Homework 4 Astro Order of Magnitude: Fall 2016

Transcription

1 1) Supernova Explosions. Type Ia supernovae are believed to be the thermonuclear explosions of accreting white dwarfs that have reached the Chandrasekhar limit. A). Mean pressure inside a white dwarf of mass M and radius R. To maintain hydrostatic equilibrium, the total outward pressure must balance gravitational pressure, which is P = E g V = 3GM 2 4πR 4 GM 2 4πR 4 = F g (1) A Illustrating that using the energy density is roughly equivalent to finding the gravitational force per unit area. B). Estimate, the speed of sound, v s P/ρ. Within how much time does the flame traverse the radius of the white dwarf, assuming R = 10 4 km, M = 1.4M? Show that this sound crossing time is comparable to the free-fall timescale. For the speed of sound: v s ( 3GM 2 4πR 4 3M 4πR 3 ) 1/2 = ( ) GM 1/2 (2) R The sound crossing time: t r ( ) R 3 1/2 v = (3) GM which is equivalent to the free-fall timescale: v ff = 1/ Gρ. Plugging in for R and M yields, t 2 s. C). Calculate the total energy output of the explosion, assuming each atom of the white dwarf is converted from carbon to nickel. Compare this energy to the gravitational binding energy of the white dwarf, to demonstrate that the white dwarf explodes completely, without leaving any remnant. There was potential for confusion with the wording of this question, since each individual atom is not converted from carbon to nickel (this would give you a huge m, and the number of nucleons would increase a lot). The number of nucleons in the WD is N nucleons = M W D = g m nucleon g 1057 nucleons. (4) The energy per nucleon of carbon is the mass-energy of carbon divided by the number of nucleons in carbon. Same for nickel. The amount of energy released per nucleon going from carbon to nickel is the difference in these energies per nucleon: mc 2 ( nucleon = mc 12 m ) Ni c ergs (5) 59 Then the total energy released in the explosion is N nucleons mc 2 nucleon 1052 ergs (6)

2 The gravitational binding energy of the WD is so there is no remnant. U GM 2 R ergs, (7) 2) Binary Stars and Supernovae. A). Starting with two stars (M 1 & M 2 ) in a circular orbit, find the maximum mass that can be lost by one of the stars for the system to remain bound. We can do this problem two ways: (1) using the CM frame, and (2) using the virial theorem. The first involves more algebra. First way: The limiting case (between bound and unbound) is simply when the total energy of the binary system, in the C.O.M. frame after the super-nova, is zero. Figure 1. Consider the parameters as indicated in Fig. 1, where after the supernova M 1 M 1 = M 1 M. Because the supernova was instantaneous and isotropic, neither velocity is altered in the explosion. The C.O.M. however, will obtain a velocity! The total energy, in the new C.O.M. frame will be E com = 1 2 (M 1v M 2 v 2 2) T com G M 1 M 2 r To find the kinetic energy of the C.O.M. (T com ), we must find its velocity (v com ). The COM velocity is v com (M 1 v 1 + M 2 v 2 )/(M 1 + M 2), where from the pre-initial binary we know that v 1 = M 2 v 2 /M 1, and from our definitions, M 1 = M 1 M. With a little algebra, we can find that v com = v 2 MM 2 M 1 (M 1 + M 2 M) Note that if M = M 1 (i.e. the entire star is destroyed), the center of mass velocity is simply that of the second star (v 2 ), as it must. (8) (9)

3 The rest is just plugging in and doing some ugly algebra, but the end result is that a final energy E com = 0 requires M = 1 2 (M 1 + M 2 ) (10) If more than half the mass of the binary is lost in the supernova, the system is inevitably unbound. Second way: Recall that the virial theorem tells us E tot = U/2. (11) Consider a general mass distribution (not necessarily a binary). If this system is bound, it will have GM 2 i R i M iv 2 i < 0. (12) Now consider our system, with M 1 and M 2. Say M 1 loses the mass. Initially, M 2 is moving at v keplerian at the initial orbit radius. After M 1 loses mass, M 2 will be going too fast and so overshoots into a larger orbit. To be unbound, we want v keplerian = v esc for this system, so that the final orbit radius is infinity. Equating the energy after the explosion (LHS) to the virial energy of the unbound system (RHS), we have GM 2 f R i M f GM i R i = GM 2 f 2R f, (13) where we assume that the radial separation does not change instantaneously when the mass does. When the system is unbound, R f, so the RHS is 0. Solving gives M f = M i /2. So if then the system is unbound. M > 1 2 M i = 1 2 (M 1 + M 2 ) (14) B). 1% anisotropy in a single direction, how much velocity could this impart upon the star? (energy released ergs.) ergs is thus bestowed upon the star lets say a 1M neutron star. 2 E 10 v = m cm s 1 = 1000 km s 1 (15) This is consistent with observations of pulsars and supernova remnants. C). Conceptually, what effects would it have on the binary? If the velocity kick adds (vectorally) with the orbital motion, it will aid in unbinding the system. If the kick subtracts from the orbital motion, the binary could tighten (one rare possibility is a Thorne-Zytkow object). Note, though, that even if the kick and orbit vectors are in opposite directions, the binary can loosen if the kick velocity is more than twice the orbit velocity. What really matters for whether the binary remains bound/unbound is whether the total energy is still negative after the kick.

4 3) Neutron Star Spin-Up. Consider a Neutron Star of mass M = 1.4M, radius R = 10 4 m, and period of 1 s. The neutron star is accreting mass through an accretion disk, at a rate of Ṁ = 10 9 M yr 1. A). Derive a differential equation for P, the rate at which the neutron star period decreases. The angular momentum of a neutron star with period P, will be: L ns = Iω = 2πI P ; where we have assumed that the moment of inertia is constant due to the slow rate of mass accretion. Rearranging and taking the derivative, we find for the change in period: P = 2πIL 2 L. ns Conservation of momentum forces the change in momentum of the NS (L ns ) to be the same as that lost by each particle (L p ) accreted: Lp = L ns. The velocity of a particle, infinitesimally above the surface of GM the NS, in a circular orbit will be v = R. Thus the momentum of an accreted particle will be given by L p = mvr = m GMR (16) and therefore L p = ṁ GMR = L ns (17) In equation 9, it has additionally been assumed that the acceleration of the keplerian circular velocity due to increasing neutron star mass, is negligible again, due to the low accretion rate. Plugging this in the for change in period of the neutron star yields: P ns = 2πI L 2 ns L ns = ṁ GMR P 2 αp 2 (18) 2πI B). What is the maximum spin rate of such a neutron star? The maximum spin rate occurs when the centrifugal force on a particle at the stars edge matches the force of gravity. At this point, in effect, the particle enters orbit around the star. By equating forces, we see that at this point Or using P = 2πR/v mv 2 R R 3 P c = 2π GM = = GMm GM R 2 v = R (19) 2π Gρ s (20) C). Solve the equation from part a to find how long it takes to reach the break-up velocity (part b ). We can rearrange equation 10 to be dt = dp αp 2 (21) where α ṁ GMR 2πI This can be integrated from initial period (P 0 ) to the final (critial) period (P c ): t = 1 ( 1 1 ) s 10 7 yrs (22) α P 0 P c

5 Multiplying this resultant time with the mass accretion rate shows that the neutron-star will only increase its mass by about 1% validating our previous assumptions. 4) Statistics and Stellar Remnants stars, initial mass function dn/dm M 2.35 in the range M M. A). Find the fraction of stars formed with M > 8M. How many NS and BHs should there be in the galaxy, with roughly how much mass? Given the initial mass distribution function, dn M 2.35 dm, the fraction of stars with mass greater than 8 solar masses is simply: M 2.35 dm f 8 = M 2.35 dm (23) 2.56 where the proportionality constants (or normalization constants) have been canceled by taking the ratio [this is equivalent to multiplying the numerator by a constant which is equal to k = 1/(the integral over all masses) 0.39]. The total number of stars above 8 solar masses, is thus N 8 = f stars. The total mass of these stars can be found by performing the integral over MdN: M 8 = M M 2.35 dm M (24) Which also suggests that the average mass of one of these objects would be about M 16M. B). Assume that every stellar core collapse distributed 0.05M of iron into the interstellar medium. If the MW started with M of gas, what is the mean interstellar mass abundance of iron in the Galaxy? From the previous problem, we can expect the number of supernovae to have been about , distributing a total of M F e 10 8 M. Comparing this with the total initial amount of gas, yields a primordial mass abundance of Z Fe This abundance is only slightly greater than that of the sun suggesting that the sun formed after most of these supernovae had already occurred. C). How many pars of stars in the MW were formed in which both companions were more massive than 8M? The probability that a given star is both in a binary system and more massive than 8M is 1 2 f 8. The probability that both stars in the binary were formed greater than 8 solar masses is thus about ( f 8 2 ) 2 ; suggesting there may be about NS-NS, NS-BH and BH-BH binaries in the galaxy. 5) Neutrinos and Opacity.

6 A). M = 1.4M and a radius R =10km, estimate the mean nucleon density If we assume the mass of a neutron star is entirely neutrons/protons (nucleons), we have and the density is thus n cm 3 N p = M M p 1033 g g = 1057 (25) B). σ νn = cm 2 ; mean free path assuming the density you found in A? l = (nσ) cm (26) C). How many seconds does it take a neutrino to emerge in a random walk This is the diffusion time-scale 1, which we found in class to be t D = R2 lc (10 6 ) 2 cm cm = 0.1 s (27) cm s 1 The actual solution is about 10s, because the radius of interest is about 10 times larger. D). Twelve electron anti-neutrinos were detected by a spherical tank with 3 kton of water. σ νw = cm 2 how many neutrinos are passing through your body a second? Water has a mass of 18 grams per mole (and thus 18 nucleons per molecule), along with a density of 1 g/cm 3 ; thus the nucleon number density is ( ) 1 g/cm 3 ( molecules/mol)(18 nucleons/molecule) (28) 18 g/mol And the mean-free path is about cm (which is about 10 light-years). We can assume that the probability of interaction of a neutrino over some distance x will be roughly proportional to x/l (you can derive this rigorously by taylor expanding the Beer-Lambert law: I = I 0 e x/l ). 3 kton is about g, and therefore about cm 3 of water. The size of the detector is then about x V 1/ cm. The number of neutrinos which interact with water nucleons (N) can thus be related to the total number of neutrinos which passed through the detector (N 0 ), as N = N 0 x l ( ) N (29) 2000 This is the number of neutrinos passing through the entire tank, in an equivalent time period (about 10 seconds actually, or 0.1 s using the result of part C ). A g person (with the same density as water) is times smaller than the tank, so finally: during the supernova, there were about (or for a 10s dispersion) neutrinos passing through your body per second. 1 Because the neutrinos are highly relativistic, we can simply use the speed of light in place of their actual velocity.

7 E). Number density of nucleons in lead is cm 3, what is the mean free path? The difference between the cross-sections of neutrinos with water and with neutron-star matter is only an order of magnitude, thus either σ νw = cm 2 or σ νn = cm 2 are good estimates. Thus, l = (nσ) cm 1 light-year (30) To have a good chance of interacting with a single neutrino, one would need one light-year of solid lead. Problem set 4 Gravitational Focusing 6) Gravitational Focusing and Accretion. A). Find the ratio of One effective of the cross-section, most important to physical concepts cross-section in pairwise accretion is the gravitation deflection of a small body by a massive body. This deflection works to increase t Consider a star moving of thethrough two bodies. a dust cloud Your(or goal anyisparticulate to reproduce region), a calculation with some velocity first done v by Safrono relative to the material no notes, far away. noifdiscussion the star werewith massless, anyone. it s effective Really. area Consider would justa be test its particle cross approach sectional area πr 2, and it would accrete all of the material along a cylinder of that size. With radius R. The impact parameter is b and the velocity at infinite distance is v gravity, however, additional material will be pulled in; we can describe this by an effective area (i. πb 2, where b is referred the to two as bodies the impact would parameter, approach Seewithin Fig. 2. athe distance impact parameter of b of each describes other and with re the distance, perpendicular absencetoof that gravity, of the the motion, cross-section which determines for collision the border would between πr particles 2. We would like that will and will not cross-section be accreted. increases due to the effects of gravity. Figure 2. Gravitational a. What Focusing. is the escape Impact velocity parameter from b, the which large is the body? maximum (you ll need this for late perpendicular distance away at which particles will be accreted. Figure from Geoff Blake s Ge/Ay b. 133: Draw The Formation a diagramandof Evolution a grazing of collision Planetary between Systems. the two bodies. (A grazing point mass had just slightly more energy it would not collide with the ma Conservation of angularattention momentumto states where that the two bodies touch. c. For the case you b v just = Rdrew, v imp use the principle of conservation(31) of angular mom for an impact velocity v imp relating. Because b, vthe particle, R, andis vjust imp, barely the velocity being accreted, of the it object is justat barely impact. (Hint: W bound i.e. its energy is zero momentum (E < 0 is bound, in the E before > 0 is unbound). picture, shown above, and the total angular mo 1 impact, and set these equal) 2 mv2 imp G Mm R = 0 v2 imp = 2GM R = v2 esc (32) d. Write an energy conservation equation relating v, v imp, R, and M. e. What is the maximum value for b for which an impact will occur? Write b 2 = R 2 (1 + f) where f is called the focusing or Safronov parameter, wh terms of the escape velocity. f. (you can start collaborating again at this point) How large does an object ne

8 Therefore, b 2 R 2 = v2 esc v 2 (33) B). In a dense globular cluster (10 6 stars, radius about 10 pc), dispersion velocity? If we apply the virial theorem, we see that mv 2 G Mm GM v = (34) d d where d is the radius of the cluster, and M = NM is the total mass of the cluster. If we plug in the given values, we see that 10 v cm s 1 (35) C). Rate of collisions? Here we put it all together: the effective cross section of a star in the cluster is while the number density of stars is σ = πb 2 = πr 2 vesc 2 v 2 π(1011 ) 2 cm 2 (5 107 ) 2 (10 6 ) cm 2 (36) thus the mean free path, l = (nσ) cm and finally, the collision rate R = N v l n N d 3 = 106 (10 19 ) 3 = (37) = hz 1 per Myr (38)