Anti-symmetric part of an equation

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Discussion Overview

The discussion revolves around the anti-symmetric part of a specific equation from "Introducing Einstein's Relativity_A Deeper Understanding Ed 2." Participants explore the properties of anti-symmetry in the context of tensor notation and the implications of symmetry in the terms of the equation.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the totally anti-symmetric parts of the terms in the equation vanish, given the symmetry properties of the terms.
  • Another participant suggests that understanding the definition of the totally anti-symmetric part is essential for the discussion.
  • There is a clarification that the totally anti-symmetric part of the equation is represented as ##X_{[a}\partial_b X_{c]}##, which should change sign with any interchange of indices.
  • Participants discuss the notation and implications of the anti-symmetric part, with one participant questioning if the notation implies that only indices "a" and "c" exchange.
  • Clarifications are made regarding the notation, indicating that all indices within the square brackets are subject to interchange, not just "a" and "c."
  • One participant provides a formula for the anti-symmetric part, which is confirmed by another participant.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the notation and the properties of the anti-symmetric part, indicating that the discussion remains unresolved with multiple competing interpretations.

Contextual Notes

There are unresolved questions regarding the implications of the symmetry properties of the terms in the equation and the specific notation used for the anti-symmetric part.

GR191511
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I'm reading "Introducing Einstein's Relativity_A Deeper Understanding Ed 2"on page 271:
##X_{a}\partial_{b}X_{c}=\lambda f_{,a}\lambda_{,b}f_{,c}+\lambda^2f_{,a}f_{,cb}##
Taking the totally anti-symmetric part of this equation and noting that the first term on the right is symmetric in a and c and the second term is symmetric in b and c,we see that their totally anti-symmetric parts vanish...
"their totally anti-symmetric parts vanish"Why?
 
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GR191511 said:
I'm reading "Introducing Einstein's Relativity_A Deeper Understanding Ed 2"on page 271:
##X_{a}\partial_{b}X_{c}=\lambda f_{,a}\lambda_{,b}f_{,c}+\lambda^2f_{,a}f_{,cb}##
Taking the totally anti-symmetric part of this equation and noting that the first term on the right is symmetric in a and c and the second term is symmetric in b and c,we see that their totally anti-symmetric parts vanish...
"their totally anti-symmetric parts vanish"Why?
May be you can start with reminding yourself what totally anti-symmetric part is.
 
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GR191511 said:
Why?
Because:
GR191511 said:
the first term on the right is symmetric in a and c and the second term is symmetric in b and c
 
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Orodruin said:
Because:
Is the" totally anti-symmetric part"of that equation##X_{[a}\partial_{b}X_{c]}+X_{a[}\partial_{b}X_{c]}+X_{[a}\partial_{b]}X_{c}##?
 
No, it is ##X_{[a}\partial_b X_{c]}##. Meaning that it should change sign whenever you interchange any of the three indices.
 
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Orodruin said:
No, it is ##X_{[a}\partial_b X_{c]}##. Meaning that it should change sign whenever you interchange any of the three indices.
any of the three indices?Doesn't the notation ##X_{[a}\partial_b X_{c]}##means that "a" exchanges with "c" only?
 
GR191511 said:
any of the three indices?Doesn't the notation ##X_{[a}\partial_b X_{c]}##means that "a" exchanges with "c" only?
No. All your indices are within the square brackets.
 
GR191511 said:
any of the three indices?Doesn't the notation ##X_{[a}\partial_b X_{c]}##means that "a" exchanges with "c" only?
No. That would be ##X_{[a|}\partial_bX_{|c]}##, where the ##|## symbols are used to exclude indices between them from round or square brackets.
 
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Ibix said:
No. That would be ##X_{[a|}\partial_bX_{|c]}##, where the ##|## symbols are used to exclude indices between them from round or square brackets.
o_O...OK!##X_{[a}\partial_bX_{c]}=\frac 1 6(X_{a}\partial_bX_{c}+X_{c}\partial_aX_{b}+X_{b}\partial_cX_{a}-X_{a}\partial_cX_{b}-X_{b}\partial_aX_{c}-X_{c}\partial_bX_{a})##?
 
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GR191511 said:
o_O...OK!##X_{[a}\partial_bX_{c]}=\frac 1 6(X_{a}\partial_bX_{c}+X_{c}\partial_aX_{b}+X_{b}\partial_cX_{a}-X_{a}\partial_cX_{b}-X_{b}\partial_aX_{c}-X_{c}\partial_bX_{a})##?
Yes.
 
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