General Relativistic Anisotropic Hydrostatic Equilibrium Equations for a Magnetized Compact

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General Relativistic Anisotropic Hydrostatic Equilibrium Equations for a Magnetized Compact

Object

Daryel Manreza Paret,¹ Aurora P ´erez Mart´ınez² Jorge Horvath³

1Universidad de la Habana, Facultad de F´ısica

2ICIMAF

3IAG/USP

October 1, 2013

(FF/UH, ICIMAF, IAG/USP, & CLAF/ICTP. D. Manreza Paret) 1 / 24

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Outline

1 Introduction.

2 TOV equation.

3 Pressures anisotropy due to a magnetic field.

4 Cylindrical coordinates.

5 Hydrostatic Equilibrium Equations for a Magnetized CO.

Numerical solution

6 Conclusions.

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Introduction

1. M

M.8 ^WD

e−

e− e−

e−

M M>8

NS

u

s s

d d

?

n

n n

M

M>20 ^BH

Figure:Diagram of the final state of stars.

Figure:Sirius A and B (Hubble telescope).

Figure:Crab nebula (Hubble and Chandra).

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Introduction

Neutron Stars. Quark Stars?

M∼1.4M R∼12 km

ρ∼10¹⁴g/cm³ B ∼10¹³−10¹⁵G

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Introduction

Mass–Radius relation Measurements of the masses or radii of CO can strongly constrain the neutron star matter equation of state and rule out theoretical models of their composition.

The observed range of neutron star masses, however, has hitherto been too narrow to rule out many predictions of exotic non-nucleonic

components. _Figure:Mass–Radius diagram^†.

†P. Demorest, T. Pennucci, S. Ransom, M. Roberts and J. Hessels, Nature467, 1081 (2010)

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TOV equation

To found the static structure of a relativistic spherical star we most solve the Einstein’s equation

G^µ_ν≡R^µ_ν−1

2Rg^µ_ν = 8πGT^µν, (1) (µ, ν= 0,1,2,3), using the Schwarzschild metric

ds²=−e^2νdt²+e^2λdr²+r²dΩ², dΩ²=dθ²+ sin²φdφ², (2) and the energy momentum tensor

T^µν = (+P)u^µuν+P g^µ_ν, (3) we obtain the TOV equations

dM

dr = 4πG, dP

dr = −G(+P)(M+ 4πPr³) r²−2rM , with boundary conditionsP(R) = 0,M(0) = 0.

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Pressures anisotropy due to a magnetic field

A magnetic field breaks the spherical symmetry and produce an anisotropy in the pressures inside the object.

NE

d u s

d

u s u s

d

B

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Pressures anisotropy due to a magnetic field

Magnetized Fermion Gas

The thermodynamical potential in the framework of the MIT Bag Model is given by:

Ω_i,v(µ_i, T) =− d_iT (2π)³

Z

p

[ln[1 + exp{−E_p,i^n,η−µ_i T }]d³p], where,

E_p,i^n,η= q

p²_z+p²_⊥+m²_i, p²_⊥=m²_i{( sB

Bi^c

(2n+ 1−η) + 1−ηy_iB)²−1},

i=e, u, d, s,B_i^c=m²_i/|ei|is the critical magnetic field,yi=|Qi|/mi accounts for the AMM andη=±1correspond to the orientations of the particle

magnetic moment, parallel or antiparallel to the magnetic field. Parameters de= 1anddu,d,s= 3are degeneracy factors.

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Pressures anisotropy due to a magnetic field

For degenerate magnetized strange quark matter, the number density, energy density and pressures are given by the expressions:

N =−X

i

∂Ωi,v

∂µi

=X

i

N_i⁰ B B_i^c

nⁱ_max

X

n

X

η=±1

p^η_F,i,

ε= Ω +µN =BX

i

M⁰i nⁱ_max

X

n

X

η=±1

xip^η_F,i+h^η_i²lnxi+p^η_F,i h^η_i

! ,

P_k=−Ω =BX

i

M⁰i nⁱ_max

X

n

X

η=±1

x_ip^η_F,i−h^η_i²lnxi+p^η_F,i h^η_i

! ,

P⊥=−Ω− MB=BX

i

M⁰i nⁱ_max

X

n

X

η=±1

2h^η_iγ_i^ηlnx_i+p^η_F,i h^η_i

! ,

The sum over the Landau levelsnis up to nⁱ_max=I

(xi+ηyiB)²−1

B_i^c/(2B)

,whereI[z]denotes the integer part of z.

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Pressures anisotropy due to a magnetic field

We have defined the dimensionless quantities:

M⁰i = e_id_im²_i

4π² , N_i⁰=d_im³_i

2π² , x_i=µ_i/m_i p^η_F,i =

q

x²_i −h^η_i ², h^η_i = sB

B_i^c(2n+ 1−η) + 1−ηyiB γ_i^η = B(2n+ 1−η)

2B_i^cp

(2n+ 1−η)B/B^c_i + 1−ηy_iB,

wherexiis the dimensionless chemical potential,pF,icorresponds to the modified Fermi momentum due to the magnetic field andh^η_i corresponds to the magnetic mass.

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Pressures anisotropy due to a magnetic field

MIT Bag model for SQM

In order to study the matter inside the star we use the MIT Bag model.The stelar equilibrium conditions are then determined by the following relations:

d↔u+e⁻+ ¯νe s↔u+e⁻+ ¯νe u+s↔d+u So that we have to solve the system of equations

µu+µe−µd= 0, µd−µs = 0 βequilibrium 2nu−nd−ns−3ne = 0 charge neutrality

nu+nd+ns−3nB = 0 baryon number conservation.

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Pressures anisotropy due to a magnetic field

3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0

2 0 4 0 6 0 8 0 1 0 0 1 2 0

1 4 0 P _⊥ B = 5 ×1 0 ^{1 8} G

P _B = 5 ×1 0 ^{1 8} G P _B = 1 ×1 0 ^{1 6} G P _⊥B = 1 ×1 0 ^{1 6} G

P [MeV fm-3 ]

ε [ M e V f m ^{- 3}] B _{b a g}= 7 5 M e V f m ^{- 3} Figure:Equations of state for magnetized strange quark matter.

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Pressures anisotropy due to a magnetic field

1 E 1 7 1 E 1 8 1 E 1 9

1 1 0 1 1 5 1 2 0 1 2 5 1 3 0 1 3 5

P _⊥ P _

P [MeV fm-3 ]

B [ G ]

B _{b a g}= 7 5 M e V f m ^{- 3}

Figure:Dependence of the pressures whit respect to the magnetic field.

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Pressures anisotropy due to a magnetic field

4 5 6 7 8 9 1 0

0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0 1 . 2 1 . 4 1 . 6 1 . 8

P _⊥ B = 5 × 1 0 ^{1 8} G P _B = 5 × 1 0 ^{1 8} G P _B = 1 ×1 0 ^{1 6} G P _⊥B = 1 ×1 0 ^{1 6} G

M/M

R [ k m ] B _{b a g}= 7 5 M e V f m ^{- 3}

Figure:Mass-Radius diagram comparing the effects of taking parallel or perpendicular pressure.

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Cylindrical symmetry

^†

For the study of the impact of pressures anisotropies in the structure of magnetized stars we will use the cylindrically symmetric metric

ds²=−e^2Φdt²+e^2Λdr²+r²dφ²+e^2Ψdz² (5) whereΦ,Λ,Ω, andΨare functions ofronly.

With this metric we have for the nonzero Einstein tensor components G^t_t = e^−2Λ(Ψ⁰⁰+ Ψ⁰²−Ψ⁰Λ⁰−1

rΛ⁰+1 rΨ⁰) G^r_r = e^−2Λ(Ψ⁰Φ⁰+1

rΦ⁰+1 rΨ⁰)

G^φ_φ = e^−2Λ(Φ⁰⁰+ Φ⁰²−Φ⁰Λ⁰+ Ψ⁰⁰+ Ψ⁰²−Ψ⁰Λ⁰+ Ψ⁰Φ⁰) G^z_z = e^−2Λ(Φ⁰⁰+ Φ⁰²−Φ⁰Λ⁰−1

rΛ⁰+1 rΦ⁰)

†C. S. Trendafilova and S. A. Fulling, “Static solutions of Einstein’s equations with cylindrical symmetry,” Eur. J. Phys.32, 1663 (2011) [arXiv:1101.4668 [gr-qc]].

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Cylindrical symmetry

^†

Energy momentum tensor for magnetized mater is^†

T^µν=







0 0 0

0 P⊥ 0 0 0 0 P⊥ 0 0 0 0 P_k







. (6)

From the Einstein field equations in natural units we get the following four differential equations:

4π = −e^−2Λ(Ψ⁰⁰+ Ψ⁰²−Ψ⁰Λ⁰−1 rΛ⁰+1

rΨ⁰) 4πP_⊥ = e^−2Λ(Ψ⁰Φ⁰+1

rΦ⁰+1 rΨ⁰)

4πP_⊥ = e^−2Λ(Φ⁰⁰+ Φ⁰²−Φ⁰Λ⁰+ Ψ⁰⁰+ Ψ⁰²−Ψ⁰Λ⁰+ Ψ⁰Φ⁰) 4πP_k = e^−2Λ(Φ⁰⁰+ Φ⁰²−Φ⁰Λ⁰−1

rΛ⁰+1 rΦ⁰)

†R. G. Felipe, H. J. Mosquera Cuesta, A. Perez Martinez and H. Perez Rojas, “Quantum instability of magnetized stellar objects,” Chin. J. Astron. Astrophys.5, 399 (2005) [astro-ph/0207150].

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Hydrostatic Equilibrium Equations for a Magnetized CO

By doing some algebra with the previous system of equation we get

P_⊥⁰ = −Φ⁰(+P_⊥)−Ψ⁰(P_⊥−P_k) 4πe^2Λ(+P_k+ 2P_⊥) = Φ⁰⁰+ Φ⁰(Ψ⁰+ Φ⁰−Λ⁰) +Φ⁰

r 4πe^2Λ(+P_k−2P_⊥) = −Ψ⁰⁰−Ψ⁰(Ψ⁰+ Φ⁰−Λ⁰)−Ψ⁰

r 4πe^2Λ(P_k−) = 1

r(Ψ⁰+ Φ⁰−Λ⁰) This is a system of differential equations in the variables

P_⊥, Φ, Λ, Ψ, (7)

provided the equations of state

→f(P_⊥), P_k→f() (8) andinitial conditions.

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Hydrostatic Equilibrium Equations for a Magnetized CO

Initial conditions

Since the differential equations involve factors of1/rwe will make power series expansions ofp,Φ,Ψ, andΛaroundr= 0to find initial condition suitable for numeric calculation

P_⊥ = P_⊥0+P_⊥1r, (9)

Λ = Λ0+ Λ1r, (10)

Φ = Φ0+ Φ1r+ Φ2r², (11) Ψ = Ψ₀+ Ψ₁r+ Ψ₂r². (12) We take also

Ψ = Φ = Λ = 0atr= 0so that the corresponding metric coefficients are equal to1at that point.

Ψ⁰ = Φ⁰ = 0to get smooth solutions at the axis.

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Hydrostatic Equilibrium Equations for a Magnetized CO

Initial conditions

By substitution of these conditions in the system of differential equations we find

P_⊥(0) = P_⊥0 Λ(0) = 0 Φ(0) = 1

2(P_k0+ 2P_⊥0+0)(r₀²−2r0) Ψ(0) = 1

2(−P_k0+ 2P_⊥0−0)(r²₀−2r0) Φ⁰(0) = 0

Ψ⁰(0) = 0 And we also take

P_k(R_k) = 0, P⊥(R⊥) = 0

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Hydrostatic Equilibrium Equations for a Magnetized CO

Mass equation

For computing the mass we will use Tolman^† generalization for the the mass of a source

mT = Z √

−g(T₀⁰−T₁¹−T₂²−T₃³)dV (13) for our cylindric metric we have

m_T = Z

re^Φ+Ψ+Λ(−2P_⊥−P_k)dV (14)

†R. C. Tolman, Relativity, Thermodynamics and Cosmology. Oxford University Press, London (1934)

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Hydrostatic Equilibrium Equations for a Magnetized CO

Numerical solution

0 1 2 3 4 5 6 7 8

0 . 0 0 . 2 0 . 4 0 . 6 0 . 8

1 . 0 B = 1 _×1 0 ^{1 6} G

M/M

R [ k m ] )

B b a g= 7 5 M e V f m ^{- 3}

Figure:Mass-Radius diagram forB= 10¹⁶G.

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Hydrostatic Equilibrium Equations for a Magnetized CO

Numerical solution

0 1 2 3 4 5 6 7 8

0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0

B = 1 0 ^{1 6} G R _⊥ B = 1 0 ^{1 8} G R _ B = 1 0 ^{1 8} G M/M

R [ k m ] )

B b a g= 7 5 M e V f m ^{- 3}

Figure:Mass-Radius diagram forB= 10¹⁶G andB= 10¹⁸.

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Conclusions

1 Taking into account pressure anisotropy due to a magnetic field leads to smaller maximum masses.

2 For high values of the magnetic field the behaviour of the mass radius relation changes.

Future work

1 Make more numerical experiments varying parameters such as the Bag in order to see if there are any configurations in which the system could reach higher values of the mass.

General Relativistic Anisotropic Hydrostatic Equilibrium Equations for a Magnetized Compact