THE ENERGY-MOMENTUM PSEUDOTENSOR
Tµν of matter satisfies the (covariant) divergenceless
equation
Tµν ;ν = 0
(3)
We know it is not a conservation law, because it cannot be written as an ordinary divergence.
In a locally inertial frame (LIF): eq. (3) becomes
∂T µν
= 0,
(4)
∂xν
this means that T µν can be written as:
∂
(5)
T µν = α η µνα ,
∂x
where η µνα is antisymmetric in ν and α; INDEED
∂ 2η µνα
= 0,
∂xν ∂xα
because the derivative operator is symmetric in ν and
α
We want to find the expression of η µνα : write Einstein
eqs.

4 
8πG
c
1
Rµν − g µν R .
Gµν = 4 T µν −→ T µν =
(6)
c
8πG
2
In a LIF Rµν is: 

2
2
2
2
∂
g
1
∂
g
∂
g
∂
g
γβ
αδ
γδ
αβ 

.
Rµν = gµα gνβ gγδ 
+
−
−
2
∂xα ∂xδ ∂xγ ∂xβ ∂xα ∂xβ ∂xγ ∂xδ
By replacing in eq. (6), T µν becomes


4




c
1
∂
∂
µν αβ
µα νβ
Tµν =
(−g)
g
g
−
g
g
(7)

 16πG (−g) ∂xβ

∂xα 
The part within { } is antisymmetric in ν and α,
symmetric in µ and ν, and it is the quantity η µνα
we were looking for.



∂ µνα
η ,
∂xα
Tµν =
η µνα = 

4
c
1 ∂
(−g) gµν gαβ − gµαgνβ
β
16πG (−g) ∂x
Since in a LIF gµν,α = 0 we can extract
write this equation as
(−g)T
µν
∂ζ µνα
=
,
∂xα
1
(−g)
and
(8)
where
ζ
µνα
= (−g)η
µνα
c4
∂ µν αβ
µα νβ
. (9)
(−g) g g − g g
=
16πG ∂xβ
EQ. (8) has been derived in a locally inertial frame.
µνα
In any other frame (−g)Tµν will not equate ∂ζ∂xα , therefore, in a generic frame
∂ζ µνα
− (−g)Tµν 6= 0.
α
∂x
We shall call this difference
(−g)t
µν
(−g)tµν , i.e.
∂ζ µνα
µν
=
−
(−g)T
.
∂xα
The quantities tµν are symmetric, because T µν and
∂ζ µνα
are symmetric in µ and ν. It follows that
∂xα
(−g) (T
µν
∂ζ µνα
+t )=
∂xα ,
µν
−→
∂
µν
µν
[(−g)
(T
+
t
)] = 0,
∂xν
this is
THE CONSERVATION LAW OF THE TOTAL ENERGY AND MOMENTUM OF MATTER + GRAVITATIONAL FIELD VALID IN ANY REFERENCE
FRAME.





(−g)t
If we express
stein’s eqs.
G
µν
µν
∂ζ µνα
=
− (−g)Tµν .
α
∂x
T µν in terms of gµν and by using Ein-
8πG
= 4 T µν
c
−→
T
µν
c4  µν 1 µν 
R − g R .
=
8πG
2


and eq. (9) it is possible to show that tµν can be written as follows
t
µν
=
+
+
+
c4 δ
µα νβ
µν αβ
σ
δ
σ
δ
σ
2Γ αβ Γ δσ − Γ ασ Γ βδ − Γ αδ Γ βσ g g − g g
16πG
g µα g βδ (Γν ασ Γσ βδ + Γν βδ Γσ ασ − Γν δσ Γσ αβ − Γν αβ Γσ δσ )
g να g βδ (Γµασ Γσ βδ + Γµβδ Γσ ασ − Γµδσ Γσ αβ − Γµαβ Γσ δσ )
αβ δσ
µ
ν
µ
ν
g g (Γ αδ Γ βσ − Γ αβ Γ δσ )
This is the stress-energy pesudotensor of the gravitazional field.
tµν it is not a tensor because :
1) it is the ordinary derivative, (not the covariant one)
of a tensor
2) it is a combination of the Γ’s that are not tensors.
However, as the Γ’s, it behaves as a tensor under
a linear coordinate transformation.
Let us consider an emitting source and the associated 3-dimensional coordinate frame O (x, y, z). Be
an observer located at P = (x1, y1, z1) at a distance
√ 2
r = x1 + y12 + z12 from the origin.
The observer wants to detect the wave coming along
r
the direction identified by the versor n = |r|
.
y
y’
P
x’
n
x
z
z’
Consider a second frame O’ (x0, y 0 , z 0 ), with origin coincident with O, and having the x0-axis aligned with n.
Assuming that the wave traveling along x0 direction is
linearly polarized and has only one polarization, the
corresponding metric tensor will be

g µ0 ν 0
 (t)


 −1


=
 0


 0


0

(x)
(y)
(z)




0
0
0


,
1
0
0



0
TT

0 [1 + h+ (t, x )]
0

TT
0 
0
0
[1 − h+ (t, x )]
The observer wants to measure the energy which flows
per unit time across the unit surface orthogonal to x0,
0
i.e. t0x , therefore he needs to compute the Christoffel
symbols i.e. the derivatives of hTµ0Tν 0 . The metric perx0
turbation has the form hT T (t, x0) = const
·
f
(t
−
0
x
c ), the
only derivatives which matter are those with respect
to time and x0
∂hTT
const ˙
≡ ḣTT =
f,
∂t
x0
1 TT
const
const 0
1 const ˙
∂hTT
TT 0
f
=
−
≡
h
=
−
f
+
f
∼
−
ḣ ,
∂x0
x02
x0
c x0
c
where we have retained only the dominant 1/x0 term.
Thus, the non-vanishing Christoffel symbols are:
Γ0y0y0 = −Γ0z0 z0 =
0
0
Γx y0y0 = −Γx z0 z0 =
1
2
1
2c
0
0
Γy 0y0 = −Γz 0z0 =
ḣT+T
0
1 TT
ḣ
2 +
0
Γy y0x0 = −Γz z0 x0 = −
ḣT+T
1 TT
ḣ .
2c +
By substituting Christoffel’s symbols in tµν , we find
t0x ≡
Thus
2

TT
0
c  dh (t, x )
dEGW
=

dx0 dS 16πG
dt
3

TT
0
dEGW
c  dh (t, x )
=

dtdS
16πG
dt
2 
 
 

2 
 
 

.
.
In general, if both polarization are present

g µ0 ν 0
0
t0x =
 (t)


 −1


=
 0


 0


0
2

(x)
(y)
(z)




0
0
0


,
1
0
0


0
0
TT
TT

0 [1 + h+ (t, x )]
h× (t, x ) 

0
0 
TT
TT
0
h× (t, x )
[1 − h+ (t, x )]
TT 2
 dh+ 



c

16πG
dt

TT 2
dh

×  


+
 
 
dt
2
=
TT 2
X  dhjk  




c


32πG jk dt
 
 
.
This is the energy per unit time which flows across a
unit surface orthogonal to the direction x0.
However, the direction x0 is arbitrary; if the observer
il located in a different position and computes the energy flux he receives, he will find formally the same
but with hTjkT referreed to the TT-gauge associated with
the new direction. Therefore, if we consider a generic
direction ~r = r~n
t0r =
2
TT
X  dhjk (t, r)  



2

c


32πG jk
 
 
dt
.
(10)
Since in GR the energy of the gravitational field cannot be defined locally, to find the GW-flux we need to
average over several wavelenghts, i.e.
3

TT 2 +
X  dhjk 

dEGW
c


= ct0r =
.
dtdS
32πG jk dt
We shall now express the energy flux directly in terms
of the quadrupole moment.
*
Since

















h̄TTµ0 = 0,
µ = 0, 3
r 
2G  d2
TT
)
(t
−
h̄ik (t, r) = 4 ·  2 QTT
ik
c r dt
c


by direct substitution we find
2
+
dEGW
c3 *X  dhTT
jk 


=
dtdS
32πG jk dt
*
" ...
G
r !#2 +
TT
X
Qjk t −
=
8πc5 r2 jk
c
*
"
...
r !#2+
G
X
Pjkmn Qmn t −
=
.
8πc5 r2 jk
c

From this formula we can compute the gravitational
GW
luminosity LGW = dEdt
LGW =
Z
Z dE
dEGW
GW 2
dS =
r dΩ
dtdS
dtdS
*
G 1 Z
X
=
dΩ
2c5 4π
jk
...
Pjkmn Qmn
r !!2+
t−
.
c
Let us compute this integral. By using the properties
of Pmnjk we find
...
=
=
...
...
...
Pmnjk Qjk Pmnrs Qrs = PjkrsQjk Qrs
...
...
1
δjr δks − njnrδks − nknsδjr + 2 njnknrns QjkQrs
... ...
... ...
... ...
QjkQjk − 2nkQkrQrsns + 21 njnknrnsQjk Qrs .
The integrals of the n’s over the solid angle are:
1 Z
1
dΩninj = δij
4π
3
1 Z
1
dΩninj nr ns =
(δij δrs + δir δjs + δis δjr )
4π
15
so that
1 R
4π
...
...
...
...
dΩ Qjk Qjk − 2nkQkrQrsns +
=
... ...
2
Q Q
5 jk jk
... ... 1
nn nnQ Q
2 j k r s jk rs
and, finally, the power emitted in gravitational waves
by an evolving source is
LGW
dEGW
G *X ...
r !+
r ! ...
Q t − Qjk t −
=
= 5
.
dt
5c jk jk
c
c
where
Qij = qij −
1
δ ij qkk
3
We shall now compute the GW-luminosity of a binary
system
We shall use the formula:
LGW
G *X ... ... +
Q Qjk .
= 5
5c jk
jk
where
µ
1
Qij = qij − δij δ kl qkl = l20
2
2








ωK t sin 2ω
ωK t 0 
cos 2ω


ωK t −cos
cos 2ω
ωK t 0 
sin 2ω


0
0
0
and
µ
Qij = l20 8 ωK 3
2
...
X
jk
...
...

ωK t
 sin 2ω


 −cos
ωK t
cos 2ω


0
2
Q Qjk = 32 µ
l40

ωK t 0 
− cos 2ω


ωK t 0 
sin 2ω
−sin


0
0
ωK
6
jk
ωK =
v
u
u
u
t
GM
l30
M3
= 32 µ G
l50
2
3
By direct substitution we find
dEGW
32 G4 µ2M3
LGW ≡
=
dt
5 c5
l50
and if m1 = m2 = m and consequently µ =
2m
64G4m5
LGW =
5c5l50
For the binary pulsar PSR 1913+16
LGW = 0.7 · 1031 erg/s
m
2
and M =
Since we know how much energy is radiated in GW,
we can compute the consequent variation of the orbital period
The total energy of the system is
1
Eorb = µωK 2 l20 − U,
2
therefore
where
Eorb
U=−
µM
Gµ
Gm1m2
=−
l0
l0
µM
1 Gµ
=−
2 l0
and
µM  1 dl0 
1 dl0 
dEorb 1 Gµ
=
= −Eorb 
dt
2 l0
l0 dt
l0 dt




Since
ωK = lnGM−3lnl0 →
→ 2lnω
ωK 2 = GMl−3
0
ωK
1 dω
3 1 dl0
=−
ωK dt
2 l0 dt
and since
ωK
1 dω
1 dT
1 dl0 2 1 dT
=−
→
=
ωK dt
T dt
l0 dt
3 T dt
and
dEorb
2 Eorb dT
dT
3 T dEorb
=−
→
=−
dt
3 T dt
dt
2 Eorb dt
The energy lost in GWs must be compensated by a
variation of orbital energy (adiabatic approximation
dEorb
+ LGW = 0
dt
→
dEorb
= −LGW
dt
we find
dT 3 T
LGW
=
dt
2 Eorb
dT 3 T
=
LGW
dt
2 Eorb
For PSR1913+16
T = 27907 s,
Eorb ∼ −1.4 · 1048 ergs,
LGW ∼ 0.7 · 1031 erg/s
with these data we find
dT
∼ −2.2 10−13
dt
By refining the calculations, using the equations of motion appropriate for an eccentric orbit with ' 0.617
one finds
dT
= −2.4 · 10−12
dt
The observed value is
dT
= −2.3 · 10−12 (±0.22 · 10−12)
dt
FIRST INDIRECT EVIDENCE OF THE
EXISTENCE OF GRAVITATIONAL WAVES
For PSRJ0737-3039
T = 8640 s,
Eorb ∼ −2.55 · 1048 ergs,
LGW ∼ 2.24 · 1032 erg/s
dT
∼ −1.2 10−12
dt
ORBITAL EVOLUTION
ωK
1 dω
1 dT
=−
ωK dt
T dt
dT 3 T
=
LGW
dt
2 Eorb
moreover,
LGW
dEGW 32 G4 µ2M3
=
≡
dt
5 c5
l50
Eorb = −
µM
1 Gµ
2 l0
therefore
ωK
1 dω
96 G3 µM2
=
.
ωK dt
5 c5 l40
When integrated it gives
ωK (t) =
3/8
ωK in tcoal
[tcoal − t]
in 4
5 c l0
256 G3 µM2
5
where
ωK in = ωK (t = 0)
3/8
tcoal =
(11)
lin
0 = l0 (t = 0) (12)
and
The orbital frequency, and consequently the frequency
of the emitted wave ν GW = ωK /π, change accordingly:
ν GW (t) =
3/8
ν in
GW tcoal
3/8
[tcoal − t]
(13)
the frequency increases with time
Since ω K =
is
s
GM
,
l30
using eq. (11) the orbital separation
l0(t) =
lin
0
1/4
tcoal
[tcoal − t]1/4
the orbital distance decreases with time.
(14)
WAVEFORM: AMPLITUDE AND PHASE
We have seen that if a binary system moves on a circular orbit and we look, for example, in the direction
orthogonal to the orbital plane, the wave we would
detect is
hTT
ij
µMG2 TT
4µ
=−
Akl
rl0c4
and ωK =
s
GM
l30
ATT
ij


ωK t sin 2ω
ωK t 0 
 cos 2ω




 sin 2ω
ω
cos
ω
= 
t
−cos
2ω
t
0
K
K




0
0
0
= π ν GW .
We shall model the waveform emitted in the inspiralling as follows:
1) an instantaneous amplitude
µMG2 ωK 2/3(t)
µMG2 4µ
4µ
=
· 1/3 1/3
h0(t) =
rl0(t)c4
rc4
G M
4π 2/3 G5/3 M5/3 2/3
=
ν GW (t)
c4 r
where
M5/3 = µ M2/3
→
M = µ3/5 M2/5 = chirp mass
Since the frequency increases in time the amplitude
increases too.
2) Since the wave frequency is changing in time, the
ωK t, will be replaced by
phase appearing in ATT
ij , i.e. 2ω
Φ(t) =
Z
t
2πνν GW (t) dt + Φ in ,
where
Φ in = Φ(t = 0)
The phase
Φ(t) =
Z
t
2πνν GW (t) dt + Φ in ,
Since
ν GW (t) =
and
Φ in = Φ(t = 0)
3/8
ν in
t
coal
GW
[tcoal − t]3/8
3/8
ν in tcoal
= 53/8
where
1
8π



5/8
3
c 

GM
then
1
ν GW (t) =
8π

3
5/8
 c



GM


5
tcoal − t
3/8

and the integrated phase will be
5/8
c3



+ Φin
Φ(t) = 2 
5GM (tcoal − t)
If we know the phase we can measure the chirp mass

Thus, the signal emitted during the inspiralling will
be
hTT
ij
4π 2/3 G5/3 M5/3 2/3
TT
=
ν
(t)A
GW
ij
c4 r
where
ATT
ij =


sin Φ(t) 0 
 cos Φ (t)



 sin Φ (t) −cos

cos Φ(t) 0 




0
0
0
LIGO[40 Hz − 1 − 2 kHz]
LISA[10−4 − 10−1] Hz
VIRGO[10 Hz − 1 − 2 kHz]
1e-21
[Hz
-1/2
]
1e-20
Sh
1/2
GEO
1e-22
VIRGO
LIGO
1e-23
10
100
1000
10000
log νGW (Hz)
Let us consider 3 binary system
a) m1 = m2 = 1.4 M
b) m1 = m2 = 10 M
c) m1 = m2 = 106 M
Let us first calculate what is the orbital distance between the two bodies on the innermost stable circular orbit (ISCO) and the corresponding emission frequency
lI0SCO ∼
6GM
,
c2
ωK =
v
u
u
u
t
v
u
u
u
u
t
GM
1 GM
ISCO
ν
ν
→
=
=
π
GW
GW
l30
π (lISCO
)3
0
ν GW = 1570.4 Hz
a) l0 ISCO = 24, 8 km
b) l0ISCO = 177, 2 km
ν GW = 219.8 Hz
ν GW = 2.2·10−3 Hz
c) l0ISCO = 17.720.415, 3 km
a) and b) are interesting for LIGO and VIRGO,
c) will be detected by LISA
Let us consider LIGO and VIRGO, and let us compute
the time a given signal spends in the detector bandwidth before coalescence.
From
3/8
ν in
GW tcoal
ν GW (t) =
[tcoal − t]3/8
we get

8/3 

in
ν GW  



t = tcoal 1 − 
.
ν GW (t)
Putting :
ν in
GW = lowest frequency detectable by the antenna,
and
ISCO
ν max
we find
GW = ν GW
(LIGO)
a) [40 − 1570.4 Hz]
∆t = 24.86 s
b)
[10 − 219.8 Hz]
∆t = 0.93 s
(VIRGO)
[10 − 1570.4 kHz]
∆t = 16.7 m
[10 − 219.8 kHz]
∆t = 37.82 s
VIRGO catches the signal for a longer time!
WHAT ABOUT LISA?
LISA [10−4 − 10−1] Hz
1e-20
Detection threshold
LISA 1-yr observation
1e-21
1e-22
1e-23
1e-24
0.0001
0.001
0.01
ν [ Hz ]
0.1
1
Let us consider 2 BH-BH binary systems
a) m1 = m2 = 102 M
b) m1 = m2 = 106 M
Orbital distance between the two bodies on the innermost stable circular orbit (ISCO) and the corresponding emission frequency
lI0SCO ∼
6GM
,
c2
ωK =
v
u
u
u
t
v
u
u
u
u
t
GM
1 GM
ISCO
ν
ν
=
π
→
=
GW
GW
l30
π (lISCO
)3
0
a) lISCO
= 1772 km
0
= 17.720.415, 3 km
b) lISCO
0
ν GW = 21.98 Hz
ν GW = 2.2 10−3 Hz
Time a given signal spends in the detector bandwidth
before coalescence.
a) m1 = m2 = 102 M
b) m1 = m2 = 106 M



t = tcoal 1 − 

ν in
ν GW (t)
8/3 





LISA
a)
[10−4 − 10−1 Hz]
∆t = 556.885 years
b)
[10−4 − 2.2 · 10−3 Hz]
∆t = 0, 12 years = 43 d 18 h 43 m 24 s