WHY ARE PULSAR PLANETS RARE?

It is worth considering whether it is possible (at least in principle) for planets to form in the usual way, in a protoplanetary disk around a young (massive) star, and survive the pulsar formation process (e.g., Bailes et al. 1991). However, there is some doubt as to whether planets can even form around a star massive enough to become a pulsar. There is a strong decline in the probability of a star hosting a giant planet for mass greater than 3 M_⊙ (e.g., Kennedy & Kenyon 2008). Currently, the most massive star to have a detected planet is 3 M_⊙ (Han et al. 2014). This dearth of planets around massive stars could be the result of several factors. First, there is intense UV radiation from the star, which can rapidly destroy the disk. Second, the lifetime of the star is much shorter. We should note, however, that selection effects may be important, since the planets (if they exist) would probably be impossible to detect. Because the star is more massive and brighter than a low-mass star, neither the transit method nor the radial velocity method is sensitive enough to detect a planet.

Supposing that such planets can still form, then there are several conditions that must be met in order for them to survive the death of the star. The orbital radii of the planets must be larger than about 4 AU in order to survive being engulfed during the red-giant phase. In the supernova explosion, half of the mass of the system is ejected. In order for planets to survive, the explosion must be asymmetric. The asymmetry results in a velocity kick to the newly formed pulsar, which must be in a similar direction to the motion of the planet at the time of the supernova (Blaauw 1961; Bhattacharya & van den Heuvel 1991). Planets that survive would be expected to be in eccentric orbits at orbital radii of at least a few AU (e.g., Thorsett et al. 1993).

This mechanism is unlikely to form a system with more than one planet (if any). For three planets to survive this process is almost certainly impossible and so the planets around PSR B1257+12 formed after the pulsar. Furthermore, the two other known pulsar planets most likely did not form in this way, since PSR B1620–26 is a circumbinary planet and PSR J1719–1438 is extremely close to its host star. None of the known pulsar planets formed in this way and we suggest that the probability for planets to form in this scenario is negligible.

When a pulsar forms in a supernova, most of the material of the star is ejected. However, some material can fall back toward the newly formed pulsar if it does not attain the escape velocity or if it is pushed back during the early hydrodynamical mixing phase (e.g., Colgate 1971; Chevalier 1989; Bailes et al. 1991). Estimates of the total mass, which can form the fall back disk are in the range 0.001–0.1 M_⊙ (e.g., Lin et al. 1991; Menou et al. 2001). This is similar to the masses observed in protoplanetary disks around young stars that lie in the range 0.001–0.1 M_⋆ (e.g., Williams & Cieza 2011), where M_⋆ is the mass of the host star. However, the angular momentum of the fallback disk may be very different. Recently, Perna et al. (2014) considered the formation of a supernova fallback disk around neutron stars (and black holes) with numerical simulations of supernova explosions. They found that if magnetic torques play a role in angular momentum transport, then supernova fallback disks do not form around neutron stars. Any material that falls back has too little angular momentum to orbit the neutron star even once. However, formation can occur when the magnetic torques are negligible. This requires the energy of the explosion to be finely tuned so that the fall back mass would on the one hand not cause the neutron star to collapse to a black hole, but on the other, would be sufficient to form a disk. Perna et al. (2014) found a typical fallback mass of 0.08 M_⊙ with the maximum specific angular momentum of the material ≲10¹⁷ cm² s⁻¹.

We can calculate the circularization radius for such a disk. The radius of a ring of material with a given specific angular momentum, j, is

$$\begin{eqnarray}&&{R}_{\mathrm{circ}}=\displaystyle \frac{{j}^{2}}{{{GM}}_{{\rm{p}}}},\end{eqnarray} \tag{ 2 }$$

where M_p is the mass of the pulsar. For typical parameters this is

$$\begin{eqnarray}&&{R}_{\mathrm{circ}}=5.4\times {10}^{7}{\left(\displaystyle \frac{j}{{10}^{17}\mathrm{erg}{\rm{s}}}\right)}^{2}{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{-1}\,\mathrm{cm}.\end{eqnarray} \tag{ 3 }$$

The upper limit to the total angular momentum of such a disk is estimated to be

$$\begin{eqnarray}&&J=2\times {10}^{49}\left(\displaystyle \frac{{M}_{{\rm{d}}}}{0.1\,{M}_{\odot }}\right)\left(\displaystyle \frac{j}{{10}^{17}\,\mathrm{erg}\,{\rm{s}}}\right)\,\mathrm{erg}\,{\rm{s}},\end{eqnarray} \tag{ 4 }$$

where M_d is the mass of the disk. This is in line with estimates by Phinney & Hansen (1993) and Menou et al. (2001). A simple estimate for the lifetime of such a disk is the viscous timescale

$$\begin{eqnarray}&&{\tau }_{\nu }=\displaystyle \frac{{R}^{2}}{\nu },\end{eqnarray} \tag{ 5 }$$

where the viscosity is

$$\begin{eqnarray}&&\nu =\alpha {\left(\displaystyle \frac{H}{R}\right)}^{2}{R}^{2}{\rm{\Omega }},\end{eqnarray} \tag{ 6 }$$

H/R is the disk aspect ratio, ${\rm{\Omega }}=\sqrt{{{GM}}_{{\rm{p}}}/{R}^{3}}$ is the Keplerian angular frequency and α is the Shakura & Sunyaev (1973) viscosity parameter. For typical values at the circularization radius we find

$$\begin{eqnarray}{\tau }_{\nu }({R}_{\mathrm{circ}}) & = & 2.8\times {10}^{4}{\left(\displaystyle \frac{\alpha }{0.01}\right)}^{-1}{\left(\displaystyle \frac{H/R}{0.01}\right)}^{-2}{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{-2}\\ & & \times {\left(\displaystyle \frac{j}{{10}^{17}\mathrm{erg}{\rm{s}}}\right)}^{3}\,{\rm{s}}.\end{eqnarray} \tag{ 7 }$$

The viscous disk spreads both inwards and outwards leading to a longer lifetime for larger radii. However, even an increase in this timescale by several orders of magnitude would not be sufficient for a reasonable timescale for planet formation. Even taking into account uncertainties in some of the physical parameters, if a supernova fallback disk does form, its lifetime is far too short for planet formation to occur.

Observations aimed at finding fallback disks around pulsars have not been successful (e.g., Wang et al. 2007). This, combined with the lack of observed planets around young pulsars (Kerr et al. 2015), suggests that planets do not form in a supernova fallback disk. The observed planets more likely formed during a later accretion phase that led to the spin-up of the pulsar.

This model involves a close binary composed of an MSP (or its progenitor) and a low-mass main-sequence star or compact companion (e.g., Podsiadlowski et al. 1991; Stevens et al. 1992). The companion is losing mass through evaporation, because it is being irradiated by the pulsar. If the evaporation timescale is shorter than the Kelvin–Helmholtz timescale, the secondary responds adiabatically to the mass loss. The companion star fills its Roche lobe and is dynamically disrupted if its radius increases faster than the Roche-lobe radius (e.g., Benz et al. 1990). This is most likely to happen when the companion is fully convective or degenerate (e.g., for a low-mass star the radius of the star is proportional to M^−1/3, where M is the mass of the star). The star is dynamically disrupted and forms a massive disk around the primary with a mass of around 0.1 M_⊙.

In the specific case of the merger of two white dwarfs, the orbit shrinks due to gravitational radiation until the less massive white dwarf fills its Roche lobe. However, in this case, there is some uncertainty whether the merger will lead to a type Ia supernova rather than to the formation of a pulsar (e.g., Livio 2000). It is most likely that a pulsar must be a component of the binary before the destruction of the companion.

During the dissolution of a companion star, the material forms a disk at a radius comparable to the initial separation of the binary, but then spreads out through viscous effects. Note that the disk formed is much larger than that expected from a supernova fallback disk. The total angular momentum of the disk is

$$\begin{eqnarray}&&J=1.1\times {10}^{52}\left(\displaystyle \frac{{M}_{{\rm{d}}}}{0.1\,{M}_{\odot }}\right){\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{2}}{\left(\displaystyle \frac{a}{1{\rm{A}}{\rm{U}}}\right)}^{\tfrac{1}{2}}\,\mathrm{erg}\,{\rm{s}}.\end{eqnarray} \tag{ 8 }$$

This is several orders of magnitude higher than the maximum estimate for the angular momentum of the supernova fallback disk given in Equation (4). With the viscous timescale given by Equation (5) and the viscosity in Equation (6), at a radius of R = 1 AU, we find the viscous timescale to be

$$\begin{eqnarray}\begin{array}{rcl}{\tau }_{\nu } & = & 1.34\times {10}^{5}{\left(\displaystyle \frac{\alpha }{0.01}\right)}^{-1}{\left(\displaystyle \frac{H/R}{0.01}\right)}^{-2}\\ & & \times \,{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{-\tfrac{1}{2}}{\left(\displaystyle \frac{R}{1{\rm{A}}{\rm{U}}}\right)}^{\tfrac{3}{2}}\,\mathrm{yr}.\end{array}\end{eqnarray} \tag{ 9 }$$

Since some material spreads outwards from its initial radius, the disk lifetime is longer than this. For example, material that reaches R = 5 AU has a viscous timescale of over 1 Myr. This is similar to the disk lifetime for a protoplanetary disk around a young star (e.g., Williams & Cieza 2011). We therefore suggest that the destruction of a companion star is the most likely scenario for forming the planetary system around PSR B1257+12. The accretion rate is typically $\dot{M}={M}_{{\rm{d}}}/{\tau }_{\nu }$ which is

$$\begin{eqnarray}\begin{array}{rcl}\dot{M} & = & 6.64\times {10}^{-8}\left(\displaystyle \frac{\alpha }{0.01}\right){\left(\displaystyle \frac{H/R}{0.01}\right)}^{2}\left(\displaystyle \frac{{M}_{{\rm{d}}}}{0.1\,{M}_{\odot }}\right)\\ & & \times \,{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{2}}{\left(\displaystyle \frac{R}{5{\rm{A}}{\rm{U}}}\right)}^{-\tfrac{3}{2}}\,{M}_{\odot }\,{\mathrm{yr}}^{-1}.\end{array}\end{eqnarray} \tag{ 10 }$$

This is similar to typical accretion rates observed around young T Tauri stars (e.g., Armitage et al. 2003; Calvet et al. 2004). Over time, as the mass of the disk decreases, the accretion rate also decreases. For a fully viscous disk, once the material has spread out into the inner regions, the disk can be modeled to be in a quasi-steady-state. In Section 4, we investigate the properties of this model further.

A companion star may be evaporated by the radiation from a pulsar (Fruchter et al. 1990; Bailes et al. 1991; Krolik 1991; Rasio et al. 1992; Tavani & Brookshaw 1992). The companion has to be almost, but not completely, evaporated by the pulsar leaving a remnant, which is planet sized. This is thought to be the explanation for the planet in PSR J1719–1438. This mechanism is only capable of forming a single planet around a pulsar and so this scenario is unlikely to have been responsible for the planetary system around PSR B1257+12. The fine tuning required for this mechanism for planet formation means that this is probably a rare event. If the companion is completely evaporated and the planets form from that debris, then the formation is somewhat similar to that described in the previous Section 2.3. However, the disk formed in this manner may have very little mass and so planet formation may be unlikely.

In summary, the most likely mechanisms for pulsar planet formation require a low-mass companion to the pulsar. In the next section, we examine the parameters required to form an extreme mass ratio binary, which is potentially capable of pulsar planet formation.

If a dead zone forms, the inner edge, closest to the star, is determined by the point at which the temperature of the disk drops below the critical temperature, T < T_crit. In this section, we assume that in the inner parts of the disk, the temperature is dominated by the irradiation from the pulsar. The heating due to the irradiation is

$$\begin{eqnarray}&&{Q}_{\mathrm{irr}}=\sigma {T}_{\mathrm{irr}}^{4}=\displaystyle \frac{L}{4\pi {R}^{2}}(1-\beta )\cos \phi ,\end{eqnarray} \tag{ 11 }$$

(e.g., Frank et al. 2002) where the albedo is β = 0.5 and

$$\begin{eqnarray}&&\cos \phi =\displaystyle \frac{{dH}}{{dR}}-\displaystyle \frac{H}{R}.\end{eqnarray} \tag{ 12 }$$

The accretion luminosity is

$$\begin{eqnarray}&&{L}_{\mathrm{acc}}=\displaystyle \frac{{{GM}}_{{\rm{p}}}\dot{M}}{{R}_{{\rm{p}}}},\end{eqnarray} \tag{ 13 }$$

where $\dot{M}$ is the accretion rate onto the pulsar. For typical values this is

$$\begin{eqnarray}&&{L}_{\mathrm{acc}}=3.1\times {10}^{4}\left(\displaystyle \frac{\dot{M}}{{10}^{-8}\,{M}_{\odot }\,{\mathrm{yr}}^{-1}}\right)\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4\,{M}_{\odot }}\right){\left(\displaystyle \frac{{R}_{{\rm{p}}}}{10\mathrm{km}}\right)}^{-1}\,{L}_{\odot }.\end{eqnarray} \tag{ 14 }$$

However, this is limited by the Eddington luminosity,

$$\begin{eqnarray}&&{L}_{\mathrm{EDD}}=4.5\times {10}^{4}\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4\,{M}_{\odot }}\right)\,{L}_{\odot }.\end{eqnarray} \tag{ 15 }$$

Therefore, the luminosity in Equation (11) is

$$\begin{eqnarray}&&L=\min ({L}_{\mathrm{acc}},{L}_{\mathrm{EDD}}).\end{eqnarray} \tag{ 16 }$$

We find the critical accretion rate above which the luminosity is limited by the Eddington Luminosity by solving ${L}_{\mathrm{acc}}={L}_{\mathrm{EDD}}$ to be

$$\begin{eqnarray}&&{\dot{M}}_{\mathrm{crit}}=1.45\times {10}^{-8}\left(\displaystyle \frac{{R}_{{\rm{p}}}}{10\,\mathrm{km}}\right)\,{M}_{\odot }\,{\mathrm{yr}}^{-1}.\end{eqnarray} \tag{ 17 }$$

We assume that the disk aspect ratio can be written as a power law in radius H ∝ Rⁿ, This simplifies Equation (12) to read

$$\begin{eqnarray}&&\cos \phi =\displaystyle \frac{H}{R}(n-1).\end{eqnarray} \tag{ 18 }$$

Assuming hydrostatic equilibrium and that the disk temperature is dominated by the irradiation, we have

$$\begin{eqnarray}&&H=\displaystyle \frac{{c}_{{\rm{s}}}}{{\rm{\Omega }}}=\displaystyle \frac{1}{{\rm{\Omega }}}\sqrt{\displaystyle \frac{{ \mathcal R }{T}_{\mathrm{irr}}}{\mu }},\end{eqnarray} \tag{ 19 }$$

where ${ \mathcal R }=8.31\times {10}^{7}\,\mathrm{erg}\,{{\rm{K}}}^{-1}\,{\mathrm{mol}}^{-1}$ is the gas constant and μ = 2.3 is the mean molecular weight. We take Equation (11) (with L = L_acc defined in Equation (13)) and combine it with Equation (19) to write the irradiation temperature for L_acc < L_EDD as

$$\begin{eqnarray}\begin{array}{rcl}{T}_{\mathrm{irr}} & = & 1643{\left(\displaystyle \frac{\dot{M}}{{10}^{-8}{M}_{\odot }{\mathrm{yr}}^{-1}}\right)}^{\tfrac{2}{7}}{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{7}}\\ & & \times \,{\left(\displaystyle \frac{{R}_{{\rm{p}}}}{10\mathrm{km}}\right)}^{-\tfrac{2}{7}}{\left(\displaystyle \frac{R}{1{\rm{A}}{\rm{U}}}\right)}^{-\tfrac{3}{7}}\,{\rm{K}}.\end{array}\end{eqnarray} \tag{ 20 }$$

Since we find T_irr ∝ R^−3/7, this gives $H={c}_{{\rm{s}}}/{\rm{\Omega }}\propto {T}_{\mathrm{irr}}^{1/2}{R}^{3/2}\propto {R}^{9/7}$. For a disk which is dominated by irradiation, n = 9/7 and we use this value for the rest of the work. Similarly, for higher accretion rates, with L_acc > L_EDD, we combine Equation (11) (with L = L_EDD in Equation (15)) with Equation (19) and we find

$$\begin{eqnarray}&&{T}_{\mathrm{irr},\mathrm{EDD}}=1825{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{7}}{\left(\displaystyle \frac{R}{1{\rm{A}}{\rm{U}}}\right)}^{-\tfrac{3}{7}}\,{\rm{K}}.\end{eqnarray} \tag{ 21 }$$

In Figure 2, we show the disk temperature as a function of radius for various accretion rates. We also show the critical temperature required for the MRI to operate. A dead zone forms when the temperature of the disk drops below this. We solve ${T}_{\mathrm{irr}}={T}_{\mathrm{crit}}$ to find the critical radius outside of which the temperature drops below that required for the MRI,

$$\begin{eqnarray}\begin{array}{rcl}{R}_{{\rm{d}},\mathrm{in}} & = & 1.45{\left(\displaystyle \frac{\dot{M}}{{10}^{-8}{M}_{\odot }{\mathrm{yr}}^{-1}}\right)}^{\tfrac{2}{3}}{\left(\displaystyle \frac{{R}_{{\rm{p}}}}{10\mathrm{km}}\right)}^{-\tfrac{2}{3}}\\ & & \times \,{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{3}}{\left(\displaystyle \frac{{T}_{{\rm{c}}}}{1400{\rm{K}}}\right)}^{-\tfrac{7}{3}}\,{\rm{A}}{\rm{U}}.\end{array}\end{eqnarray} \tag{ 22 }$$

If, at this radius, the surface density is sufficiently high, then a dead zone forms outside of this radius. Above the Eddington limit, we solve ${T}_{\mathrm{irr},\mathrm{EDD}}={T}_{\mathrm{crit}}$ to find the critical inner dead zone radius

$$\begin{eqnarray}&&{R}_{{\rm{d}},\mathrm{in},\mathrm{EDD}}=1.85{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{3}}{\left(\displaystyle \frac{{T}_{{\rm{c}}}}{1400{\rm{K}}}\right)}^{-\tfrac{7}{3}}\,{\rm{A}}{\rm{U}}.\end{eqnarray} \tag{ 23 }$$

The lower line in Figure 4 shows the inner edge of the dead zone as a function of the accretion rate. This is a combination of Equation (22) for $\dot{M}\lt {\dot{M}}_{\mathrm{crit}}$ and Equation (23) for $\dot{M}\gt {\dot{M}}_{\mathrm{crit}}$. Since the inner dead zone radius does not depend upon the critical surface density, it is the same in both plots. In the next section, we consider the radial location of the outer edge of the dead zone.

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If a dead zone is present, its outer edge is determined by the location at which the surface density drops below the critical, Σ < Σ_crit. The viscosity of the disk is

$$\begin{eqnarray}&&\nu =\alpha \displaystyle \frac{{c}_{{\rm{s}}}^{2}}{{\rm{\Omega }}}.\end{eqnarray} \tag{ 24 }$$

For a steady-state disk, the surface density is given by

$$\begin{eqnarray}&&{\rm{\Sigma }}=\displaystyle \frac{\dot{M}}{3\pi \nu }\mathrm{.}\end{eqnarray} \tag{ 25 }$$

(Pringle 1981). For our typical parameters, for an accretion rate which is not Eddington limited, this is

$$\begin{eqnarray}\begin{array}{rcl}{\rm{\Sigma }} & = & 26.7{\left(\displaystyle \frac{\dot{M}}{{10}^{-8}{M}_{\odot }{\mathrm{yr}}^{-1}}\right)}^{\tfrac{5}{7}}{\left(\displaystyle \frac{{R}_{{\rm{p}}}}{10\mathrm{km}}\right)}^{\tfrac{2}{7}}{\left(\displaystyle \frac{\alpha }{0.01}\right)}^{-1}{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{5}{14}}\\ & & \times \,{\left(\displaystyle \frac{R}{1{\rm{A}}{\rm{U}}}\right)}^{-\tfrac{15}{14}}\,{\rm{g}}\,{\mathrm{cm}}^{-2}.\end{array}\end{eqnarray} \tag{ 26 }$$

Similarly, for the case in which the luminosity is Eddington limited, we can find the surface density

$$\begin{eqnarray}\begin{array}{rcl}{{\rm{\Sigma }}}_{\mathrm{EDD}} & = & 24.0\left(\displaystyle \frac{\dot{M}}{{10}^{-8}\,{M}_{\odot }\,{\mathrm{yr}}^{-1}}\right){\left(\displaystyle \frac{\alpha }{0.01}\right)}^{-1}{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{5}{14}}\\ & & \times \,{\left(\displaystyle \frac{R}{1{\rm{A}}{\rm{U}}}\right)}^{-\tfrac{15}{14}}\,{\rm{g}}\,{\mathrm{cm}}^{-2}.\end{array}\end{eqnarray} \tag{ 27 }$$

In Figure 3, we plot the surface density as a function of radius for various accretion rates.

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We can find the radius of the outer edge of the dead zone by solving Σ = Σ_crit. Below the Eddington accretion rate we find

$$\begin{eqnarray}\begin{array}{rcl}{R}_{{\rm{d}},\mathrm{out}} & = & 1.3{\left(\displaystyle \frac{\dot{M}}{{10}^{-8}{M}_{\odot }{\mathrm{yr}}^{-1}}\right)}^{\tfrac{2}{3}}{\left(\displaystyle \frac{\alpha }{0.01}\right)}^{-\tfrac{14}{15}}{\left(\displaystyle \frac{{R}_{{\rm{p}}}}{10\mathrm{km}}\right)}^{\tfrac{4}{15}}\\ & & \times \,{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{3}}{\left(\displaystyle \frac{{{\rm{\Sigma }}}_{\mathrm{crit}}}{20{\rm{g}}{\mathrm{cm}}^{-2}}\right)}^{-\tfrac{14}{15}}\,{\rm{A}}{\rm{U}}\end{array}\end{eqnarray} \tag{ 28 }$$

and the corresponding critical radius above the Eddington accretion rate is

$$\begin{eqnarray}\begin{array}{rcl}{R}_{{\rm{d}},\mathrm{out},\mathrm{EDD}} & = & 1.18{\left(\displaystyle \frac{\dot{M}}{{10}^{-8}{M}_{\odot }{\mathrm{yr}}^{-1}}\right)}^{\tfrac{14}{15}}{\left(\displaystyle \frac{\alpha }{0.01}\right)}^{-\tfrac{14}{15}}\\ & & \times \,{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{1}{3}}{\left(\displaystyle \frac{{{\rm{\Sigma }}}_{\mathrm{crit}}}{20{\rm{g}}{\mathrm{cm}}^{-2}}\right)}^{-\tfrac{14}{15}}\,{\rm{A}}{\rm{U}}.\end{array}\end{eqnarray} \tag{ 29 }$$

In Figure 4, the upper lines show the outer edge of the dead zone, which depends on both the accretion rate and the critical active layer surface density. In the next section, we determine whether a dead zone can exist, since if R_d,in > R_d,out, then there is no region of parameter space for a dead zone and the disk is fully turbulent.

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The condition which must be satisfied for a dead zone to exist is that the surface density at the radius of the inner dead zone boundary must be larger than the critical surface density that is MRI active,

$$\begin{eqnarray}&&{\rm{\Sigma }}({R}_{{\rm{d}},\mathrm{in}})\gt {{\rm{\Sigma }}}_{\mathrm{crit}}.\end{eqnarray} \tag{ 30 }$$

Since the surface density in the disk increases with radius, the larger the Σ_crit the less likely it is that a dead zone may form. This condition is equivalent to

$$\begin{eqnarray}&&{R}_{{\rm{d}},\mathrm{in}}\lt {R}_{{\rm{d}},\mathrm{out}}.\end{eqnarray} \tag{ 31 }$$

We consider here disk parameters for which a dead zone exists.

In Figure 4, we show the inner and outer dead zone radii for an example with T_crit = 1400 K and ${{\rm{\Sigma }}}_{\mathrm{crit}}=2\,{\rm{g}}\,{\mathrm{cm}}^{-2}$ (left) and ${{\rm{\Sigma }}}_{\mathrm{crit}}=0.2\,{\rm{g}}\,{\mathrm{cm}}^{-2}$ (right). The shaded regions show the accretion rates for which a dead zone exists and the range of radii. We find that for Σ_crit ≳ 10 g cm⁻² a dead zone does not exist for typical accretion rates. However, for a smaller critical surface density, Σ_crit ≲ 10 g cm⁻², there is a dead zone for all accretion rates. We suggest that the formation of planets around a pulsar requires a very small critical surface density. Such a low critical surface density may be difficult to achieve, because cosmic rays are expected to penetrate ${{\rm{\Sigma }}}_{\mathrm{crit}}=200\,{\rm{g}}\,{\mathrm{cm}}^{-2}$. There are several possibilities for lowering the active layer surface density. First, as described in Section 4, cosmic rays may be swept away, thus allowing a large dead zone to form. Second, the presence of dust or polycyclic hydrocarbons can suppress the MRI. MHD simulations, which include these effects, find small critical surface densities (e.g., Bai 2011; Perez-Becker & Chiang 2011). Third, the inclusion of non-ideal MHD effects may decrease the amount of turbulence (e.g., Simon et al. 2013). There remains much uncertainty in the value of the active layer surface density.

We note that our model represents a steady-state disk, but in a real disk around the pulsar there will initially be a spreading phase before the steady state is reached. If the disk is made by the destruction of a binary companion, then, initially the material will lie at the binary orbital separation. The material spreads both inwards and outwards from there. The steady-state solutions calculated in this work may only be appropriate inside of the initial binary separation radius. Depending upon the spreading rate of the disk, the dead zone may extend to the outer edge of the disk. This is especially likely for high accretion rates, early in the disk evolution. The spreading of the disk may be slowed, because of the dead zone. We discuss this further in Section 5.

In this work, so far we have not included the effect of viscous heating on the steady-state disk model. We can calculate the temperature from the viscous heating

$$\begin{eqnarray}&&\sigma {T}_{\mathrm{visc}}^{4}=\displaystyle \frac{9}{8}\nu {\rm{\Sigma }}{{\rm{\Omega }}}^{2}.\end{eqnarray} \tag{ 32 }$$

With this, we calculate (for accretion rates below the Eddington limit) the relative temperature compared with the irradiation temperature

$$\begin{eqnarray}\begin{array}{rcl}\displaystyle \frac{{T}_{\mathrm{visc}}}{{T}_{\mathrm{irr}}} & = & 0.056{\left(\displaystyle \frac{\dot{M}}{{10}^{-8}{M}_{\odot }{\mathrm{yr}}^{-1}}\right)}^{-\tfrac{1}{28}}{\left(\displaystyle \frac{{R}_{{\rm{p}}}}{10\mathrm{km}}\right)}^{\tfrac{2}{7}}{\left(\displaystyle \frac{{M}_{{\rm{p}}}}{1.4{M}_{\odot }}\right)}^{\tfrac{3}{28}}\\ & & \times \,{\left(\displaystyle \frac{R}{1{\rm{A}}{\rm{U}}}\right)}^{-\tfrac{9}{28}}.\end{array}\end{eqnarray} \tag{ 33 }$$

In Figure 5, we show this ratio as a function of radius for an accretion rate of 10⁻⁸ M_⊙ yr⁻¹. As can be seen from the above equation, this is not very sensitive to the accretion rate. Furthermore, the heating from viscous effects is much smaller than the heating due to the irradiation from the pulsar. If we were to include this heating into our steady-state disk models, the temperature of the disk would increase slightly. The models we have shown here represent the maximum size of the dead zone in the absence of viscous heating.

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