Deriving a unique derivative operator with torsion such that $triangledown g = 0$
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From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:
$triangledown g = 0$ becomes again
$$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$
If we add the permutation (ab) and subtract the permutation (cab) as before we get
$$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$
How have we gone form C's to T's?
derivatives differential-geometry general-relativity special-relativity
asked Dec 5 at 12:32
Permian
2,0811034
add a comment |
up vote
1
down vote
favorite
From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:
$triangledown g = 0$ becomes again
$$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$
If we add the permutation (ab) and subtract the permutation (cab) as before we get
$$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$
How have we gone form C's to T's?
derivatives differential-geometry general-relativity special-relativity
asked Dec 5 at 12:32
Permian
2,0811034
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:
$triangledown g = 0$ becomes again
$$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$
If we add the permutation (ab) and subtract the permutation (cab) as before we get
$$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$
How have we gone form C's to T's?
derivatives differential-geometry general-relativity special-relativity
asked Dec 5 at 12:32
Permian
2,0811034
From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:
$triangledown g = 0$ becomes again
$$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$
If we add the permutation (ab) and subtract the permutation (cab) as before we get
$$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$
How have we gone form C's to T's?
derivatives differential-geometry general-relativity special-relativity
derivatives differential-geometry general-relativity special-relativity
asked Dec 5 at 12:32
Permian
2,0811034
asked Dec 5 at 12:32
Permian
2,0811034
asked Dec 5 at 12:32
Permian
2,0811034
asked Dec 5 at 12:32
Permian
2,0811034
asked Dec 5 at 12:32
Permian
2,0811034
2,0811034
add a comment |
add a comment |
1 Answer
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up vote
1
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accepted
Remember that the torsion in basically the antisymmetric part of $C$
$$
T_{abc} = C_{a[bc]} tag{1}
$$
So that
begin{eqnarray}
color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
end{eqnarray}
With this in mind ($nabla g = 0$)
begin{eqnarray}
tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
&& -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
&=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
&stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
&stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
end{eqnarray}
answered Dec 5 at 13:38
caverac
12.8k21028
Ill have another look but I think statement is the bit im not seeing
– Permian
Dec 5 at 14:07
Yeah I cant see how (1) is so obvious to you
– Permian
Dec 5 at 20:43
@Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
– caverac
Dec 5 at 20:45
where are the C's in 1a)?
– Permian
Dec 5 at 20:47
1
It is a very fascinating subject, congratulations!
– caverac
Dec 7 at 20:02
|
show 7 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Remember that the torsion in basically the antisymmetric part of $C$
$$
T_{abc} = C_{a[bc]} tag{1}
$$
So that
begin{eqnarray}
color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
end{eqnarray}
With this in mind ($nabla g = 0$)
begin{eqnarray}
tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
&& -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
&=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
&stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
&stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
end{eqnarray}
answered Dec 5 at 13:38
caverac
12.8k21028
Ill have another look but I think statement is the bit im not seeing
– Permian
Dec 5 at 14:07
Yeah I cant see how (1) is so obvious to you
– Permian
Dec 5 at 20:43
@Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
– caverac
Dec 5 at 20:45
where are the C's in 1a)?
– Permian
Dec 5 at 20:47
1
It is a very fascinating subject, congratulations!
– caverac
Dec 7 at 20:02
|
show 7 more comments
up vote
1
down vote
accepted
Remember that the torsion in basically the antisymmetric part of $C$
$$
T_{abc} = C_{a[bc]} tag{1}
$$
So that
begin{eqnarray}
color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
end{eqnarray}
With this in mind ($nabla g = 0$)
begin{eqnarray}
tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
&& -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
&=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
&stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
&stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
end{eqnarray}
answered Dec 5 at 13:38
caverac
12.8k21028
Ill have another look but I think statement is the bit im not seeing
– Permian
Dec 5 at 14:07
Yeah I cant see how (1) is so obvious to you
– Permian
Dec 5 at 20:43
@Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
– caverac
Dec 5 at 20:45
where are the C's in 1a)?
– Permian
Dec 5 at 20:47
1
It is a very fascinating subject, congratulations!
– caverac
Dec 7 at 20:02
|
show 7 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Remember that the torsion in basically the antisymmetric part of $C$
$$
T_{abc} = C_{a[bc]} tag{1}
$$
So that
begin{eqnarray}
color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
end{eqnarray}
With this in mind ($nabla g = 0$)
begin{eqnarray}
tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
&& -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
&=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
&stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
&stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
end{eqnarray}
answered Dec 5 at 13:38
caverac
12.8k21028
Remember that the torsion in basically the antisymmetric part of $C$
$$
T_{abc} = C_{a[bc]} tag{1}
$$
So that
begin{eqnarray}
color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
end{eqnarray}
With this in mind ($nabla g = 0$)
begin{eqnarray}
tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
&& -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
&=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
&stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
&stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
end{eqnarray}
answered Dec 5 at 13:38
caverac
12.8k21028
answered Dec 5 at 13:38
caverac
12.8k21028
answered Dec 5 at 13:38
caverac
12.8k21028
answered Dec 5 at 13:38
caverac
12.8k21028
12.8k21028
Ill have another look but I think statement is the bit im not seeing
– Permian
Dec 5 at 14:07
Yeah I cant see how (1) is so obvious to you
– Permian
Dec 5 at 20:43
@Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
– caverac
Dec 5 at 20:45
where are the C's in 1a)?
– Permian
Dec 5 at 20:47
1
It is a very fascinating subject, congratulations!
– caverac
Dec 7 at 20:02
|
show 7 more comments
Ill have another look but I think statement is the bit im not seeing
– Permian
Dec 5 at 14:07
Yeah I cant see how (1) is so obvious to you
– Permian
Dec 5 at 20:43
@Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
– caverac
Dec 5 at 20:45
where are the C's in 1a)?
– Permian
Dec 5 at 20:47
1
It is a very fascinating subject, congratulations!
– caverac
Dec 7 at 20:02
Ill have another look but I think statement is the bit im not seeing
– Permian
Dec 5 at 14:07
Ill have another look but I think statement is the bit im not seeing
– Permian
Dec 5 at 14:07
Yeah I cant see how (1) is so obvious to you
– Permian
Dec 5 at 20:43
Yeah I cant see how (1) is so obvious to you
– Permian
Dec 5 at 20:43
@Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
– caverac
Dec 5 at 20:45
@Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
– caverac
Dec 5 at 20:45
where are the C's in 1a)?
– Permian
Dec 5 at 20:47
where are the C's in 1a)?
– Permian
Dec 5 at 20:47
1
1
It is a very fascinating subject, congratulations!
– caverac
Dec 7 at 20:02
It is a very fascinating subject, congratulations!
– caverac
Dec 7 at 20:02
|
show 7 more comments
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