Chapter 18
The far field limit of an isolated,
stationary object
In this chapter we will derive the metric which describes the gravitational field generated by
an isolated, stationary object. Since the source is isolated, in the exterior Tµν = 0 and the
spacetime is vacuum. Therefore, it is reasonable to assume that far away from the source
the metric tends to Minkowski’s metric, i.e. the metric must satisfy the asymptotic flatness
condition.
If the spacetime is asymptotically flat, we can define, in an appropriate coordinate frame,
a space coordinate r such that
lim gµν = ηµν .
(18.1)
r→∞
We call far field limit the region of spacetime where r ≫ R, being R a lengthscale characteristic of the source. In the far field limit,
gµν = ηµν + O
! "
1
r
.
(18.2)
We also assume the metric is stationary, i.e. that it admits a timelike Killing vector so that
(see chapter 9), by a suitable choice of coordinates, the metric can be made independent of
time; this also implies that the source stress-energy tensor is independent of time.
We shall now show that the metric of a stationary axisymmetric source in the far field
limit, up to terms of order 1/r in the expansion (18.2), is
!
"
!
"
2M
2M
dt2 + 1 +
dr 2
r
r
4J
sin2 θdtdφ
+r 2 (dθ2 + sin2 θdφ2 ) −
r
+ higher order terms in 1/r ,
ds2 = − 1 −
(18.3)
where M is the source mass and J its angular momentum.
Let us write the expansion (18.2), which holds at large distance from the source, in a
perturbative form
gµν = ηµν + hµν
(18.4)
270
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT271
with |hµν | ≪ 1. The perturbation hµν is a solution of the equations of the linearized Einstein
equations in vacuum (see Chapter 13, eq. (13.26))
✷F h̄µν = 0
h̄µν,µ = 0
(18.5)
(18.6)
where we remind that ✷F is the d’Alambertian of the flat spacetime
✷F = η αβ
∂ ∂
∂2
=
−
+ ∇2 ,
∂xα ∂xβ
c2 ∂t2
and
1
h̄µν ≡ hµν − ηµν hαα .
(18.7)
2
Since we are assuming the spacetime to be stationary, therefore (18.5), (18.6) become
∇2 h̄µν = 0,
h̄i ν,i = 0 .
(18.8)
(18.9)
We stress that eqs. (18.8) and (18.9) hold only in the far field limit r ≫ R.
We shall now derive eq. (18.3) in the simple case when the gravitational field generated
by the source is weak everywhere, i.e. also inside the source and on its boundary. We shall
subsequently show that the metric (18.3) holds in the more general case when the field near
the source is strong.
18.1
The weak field case.
If the gravitational field of the source is weak everywhere, as shown in Chapter 12.1, Einstein’s equations for the metric perturbation h̄µν inside the source become
✷F h̄µν = −16πTµν
h̄µν,µ = 0 ,
(18.10)
and the general solution is (13.27)
h̄µν (t, x) = 4
#
V
Tµν (t − |x-x′ |, x′ ) 3 ′
dx
|x-x′ |
(18.11)
where V is the source three-volume. On the source boundary, ∂V , by definition the stressenergy tensor vanishes, Tµν = 0.
In Chapters 12.1 and 14 we were interested in the time-dependent part of the solution
(18.11), since we were interested in gravitational waves. Here, instead, we are considering a
stationary source, for which Tµν = Tµν (x′ ), therefore the solution (18.11) becomes
h̄µν (x) = 4
#
V
Tµν (x′ ) 3 ′
dx.
|x − x′ |
(18.12)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT272
In the following space indices 1, 2, 3 will be denoted by latin letters i, j. As in Chapter 12.1
the indices if hµν will be raised by using Minkowski’s metric, thus hiµ = hi µ .
Let us consider a reference frame centered on the source center of mass. Be x a position
vector pointing far away from the source, and x′ the position vector of a generic source point;
be |x| ≫ |x′ |. If we define r ≡ |x|, and Taylor expand the quantity 1/|x − x′ | (this expansion
is commonly named multipolar expansion) we find
1
1
= +
′
|x − x |
r
$
i
!
xi x′i
1
+O 3
3
r
r
"
i = 1, 3.
(18.13)
Tµν x′i d3 x′ .
(18.14)
Substituting in (18.12) we find
4
h̄µν (x) =
r
#
3 ′
Tµν d x +
V
4
$
i
ix
r3
#
V
Let us evaluate the 00-component of h̄µν In the weak field limit T00 ∼ ρc2 is the source
mass-density, therefore the first integral in eq. (18.14) gives
#
T00 d3 x′ = M .
V
(18.15)
The 00 component of the second integral in eq. (18.14) gives the position of the source center
of mass, which coincides with the origin of the coordinates frame; thus
#
T00 x′i d3 x′ = Mx′icdm = 0 .
V
(18.16)
From eqs. (18.14), (18.15) and (18.16) we find
h̄00 =
!
1
4M
+O 3
r
r
"
.
(18.17)
We shall now compute the µi components of h̄µν . To compute the first integral we shall use
the divergenceless equation satisfied by Tµν which, for a stationary source, becomes
T µν,ν = T µ0,0 + T µi,i = T µi,i = 0
Using (18.18), and the property
#
V
µi 3 ′
T dx =
#
T µk δki d3 x′
V
∂xi
∂xj
=
µ = 0, 3, i = 1, 3.
(18.18)
= δji , we find
#
V
T
µk
∂x′i 3 ′
d x =−
∂x′k
# %
V
&
∂T µk
x′i d3 x′ = 0
′k
∂x
(18.19)
where we have integrated by parts. Remind: the surface terms do not contribute, because
on the boundary of V , Tµν = 0. Thus,
#
V
T µi d3 x′ = 0 .
(18.20)
We shall compute the second integral using the following property:
# '
V
=
#
µi ′j
µj ′i
T x +T x
V
T µk
(
3 ′
dx =
#
V
T
µk
%
&
∂x′i ′j ∂x′j ′i 3 ′
x + ′k x d x
∂x′k
∂x
#
∂ ' ′i ′j ( 3 ′
x
x
d
x
=
−
x′i x′j T µk,k d3 x′ = 0;
∂x′k
V
(18.21)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT273
thus
# '
V
(
T µi x′j + T µj x′i d3 x′ = 0,
which implies that the second integral in eq. (18.14) is antisymmetric in the last two indices:
#
V
µi ′j 3 ′
T x d x =−
#
T µj x′i d3 x′ .
V
(18.22)
Let us consider now the space components of the second integral in eq. (18.14)
#
V
T jk x′i d3 x′ .
(18.23)
this expression is symmetric in the first two indices, and antisymmetric in the last two indices
because of eq. (18.22); consequently
#
T ki x′j d3 x′ = −
V#
=
i.e.
V
ji ′k 3 ′
T x dx =
#
V
#
#V
T kj x′i d3 x′ = −
ij ′k 3 ′
T x d x =−
V
ki ′j 3 ′
T x d x =−
#
#
#V
V
T jk x′i d3 x′
T ik x′j d3 x′ ,
(18.24)
T ik x′j d3 x′ .
V
The only possibility for this equality to be satisfied is that
#
V
T ki x′j d3 x′ = 0 .
(18.25)
Consequently, from eqs. (18.14), (18.20) and (18.25) we find
!
1
h̄ik = O 3
r
"
.
(18.26)
The last metric components we need to compute are h̄0i
4
h̄0i (x) =
r
#
V
3 ′
T0i d x +
4
$
j
jx
r3
#
V
T0i x′j d3 x′ .
(18.27)
From eq. (18.20) we know that the first term is zero, whereas eq. (18.22) implies that
#
V
0i ′j 3 ′
T x d x =−
#
V
T 0j x′i d3 x′ .
(18.28)
We shall now show that this integral is related to the source angular momentum. The
components T 0i are the density of the i-th component of momentum of the source
T 0i = P i ;
(18.29)
a matter element in the volume d3 x′ , at a distance x′ from the origin, has momentum Pd3 x′ .
According to Newtonian theory, its angular momentum is dJ = x′ ×Pd3 x′ , where × indicates
the vector product. Thus the source angular momentum is
J=
#
x′ × Pd3 x′ .
(18.30)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT274
The components of J can be written as follows
i
J =−
#
V
ϵijk T 0j x′k d3 x′ ,
(18.31)
where ϵijk is the 3-D Levi-Civita tensor density we introduced in section 9.7. We remind that
it is completely antisymmetric, since its components change sign under interchange of any
pair of indices. Since it is completely antisymmetric, the components with two equal indices
are zero, and the only non-vanishing components are those for which the three indices are
different. Moreover
e123 = 1.
(18.32)
Equation (18.31) can be written as
#
V
1
T 0i x′j d3 x′ = − ϵijk J k
2
(18.33)
***********************************************
PROOF
Be B ij = −B ji an antisymmetric tensor and
Ak = ϵklm B lm .
(18.34)
1
1
ϵijk Ak = ϵijk ϵklm B lm ;
2
2
(18.35)
Let us multiply bot members by 12 ϵijk
the following equality is easy to prove
1
ϵijk ϵklm = δil δjm − δim δjl
(18.36)
1
1
1
ϵijk Ak = ϵijk ϵklm B lm = (δil δjm − δim δjl ) B lm = B ij
2
2
2
(18.37)
Consequently,
Using this property, eq. (18.33) follows immediately.
***********************************************
Thus, using eq. (18.33) the terms in the sum appearing in h̄0i (see eq. (18.27) can be written
as
4 j # 0i ′j 3 ′
2
4 j#
′j 3 ′
x
T
x
d
x
=
−
x
T
x
d
x
=
ϵijk xj J k .
(18.38)
0i
3
3
3
r
r
r
V
V
1
ϵijk ̸= 0 only if its three indices are all different, thus ı ̸= k and j ̸= k; similarly for ϵlmk . Therefore
ϵijk ϵlmk ̸= 0 only if the indices ij and lm are the same. If they have the same order, i.e. ij = lm,
then ϵijk ϵlmk = 1; if they have the opposite order, i.e. ij = ml, then ϵijk ϵlmk = −1. Consequently,
ϵijk ϵlmk = δ il δ jm − δ im δ jl .
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT275
From eqs. (18.27), (18.20) and (18.38) we find
h̄0i =
!
1
2
ϵijk xj J k + O 3
3
r
r
"
(18.39)
In summary, the multipolar expansion (18.14) gives
h̄00
h̄0i
h̄ij
In terms of hµν
!
"
1
4M
+O 3
=
r
r
! "
1
2
= 3 ϵijk xj J k + O 3
r
r
! "
1
= O 3 .
r
(18.40)
2
1
hµν = h̄µν − ηµν h̄αα ,
2
(18.41)
we find3
h00
h0i
hij
18.1.1
!
"
1
2M
+O 3
=
r
r
! "
1
2
j k
= 3 ϵijk x J + O 3
r
r
! "
1
2M
δij + O 3 .
=
r
r
(18.42)
The far field limit metric in polar coordinates
Let us transform the solution (18.42) in polar coordinates
x1 = r sin θ cos φ
x2 = r sin θ sin φ
x3 = r cos θ .
Since
)
(dxi )2 = dr 2 + r 2 dθ2 + r 2 sin2 θdφ2 ,
(18.43)
(18.44)
i
then
(
2M ' 2
dr + r 2 dθ2 + r 2 sin2 θdφ2 .
(18.45)
r
The transformation of h0i dx0 dxi is less trivial. If we choose the frame orientation such that
the angular momentum is directed along the z axis, i.e.
hij dxi dxj =
J = (0, 0, J) ,
2
(18.46)
To invert (18.7) we first take the trace of (18.7), finding h̄λλ = −hλλ , then substitute into (18.7).
* +
j
3
Notice that xr3 is an O r12 term, because in the far field limit xj ∼ r.
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT276
then
0
i
h0i dx dx
!
"
!
"
2 0
2
=
dx ϵijk xj J k dxi = − 3 dx0 J(x1 dx2 − x2 dx1 )
3
r
r
!
"
2J
2 0
= − 3 dx Jr 2 sin2 θdφ = − sin2 θdtdφ,
r
r
(18.47)
where the equality x1 dx2 − x2 dx1 = r 2 sin2 θ can be found by differentiating eq. (18.43).
In conclusion, the line element is
!
!
""
1
2M
+O 3
dt2
r
r
!
! "" ,
2M
1
+ 1+
dr 2 + r 2 (dθ2 + sin2 θdφ2 )
+O 3
r
r
!
! ""
1
4J
2
+ − sin θ + O 2
dtdφ .
r
r
ds2 = − 1 −
(18.48)
This is the solution of the linearized Einstein equations in the weak field limit. If we consider
the full, non linear Einstein equations, we have terms of order O(|hµν |2 ), which produce
terms of order ∼ M 2 /r 2 , ∼ J 2 /r 2 and, due to the non linearity, also to higher order terms.
Therefore, with respect to the fully non linear solution, our expansion does not include terms
of order O(1/r 2), i.e. it is more correct to write
!
!
""
1
2M
dt2
ds = − 1 −
+O 2
r
r
!
! "" ,
2M
1
+O 2
+ 1+
dr 2 + r 2 (dθ2 + sin2 θdφ2 )
r
r
!
! ""
1
4J
+ −
sin2 θdtdφ .
+O 2
r
r
Finally, we redefine the radial coordinate as follows:
2
r → r−M.
(18.49)
(18.50)
Neglecting contributions of order O(1/r 2), the only term which produces a change in the
metric is
!
"
!
"
2M 2 2
2M
2
2
1+
r (dθ + sin θdφ ) → 1 +
(r − M)2 (dθ2 + sin2 θdφ2 )
r
r
!
! ""
1
(dθ2 + sin2 θdφ2 ) .
(18.51)
= r2 1 + O
r
With this coordinate definition, we finally reduce the metric (18.49) to the following form:
ds
2
!
"
!
"
2M
2M
= − 1−
dt2 + 1 +
dr 2
r
r
4J
sin2 θdtdφ
+r 2 (dθ2 + sin2 θdφ2 ) −
r
+ higher order terms in 1/r .
(18.52)
which coincides with eq. (18.3).
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT277
18.2
The strong field case
In this section we shall drop the weak field assumption, and we shall assume that near and
inside the source the field can be strong. However far away, where we want to find the
solution of Einstein’s equations, the field is still weak weak, the metric can be written as
gµν = ηµν + hµν ,
(18.53)
and we shall neglect terms of order O(|hµν |2 ), and terms that decay with powers larger than
of 1/r, where r is the distance from the source. We seek a solution of the form
h̄µν
!
1
aµν (θ, φ) bµν (θ, φ)
=
+O 3
+
2
r
r
r
"
.
(18.54)
The coefficients aµν , bµν depend only on the angular variables θ, φ, so that they remain finite
for r → ∞. The metric perturbation, which we assume to be stationary, satisfies equation
(18.8),
∇2 h̄µν = 0 .
(18.55)
The Laplace operator in spherical coordinates has the form4
∇2 =
1
IL
∂r r 2 ∂r + 2
2
r
r
(18.56)
where IL is an operator acting on the angular variables:
IL ≡ ∂θ2 + cot θ∂θ + sin−2 θ∂φ2 .
(18.57)
By substituting eq. (18.54) in (18.56) we easily find
ILaµν (θ, φ) = 0
ILbµν (θ, φ) = −2bµν (θ, φ) .
(18.58)
(18.59)
The eigenfunctions of the operator IL are the spherical harmonics Ylm (θ, φ), with l = 0, 1, . . .
and m = −l, −l + 1, . . . , l − 1, l. They are defined by the property
ILYlm = −l(l + 1)Ylm .
(18.60)
Equation (18.59) tells us that bµν is a linear combination of the spherical harmonics with
l = 1, which are
.
3
sin θeiφ
8π
.
3
cos θ
=
4π
Y11 = −
Y11
Y1−1 =
.
3
sin θe−iφ .
8π
(18.61)
4
The theory of Laplace equation and the properties of spherical harmonics are extensively discussed in
the literature. See for instance Jackson’s book Electromagnetism, Chapter 3.
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT278
This is equivalent to say that bµν is a linear combination of the direction cosines ni = xi /r,
because
.
8π −Y11 + Y1−1
x1
n =
= sin θ cos φ =
r
3
2
.
2
x
8π −Y11 − Y1−1
n2 =
= sin θ sin φ =
r
3
2i
1
x3
n =
= cos θ =
r
3
.
4π
Y10 .
3
(18.62)
Therefore, while aµν does not depend on the angular variables ni , bµν can be written as a
linear combination of the ni ’s
i
bµν (ni ) = bµν
(18.63)
i n .
Consequently, the expansion (18.54) can be written as
h̄µν =
!
xi
1
aµν bµν
+ i 3 +O 3
r
r
r
"
,
(18.64)
with aµν , bµν
i constant coefficients.
We now impose on (18.64) the gauge condition (18.6)
h̄µν,ν = 0
(18.65)
which, in the case of stationary perturbations, becomes
h̄µi,i = 0 .
(18.66)
We get (remember that in linearized gravity it is irrelevant if a space index i is up or down)
h̄µj,j = −
aµj xj bµji (δ ij r 2 − 3xi xj )
+
=0
r3
r5
(18.67)
which has to be satisfied for all (large) values of r and for all values of ni = xi /r. Eq. (18.67)
is satisfied only if the coefficients of different powers of r vanish, i.e.
'
ij
i j
δ − 3n n
(
aµj = 0
bµij = 0 .
(18.68)
These equations do not involve a00 , b00i , which are in general nonvanishing, free constants;
to simplify the notation, we rewrite them as
a ≡ a00
bi ≡ b00i .
(18.69)
The first of eqs. (18.68) says that all the constant aµν different from a00 vanish. The second
equation can be rewritten as
H ij b0ij = 0
H ij bkij = 0
(18.70)
(18.71)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT279
where we have defined
H ij ≡ δ ij − 3ni nj .
(18.72)
b0ij = bδij + ck ϵijk
bkij = dk δij + di δkj − dj δki
(18.73)
(18.74)
The general solution of (18.70), (18.71) is
where b, ck , dk are constants. A rigorous proof of (18.73), (18.74) would require the use of
the structures of Group Theory, which goes beyond the scope of this book. Here we will
only give an intuitive, non-rigorous proof of the first solution, (18.73).
Equation (18.70) must be satisfied for any value of ni , i.e. for any value of the angular
variables θ, φ. While H ij depends on the angles, bµij cannot depend on the angles, therefore
equation (18.70) can be satisfied only because of the symmetry properties of H ij , which is
symmetric and traceless
H ij = H ji
δij H ij = 0 .
(18.75)
All quantities considered here are tensors in the euclidean three-dimensional space. The only
constant tensors which vanish when contracted with H ij are the Kronecker delta, δij , and
the completely antisymmetric tensor, ϵijk : the former vanishes because H ij is traceless, the
latter because H ij is symmetric. Thus b0ij must be a combination of these tensors, as shown
in eq. (18.73).
Summarizing, by imposing the gauge condition (18.6) on the expansion (18.64) we get
h̄00
h̄0i
h̄ij
!
"
a bi xi
1
+ 3 +O 3
=
r
r
r
! "
xj ck
1
bxi
=
+
ϵ
+
O
ijk
3
3
r
r
r3
! "
(
1 '
1
= 3 −δij dk xk + di xj + dj xi + O 3 ;
r
r
(18.76)
this solution depends on the constants a, bi , b, ck , dk .
The constants bi , dk can be eliminated by a (position dependent) infinitesimal diffeomorphism xµ → xµ + ξ µ with parameter
%
b di
ξ = − ,−
r
r
µ
&
(18.77)
(it is infinitesimal in the sense that r is large and ξ ∼ 1/r).
The change in the metric is (see Chapter ??)
gµν = ηµν + hµν → gµν + gµα ξ α,ν + gνα ξ α,µ + gµν ξ α,α
= ηµν + hµν + gµα ξ α,ν + gνα ξ α,µ + gµν ξ α,α ,
(18.78)
therefore, there is a change in the perturbation hµν given by
δhµν = gµα ξ α,ν + gνα ξ α,µ + gµν ξ α,α .
(18.79)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT280
Since ξ µ = O(|hµν |), by neglecting terms quadratic in hµν we find
Changing to h̄µν we find
δhµν = ηµα ξ α,ν + ηνα ξ α,µ .
(18.80)
1
h̄µν = hµν − ηµν η αβ , hαβ
2
(18.81)
thus
1
δ h̄µν = δhµν − ηµν η αβ δhαβ
2
= ηµα ξ α,ν + ηνα ξ α,µ − ηµν ξ α,α .
(18.82)
Since
ξ µ,0 = 0
bxi
ξ 0,i =
r3
dk xi
ξ k,i =
,
r3
(18.83)
then
δ h̄0i
δ h̄ij
!
"
!
"
1
dk xk
+
O
r3
r3
! "
! "
i
1
1
bx
= η00 ξ,i0 + O 3 = − 3 + O 3
r
r
r
,
1
= ηik ξ k,j + ηjk ξ k,i − ηij ξ k,k = 3 di xj + dj xj − ηij dk xk ;
r
δ h̄00 = −η00 ξ k,k + O
1
r3
=
(18.84)
thus, after the diffeomorphism,
h̄00
h̄0i
h̄ij
!
"
1
a b̃i xi
+ 3 +O 3
=
r
r
r
! "
xj ck
1
= ϵijk 3 + O 3
r
r
! "
1
= O 3 ,
r
(18.85)
where we have defined
b̃i ≡ bi + di .
(18.86)
Furthermore, we can get rid of b̃i by performing a (rigid) translation
xi → xi +
b̃i
a
(18.87)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT281
which produces the following change in the a/r term:
⎛%
'
(−1/2
a
b̃i
→ a ⎝ xi +
= a (xi )2
r
a
%
= a r
=
%
2
%
&2 ⎞−1/2
⎠
b̃i xi
1+2 2
r a
a
b̃i xi
1− 2
r
r a
&
&&−1/2
+O
!
1
r3
"
!
1
+O 3
r
=
"
!
1
a b̃i xi
− 3 +O 3
r
r
r
"
.
(18.88)
Therefore,
h̄00
h̄0i
h̄ij
!
a
=
+O
r
xj ck
= ϵijk 3
r
! "
1
= O 3
r
1
r3
"
!
1
+O 3
r
"
.
(18.89)
Finally, we compute
1
hµν = h̄µν − ηµν η αβ h̄αβ
(18.90)
2
(which follows from the definition (18.7) because η µν hµν = −η µν h̄µν , as can easily be seen
by taking the trace of (18.7)). We have
a
1 αβ
η h̄αβ = −
2
2r
(18.91)
therefore
h00
h0i
hij
!
"
a
1
=
+O 3
2r
r
! "
xj ck
1
= ϵijk 3 + O 3
r
r
! "
a
1
= δij + O 3 .
2r
r
(18.92)
With the identifications
a = 4M
ck = 2J k
(18.93)
the solution (18.92) coincides with the solution (18.42), which we derived in the case of a
weak field source, and that we have already shown to coincide with the solution (18.3).
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT282
18.3
Mass and angular momentum of an isolated object
As we have seen, the metric
ds
2
!
"
!
"
2M
2M
= − 1−
dt2 + 1 +
dr 2
r
r
4J
sin2 θdtdφ
+r 2 (dθ2 + sin2 θdφ2 ) −
r
+ higher order terms in 1/r .
(18.94)
describes the far field limit of an isolated, stationary source. If the source is weakly gravitating, we have seen that, according to Newtonian physics, M and J have a simple interpretation: they are, respectively, the mass and the angular momentum of the source.
If the source is not weakly gravitating M and J arise as integration constants of the
general far field solution (18.94), and their physical interpretation needs to be found.
One possibilty is through the study of the motion of test bodies in the metric (18.94).
If the test body is far away from a strongly gravitating source characterized by the two
constanst, say M̄ and J¯, its motion cannot be distinguished from that it would have if
moving around a weakly gravitating source with mass M = M̄ and angular momentum
¯ Thus, an operational definition of the mass and angular momentum of the strongly
J = J.
gravitating source can be given by studying geodesic motion far away from the source. The
mass will be measured from the orbital frequency of the test mass through Kepler’s third
law, and the angular momentum by measuring the precession of gyroscopes orbiting around
the source.
A different answer is based on the stress-energy pseudotensor, which we have defined
in Capter 14. We remind that the stress-energy pseudotensor tµν describes the energy and
momentum carried by the gravitational field, and satisfies, together with the stress-energy
tensor Tµν , a conservation law:
[(−g)(T µν + tµν )],ν = 0 .
(18.95)
It can be expressed as a divergence:
(−g)(T µν + tµν ) =
∂ζ µνα
∂xα
(18.96)
where
1 ∂ ,
µν αβ
µα νβ
(−g)(g
g
−
g
g
)
.
16π ∂xβ
Since we are considering a stationary spacetime, eq. (18.96) becomes
ζ µνα =
(−g)(T µν + tµν ) =
∂ζ µνk
,
∂xk
k = 1, 3.
(18.97)
(18.98)
Let us now consider a spherical three-dimensional volume V centered on the source, with
radius r much larger than the source size. WARNING: V is not the source volume, it is
much larger!
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT283
The total four-momentum P µ , enclosed in the volume V , is due both to the source
and to the gravitational field:
µ
P =
#
d3 x(−g)(T 0µ + t0µ ) .
V
(18.99)
Substituting (18.98) in (18.99) we find
Pµ =
#
V
d3 x
∂ζ 0µk
.
∂xk
(18.100)
By Gauss’ theorem, we write this integral as an integral over the spherical surface S surrounding the volume V :
#
P µ = ζ 0µk dSk .
(18.101)
S
Thus, for instance, the total mass-energy of the system is
Mtot = P 0 =
3
S
ζ 00k dSk .
(18.102)
As explained in Section 18.1, the three-momentum of the matter element of volume d3 x,
located at a point of coordinates xi , is
P i d3 x ,
(18.103)
and the angular momentum of the matter element is
dJ i = (x × P)i d3 x = −ϵijk P j xk d3 x .
(18.104)
Therefore, the total angular momentum which generalizes eq. (18.33) and includes the
contribution of the gravitational field, is
J i = −ϵijk
#
V
d3 x(−g)(T 0j + t0j )xk .
(18.105)
Using eq. (18.96), eq. (18.105) gives
Ji
4
#
#
0jl
k
∂(ζ 0jl xk )
3 ∂ζ
k
3
0jl ∂x
= −ϵijk d x
x
=
−ϵ
d
x
−
ζ
ijk
∂xl
∂xl
∂xl
V
V
4
5
#
∂(ζ 0jl xk )
= −ϵijk d3 x
− ζ 0jk .
∂xl
V
5
(18.106)
We now introduce the quantity λµναβ defined as (see eq. (18.97))
λµναβ ≡
1 ,
(−g)(g µν g αβ − g µα g νβ ) ,
16π
(18.107)
related to ζ µνα by the following equation
ζ µνα =
∂λµναβ
,
∂xβ
(18.108)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT284
which, for a stationary spacetime becomes
ζ
µνα
∂λµναi
=
.
∂xi
(18.109)
By replacing the quantity ζ 0jk in terms of λ0jkl as given by eq. (18.109), eq. (18.106) becomes
J i = −ϵijk
= −ϵijk
#
d3 x
#V '
S
(
∂ ' 0jl k
0jkl
ζ
x
−
λ
∂xl
(
ζ 0jl xk − λ0jlk dSl .
(18.110)
(18.111)
In conclusion, the total angular momentum of the source is
J i = −ϵijk
3 ' 0jl k
ζ x
S
(
− λ0jlk dSl .
(18.112)
Thus, given the metric, using eqs. (18.102) and (18.112) we can evaluate the total massenergy and the total angular momentum of the source.
It is possible to show that, using the metric of the far field limit gµν = ηµν + hµν , where
hµν is given by eqs. (18.92), i.e.
!
h0i
hij
"
1
2M
+O 2
r
r
! "
2
1
= 3 ϵijk xj J k + O 3
r
r
! "
1
2M
δij + O 2 ,
=
r
r
h00 =
(18.113)
the 4-momentum and the angular momentum of the stationary source are
P µ = (M, 0, 0, 0) ,
J i = (0, 0, J) .
(18.114)
We show explicitly the calculations to find P 0 . From eq. (18.102) we have
0
P =
#
S
ζ
00i
dSi ≡
#
S
ζ 00i ni dS,
(18.115)
where dS = r 2 dΩ and ni is the unit vector orthogonal to the surface element dS. Being
gµν = ηµν + hµν + O(|hµν |2 ), the property g µν gνρ = δρµ implies
g µν = η µν − hµν + O(|hµν |2 )
(18.116)
where the indices of hµν have been raised with Minkowski’s metric. Indeed,
(η µν − hµν )(ηνρ + hνρ ) = δρµ + O(|hµν |2 ) .
(18.117)
CHAPTER 18. THE FAR FIELD LIMIT OF AN ISOLATED, STATIONARY OBJECT285
Therefore,
2M
+ O(|hµν |2 )
r
!
"
2M ij
= δ ij − hij + O(|hµν |2 ) = 1 −
δ + O(|hµν |2 ) .
r
g 00 = −1 − h00 + O(|hµν |2 ) = −1 −
g ij
(18.118)
(18.119)
The determinant of gµν is
g = (−1 + h00 )(1 + hii ) = −(1 +
4M
) + O(|hµν |2 ) .
r
(18.120)
Note that in this expression we neglect the term J/r 3 with respect to M/r, since we are in
the far field limit.
From eq. (18.97), neglecting terms O(|hµν |2 ) (like the terms ∼ M 2 , ∼ J 2 ), we find
'
(1 i ∂ ,
1 i ∂ ,
00 ij
0i 0j
00 ij
(−g)
g
g
−
g
g
∼
(−g)g
g
n
n
16π ∂xj 6!
16π ∂x"j 7
"!
"!
1 i ∂
4M
2M
2M
n
=
1+
−1 −
1−
δij + O(|hµν |2 )
j
16π ∂x
r
r
r
1 i ∂ 4M
(18.121)
= −
n
δij + O(|hµν |2 ).
16π ∂xj r
ζ 00i ni =
Since
nj
∂ 1
=
−
,
∂xj r
r2
then
ζ 00i ni =
and
0
P =
1 M ) i i
1 M
n
n
=
4π r 2 i=1,3
4π r 2
(18.122)
#
(18.123)
S
ζ 00i ni r 2 dΩ = M .
The calculation for the angular momentum are similar.
We can conclude that the integration constants M and J appearing in the far field limit
metric of an isolated source (18.3) can be correctly interpreted as the mass-energy and the
angular momentum of the system. In the case of a weakly gravitating source, the contribution
of the gravitational field to the mass and to the angular momentum are negligible; if the
source has a strong gravitational field, the field contributes to the total mass and angular
momentum, through the stress-energy pseudotensor tµν .
We stress again that, being the source isolated, at large distance the metric tends to
Minkowski’s metric; this allows us to assume that, for r sufficiently large, gµν = ηµν +hµν with
hµν small. Furthermore, hµν can be expanded in powers of 1/r. The dominant contribution
in this expansion gives the total mass-energy and angular momentum of the system.