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Archived: 07 Aug 2014, 09:55
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jbeckmann
Joined: 22 Apr 2010, 15:52 Posts: 43 Location: Olpe, Germany
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 Exercise [12.16]
Here is my (WRONG!) proposal: THIS PROOF COMPRISES AN ERROR (cf. Roberto, below). Better try this one:
Last edited by jbeckmann on 06 Dec 2010, 20:44, edited 1 time in total.
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05 Dec 2010, 09:13 |
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Roberto
Joined: 03 Jun 2010, 15:18 Posts: 136
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 Re: Exercise [12.16]
I think your proof of "a_[rs a_u]v implies that a_rs is simple " is not correct.
The proof is based on the assumption that there is a vector x^v satisfying condition x^m x^n a_mn =1
This is never possible, because being a_mn antisymmetric and x^m x^n symmetric their contraction is always 0. This can easily be seen in the simple 2 dimensional case where a_11=a_22=0, a_12 = -a_21=a then x^m x^n a_mn= x_1 x_2 a - x_2 x_1 a = 0.
The formal proof for the general case is: x^m x^n a_mn = x^n x^m a_nm = - x^n x^m a_mn = - x^m x^n a_mn (where in the first step I just renamed the contracted indices and in second step antisymmetry of a was exploited) therefore x^m x^n a_mn =0.
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06 Dec 2010, 14:27 |
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jbeckmann
Joined: 22 Apr 2010, 15:52 Posts: 43 Location: Olpe, Germany
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 Re: Exercise [12.16]
You are right, Roberto! Thanks for the hint and sorry for the error. I hope this fixes the bug:
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06 Dec 2010, 20:42 |
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