Astro 2299 The Search for Life in the Universe Lecture 8 Last time: Formation and function of stars This time (and probably next): The Sun, hydrogen fusion Virial theorem and internal temperatures of stars Jeans mass Formation of protostars and planetary systems Assignment 2 posted Summary about stars to be posted today Instructors: Jim Cordes & Shami Chatterjee http://www.astro.cornell.edu/academics/courses/astro2299/ 1
Papers on the NS-NS/GW/kilonova Gravitational waves: A golden binary Nature, 551, 36, 2017 https://www.nature.com/articles/nature24153 Spectroscopic identification of r-process nucleosynthesis in a double neutron-star merger Nature, 551, 67, 2017 https://www.nature.com/articles/nature24298 Optical emission from a kilonova following a gravitational-wavedetected neutron-star merger Nature, 551, 64, 2017 https://www.nature.com/articles/nature24291 2
Stars: from cradle to grave
100 75 50 25 The Milky Way 75% of the Milky Way s luminosity arise from the rarest stars. Percentage of Galactic Luminosity Percentage in Number Percentage of Galactic Stellar Mass K & M stars account for ¾ s of the stars in the galaxy but contribute less than 5% of its luminosity. 0 O-M F-M O B A F G K M B-F Supergiant (I & II) Red Giant (III) Main Sequence (V) White Dwarf Luminosity Class and Spectral Type
8
Radius of Sun s photosphere: R ~ 7 x 10 10 cm ~ 7 x 10 5 km 9
FIG. 3. Elemental composition of the Sun (a typical star) by mass. The word metal here is used in the way commonly applied to stellar compositions and includes all elements heavier than H and He. (Credit: Martin Asplund) From Astrobiology Primer FIG. 4. The abundance of elements (by number of atoms normalized to silicon) in Earth s crust. The crust is depleted in siderophilic ( iron-loving ) elements relative to the bulk Earth as siderophiles were partitioned into the Earth s core. Compared with the mantle, Earth s crust is enriched in lithophilic (rock-forming) elements. (Credit: Aditya Chopra) 568 10
The Core The core temperature of the sun is about 15 10 6 K. At this temperature, hydrogen is fused into helium The sun has been on the main sequence for 4.6 Gyr and will continue for approximately another 5 Gyr. Once the inner 10% of hydrogen has been expended, the sun will move off the main sequence and ultimately form a red giant and planetary nebula with a white dwarf at its interior
Solar Interior Radiative zone: Energy is transported by electromagnetic radiation Convection zone: Energy carried by convection A computer-generated image showing the pattern of a p-mode solar acoustic oscillation both in the interior and on the surface of the sun. (l=20, m=16, and n=14.) Note that the increase in the speed of sound as waves approach the center of the sun causes a corresponding increase in the acoustic wavelength. https://en.wikipedia.org/wiki/helioseismology
Proton-proton chain CNO cycle 12 C acts as a catalyst because there is no net production of it in the CNO cycle 13
The Proton-Proton Chain Reaction Three steps complete this fusion reaction: Net reaction: 4p è 4 He + energy (about 25 MeV per 4 He produced) The release of energy is about 0.007 times the rest mass of the input hydrogen The main sequence lifetime then corresponds to releasing 0.1 x 0.007 x M c 2 of energy as heat and radiation You can check this by calculating this total energy and dividing by the current luminosity of the sun
The CNO Cycle Six steps complete this fusion reaction: The CNO cycle requires higher temperatures than the proton-proton chain because C and N nuclei have larger positive charge that the proton needs to push against This requires higher thermal velocities for the protons
Solar Neutrinos https://en.wikipedia.org/wiki/solar_neutrino Solar neutrinos have been measured and are consistent with the rate expected (after including neutrino oscillations) 18
Virial Theorem and Stellar Temperatures (For discussion of the VT, see star notes on web page and wikipedia) The virial theorem says that in a stable object the internal and gravitational energy are balanced: pressure vs. gravity 2 x KE + PE = 0 Example: a planet of mass m orbiting a star of mass M The KE and PE are: mv 2 KE = 1 2 mv2 = GMm 2r r = GMm r 2 PE = Force balance for a circular orbit GMm r So the VT is satisfied. The same is true for any stable object that is held together by gravity
Virial theorem and Stars M = mass, R = radius In a star, the kinetic energy is thermal (possibly combined with convection and turbulence, which we ignore here) The gravitational potential energy is (sphere with uniform density) PE = 3GM 2 5R The thermal energy is (if uniform temperature) N = number of particles (nuclei) in star of mass M m = mean mass of each nucleus Using the VT we can solve for temperature T = 1 GMm 5 kr N ~ M / m k = Boltzmann s constant
Use Internal Temperatures of Stars T = 1 5 GMm kr G = 6.67 x 10-8 cgs M = 2 x 10 33 g R = 7 x 10 10 cm m = m H = 1.67 x 10-24 k = 1.38 x 10-16 cgs to estimate T ~ 4.6 x 10 6 K for the sun This is an average temperature but is comparable to what is needed to drive nuclear reactions The estimated energies (KE, PE) are dominated by the core of the sun where the mass density is highest
Kelvin-Helmholtz Contraction Time The VT says that KE and PE are balanced. In order for a star like the Sun to contract, it must lose energy. The VT further implies that while ½ of the PE goes into an increase in KE, the other half must be dissipated in some other way: radiation. The measured luminosity of the Sun is L ~ 4 x 10 33 erg s -1. If this luminosity were due solely to radiation of GPE, the lifetime of the Sun would be only about 3x10 14 s ~ 10 Myr = the K-H contraction time. See star notes on web page for details. What gives? Either the solar system is very young or there is another source of energy, i.e. nuclear reactions. That was the realization in the early 20 th century
What is needed to form a star? A star typically means an object that shines because nucleosynthesis occurs in its core. A star needs to have > ~ 0.1 M of mass Initial reactions: 4H à He (CNO cycle, proto-proton chain) Later: He à C à Fe Requirements: Collapse of a gas cloud that contains H Sufficient mass in protostar so that central temperature is high enough to drive nuclear reactions Collapse of gas clouds is constrained by the temperature and density of the gas cloud, also magnetic fields, and angular momentum
Jeans Scale and Mass Approach 1: Equation of hydrostatic equilibrium F P FG dp dr = GM(r) (r) r 2 r A gradient in pressure dp/dr is needed to counteract the gravitational force Stars and planets can be stable for Gyr and this equation needs to be satisfied with the appropriate density and pressure Dimensional analysis: Pressure: P = nkt, n = (ρ/m)kt à dp/dr ~ P/R ~ ρkt/mr Similarly GMρ/r 2 ~ GMρ/R 2 Set equal ρkt/mr ~ GMρ/R 2 k = Boltzmann s constant G = Gravitational constant à M/R ~ kt/gm Use M ~ (4π/3) ρ R 3 à R ~ [ (3/4π) kt/gρm ] 1/2
Jeans Scale and Mass Approach 1: (continued) call this the Jeans length F P FG r R J ~ [kt/gρm ] 1/2 Can be written using the sound speed C s : C s ~ (kt/m) 1/2 à R J ~ C s / (Gρ) 1/2
R 0 Jeans Scale and Mass Approach 2: Free fall vs pressure wave Suppose we squeeze on the outer shell of a region of size R 0 that has mass M internal to the shell The gravitational force on the outer shell (directed inward) is If there is no counteracting pressure, the shell will free fall and its radius vs time is (see next slide) R 0 = initial radius M = mass inside R 0 The time taken to reach r = 0 is the free-fall time: All shells collapse with the same free-fall time if density! = constant We can compare this time to the time it takes sound at speed C s to propagate across the region, t P = R/C s If t ff < t P then the region will collapse faster than pressure can push back For equal values of the time scales, the same length scale R J is gotten
Consider the spherical region to consist of many spherical shells, all of which collapse. The mass inside a shell at t=0 determines the acceleration of the shell; that mass stays constant during the collapse because all shells collapse and they do not cross at any time. Derivation of r(t)
Example In the Milky Way there are cold dense clouds that are actively forming stars today Typical temperatures are 10 K and densities 10-22 g/cm 3 Evaluating R J ~ (kt/m) 1/2 (1 / (Gρ) 1/2 ~ 3 pc We can also calculate the Jeans mass as M J ~ ρ R J 3 and we get about 50 M Interpretation: Relatively large mass regions collapse. Subregions inside them fragment as their temperatures fall and their densities increase. à A wide range of stellar masses
Initial mass function of stars = distribution of masses at birth
Gravitational stability: The case of B68 Optical Near-Infrared Starless Bok Globule Gravitational stable or at the verge of collapse