- #1
unscientific
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Contracting \mu & \alpha - What Does It Mean?
- Context: Graduate
- Thread starter unscientific
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Discussion Overview
The discussion centers around the operation of contracting indices in tensor calculus, specifically focusing on the contraction of the indices ##\mu## and ##\alpha##. Participants explore the implications of this operation within the context of the Bianchi identity and the Riemann tensor, discussing both theoretical aspects and specific mathematical steps involved in the contraction process.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- Some participants describe contraction as an operation applicable to tensors with upper and lower indices, emphasizing the summation over components where the indices match.
- There is a detailed explanation of how contracting the indices in the Bianchi identity leads to a more complex expression, highlighting the importance of the Einstein summation convention.
- A participant requests clarification on the steps leading to a specific result involving the Riemann tensor and Ricci tensor, indicating a desire for a deeper understanding of the contraction process.
- Another participant outlines the steps of contraction, demonstrating how the indices are manipulated and how the antisymmetry of the Riemann tensor affects the signs in the resulting equations.
- There is mention of the Ricci scalar and its relationship to the contracted indices, suggesting a connection between the operations discussed and broader concepts in differential geometry.
Areas of Agreement / Disagreement
Participants generally agree on the mechanics of tensor contraction and its application to the Bianchi identity, but there are varying interpretations of the implications and specific steps involved in the process. The discussion remains unresolved regarding the clarity and correctness of the final expressions derived from the contractions.
Contextual Notes
Some participants note the importance of the Einstein summation convention and the antisymmetry properties of the Riemann tensor, which may introduce complexities in the manipulation of indices. There are also unresolved aspects related to the interpretation of the final results and their implications in the context of general relativity.
- #2
PeterDonis
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Contraction is an operation that can be applied to any tensor or product of tensors with an upper and a lower index free. (In this case the upper index is ##\mu## and the lower index is ##\alpha##.) The contraction is just a sum over all tensor components for which ##\mu## and ##\alpha## take the same value. So, for example, a tensor ##T^{\mu}{}_{\alpha}## with one upper and one lower index can be contracted to a scalar ##T = T^0{}_0 + T^1{}_1 + T^2{}_2 + T^3{}_3##.
The contraction of the Bianchi identity has more terms because the identity itself has three terms, and each one becomes a sum of four terms when contracted. Note that the second step, multiplying by ##g^{\nu \gamma}##, is also a contraction, because the indexes ##\nu## and ##\gamma## appear as lower indexes in the Bianchi identity.
The contraction of the Bianchi identity has more terms because the identity itself has three terms, and each one becomes a sum of four terms when contracted. Note that the second step, multiplying by ##g^{\nu \gamma}##, is also a contraction, because the indexes ##\nu## and ##\gamma## appear as lower indexes in the Bianchi identity.
- #3
unscientific
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PeterDonis said:
Do you mind showing the steps leading to the final result ## 2\nabla_v R_\beta^v -\nabla_\beta R = 0##?
- #4
PeterDonis
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We start with
$$
\nabla_{\gamma} R^{\mu}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\mu}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\mu}{}_{\nu \beta \gamma} = 0
$$
Contracting ##\mu## and ##\alpha## gives (note that an index that is repeated, once upper and once lower, is summed over as I described; this is called the "Einstein summation convention" and is extremely useful):
$$
\nabla_{\gamma} R^{\alpha}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\alpha}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$
Completing the contractions (meaning, collapsing the contracted indexes and using the fact that contracting the Riemann tensor on the upper and the middle lower index gives the Ricci tensor) gives:
$$
\nabla_{\gamma} R_{\nu \beta} - \nabla_{\beta} R_{\nu \gamma} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$
where the minus sign in the second term comes in because the ##\alpha## index was the last lower index, not the middle one; swapping the indexes flips the sign (because the Riemann tensor is antisymmetric in the last two lower indexes). Now we contract with ##g^{\nu \gamma}## to give:
$$
g^{\nu \gamma} \nabla_{\gamma} R_{\nu \beta} - g^{\nu \gamma} \nabla_{\beta} R_{\nu \gamma} + g^{\nu \gamma} \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$
Completing the contractions gives (note that ##R = g^{\nu \gamma} R_{\nu \gamma}## is the Ricci scalar):
$$
\nabla^{\nu} R_{\nu \beta} - \nabla_{\beta} R + \nabla_{\alpha} R^{\alpha}{}_{\beta} = 0
$$
The first and third terms are really the same thing, because the contracted index is a "dummy" index and we can relabel it freely, and we can also freely "flip" the indexes in the contraction (to put the upper index on ##R## and the lower index on ##\nabla## in the first term). This gives what we were looking for:
$$
2 \nabla_{\nu} R^{\nu}{}_{\beta} - \nabla_{\beta} R = 0
$$
$$
\nabla_{\gamma} R^{\mu}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\mu}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\mu}{}_{\nu \beta \gamma} = 0
$$
Contracting ##\mu## and ##\alpha## gives (note that an index that is repeated, once upper and once lower, is summed over as I described; this is called the "Einstein summation convention" and is extremely useful):
$$
\nabla_{\gamma} R^{\alpha}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\alpha}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$
Completing the contractions (meaning, collapsing the contracted indexes and using the fact that contracting the Riemann tensor on the upper and the middle lower index gives the Ricci tensor) gives:
$$
\nabla_{\gamma} R_{\nu \beta} - \nabla_{\beta} R_{\nu \gamma} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$
where the minus sign in the second term comes in because the ##\alpha## index was the last lower index, not the middle one; swapping the indexes flips the sign (because the Riemann tensor is antisymmetric in the last two lower indexes). Now we contract with ##g^{\nu \gamma}## to give:
$$
g^{\nu \gamma} \nabla_{\gamma} R_{\nu \beta} - g^{\nu \gamma} \nabla_{\beta} R_{\nu \gamma} + g^{\nu \gamma} \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
$$
Completing the contractions gives (note that ##R = g^{\nu \gamma} R_{\nu \gamma}## is the Ricci scalar):
$$
\nabla^{\nu} R_{\nu \beta} - \nabla_{\beta} R + \nabla_{\alpha} R^{\alpha}{}_{\beta} = 0
$$
The first and third terms are really the same thing, because the contracted index is a "dummy" index and we can relabel it freely, and we can also freely "flip" the indexes in the contraction (to put the upper index on ##R## and the lower index on ##\nabla## in the first term). This gives what we were looking for:
$$
2 \nabla_{\nu} R^{\nu}{}_{\beta} - \nabla_{\beta} R = 0
$$