3+1 formalism and bases of numerical relativity - LUTh ...

7.6 Komar mass and angular momentum 123
The extrinsic curvature component Krϕ can be evaluated via formula (4.63), which reduces
to 2NKij = Lβ γij since ∂γij/∂t = 0. From the Boyer-Lindquist line element (see e.g.
Eq. (5.29) in Ref. [156]), we read the components of the shift:
(β r ,β θ ,β ϕ
2amr
) = 0, 0, −
(r2 + a2 )(r2 + a2 cos2 θ) + 2a2mr sin2
, (7.107)
θ
where m and a are the two parameters of the Kerr solution. Then, using Eq. (A.6),
Krϕ = 1
2N Lβ γrϕ = 1
ϕ ∂γrϕ ∂β
β +γϕϕ
2N ∂ϕ
=0
ϕ ∂β
+ γrϕ
∂r ϕ
=
∂ϕ
=0
1
2N γϕϕ
∂βϕ .
∂r
(7.108)
Hence
JK = 1
s
16π r=const
r
N γϕϕ
∂βϕ √
q dθ dϕ. (7.109)
∂r
The values of s r , N, γϕϕ and √ q can all be read on the Boyer-Lindquist line element.
However this is a bit tedious. To simplify things, let us evaluate JK only in the limit
r → ∞. Then s r ∼ 1, N ∼ 1, γϕϕ ∼ r 2 sin 2 θ, √ q ∼ r 2 sin θ and, from Eq. (7.107),
βϕ ∼ −2am/r3 , so that
JK = 1
16π
r 2 2 6am
sin
r4 r2 sin θ dθ dϕ = 3am
π
× 2π × sin
8π 0
3 θ dθ. (7.110)
Hence, as expected,
r=const
JK = am. (7.111)
Let us now find the 3+1 expression of the volume version (7.101) of the Komar angular
momentum. We have n · φ = 0 and, from the 3+1 decomposition (4.10) of T:
Hence formula (7.101) becomes
with
JK =
T(n,φ) = −〈p,φ〉. (7.112)
Σt
〈p,φ〉 √ γ d 3 x + J H K , (7.113)
J H K = 1
Kijs
8π Ht
i φ j√ q d 2 y . (7.114)
Example : Let us consider a perfect fluid. Then p = (E + P)U [Eq. (5.61)], so that
JK = (E + P)U · φ √ γ d 3 x + J H K . (7.115)
Σt
Taking φ = −y∂x +x∂y (symmetry axis = z-axis), the Newtonian limit of this expression
is then
JK = ρ(−yU x + xU y )dxdy dz, (7.116)
Σt
i.e. we recognize the standard expression for the angular momentum around the z-axis.