(View Cosmos DVD 5, episode 9, on solar activity; mov of solar surface (source) and Hinode mpg of same (source); also SOHO EUIT mov of solar activity from 10/21/03 to 11/06/03 (source); also Hinode mpg and magnetic field images from 12/13/06 flare (source) and Hinode mov of solar X-ray jets (source).)The most recent complete solar cycle (seen here in 284 Angstroms) began in 1996:
(source).
Sunspot activity defines the solar cycle; note the Maunder Minimum between 1645 and 1715:
(source)
At the start of each new cycle, the polarity of sunspot "pairs" reverse. For instance, during the cycle which is beginning in 2008, active regions in the northern hemisphere are oriented so that their south poles are west of their north poles (the opposite occurs in the southern hemisphere). In the previous cycle, the order was reversed.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:(View SOHO mpeg from May 1998 (source))
(source)
The solar wind speed is around 400 km/s, which is about 2/3 of the escape velocity at the surface of the Sun. The wind, however, starts from the corona, which extends several solar radii; the escape velocity at 2.4 solar radii is 400 km/s. The current solar wind velocity is available from SOHO.Even radiation pressure sculpts the interstellar medium.The solar wind carries away about 10-14 solar masses annually. This is about 634 million kg per second. If the wind was entirely made up of protons, the outgoing flux would be 3.79 * 1035 protons per second. Assuming that they are emitted isotropically, we should detect an average of about 3.37 per cubic centimeter at the Earth. The measured value is actually about twice that.
The spectral data varies in resolution and coverage among the stars: some spectra include ultraviolet and/or infrared wavelengths while others cover only part of the optical spectrum. All are flux-calibrated, meaning that the flux values have calibrated physical units of Watts per square meter per Angstrom. Some of the original spectra were compiled at resolutions greater than the applet's finest resolution of one value per Angstrom, and those spectra have been smoothed for our use here. In addition, some spectra have been truncated because of the applet's limited color resolution: it can show only 256 levels of intensity for any given hue.
Begin by choosing a star from the pop-up menu. An image of the star will be displayed and, for most stars, the measured annual parallax angle and absolute error in that measurement will be displayed in the text box to the right of the image.
Note that the image is not, except in the case of Sol, an image of the stellar surface. It is actually a diffraction pattern with a strong, wide central maximum (the central circular "image") and in most cases indiscernible secondary maxima.For all stars, the text box will also contain the minimum and maximum flux values over the range of the displayed spectrum, along with the total flux over that range.
Below this is another pop-up menu which selects an element (or ion or molecule) for spectral line identification, followed by a button and scroll bar enabling a black body temperature fit to the spectrum. Beneath those are three displays. The top display shows the spectral lines associated with the element selected (they are displayed as emission lines, that is, bright against a black background). Below that is a reconstructed color image of the actual spectra (wavelengths in the ultraviolet and infrared are displayed in shades of gray). Finally there is a graph of the flux as a function of wavelength:
You may click on either the spectrum image or graph to mark a central wavelength value on which to focus for detailed inquiries. The "Zoom In" and "Zoom Out" buttons allow you to quickly change the range of wavelengths displayed about that central value, and the values of the wavelength at the left and right edges of the graphs are displayed to the right of those buttons. The scroll bar labeled "Central lambda" allows you to smoothly vary the central value, and the value of the flux at the central wavelength is displayed to the right of the central lambda value. Finally, the "Gaussian Fit" button draws a Gaussian fit to the spectrum graph centered on the central lambda value, with a standard deviation as specified by the scroll bar to the right.
(source, source, source, source, source, source, source)
energy flux = luminosity / (4 p distance2)Using values from the Spectrum Viewer for the star 108 Virginis, we see from its parallax (5.31 mas) that its distance from Earth is
r = 1000 / 5.31 = 188.3 parsecs, times 3.086 * 1016 m/parsecThe Viewer also tells us that its total flux is 9.9775 * 10-11 W/m2, so its luminosity is= 5.81 * 1018 m
L = 4 p * (5.81 * 1018 m)2 * 9.9775 * 10-11 W/m2Note that we do not have a "complete" spectrum; there appears to be additional flux in the ultraviolet that is missing from the data for this star. Because of this, we have really underestimated the luminosity of this star.= 4.23 * 1028 W, divided by 3.85 * 1026 W for our Sun= 110 Lsolar
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.
Here are sample optical spectra arranged by spectral class; can you identify the prominent lines?
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 temperature4Using our temperature of 9300 K for 108 Virginis (and 5.67 * 10-8 for s), we obtain
r = (4.23 * 1028 / 4 p s 93004)1/2Since our estimate for the luminosity was low, this is an underestimate for the radius. But because of the square root, we are closer here than in our luminosity calculation (if the luminosity was off by a factor of 2, the radius will only be off by a factor of 21/2 = 1.414).= 2.82 * 109 m, divided by the radius of our Sun (7 * 108 m)= 4 rsolar
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 three 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.
For one or more of your stars, the applet may not have parallax information; for these you will only be able to find the black body temperature. Several of the spectra show strong emission lines (associated with flares) which alter the scale of the graph and make black body temperature estimation more difficult.
apparent wavelength / actual wavelength = 1 + recession velocity / cIf the apparent wavelength is shorter than the actual, the motion is toward us and the recession velocity is taken to be negative. The Spectrum Viewer does not have sufficient resolution to enable us to compute recession shifts for any of the stars it has data for. But we will return to this computation when we discuss Cosmology.
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 / cThe line width is twice the standard deviation reported by the Spectrum Viewer.
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.If the lines are not clearly defined and relatively deep, it will not be possible to do this. This will occur if one or more of your stars have low-resolution spectra, or those lines are located in very low-flux wavelengths. List those stars which you are unable to evaluate.Are the speeds consistent for each star? What does the range tell you about this computation?
Using our temperature of 9300 K, we see that 108 Virginis is likely close to a class A1 (interpolating between the temperatures for A0 and A2 in the table).
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
absolute magnitude = 4.83 - 2.5 * log10 luminositysolarFor 108 Virginis, at a distance of 188.3 parsecs and a luminosity of 110 Lsolar, we obtain an absolute magnitudeabsolute magnitude = apparent magnitude + 5 - 5 * log10 distancepc
M = 4.83 - 2.5 * log10 110 = - 0.27and an apparent magnitude of
m = - 0.27 - 5 + 5 * log 10 188.3 = 6.1Since the absolute magnitude is the magnitude a star would have at a fixed distance of 10 parsecs, m will be larger than M if the star is more than 10 parsecs distant, and less than M if it is closer. Note that because of our underestimate of the luminosity, M is larger (less negative) than it should be, and m is also larger than it should be. Here the error is even less than in our radius calculation because the log function increases more slowly than the square root.
You can barely see a star of apparent magnitude 6 with your eyes, if you are in a region of dark skies. Under the same conditions, binoculars takes you to magnitude 10 and our 8 inch telescopes to about magnitude 13. The Hubble can just see objects of apparent magnitude 30.
The following table relates the absolute magnitude to the luminosity class for each spectral class (the magnitudes given are the most likely values for each class):
Using SIMBAD, you can look up the accepted spectral class for 108 Virginis. You will find it listed as "B9.5V". Our choice of A1 is very close to B9.5, and the "V" indicates a main sequence star, so our choice of luminosity class was correct (if a little lucky).
Spectral Class M (Main Sequence) M (Giants) M (Supergiants) M (White Dwarfs) O5 -5.8 O6 -4.8 O8 -4.1 B0 -3.3 -6.4 10.2 B1 -2.9 B2 -2.5 B3 -2.0 B4 -1.5 B5 -1.1 B8 0.0 A0 0.7 -5.0 A2 1.3 -5.0 A5 1.9 -5.0 A7 2.3 -4.9 F0 2.7 1.0 -4.8 12.9 F2 3.0 0.9 -4.8 F5 3.5 0.8 -4.7 F8 4.0 0.7 -4.6 G0 4.4 0.6 -4.6 G2 4.7 0.5 -4.6 G5 5.1 0.4 -4.5 G8 5.6 0.3 -4.5 K0 6.0 0.2 -4.5 K2 6.5 0.1 -4.5 K3 6.8 0.1 -4.5 K5 7.5 0.0 -4.5 K7 8.0 -0.1 -4.5 M0 8.8 -0.2 M2 9.8 -0.2 M5 12.0 -0.2 M8 16.0 A white dwarf is a star which has collapsed to a small, hot cinder after fusion processes have stopped.Using the estimated class A1 for 108 Virginis and our absolute magnitude of -0.27, we see that the value in the main sequence column (1) is closer than the value in the supergiant column (-5), so we suspect that 108 Virginis is a main sequence star. Since we have no data in the giant column, it could be a giant star, but it definitely is not a supergiant or white dwarf.Standard luminosity classes are
Ia - Bright Supergiants
- Ib - Supergiants
- II - Bright Giants
- III - Giants
- IV - Subgiants
- V - Dwarfs (including Main Sequence Stars)
A spectral class beginning with "D" indicates a degenerate star, or white dwarf. Additional designations include e (has emission lines), n (has diffuse lines), p (has a peculiar spectrum; see, even physicists don't understand it all yet!), s (has sharp lines), and v (variable star).
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 = (mobserved - mexpected) / distancehere measured in magnitudes per kiloparsec; note that it is a function of wavelength.
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 = 8.36However, the observed apparent magnitude (measured near 5448 Angstroms, the effective wavelength for the V band) is 11.4 (source), meaning that the extinction is
(11.4 - 8.36) / 0.236 = 12.86 magnitudes / kpcIn this case, the extinction is due largely to an obvious cloud:
and it is more reasonable to quote the extinction simply as 3.04 magnitudes.
(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 Msolar, while those of type B0 are about 15 Msolar (stars begin their main sequence lives with masses varying from 0.04 to about 200 Msolar (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 Msolar.The first thing to notice is that by far, most of the data points lie on the main sequence. Since we are looking at a population of stars at a different time in each star's life, that tells us that stars spend most of their lives on the main sequence. As a consequence, the (super)giant and white dwarf regions of the diagram must represent stars at various stages near the ends of their lives (assuming that most of the data points represent stars and not protostars).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 Rsolar. 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.
stellar lifetimeyears = 1010 masssolar / luminositysolarSince luminosity is approximately equal to mass3.5 (in solar units), this is equivalent to
stellar lifetimeyears = 1010 / masssolar2.5leading us to the conclusion that more massive stars burn out more quickly.
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
What happens next is a tug of war between the opposing pressures due to fusion and gravity: as fusion products build up in the core, its density increases. The core shrinks, the pressure increases, the temperature increases, heavier nuclei begin to fuse in the core and previous 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, with multiple shells supporting different reactions simultaneously.
The fusion of successively heavier nuclei is an accelerating process, due to the drop in the number of nuclei available to fuse. Carbon fusion might last for hundreds of years before Oxygen fusion begins, but Oxygen burning might last only a year or so before Silicon fusion begins. It takes about a day for the core to fuse into Iron.
Numerical values in this section are approximate. Stellar life cycles are based on models described in 1965 and 1967 by Iben (Astrophysical Journal 141:993 and Annual Reviews of Astronomy and Astrophysics 5:571, respectively).
In these nuclear reactions, the difference between the masses of the products and reactants can be used to compute the energy released by using
- 1H + 1H -> 2H + e+ + ne
- 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 * c2The mass of a proton (1H) is 1.673 * 10-27 kg; the mass of a Helium nucleus (4He) is 6.647 * 10-27 kg, and the mass of a positron (e+) is 9.109 * 10-31 kg. The energy released in neutrinos (ne) and gamma rays (g) is thenMost 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)2Using the solar constant (1365 W/m2) as the energy flux and 1 AU as the distance in the equation below, we can compute the luminosity of the Sun, obtaining LS = 3.85 * 1026 W. Dividing the luminosity by the energy per reaction gives the number of reactions per second, 9.92 * 1037. Since each reaction consumes 4 protons, the rate of Hydrogen consumption is= 3.881 * 10-12 J(= 24.225 MeV)
9.92 * 1037 * 4 * 1.673 * 10-27The photons released in these reactions are gamma rays, but the bulk of the electromagnetic radiation leaving the surface of the Sun is in the visible spectrum. The average travel distance ("mean free path") between collisions for a photon in the center of the Sun is about 1 cm. On average, each photon leaving the surface has interacted with matter about 1021 times since being emitted in the core. Since many of these interactions involve the absorption and re-emission of the photon, the photons emitted at the surface are really distant ancestors of the ones created during fusion. Most interactions involve some energy loss (they are "inelastic"), and this accounts for the shift in spectrum from gamma to visible. The average time for a photon to travel to the surface from the core is= 6.64 * 1011 kg / s.
1021 * .01 m / c = 33 billion secondsor over a thousand years.
In the following list of nuclear reactions occurring in stars, numbers in parentheses are approximate energy yields in MeV; negative values indicate additional energy is required to make these reactions occur. Note that these are not all of the possible reactions; any reaction which conserves energy, momentum, angular momentum, electric charge, baryon number (protons and neutrons are baryons) and lepton number (electrons, positrons and neutrinos are leptons) can occur. Each reaction occurs with different frequency based on how much energy is required to overcome the electric repulsion between nuclei before the strong force (which binds nuclei together) takes over.
- 1H + 1H -> 2H + e+ + ne (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+ + ne (1.1)*
- 13C + 1H -> 14N + g (11.2)
- 14N + 1H -> 15O + g (3.6)
- 15O -> 15N + e+ + ne (1.8)*
- 15N + 1H -> 12C + 4He (4.9)
- 4He + 4He -> 8Be + g (-0.09)
- 4He + 8Be ->12C + g (7.4)
- 12C + 4He -> 16O + g (7.2)
- 12C + 12C -> 20Ne + 4He (4.7)
- 12C + 12C -> 23Na + 1H (2.2)
- 12C + 12C -> 23Mg + 1n (-2.6)
- 12C + 12C -> 24Mg + g (14)
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 spectra lack Hydrogen lines; Type Ia typically have Silicon lines).
The second to last reaction in this set is included to illustrate reactions which release neutrons. These free neutrons will enter into "neutron capture" reactions below.
- 16O + 4He -> 20Ne + g (4.8)
- 16O + 16O -> 28Si + 4He (0.47)
- 28Si + 4He -> 32S + g (16)
- 32S + 4He -> 36Ar + g (6.7)
- 36Ar + 4He -> 40Ca + g (7)
- 40Ca + 4He -> 44Ti + g (5.1)
- 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+ + ne (1.1)*
- 56Co -> 56Fe + e+ + ne (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.Neutron capture can form atoms with atomic masses larger than Iron, for instance in the series of processes:
n 1.0087 14N 13.9954 32S 31.9633 1H 1.00725 15N 14.9963 36Ar 35.9576 2H 2.01355 15O 14.9987 40Ca 39.9516 3He 3.0149 16O 15.9905 44Ti 43.9476 4He 4.0015 20Ne 19.9869 48Cr 47.9408 8Be 8.00311 23Na 22.9838 52Fe 51.9338 12C 11.9967 23Mg 22.9875 56Fe 55.9206 13C 13.0001 24Mg 23.9784 56Co 55.925 13N 13.0019 28Si 27.979 56Ni 55.9267 You can find the atomic numbers of these atoms from the Periodic Table of the Elements.
56Fe + 1n -> 57FeBut nuclear fusion ends with Iron production. This is because 56Fe fusion is an endothermic process: it takes more energy than it releases. For instance,57Fe + 1n -> 58Fe
58Fe + 1n -> 59Fe
59Fe -> 59Co + e- + ne
59Co + 1n -> 60Co
60Co -> 60Ni + e- + ne
56Fe + 56Fe -> 112Terequires over 44 MeV to take place. While we have seen several such reactions above, they have proceeded because excess energy is available in the active core of the star. Once an Iron core forms, there are no reactions available to provide the additional energy necessary for Iron fusion to occur.
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 (-8.8 MeV / baryon).
Portfolio Exercise 4: For each of the fusion reactions above, compute the energy yield.
IC 4406 (source); note that in this image, red, green and blue colors represent concentrations of Nitrogen, Hydrogen and Oxygen, respectively. We cannot, however, use these colors to distinguish between isotopes of Oxygen.
NGC 6543: The Cat's Eye Nebula (source)
NGC 2392: The Eskimo Nebula (source); red, green, blue and violet colors represent concentrations of Nitrogen, Hydrogen, Oxygen and Helium, respectively.These stars typically shed enough mass:
Eta Carinae (source)so that their final mass is less than the Chandrasekhar Limit of 1.44 solar masses to become a white dwarf: the compressed core of the final fusion product:
NGC 2440 (source); in this image, red, blue-green and blue colors represent Nitrogen and Hydrogen, Oxygen and Helium, respectively.
A white dwarf is "degenerate" because its atoms are as close as possible within the constraints of the Pauli Exclusion Principle, which tells us that no two electrons can be in the same state. In this condition, every atom is in the lowest state possible, so that the number of available states, or degeneracy, is as low as possible.
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)/2.5Its black body temperature is around 14800 K, so its radius is= 10-2.548 = 0.0028 Lsolar
r = (L / (4 p s T4))1/2From this we can compute its average density= (0.0028 * 3.85 * 1026 / (4 p s 148004))1/2= 5.62 * 106 m = 0.008 Rsolar
r = m / (4/3 p r3)its escape velocity= 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)1/2and its surface gravity= (2 * G * 2 * 1030 / 5.62 * 106)1/2= 6.89 * 106 m/s
= 0.023 c,
surface gravity = G * mass / radius2which means that if you could survive on its surface, you would weight about 430 thousand times as much as you do on Earth! These values are typical orders of magnitude for white dwarfs:= G * 2 * 1030 / (5.62 * 106)2= 4.23 * 106 m / s2,
Note that at 8.6 light years, Sirius B is the closest white dwarf to Earth. That, coupled with the fact that it is a binary partner to Sirius A, allows us to understand it much better than many other white dwarfs. Careful observation of their orbital period and separation was the key to computing its mass.
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)
For Type II supernovae, the "-P" and "-L" designations stand for "plateau" and "linear". The timings of the light curve decays seem correlated with the radioactive decay rates of 56Ni and 56Ni.
If the final mass at core collapse (after the shells have been shed) is less than 3-3.2* solar masses, the remnant will be a neutron star. In a neutron star, the atoms have come so close to each other that the electrons have been absorbed into the nuclei, releasing neutrinos. The remaining neutrons are now degenerate: they are as close as possible within the constraints of the Pauli Exclusion Principle. A neutron star is essentially one large, extremely dense nucleus, with no net electric charge.
*Most neutron star masses are close to 1.4 Msolar; the largest measured so far is about 2.1 Msolar. The actual limit depends on how neutron matter behaves, and is the subject of ongoing research.If the neutron star is rotating, and has a sufficiently large magnetic field whose axis is not along the rotation axis, we may observe it as a pulsar. It is assumed that most if not all stars rotate, so by conservation of angular momentum, a collapsed core such as a white dwarf or neutron star will rotate much faster than its progenitor, by a factor of (rbefore/rafter)2.
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)
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 Msolar and its luminosity is about 50000 Lsolar. That means that its radius is
r = (50000 * 3.85 *1026 / (4 p s 40004))1/2Since its luminosity is varying, we know it is in the horizontal branch, and it is estimated that within 10000 years it will explode as a supernova. Assuming the supernova has a peak luminosity of 1010 Lsolar, the energy flux we should experience here on Earth is= 3.25 * 1011 m = 464 Rsolar
energy flux = luminosity / (4 p distance2)This may not seem like much compared to the total flux we receive from the Sun, but consider that much of this is in gamma radiation. If we assume the normal gamma flux is 5 MeV/cm2/s, or around 8 * 10-9 W/m2, this is a lot of radiation. The lesson here is that the neighborhood affected by a supernova is very large. In any given galaxy, every 30 to 50 years another star explodes as a supernova:= 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)This makes the universe seem like a violent place, and it can be. But it is important to remember that the explosion of a supernova not only disrupts its neighborhood, but it also seeds the interstellar medium with heavier elements, and the shock waves compress that medium so that more stars can develop in the future:
(source).
Star forming regions in galaxies are identified by the hydrogen gas glowing in infrared and by the ultraviolet light of young, bright stars (see M31).
(View Cosmos DVD 6, episode 9, on Flatland and curved space.)The Special Relativity Primer may be a good place to start trying to understand the idea of spacetime. The Relativity Playground may help you better understand the principles of General Relativity. Note that since mass is a scalar (having no directionality), it is a constant in every reference frame.
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)from the center of the black hole. This is a complicated function, so we will specialize for a moment to the (probably) unphysical case of a static black hole, which does not spin. Its horizon is then at
r = 2 G M / c2Using this expression for r, the escape velocity at the horizon is= 2953 meters * M / MsolarNote that this implies that G / c2 is a conversion factor for converting mass to length.
ev = (2 G M / r)1/2which is consistent with our notions a) of nothing (particularly light) being able to escape from inside the horizon of a black hole, and b) all light cones are tangent to the horizon of a black hole.= c
The surface gravity at the horizon is
g = c4 / (4 G M)For M = 2 Msolar, this is almost 2 million times the surface gravity of Sirius B, and 4 times that of the neutron star in the center of the Crab Nebula. But there is a caveat for the neutron star: it is a pulsar, so we know it spins, and therefore using our simplistic version of the equation for the horizon is not exactly a valid comparison. However, even though the pulsar is spinning 30 times each second, the angular momentum term is less than 0.03% of the mass term, so this result is very close.= 1.52 * 1013 m/s2 / (M / Msolar)
Consider now the tidal acceleration experienced by a 2-meter object at the horizon:
G M / r2 - G M / (r+2)2Since M is a multiple of Msolar, G M is much greater than c2, and this reduces to approximately= (c6 / (4 G M)) (c2 + 2 G M) / (c2 + G M)2
c6 / (2 G2 M2)For M = 2 Msolar, this is about 5 * 109 m/s2. This means that our 2-meter object experiences a tidal acceleration over 500 million times Earth's surface gravity! But for M = 106 Msolar, the tidal acceleration is 0.02 m/s2: unnoticeable.= 2 * 1010 / (M / Msolar)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))The disc-like disturbances around the pulsar and the accretion discs around the black holes are related in that the curvature of spacetime is essentially the same (except in magnitude) around all spinning masses. It is probable that most if not all of the violent events we see in the universe, from novae to gamma ray bursts, are powered by either stellar collapse, the accretion of matter onto a massive compact object like a white dwarf, neutron star or black hole, or mergers of same.
(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 Msolar. 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?
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