Separability of Dirac equation in higher dimensional Kerr-NUT-de ...

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Jan 16, 2008 - arXiv:0711.0078v2 [hep-th] 16 Jan 2008. November, 2007. OCU-PHYS 280. Separability of Dirac equation in higher dimensional Kerr-NUT-de ...
November, 2007 OCU-PHYS 280

arXiv:0711.0078v2 [hep-th] 16 Jan 2008

Separability of Dirac equation in higher dimensional Kerr-NUT-de Sitter spacetime

Takeshi Ootaa1 and Yukinori Yasuib2

a Osaka

City University Advanced Mathematical Institute (OCAMI) 3-3-138 Sugimoto, Sumiyoshi, Osaka 558-8585, JAPAN

b Department

of Mathematics and Physics, Graduate School of Science, Osaka City University 3-3-138 Sugimoto, Sumiyoshi, Osaka 558-8585, JAPAN

Abstract It is shown that the Dirac equations in general higher dimensional Kerr-NUT-de Sitter spacetimes are separated into ordinary differential equations.

1 2

[email protected] [email protected]

1

Recently, the separability of Klein-Gordon equations in higher dimensional Kerr-NUT-de Sitter spacetimes [1] was shown by Frolov, Krtouˇs and Kubizˇ na´k [2]. This separation is deeply related to that of geodesic Hamilton-Jacobi equations. Indeed, a geometrical object called conformal KillingYano tensor plays an important role in the separability theory [3, 4, 5, 2, 6, 7, 8, 9]. However, at present, a similar separation of the variables of Dirac equations is lacking, although the separability in the four dimensional Kerr geometry was given by Chandrasekhar [10]. In this paper we shall show that Dirac equations can also be separated in general Kerr-NUT-de Sitter spacetimes. The D-dimensional Kerr-NUT-de Sitter metrics are written as follows [1]: (a) D = 2n !2 n−1 n n X X X dx2µ (k) (2n) (1) Aµ dψk , + Qµ (x) g = Q (x) µ=1 µ=1 µ k=0 (b) D = 2n + 1 g

(2n+1)

=

n X dx2µ

µ=1

Qµ (x)

+

n X

Qµ (x)

µ=1

n−1 X

A(k) µ dψk

k=0

!2

c + (n) A

n X

(k)

A

dψk

k=0

!2

.

(2)

The functions Qµ (µ = 1, 2, · · · , n) are given by Xµ , Uµ

Qµ (x) =

Uµ =

n Y

ν=1 (ν6=µ)

(x2µ − x2ν ),

(3)

(k)

where Xµ is a function depending only on the coordinate xµ , and A(k) and Aµ are the elementary symmetric functions of {x2ν } and {x2ν }ν6=µ respectively: n Y

(t − x2ν ) = A(0) tn − A(1) tn−1 + · · · + (−1)n A(n) ,

(4)

ν=1 n Y

ν=1 (ν6=µ)

n−1 n−2 (t − x2ν ) = A(0) − A(1) + · · · + (−1)n−1 Aµ(n−1) . µ t µ t

(5)

The metrics are Einstein if Xµ takes the form [1, 11] (a) D = 2n n X

Xµ =

c2k x2k µ + bµ xµ ,

(6)

k=0

(b) D = 2n + 1 Xµ =

n X

c2k x2k µ + bµ +

k=0

where c, c2k and bµ are free parameters.

(−1)n c , x2µ

(7)

1. D=2n For the metric (1) we introduce the following orthonormal basis {ea } = {eµ , en+µ } (µ = 1, 2, · · · , n): dxµ , eµ = p Qµ

en+µ =

q



n−1 X

A(k) µ dψk .

(8)

k=0

The dual vector fields are given by eµ =

q



∂ , ∂xµ

en+µ =

2

n−1 X

2(n−1−k)

(−1)k xµ p Qµ Uµ k=0

∂ . ∂ψk

(9)

The spin connection is calculated as [11] p √ xν Qν µ xµ Qµ ν ωµν = − 2 e − 2 e , xµ − x2ν xµ − x2ν q

n+µ

ωµ,n+µ = −(∂µ Qµ ) e



X xµ

x2 ρ6=µ ρ

(µ 6= ν)

p

Qρ n+ρ e , − x2µ

(no sum over µ),

(10)

p √ xµ Qν n+µ xµ Qµ n+ν = e − 2 e , (µ 6= ν) x2µ − x2ν xµ − x2ν p √ xµ Qν µ xν Qµ ν = − 2 e − 2 e , (µ 6= ν). xµ − x2ν xµ − x2ν

ωµ,n+ν ωn+µ,n+ν

Then, the Dirac equation is written in the form (γ a Da + m)Ψ = 0,

(11)

1 Da = ea + ωbc (ea )γ b γ c . 4

(12)

where Da is a covariant differentiation,

From (9),(10) and (12), we obtain the explicit expression for the Dirac equation n X

γ

µ=1

+

n X

µ=1

γ

µ

q



n+µ

!

′ n xµ 1 Xµ 1 X ∂ + + Ψ 2 ∂xµ 2 Xµ 2 xµ − x2ν

ν=1 (ν6=µ)

2(n−1−k)

n−1 X

(−1)k xµ Xµ k=0

q



(13) !

n ∂ 1 X xν + (γ ν γ n+ν ) Ψ + mΨ = 0. 2 ∂ψk 2 xµ − x2ν ν=1 (ν6=µ)

Let us use the following representation of γ-matrices: {γ a , γ b } = 2δab , γ µ = σ3 ⊗ σ3 ⊗ · · · ⊗ σ3 ⊗σ1 ⊗ I ⊗ · · · ⊗ I, γ

n+µ

|

{z

}

µ−1

(14)

= σ3 ⊗ σ3 ⊗ · · · ⊗ σ3 ⊗σ2 ⊗ I ⊗ · · · ⊗ I, |

{z

}

µ−1

where I is the 2 × 2 identity matrix and σi are the Pauli matrices. In this representation, we write the 2n components of the spinor field as Ψǫ1 ǫ2 ···ǫn (ǫµ = ±1), and it follows that (γ µ Ψ)ǫ1 ǫ2 ···ǫn (γ n+µ Ψ)ǫ1 ǫ2 ···ǫn



= 

µ−1 Y ν=1



ǫν  Ψǫ1···ǫµ−1 (−ǫµ )ǫµ+1 ···ǫn ,



= −iǫµ 

µ−1 Y ν=1

By the isometry the spinor field takes the form



ǫν  Ψǫ1 ···ǫµ−1 (−ǫµ )ǫµ+1 ···ǫn .

ˆ ǫ ǫ ···ǫn (x) exp i Ψǫ1 ǫ2 ···ǫn (x, ψ) = Ψ 1 2

n−1 X k=0

3

(15)

Nk ψk

!

(16)

with arbitrary constants Nk . Substituting (15) into (13), we obtain n X

µ=1

µ−1 Y

p



+

ǫρ

ρ=1

!

!



n ∂ 1 1 Xµ 1 ǫµ Yµ 1 X ˆ ǫ ···ǫ (−ǫ )ǫ ···ǫ + + + Ψ µ µ+1 n 1 µ−1 ∂xµ 2 Xµ 2 Xµ 2 xµ − ǫ µ ǫ ν xν ν=1 (ν6=µ)

ˆ ǫ ǫ ···ǫn = 0, mΨ 1 2

(17)

where we have introduced the function Yµ =

n−1 X

(−1)k xµ2(n−1−k) Nk ,

(18)

k=0

which depends only on xµ . Consider now the region xµ − xν > 0 for µ < ν and xµ + xν > 0. Let us define Φǫ1ǫ2 ···ǫn (x) =

Y

1≤µ