The ADM energy of gravitational waves?
Answers
Answered by
2
First of all, we let
g
ab
=
η
ab
+
γ
ab
.
gab=ηab+γab.
Then use the linear Einstein's Field Equation,
R
ab
=0 ⇒
□
2
γ
ab
=0.
Rab=0 ⇒ ◻2γab=0.
For a plane wave propagating along the
x
3
x3
-axis,we know that the only components of
γ
μν
γμν
that are different from zero are
γ
11
=−
γ
22
,
γ
12
=
γ
21
γ11=−γ22,γ12=γ21
So,
γ
jj
=
γ
11
+
γ
22
+
γ
33
=0
γjj=γ11+γ22+γ33=0
Consider the ADM Energy
E=
c
4
16πG
lim
r→∞
∬
S
r
⊂⊃(
∂
j
h
ij
−
∂
i
h
jj
)d
S
i
E=c416πGlimr→∞∬Sr⊂⊃(∂jhij−∂ihjj)dSi
Some calculation about
h
ab
hab
,
h
ab
=
g
ab
∓
n
a
n
b
⇒
h
ij
=
g
ij
=
η
ij
+
γ
ij
hab=gab∓nanb ⇒ hij=gij=ηij+γij
h
jj
=
η
jj
+
γ
jj
=
η
jj
=const
hjj=ηjj+γjj=ηjj=const
∂
1
h
ij
=
∂
2
h
ij
=0
∂1hij=∂2hij=0
Finally, we have
E=0
E=0
g
ab
=
η
ab
+
γ
ab
.
gab=ηab+γab.
Then use the linear Einstein's Field Equation,
R
ab
=0 ⇒
□
2
γ
ab
=0.
Rab=0 ⇒ ◻2γab=0.
For a plane wave propagating along the
x
3
x3
-axis,we know that the only components of
γ
μν
γμν
that are different from zero are
γ
11
=−
γ
22
,
γ
12
=
γ
21
γ11=−γ22,γ12=γ21
So,
γ
jj
=
γ
11
+
γ
22
+
γ
33
=0
γjj=γ11+γ22+γ33=0
Consider the ADM Energy
E=
c
4
16πG
lim
r→∞
∬
S
r
⊂⊃(
∂
j
h
ij
−
∂
i
h
jj
)d
S
i
E=c416πGlimr→∞∬Sr⊂⊃(∂jhij−∂ihjj)dSi
Some calculation about
h
ab
hab
,
h
ab
=
g
ab
∓
n
a
n
b
⇒
h
ij
=
g
ij
=
η
ij
+
γ
ij
hab=gab∓nanb ⇒ hij=gij=ηij+γij
h
jj
=
η
jj
+
γ
jj
=
η
jj
=const
hjj=ηjj+γjj=ηjj=const
∂
1
h
ij
=
∂
2
h
ij
=0
∂1hij=∂2hij=0
Finally, we have
E=0
E=0
Answered by
0
letgab=ηab+γab.gab=ηab+γab.Then use the linear Einstein's Field Equation,Rab=0 ⇒ 2γab=0.Rab=0 ⇒ 2γab=0.
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