Question

Traceless-transverse gauge vs traceless-transverse part?

Answer 1

In Hobson et al (2006) pg514 they write the formula for the flux in the traceless-transverse gauge:

F(n⃗ )=−c432πG⟨(∂thTTij)(n⃗ ⋅∇⃗ )hijTT⟩F(n→)=−c432πG⟨(∂thijTT)(n→⋅∇→)hTTij⟩

They then take the quadrupole formula:

h¯ij=−2Gc6rI¨ijh¯ij=−2Gc6rI¨ij

and take the traceless-transverse part which from what I can tell means they do the following:

h¯ij−13δijh¯=−2Gc6r(I¨ij−13δijI¨)h¯ij−13δijh¯=−2Gc6r(I¨ij−13δijI¨)

≡−2Gc6rJ¨ij≡−2Gc6rJ¨ij

They then project this onto 'a spatial surface orthogonal to the radial direction at any point' using the operator:

PkiPlj−12PijPklPikPjl−12PijPkl

to get:

(PkiPlj−12PijPkl)(h¯ij−13δijh¯)=−2Gc6rJ¨ijTT(PikPjl−12PijPkl)(h¯ij−13δijh¯)=−2Gc6rJ¨TTij

They then call (implicitly) the left hand side of this hijTThTTij and start treating it as the metric tensor in the traceless-transverse gauge. Please can someone explain to me why this is valid? Since to me

(PkiPlj−12PijPkl)(h¯ij−13δijh¯)(PikPjl−12PijPkl)(h¯ij−13δijh¯)

is simply the traceless-transverse PART of hijhijused in the gauge used in the quadrapole formula rather then hijhij in the traceless-transverse gauge (i.e. hTTijhijTT).

Edit

Just to make what I am asking clearly; If we have h¯klh¯kl in an arbitrary gauge can we always retrieve the TT-gauge by using

h¯TTij=(PkiPlj−12PijPkl)h¯klh¯ijTT=(PikPjl−12PijPkl)h¯kl

and does it matter what unit vector nini we use in the projection operator?

Answer 2

In Hobson et al (2006) pg514 they write the formula for the flux in the traceless-transverse gauge:
F(
n
⃗
)=−
c
4
32πG
⟨(
∂
t
h
TT
ij
)(
n
⃗
⋅
∇
⃗
)
h
ij
TT
⟩
F(n→)=−c432πG⟨(∂thijTT)(n→⋅∇→)hTTij⟩
They then take the quadrupole formula:
h
¯
ij
=−
2G
c
6
r
I¨
ij
h¯ij=−2Gc6rI¨ij
and take the traceless-transverse part which from what I can tell means they do the following:
h
¯
ij
−
1
3
δ
ij
h
¯
=−
2G
c
6
r
(
I¨
ij
−
1
3
δ
ij
I¨
)
h¯ij−13δijh¯=−2Gc6r(I¨ij−13δijI¨)
≡−
2G
c
6
r
J
¨
ij
≡−2Gc6rJ¨ij
They then project this onto 'a spatial surface orthogonal to the radial direction at any point' using the operator:
P
k
i
P
l
j
−
1
2
P
ij
P
kl
PikPjl−12PijPkl
to get:
(
P
k
i
P
l
j
−
1
2
P
ij
P
kl
)(
h
¯
ij
−
1
3
δ
ij
h
¯
)=−
2G
c
6
r
J
¨
ij
TT
(PikPjl−12PijPkl)(h¯ij−13δijh¯)=−2Gc6rJ¨TTij
They then call (implicitly) the left hand side of this
h
ij
TT
hTTij
and start treating it as the metric tensor in the traceless-transverse gauge. Please can someone explain to me why this is valid? Since to me
(
P
k
i
P
l
j
−
1
2
P
ij
P
kl
)(
h
¯
ij
−
1
3
δ
ij
h
¯
)
(PikPjl−12PijPkl)(h¯ij−13δijh¯)
is simply the traceless-transverse PART of
h
ij
hij
used in the gauge used in the quadrapole formula rather then
h
ij
hij
in the traceless-transverse gauge (i.e.
h
TT
ij