ASTR 304 - 2003W [Answers for Week 12]

Navigation BarHome PageSyllabusBibliographyProblem SetsNewsContact Info

By week:

Week 12 - Understanding Cosmological Gamma-Ray Bursts

Problem Set

Problem 1 - Burst Rates Top

We are going to calculate the fraction of stars that are born that will end up as binary neutron stars or collapsars. We will use the Salpeter IMF: dN = A M-2.25 dM for M>0.1 .
  1. Black holes: what fraction of stars has initial masses greater than 20 ?
  2. Neutron stars: what fraction of stars has initial masses greater than 8 and less than 20 ?
  3. In steady state, if a galaxy has a star-formation-rate of 1 /yr, what is the rate of supernova that produce black holes and neutron stars?
  4. Binary neutron stars: let's assume that half of all stars are in binaries and that stars pick their companions randomly, what fraction of neutron stars will have companions that will become neutron stars?
  5. Finally, let us assume that only half of binaries survive a supernova explosion, what is the rate of binary-neutron-star mergers in the galaxy?
  6. What is the mean star-formation rate of the universe? Take Ω*=0.00245. You want a round number. What is the rate of GRBs in the two models?

Answer for Problem 1

  1. First let's calculate the total number of stars
    Ntot = /∞,
    |
    /0.1     		A M-2.25 dM = 14.23 A -5/4
    The number above 20 is 0.0189 A, yielding a fraction of 1.3 x 10-3.
  2. 2.9 x 10-3
  3. This is actually a bit more complicated that it looks. The star-formation rate is in solar masses per year, not stars per year, so you have to calculate the total mass of stars in the IMF
    Mtot = /∞,
    |
    /0.1     		A M M-2.25 dM = 7.11 A -1/4
    So a star-formation rate of 1 solar mass per year corresponds to the formation of two stars per year, so we get 2.6 x 10-3 black-hole supernovae per year and 5.7 x 10-3 neutron-star supernovae per year.
  4. 2.9 x 10-3 / 2 = 1.4 x 10-3
  5. There are two supernovae, so only one quarter of the binaries survive, so the answer is
    0.25 ( 1.4 x 10-3 ) 5.7 x 10-3/yr = 2 x 10-6/yr.
  6. Let's start with ρcrit, the critical density to close the universe.
    ρcrit = 3 H02
    8 π G
    The mean star-formation rate is
    Ω*ρcrit Volume of Universe
    Age of Universe     		= Ω* 3 H02
    8 π G     		(c / H0)3
    1/H0     		=Ω* 3 c3
    8 π G
    The mean star-formation rate is Ω* (5 x 1037) g s-1 = 2 x 109 /yr. Yielding a rate of black-hole GRBs of 4 x 106/yr and neutron-star-inspiral GRBs of 3000/yr.

    These rates of GRBs are an overestimate for both cases. First, only a fraction of the supernovae that result in black holes produce relativistic jets. For the second model, only a small fraction of neutron-star binaries are tight enough to spiral in over the age of the universe.

Problem 2 - Neutrino-Eddington Limit Top

  1. Use the cross section for neutrino pair-production as an estimate of the cross for a neutrino to scatter of an electron. What is the neutrino Eddington limit to the luminosity as a function of the mass of the star in solar masses and the energy of the neutrinos?
  2. What is the Eddington limit to the accretion rate?
  3. Use this Eddington-limited accretion rate to estimate the maximum value of Γ for a gamma-ray burst.

Answer for Problem 2

  1. Let's start with the formula for the Eddington luminosity
    LEdd = 4 π c G M mp
    σT
    Instead of the Thomson cross-section we use the neutrino cross section, 4 x 10-38 E2 GeV-2 cm2 to get
    LEdd = 2 x 1051 erg/s GeV2 M
    E2     		8 x 1057 erg/s Γ-2 M
  2. Let's use the efficiency of a Schwarschild hole, 0.057 c2 to get
    MdotEdd = 1.5 x 1038 g/s Γ-2 M
    		= 7.7 x 104 /s Γ-2 M
  3. Finally let's add the formula in class to estimate the Eddington-limited value of Γ. We have
    MdotEdd = 7.7 x 104 /s Γ-2 M
    		=10-6 /s Γ4 /
    |
    \    		 M
    		\2
    |
    /
    This gives an Eddington-limited value of Γ of 65 (M/)-1/6. This is a really rough calculation but it may explain why you don't find values of Γ as large as 103 or so. There is a way to evade this limit which is that neutrinos produced in the disk could gain energy before forming pairs by scattering off of high-energy electrons that had already been produced.

Problem 3 - The Millisecond Magnetar Top

A neutron star is born spinning with a period of 1.6 milliseconds. It has a magnetic field of 1016 G.
  1. What is P-dot for the magnetar when it is born?
  2. What is the initial spin-down luminosity of the magnetar?
  3. Does the spin-down luminosity of the magnetar increase or decrease with time? What does this mean in the context of the internal shocks model for gamma-ray burst emission?

Answer for Problem 3

  1. Using the dipole formula for the magnetic field of a neutron star,
    Bp sinα = 6.4 x 1019 I451/2 R6-3 (P1 Pdot)1/2 G
    We have Bp=1016 G and P=1.6 ms, so Pdot=1.52e-5 s/s. The spin-down timescale is around 100s.
  2. The spindown luminosity is -IΩdΩ/dt or 4π2I dP/dt P-3=1.5 x 1050 erg/s.
  3. The spindown luminosity will typically decrease with time, so the later bits of the burst might have lower values of Γ so it might be difficult to have the internal shocks collide.
Last modified: Tuesday, 06 April 2004 07:28:05