“Anti-glitches” in the Quark-Nova model for AXPs II

Abstract

We investigate the “antiglitch” AXP 1E 2259+586 experienced between MJD=56031 and 56045 in the context of the Quark-Nova model in which an AXP is a quark star surrounded by a degenerate Keplerian disk. In a companion paper we assumed the “anti-glitch” to be instantaneous, whereas in this paper we consider the quark star to undergo a period of enhanced spin-down over several days. We find that the Quark-Nova model can account for the spin-down and at the same time the enhanced 2–10 keV observed flux without introducing any new physics to the model.

Appendix: Mass loss due to wind

The mass loss from the ring atmosphere is given by equation B.6 in Ouyed et al. (2007b):

$$ \dot{m}(t) \approx1.66 \times10^{18}\mbox{ gm}\,\mbox{s}^{-1} \frac{R_{\mathrm{in},25}^3 T_{\mathrm{ring}, \mathrm{keV}}^3}{\mu_{\mathrm {{atm}}}(t)^{3/2} M_{\mathrm{QS},1.5}} $$

(A1)

where R _in,25 is the inner radius of the ring in units of 25 km, T _ring,keV is the temperature of the ring in units of keV, and M _QS,1.5 is the mass of the QS in units of 1.5 M_⊙. The mean molecular weight of the atmosphere, μ _atm(t), evolves in time after a burst according to equation 27 in Ouyed et al. (2007b):

$$ \mu_{\mathrm{{atm}}}(t) = \biggl[\frac{1}{\mu_q} + \biggl(\frac{1}{\mu_b}- \frac{1}{\mu_q} \biggr)e^{-t/\tau} \biggr]^{-1} $$

(A2)

where μ _q and μ _b are the mean molecular weight of the atmosphere in quiescence and during the burst respectively. τ is given by equation 26 in Ouyed et al. (2007b):

$$ \tau= 2.52 \times10^{7} \mbox{ s} \frac{M_{\mathrm{QS},1.5}^4}{\eta _{0.1}^3 R_{\mathrm{in},25}^6} $$

(A3)

where η _0.1 is the wall accretion efficiency in unis of 0.1. If we assume a fraction of the mass-loss, β, is removed as the wind then (1−β) is available for accretion onto the QS giving an accretion luminosity of:

$$\begin{aligned}& L_{\mathrm{{acc}}}(t) = 1.5 \times10^{38} \mbox{ erg}\,\mbox{s}^{-1} \\& \hphantom{L_{\mathrm{{acc}}}(t) =} {}\times \eta_{0.1} (1-\beta) \biggl[\frac{R_{\mathrm{in},25}^3 T_{\mathrm {ring}, \mathrm{keV}}^3}{\mu_{\mathrm{{atm}}}(t)^{3/2} M_{\mathrm{QS}, 1.5}} \biggr] \end{aligned}$$

(A4)

Equation B.13 in Ouyed et al. (2007b) then gives the equilibrium temperature of the ring:

$$ T_{\mathrm{ring}, \mathrm{keV}}(t) = 0.8\mbox{ keV} \frac{(1-\beta)\eta_{0.1} R_{\mathrm{in},25}}{\mu_{\mathrm{{atm}}}(t)^{3/2} M_{\mathrm {QS},1.5}} $$

(A5)

Substituting Eq. (A5) into (A1) (and multiplying by β) we get the mass loss rate due to the wind:

$$ \dot{m}_{\mathrm{{wind}}}(t) \approx3.9 \times10^{17}\mbox{ gm}\,\mbox{s}^{-1} \beta(1-\beta)^3 \frac{R_{\mathrm{in},25}^6 \eta_{0.1}^3}{\mu _{\mathrm{{atm}}}(t)^6 M_{\mathrm{QS},1.5}^4} $$

(A6)

The total mass lost due to the wind from the time of the burst, t _b , to time t is:

$$\begin{aligned}& \Delta m_{\mathrm{{wind}}}(t) \approx3.9 \times10^{17}\mbox{ gm}\,\beta (1-\beta)^3 \\& \hphantom{\Delta m_{\mathrm{{wind}}}(t) \approx} {}\times \frac{R_{\mathrm{in},25}^6 \eta_{0.1}^3}{\mu_{\mathrm {{atm}}}(t)^6 M_{\mathrm{QS},1.5}^4} \biggl[\int_{t_b}^{t} \mu_{\mathrm{{atm}}}(t)^{-6} dt \biggr] \end{aligned}$$

(A7)