Anti-symmetric part of an equation

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SUMMARY

The discussion centers on the anti-symmetric part of the equation presented in "Introducing Einstein's Relativity_A Deeper Understanding Ed 2," specifically on page 271. The equation, ##X_{a}\partial_{b}X_{c}=\lambda f_{,a}\lambda_{,b}f_{,c}+\lambda^2f_{,a}f_{,cb}##, reveals that the totally anti-symmetric parts vanish due to the symmetry properties of the terms involved. The first term is symmetric in indices a and c, while the second term is symmetric in indices b and c, leading to the conclusion that their anti-symmetric components do not contribute. The notation for the totally anti-symmetric part is clarified as ##X_{[a}\partial_b X_{c]}##, which changes sign upon the interchange of any of the three indices.

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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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