Using symmetry of Riemann tensor to vanish componentsRiemann tensor in 2d and 3dRiemann Curvature Tensor Symmetries ProofFinding the Riemann tensor for the surface of a sphere with sympy.diffgeomVariation with respect to $R_abcd$? How to compute$fracpartial Rpartial R_abcd=frac12(g^ac g^bd - g^ad g^bc)$?Variation of term like $fracpartial R_ab R^abpartial R_abcd$, $fracpartial R_abcd R^abcdpartial R_efgh$Variation of products of Riemann tensor $delta (sqrt-g RR epsilon epsilon)$Is there any additional symmetry in Riemann curvature tensor to tell which components are zero?Intuition for Interchange Symmetry of the Riemann TensorShow that the symmetry properties of a tensor are invariantWhy does the lowered Riemann tensor only have 20 independent components for the Schwarzschild metric?
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Using symmetry of Riemann tensor to vanish components
Riemann tensor in 2d and 3dRiemann Curvature Tensor Symmetries ProofFinding the Riemann tensor for the surface of a sphere with sympy.diffgeomVariation with respect to $R_abcd$? How to compute$fracpartial Rpartial R_abcd=frac12(g^ac g^bd - g^ad g^bc)$?Variation of term like $fracpartial R_ab R^abpartial R_abcd$, $fracpartial R_abcd R^abcdpartial R_efgh$Variation of products of Riemann tensor $delta (sqrt-g RR epsilon epsilon)$Is there any additional symmetry in Riemann curvature tensor to tell which components are zero?Intuition for Interchange Symmetry of the Riemann TensorShow that the symmetry properties of a tensor are invariantWhy does the lowered Riemann tensor only have 20 independent components for the Schwarzschild metric?
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$begingroup$
The Riemann tensor is skew-symmetric in its first and last pair of indices, i.e.,
beginalign
R_abcd = -R_abdc = -R_bacd
endalign
Can I simply use this to say that, for example, the component $R_0001 = 0$ because $R_0001 = -R_0001$?
general-relativity tensor-calculus curvature
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
$endgroup$
add a comment |
$begingroup$
The Riemann tensor is skew-symmetric in its first and last pair of indices, i.e.,
beginalign
R_abcd = -R_abdc = -R_bacd
endalign
Can I simply use this to say that, for example, the component $R_0001 = 0$ because $R_0001 = -R_0001$?
general-relativity tensor-calculus curvature
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
$endgroup$
add a comment |
$begingroup$
The Riemann tensor is skew-symmetric in its first and last pair of indices, i.e.,
beginalign
R_abcd = -R_abdc = -R_bacd
endalign
Can I simply use this to say that, for example, the component $R_0001 = 0$ because $R_0001 = -R_0001$?
general-relativity tensor-calculus curvature
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
$endgroup$
The Riemann tensor is skew-symmetric in its first and last pair of indices, i.e.,
beginalign
R_abcd = -R_abdc = -R_bacd
endalign
Can I simply use this to say that, for example, the component $R_0001 = 0$ because $R_0001 = -R_0001$?
general-relativity tensor-calculus curvature
general-relativity tensor-calculus curvature
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
asked 8 hours ago
Elismar LöschElismar Lösch
154 bronze badges
154 bronze badges
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
$endgroup$
$begingroup$
Wow, this tip was very nice!
$endgroup$
– Elismar Lösch
7 hours ago
$begingroup$
Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks!
$endgroup$
– Elismar Lösch
6 hours ago
$begingroup$
The algebraic Bianchi identity is $R_abcd+R_acdb+R_adbc=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations.
$endgroup$
– G. Smith
6 hours ago
add a comment |
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$begingroup$
Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
$endgroup$
$begingroup$
Wow, this tip was very nice!
$endgroup$
– Elismar Lösch
7 hours ago
$begingroup$
Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks!
$endgroup$
– Elismar Lösch
6 hours ago
$begingroup$
The algebraic Bianchi identity is $R_abcd+R_acdb+R_adbc=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations.
$endgroup$
– G. Smith
6 hours ago
add a comment |
$begingroup$
Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
$endgroup$
$begingroup$
Wow, this tip was very nice!
$endgroup$
– Elismar Lösch
7 hours ago
$begingroup$
Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks!
$endgroup$
– Elismar Lösch
6 hours ago
$begingroup$
The algebraic Bianchi identity is $R_abcd+R_acdb+R_adbc=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations.
$endgroup$
– G. Smith
6 hours ago
add a comment |
$begingroup$
Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
$endgroup$
Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
edited 7 hours ago
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
answered 8 hours ago
G. SmithG. Smith
16.5k1 gold badge27 silver badges54 bronze badges
16.5k1 gold badge27 silver badges54 bronze badges
$begingroup$
Wow, this tip was very nice!
$endgroup$
– Elismar Lösch
7 hours ago
$begingroup$
Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks!
$endgroup$
– Elismar Lösch
6 hours ago
$begingroup$
The algebraic Bianchi identity is $R_abcd+R_acdb+R_adbc=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations.
$endgroup$
– G. Smith
6 hours ago
add a comment |
$begingroup$
Wow, this tip was very nice!
$endgroup$
– Elismar Lösch
7 hours ago
$begingroup$
Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks!
$endgroup$
– Elismar Lösch
6 hours ago
$begingroup$
The algebraic Bianchi identity is $R_abcd+R_acdb+R_adbc=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations.
$endgroup$
– G. Smith
6 hours ago
$begingroup$
Wow, this tip was very nice!
$endgroup$
– Elismar Lösch
7 hours ago
$begingroup$
Wow, this tip was very nice!
$endgroup$
– Elismar Lösch
7 hours ago
$begingroup$
Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks!
$endgroup$
– Elismar Lösch
6 hours ago
$begingroup$
Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks!
$endgroup$
– Elismar Lösch
6 hours ago
$begingroup$
The algebraic Bianchi identity is $R_abcd+R_acdb+R_adbc=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations.
$endgroup$
– G. Smith
6 hours ago
$begingroup$
The algebraic Bianchi identity is $R_abcd+R_acdb+R_adbc=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations.
$endgroup$
– G. Smith
6 hours ago
add a comment |
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