reactions marked with an asterisk above are all beta decays. Fusion stops with Iron because ⁵⁶Fe has the lowest energy per baryon of any nucleus.

Portfolio Exercise 4: For each of the nuclear reactions above, compute the energy yield.

Stellar Evolution (II)

IC 4406 (source)

Eta Carinae (source)

NGC 2440 (source)

*Most neutron star masses are close to 1.4 M_solar; the largest measured so far is about 2.1 M_solar. The actual limit depends on how neutron matter behaves, and is the subject of ongoing research.

Sirius B is a white dwarf whose absolute magnitude is 11.2 and whose mass is approximately equal to that of our Sun. Its luminosity is then

L = 10^{(4.83 - M)/2.5}

= 10^-2.548 = 0.0028 L_solar

r = (L / (4 p s T⁴))^1/2

= (0.0028 * 3.85 * 10²⁶ / (4 p s 14800⁴))^1/2

= 5.62 * 10⁶ m = 0.008 R_solar

r = m / (4/3 p r³)

= 2 * 10³⁰ / (4/3 p (5.62 * 10⁶)³), times 0.001 to convert kg/m³ to g/cm³,

= 2.7 * 10⁶ g / cm³ (!)

escape velocity = (2 * G * mass / radius)^1/2

= (2 * G * 2 * 10³⁰ / 5.62 * 10⁶)^1/2

= 6.89 * 10⁶ m/s

= 0.023 c,

surface gravity = G * mass / radius²

= G * 2 * 10³⁰ / (5.62 * 10⁶)²

= 4.23 * 10⁶ m / s²,

White Dwarf	Absolute Magnitude	Masssolar	Radiussolar
40 Eridani B	11	0.501	0.0136
Procyon B	13.2	0.604	0.0096
Sirius B	11.2	1	0.0084

(source)

The neutron star at the heart of the Crab Nebula has about 1.4 times the mass of Sirius B but its radius is only about 10 kilometers. This is 560 times smaller, so its density is 250 million times larger, its escape velocity is 28 times larger (0.64 c) and its surface gravity is 440 thousand times greater! We have already seen the Crab Nebula in multiple wavelengths; here is an x-ray image of the neutron star and supernova remnant Cassiopeia A:

(source)

The planetary nebulas glow from irradiation by the remnant white dwarf or neutron star.

Now consider our old friend Betelgeuse. As we mentioned in the Introduction, it is an M2Ib star located 427.3 light years from us. Its temperature is about 4000 K, its mass is about 20 M_solar and its luminosity is about 50000 L_solar. That means that its radius is

r = (50000 * 3.85 *10²⁶ / (4 p s 4000⁴))^1/2

= 3.25 * 10¹¹ m = 464 R_solar

energy flux = luminosity / (4 p distance²)

= 10¹⁰ * 3.85 *10²⁶ W / (4 p (427.3 ly * 9.461 * 10¹⁵ m/ly)²)

= 0.0187 W/m²

SN 1994D in NGC 4526 (source)

(source).

Relativity and Black Holes

(View Cosmos DVD 6, episode 9, on Flatland and curved space.)

We define the future light cone of a particle to be the collection of all the places that particle might ever be at any given time in its future. The surface of the cone is where it could be if it travels at the speed of light, and the interior of the cone is where it could be if it travels more slowly. Therefore photons are destined to stay on the surface of their light cones and massive particles are destined to stay in the interior of their light cones.

All this means that you will indeed expire when you cross the horizon of a black hole. Assuming you cross feet first, the future light cone of your feet is pointing into the center of the black hole, so no nerve impulses can reach your head from there. When your heart crosses, no blood can flow from it to your head. And when your head crosses the horizon, the parts of your brain cannot communicate with each other and your consciousness ceases.

For a black hole of mass M and angular momentum L, the horizon is a spherical surface located at a distance

r = (G M² + (G² M⁴ - c² L²)^1/2) / (M c²)

r = 2 G M / c²

= 2953 meters * M / M_solar

Note that this implies that G / c² is a conversion factor for converting mass to length.

ev = (2 G M / r)^1/2

= c

The surface gravity at the horizon is

g = c⁴ / (4 G M)

= 1.52 * 10¹³ m/s² / (M / M_solar)

Consider now the tidal acceleration experienced by a 2-meter object at the horizon:

G M / r² - G M / (r+2)²

= (c⁶ / (4 G M)) (c² + 2 G M) / (c² + G M)²

c⁶ / (2 G² M²)

= 2 * 10¹⁰ / (M / M_solar)²

It is expected that many if not most galaxies harbor a supermassive black hole in their cores, probably surrounded by an accretion disc, whose matter is accelerated to relativistic speeds and radiates tremendous amounts of energy as it falls into the horizon:

(source)

(View movie of crab pulsar (source))

(View simulation of blazar BL Lac (source))

Portfolio Exercise 5: Find a white dwarf other than Sirius B and a neutron star other than the one at the heart of the Crab Nebula, and compute the following: its density, the escape velocity at its surface, and its surface gravity. For the white dwarf, you will need to know its radius and mass; you will have to find its mass (which means it will probably have to be a binary companion), and you can compute its radius if you find its absolute magnitude and temperature. For the neutron star, your source will have to provide its mass and radius.

Now compute the horizon radius for a black hole of mass equal to the masses of the white dwarf and neutron star you found. Using that radius, compute the escape velocity and surface gravity of the "equivalent" black hole. Express all escape velocities in terms of the speed of light, and all surface gravities in terms of Earth's surface gravity (9.8 m/s²).

Portfolio Exercise 6: Using the Relativity Playground: vary the initial position and speed to see how the escape velocity depends on distance from the horizon. Do this first for a = 0 and masses of 2, 2000 and 2000000 M_solar. Then set a = .9 M and do the same. Does the escape velocity depend on the mass? Does it matter whether the probes follow an azimuthal or polar orbit?

• Introduction
• Astronomical Observation
• The Solar System
• Stars
• Galaxies
• Cosmology
• Postscript: Extraterrestrial Life
• References
• Useful Equations and Constants
• Glossary

©2008, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

Please send comments or suggestions to the author.