Deriving Riemann tensor with four lower indices
$begingroup$
In terms of the connection coefficients
$$R_{abc}^d = partial_a Gamma_{bc}^d-partial_b Gamma_{ac}^d - Gamma_{be}^dGamma_{ac}^e+Gamma_{ae}^dGamma_{bc}^e$$
Pick an event $A$ and choose coordinates such that $partial_a g_{bc} = 0$ at $A$... So at the event $A$, but not elsehwere in general,
$$R_{abcd} = g_{de}partial_a(Gamma_{bc}^e) - g_{de}partial_b(Gamma_{ac}^e)$$
Where has this last equation (RHS) come from?
differential-geometry riemannian-geometry tensors general-relativity
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
$endgroup$
add a comment |
$begingroup$
In terms of the connection coefficients
$$R_{abc}^d = partial_a Gamma_{bc}^d-partial_b Gamma_{ac}^d - Gamma_{be}^dGamma_{ac}^e+Gamma_{ae}^dGamma_{bc}^e$$
Pick an event $A$ and choose coordinates such that $partial_a g_{bc} = 0$ at $A$... So at the event $A$, but not elsehwere in general,
$$R_{abcd} = g_{de}partial_a(Gamma_{bc}^e) - g_{de}partial_b(Gamma_{ac}^e)$$
Where has this last equation (RHS) come from?
differential-geometry riemannian-geometry tensors general-relativity
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
$endgroup$
$begingroup$
I couldnt think of a more precise wording for the title (possible edit here)
$endgroup$
– Permian
Dec 20 '18 at 21:12
1
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 21:29
add a comment |
$begingroup$
In terms of the connection coefficients
$$R_{abc}^d = partial_a Gamma_{bc}^d-partial_b Gamma_{ac}^d - Gamma_{be}^dGamma_{ac}^e+Gamma_{ae}^dGamma_{bc}^e$$
Pick an event $A$ and choose coordinates such that $partial_a g_{bc} = 0$ at $A$... So at the event $A$, but not elsehwere in general,
$$R_{abcd} = g_{de}partial_a(Gamma_{bc}^e) - g_{de}partial_b(Gamma_{ac}^e)$$
Where has this last equation (RHS) come from?
differential-geometry riemannian-geometry tensors general-relativity
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
$endgroup$
In terms of the connection coefficients
$$R_{abc}^d = partial_a Gamma_{bc}^d-partial_b Gamma_{ac}^d - Gamma_{be}^dGamma_{ac}^e+Gamma_{ae}^dGamma_{bc}^e$$
Pick an event $A$ and choose coordinates such that $partial_a g_{bc} = 0$ at $A$... So at the event $A$, but not elsehwere in general,
$$R_{abcd} = g_{de}partial_a(Gamma_{bc}^e) - g_{de}partial_b(Gamma_{ac}^e)$$
Where has this last equation (RHS) come from?
differential-geometry riemannian-geometry tensors general-relativity
differential-geometry riemannian-geometry tensors general-relativity
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
asked Dec 20 '18 at 21:11
PermianPermian
2,2061135
2,2061135
$begingroup$
I couldnt think of a more precise wording for the title (possible edit here)
$endgroup$
– Permian
Dec 20 '18 at 21:12
1
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 21:29
add a comment |
$begingroup$
I couldnt think of a more precise wording for the title (possible edit here)
$endgroup$
– Permian
Dec 20 '18 at 21:12
1
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 21:29
$begingroup$
I couldnt think of a more precise wording for the title (possible edit here)
$endgroup$
– Permian
Dec 20 '18 at 21:12
$begingroup$
I couldnt think of a more precise wording for the title (possible edit here)
$endgroup$
– Permian
Dec 20 '18 at 21:12
1
1
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 21:29
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 21:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
$endgroup$
add a comment |
$begingroup$
Multiplying $R_{abc}^e=(partial_aGamma_{bc}^e+Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$ by $g_{de}$ with contraction over $e$ gives $R_{abcd}=(g_{de}partial_aGamma_{bc}^e+g_{de}Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$, so you're asking why $Gamma_{af}^eGamma_{bc}^f$ is $aleftrightarrow b$-symmetric at $A$. As Ted Shifrin notes, this expression is locally $0$ in the Riemann normal coordinates at $A$.
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
$endgroup$
add a comment |
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
$endgroup$
add a comment |
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
$endgroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
answered Dec 21 '18 at 0:55
Ted ShifrinTed Shifrin
63.5k44591
63.5k44591
add a comment |
add a comment |
$begingroup$
Multiplying $R_{abc}^e=(partial_aGamma_{bc}^e+Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$ by $g_{de}$ with contraction over $e$ gives $R_{abcd}=(g_{de}partial_aGamma_{bc}^e+g_{de}Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$, so you're asking why $Gamma_{af}^eGamma_{bc}^f$ is $aleftrightarrow b$-symmetric at $A$. As Ted Shifrin notes, this expression is locally $0$ in the Riemann normal coordinates at $A$.
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
$begingroup$
Multiplying $R_{abc}^e=(partial_aGamma_{bc}^e+Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$ by $g_{de}$ with contraction over $e$ gives $R_{abcd}=(g_{de}partial_aGamma_{bc}^e+g_{de}Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$, so you're asking why $Gamma_{af}^eGamma_{bc}^f$ is $aleftrightarrow b$-symmetric at $A$. As Ted Shifrin notes, this expression is locally $0$ in the Riemann normal coordinates at $A$.
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
$begingroup$
Multiplying $R_{abc}^e=(partial_aGamma_{bc}^e+Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$ by $g_{de}$ with contraction over $e$ gives $R_{abcd}=(g_{de}partial_aGamma_{bc}^e+g_{de}Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$, so you're asking why $Gamma_{af}^eGamma_{bc}^f$ is $aleftrightarrow b$-symmetric at $A$. As Ted Shifrin notes, this expression is locally $0$ in the Riemann normal coordinates at $A$.
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
$endgroup$
Multiplying $R_{abc}^e=(partial_aGamma_{bc}^e+Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$ by $g_{de}$ with contraction over $e$ gives $R_{abcd}=(g_{de}partial_aGamma_{bc}^e+g_{de}Gamma_{af}^eGamma_{bc}^f)-aleftrightarrow b$, so you're asking why $Gamma_{af}^eGamma_{bc}^f$ is $aleftrightarrow b$-symmetric at $A$. As Ted Shifrin notes, this expression is locally $0$ in the Riemann normal coordinates at $A$.
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
answered Dec 20 '18 at 22:28
J.G.J.G.
25.2k22539
25.2k22539
add a comment |
add a comment |
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$begingroup$
I couldnt think of a more precise wording for the title (possible edit here)
$endgroup$
– Permian
Dec 20 '18 at 21:12
1
$begingroup$
Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 21:29