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Generalized divergence of tensor in GR
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Although I've forgotten the proof (and cannot find it in, say, Carroll's book), the following formula holds for the covariant divergence in general relativity:
$$nabla_mu A^mu = frac1sqrt partial_mu left( sqrt A^muright),$$
where $g = det(g_alphabeta)$. I was wondering if this formula holds if $A^mu$ is replaced with a general rank $(n,m)$ tensor
$$T^mu mu_1mu_2 cdots mu_n-1_,,,,,,,,,,,,,,,,,,,nu_1cdots nu_m?$$
If not, could you point me to any references that have divergence formulas for higher rank tensors?
general-relativity differential-geometry tensor-calculus definition differentiation
asked 2 hours ago
Dwagg
501110
add a comment |Â
up vote
2
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Although I've forgotten the proof (and cannot find it in, say, Carroll's book), the following formula holds for the covariant divergence in general relativity:
$$nabla_mu A^mu = frac1sqrt partial_mu left( sqrt A^muright),$$
where $g = det(g_alphabeta)$. I was wondering if this formula holds if $A^mu$ is replaced with a general rank $(n,m)$ tensor
$$T^mu mu_1mu_2 cdots mu_n-1_,,,,,,,,,,,,,,,,,,,nu_1cdots nu_m?$$
If not, could you point me to any references that have divergence formulas for higher rank tensors?
general-relativity differential-geometry tensor-calculus definition differentiation
asked 2 hours ago
Dwagg
501110
It does not. It holds when $A^mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor.
– Prahar
1 hour ago
add a comment |Â
up vote
2
down vote
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up vote
2
down vote
favorite
Although I've forgotten the proof (and cannot find it in, say, Carroll's book), the following formula holds for the covariant divergence in general relativity:
$$nabla_mu A^mu = frac1sqrt partial_mu left( sqrt A^muright),$$
where $g = det(g_alphabeta)$. I was wondering if this formula holds if $A^mu$ is replaced with a general rank $(n,m)$ tensor
$$T^mu mu_1mu_2 cdots mu_n-1_,,,,,,,,,,,,,,,,,,,nu_1cdots nu_m?$$
If not, could you point me to any references that have divergence formulas for higher rank tensors?
general-relativity differential-geometry tensor-calculus definition differentiation
asked 2 hours ago
Dwagg
501110
Although I've forgotten the proof (and cannot find it in, say, Carroll's book), the following formula holds for the covariant divergence in general relativity:
$$nabla_mu A^mu = frac1sqrt partial_mu left( sqrt A^muright),$$
where $g = det(g_alphabeta)$. I was wondering if this formula holds if $A^mu$ is replaced with a general rank $(n,m)$ tensor
$$T^mu mu_1mu_2 cdots mu_n-1_,,,,,,,,,,,,,,,,,,,nu_1cdots nu_m?$$
If not, could you point me to any references that have divergence formulas for higher rank tensors?
general-relativity differential-geometry tensor-calculus definition differentiation
general-relativity differential-geometry tensor-calculus definition differentiation
asked 2 hours ago
Dwagg
501110
asked 2 hours ago
Dwagg
501110
edited 7 mins ago
asked 2 hours ago
Dwagg
501110
asked 2 hours ago
Dwagg
501110
asked 2 hours ago
Dwagg
501110
501110
It does not. It holds when $A^mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor.
– Prahar
1 hour ago
add a comment |Â
It does not. It holds when $A^mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor.
– Prahar
1 hour ago
It does not. It holds when $A^mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor.
– Prahar
1 hour ago
It does not. It holds when $A^mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor.
– Prahar
1 hour ago
add a comment |Â
1 Answer
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No, this does not hold in general for higher-rank tensors. The general equation for the divergence of a completely contravariant tensor in terms of a coordinate derivative operator $partial_mu$ is
$$
nabla_mu T^mu nu_1 dots nu_n = partial_mu T^mu nu_1 dots nu_n + Gamma^mu _mu rho T^rho nu_1 dots nu_n + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
We also have the fact that
$$
Gamma^mu _mu rho = frac1sqrt partial_mu sqrt.
$$
Thus,
$$
nabla_mu T^mu nu_1 dots nu_n = frac1sqrt partial_mu left( sqrt T^mu nu_1 dots nu_n right) + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
This last sum will not vanish for a general tensor. However, some or all of the terms may vanish for tensors with a particular symmetry structure. In particular, if $T^mu nu_1 dots nu_n$ is antisymmetric in all of its indices, then any contraction of two of its indices with the symmetric indices of the Christoffel symbols automatically vanishes; and thus the entire sum goes away.
For references as to how to take the covariant derivative of a general tensor, see Chapter 5 of Schutz's A First Course in General Relativity for a coordinate-based approach, or Chapter 3 of Wald's General Relativity for a more general approach.
answered 1 hour ago
Michael Seifert
13.8k12651
I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.
– Void
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
No, this does not hold in general for higher-rank tensors. The general equation for the divergence of a completely contravariant tensor in terms of a coordinate derivative operator $partial_mu$ is
$$
nabla_mu T^mu nu_1 dots nu_n = partial_mu T^mu nu_1 dots nu_n + Gamma^mu _mu rho T^rho nu_1 dots nu_n + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
We also have the fact that
$$
Gamma^mu _mu rho = frac1sqrt partial_mu sqrt.
$$
Thus,
$$
nabla_mu T^mu nu_1 dots nu_n = frac1sqrt partial_mu left( sqrt T^mu nu_1 dots nu_n right) + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
This last sum will not vanish for a general tensor. However, some or all of the terms may vanish for tensors with a particular symmetry structure. In particular, if $T^mu nu_1 dots nu_n$ is antisymmetric in all of its indices, then any contraction of two of its indices with the symmetric indices of the Christoffel symbols automatically vanishes; and thus the entire sum goes away.
For references as to how to take the covariant derivative of a general tensor, see Chapter 5 of Schutz's A First Course in General Relativity for a coordinate-based approach, or Chapter 3 of Wald's General Relativity for a more general approach.
answered 1 hour ago
Michael Seifert
13.8k12651
I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.
– Void
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
No, this does not hold in general for higher-rank tensors. The general equation for the divergence of a completely contravariant tensor in terms of a coordinate derivative operator $partial_mu$ is
$$
nabla_mu T^mu nu_1 dots nu_n = partial_mu T^mu nu_1 dots nu_n + Gamma^mu _mu rho T^rho nu_1 dots nu_n + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
We also have the fact that
$$
Gamma^mu _mu rho = frac1sqrt partial_mu sqrt.
$$
Thus,
$$
nabla_mu T^mu nu_1 dots nu_n = frac1sqrt partial_mu left( sqrt T^mu nu_1 dots nu_n right) + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
This last sum will not vanish for a general tensor. However, some or all of the terms may vanish for tensors with a particular symmetry structure. In particular, if $T^mu nu_1 dots nu_n$ is antisymmetric in all of its indices, then any contraction of two of its indices with the symmetric indices of the Christoffel symbols automatically vanishes; and thus the entire sum goes away.
For references as to how to take the covariant derivative of a general tensor, see Chapter 5 of Schutz's A First Course in General Relativity for a coordinate-based approach, or Chapter 3 of Wald's General Relativity for a more general approach.
answered 1 hour ago
Michael Seifert
13.8k12651
I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.
– Void
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
No, this does not hold in general for higher-rank tensors. The general equation for the divergence of a completely contravariant tensor in terms of a coordinate derivative operator $partial_mu$ is
$$
nabla_mu T^mu nu_1 dots nu_n = partial_mu T^mu nu_1 dots nu_n + Gamma^mu _mu rho T^rho nu_1 dots nu_n + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
We also have the fact that
$$
Gamma^mu _mu rho = frac1sqrt partial_mu sqrt.
$$
Thus,
$$
nabla_mu T^mu nu_1 dots nu_n = frac1sqrt partial_mu left( sqrt T^mu nu_1 dots nu_n right) + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
This last sum will not vanish for a general tensor. However, some or all of the terms may vanish for tensors with a particular symmetry structure. In particular, if $T^mu nu_1 dots nu_n$ is antisymmetric in all of its indices, then any contraction of two of its indices with the symmetric indices of the Christoffel symbols automatically vanishes; and thus the entire sum goes away.
For references as to how to take the covariant derivative of a general tensor, see Chapter 5 of Schutz's A First Course in General Relativity for a coordinate-based approach, or Chapter 3 of Wald's General Relativity for a more general approach.
answered 1 hour ago
Michael Seifert
13.8k12651
No, this does not hold in general for higher-rank tensors. The general equation for the divergence of a completely contravariant tensor in terms of a coordinate derivative operator $partial_mu$ is
$$
nabla_mu T^mu nu_1 dots nu_n = partial_mu T^mu nu_1 dots nu_n + Gamma^mu _mu rho T^rho nu_1 dots nu_n + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
We also have the fact that
$$
Gamma^mu _mu rho = frac1sqrt partial_mu sqrt.
$$
Thus,
$$
nabla_mu T^mu nu_1 dots nu_n = frac1sqrt partial_mu left( sqrt T^mu nu_1 dots nu_n right) + sum_i = 1^n Gamma^nu_i _mu rho T^mu nu_1 dots rho dots nu_n.
$$
This last sum will not vanish for a general tensor. However, some or all of the terms may vanish for tensors with a particular symmetry structure. In particular, if $T^mu nu_1 dots nu_n$ is antisymmetric in all of its indices, then any contraction of two of its indices with the symmetric indices of the Christoffel symbols automatically vanishes; and thus the entire sum goes away.
For references as to how to take the covariant derivative of a general tensor, see Chapter 5 of Schutz's A First Course in General Relativity for a coordinate-based approach, or Chapter 3 of Wald's General Relativity for a more general approach.
answered 1 hour ago
Michael Seifert
13.8k12651
answered 1 hour ago
Michael Seifert
13.8k12651
answered 1 hour ago
Michael Seifert
13.8k12651
answered 1 hour ago
Michael Seifert
13.8k12651
13.8k12651
I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.
– Void
1 hour ago
add a comment |Â
I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.
– Void
1 hour ago
I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.
– Void
1 hour ago
I believe that for general Christoffel symbols and non-zero $T$, total antisymmetry is the only case where the extra sum vanishes.
– Void
1 hour ago
add a comment |Â
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It does not. It holds when $A^mu$ is replaced with a general $p$-form (which is a totally antisymmetric $(p,0)$ tensor.
– Prahar
1 hour ago