ASTR 304 - 2009W [Answers for Week 5]

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Week 5 - Accreting Neutron Stars

Problem Set

Problem 1 - Accretion Top

  1. Let's use Newtonian gravity for simplicity here. How much kinetic energy does a gram of material have if it falls freely from infinity to the surface of a star of mass M and radius R?
  2. How much energy is released if a gram of material falls from a circular orbit just above the stellar surface onto the stellar surface? To put it another way, what is the kinetic energy of the material in the circular orbit?
  3. Hydrogen burning releases about 6 x 1018 erg/g. How does accretion of hydrogen onto a neutron star (R=10km, M=1.4) differ from accretion onto a white dwarf (R=10000 km, M=0.6)?
  4. What is the total about of energy released per gram of material as it falls from infinity to the surface of a neutron star? How many grams of material would have to fall each second on the neutron star to generate an Eddington luminosity through accretion? This is called the Eddington accretion rate.

Answer for Problem 1

  1. E/m=G M/R
  2. E/m=G M/(2 R)
  3. E/m=1.8 x 1020 erg/g for a neutron star, and E/m=8 x 1016 erg/g for a white dwarf. For a white dwarf the nuclear energy of the infalling material exceeds the gravitational energy, so a nuclear explosion of the material could eject the accumulated material. For a neutron star the material is stuck on the surface.
  4. E/m=1.8 x 1020 erg/g while LEdd=1.8 x 1038 erg/s so the Eddington accretion rate is 1018 g/s or 1.6 x 10-8 /yr.

Problem 2 - Bursts Top

We will try to model Type-I X-ray bursts using a simple model for the instability. We will calculate how much material will accumulate on a neutron star before it bursts.

  1. Let us assume that the star accretes pure helium, that the temperature of the degenerate layer is constant down to the core (Tc), how much luminosity emerges from the surface of the star? (You shouldn't have to derive this formula (I gave it to you in class).
  2. Let us assume that the helium layer has a mass, dM, and that the enregy generation rate for helium burning is given by
    ε3α = 3.5 x 1020 T9-3 exp(-4.32/T9) erg s-1 g-1
    where T9=T/109K. The energy generation rate is a function of density too, but let's forget about that to keep things simple. How much power does the helium layer generate as a function of dM?
  3. Equate your answer to (1) to the answer to (2) and solve for dM. This is the thickness of a layer in thermal equilibrium.
  4. Let's assume that the potential burst starts by the temperature in the accreted layer jiggling up by a wee bit. If the surface luminosity increases faster with temperature than the helium burning rate, then the layer is stable. Calculate dLsurface/dT and dPhelium/dT.
  5. Calculate the value of dM for which dPhelium/dT exceeds dLsurface/dT and the layer bursts.
  6. Equate your value of dM in (3) and (5) and solve for T. What is dM? How long will it take for such a layer to accumulate if the star is accreting at one-tenth of the Eddington accretion rate?

Answer for Problem 2

  1. Using the formula from Week 3 gives
    Lγ,1= 2.35 x 108 erg/s (T/1 K)7/2.

    If you used the black body formula you would get
    Lγ,2= 7 x 108 erg/s (T/1 K)4.

  2. PHelium = ε3α dM = 3.5 x 1020 T9-3 exp(-4.32/T9) erg s-1 g-1 dM
  3. For the envelope calculation you get
    d Lγ,1
    d T    		 = 8.2 x 108 erg/s/K (T/1 K)5/2.
    For the blackbody calculation you get
    d Lγ,2
    d T    		 = 2.8 x 109 erg/s/K (T/1 K)3.
    For the helium burning we get
    4. d PHelium
    d T    		 = 4.2 x 1010 erg/s/g/K T9-5 exp(-4.32/T9) (36 - 25 T9)
  4. For the envelope formula we get

    dM = 6.19 x 1020 T915/2 exp(4.32/T9) (36 - 25 T9)-1 g

    And for the blackbody formula we get

    dM = 6.67 x 1025 T98 exp(4.32/T9) (36 - 25 T9)-1 g

  5. Using T=108 K, yields dM=3.4 x 1030 g (106 yr) and 1.1 x 1035 g (3.6 x 1010 yr).
  6. Something is clear wrong here. Argh!! The problem is the estimate of the temperature. The thickness of the mass layer is very sensitive to the temperature so we should try to get it right.

    The answer to part (4) is the thickness of the marginally stable layer as a function of the temperature. We can also calculate the thickness of a layer in equilibrium by equating the answers to part (1) and (2). Here goes

    For the envelope formula we get
    dM = 2.12 x 1019 T913/2 exp(4.32/T9) g

    For the blackbody formula we get
    dM = 2 x 1024 T97 exp(4.32/T9) g

    Next let's set the equilibrium thickness derived above to be equal to the marginally unstable thickness in part (4) and solve for T9. To keep things compact, we'll call the constant in front of the unstable thickness A and the one in front of the equilibrium thickness B. Believe it or not, we can do this analytically.
    T9 = 36 B
    A + 25 B     		T9 = 0.664 and 0.617 for the two cases.

    This gives a more reasonable layer thickness of 1021 g and 7 x 1025 g and accretion timescales of 2.8 hours and 24 years. The insulation of the envelope makes a big difference. Type-I bursts typically recur on a timescale of hours at one-tenth of the Eddington accretion rate.

Last modified: Thursday, 08 April 2010 14:15:28