Levi-Civita Symbol multiplied by itself

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    Levi-civita Symbol
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Homework Help Overview

The discussion revolves around evaluating the expression \(\epsilon_{ijk}\epsilon_{ijk}\), where \(\epsilon\) represents the antisymmetric Levi-Civita symbol in three dimensions. Participants are exploring the implications of this evaluation in the context of determinants and Kronecker deltas.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the Levi-Civita symbol and the Kronecker delta, with one participant attempting to derive the expression and another clarifying notation and summation indices. Questions arise regarding the properties of the delta function and the validity of swapping indices.

Discussion Status

The discussion is active, with participants providing clarifications and corrections to notation. There is an acknowledgment of the properties of the delta function, and some participants are exploring the implications of their calculations without reaching a definitive conclusion.

Contextual Notes

Participants mention the stress of impending exams, which may influence the clarity and thoroughness of their discussions. There is a focus on notation and the proper application of mathematical properties.

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Homework Statement



evaluate \epsilon_{ijk}\epsilon_{ijk} where \epsilon is is the antisymetric levi-civita symbol in 3D

Homework Equations



determinant of deltas = product of levi-civita -> would take ages to write out.

The Attempt at a Solution



\epsilon_{ijk}\epsilon_{ijk}=\delta_{kk}\delta_{ll}-\delta_{lk}\delta_{kl}=1-2\delta_{lk}

But i have a feeling the answer is 3? (because of this)

263e9dccaced8adf5ba6d68403150f47.png
 
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Well, firstly clean your notation. It should be
<br /> \epsilon_{ijk}\epsilon_{ijk}=\delta_{kk}\delta_{jj}-\delta_{jk}\delta_{jk}<br />
And there is a summation over j and k.
So that gives
<br /> \sum\limits_{j,k}\delta_{kk}\delta_{jj}-\delta_{jk}\delta_{jk} = 3.3-3 = 6 = 3!<br />
Also 3! = 6, not 3.
 
Thanks a lot. I was in a rush and I am pretty stressed at the moment as its only a few days before exams, the \epsilon [\tex]was supposed to read \epsilon_{jlk}[\tex].&lt;br /&gt; &lt;br /&gt; How is it okay to swap the order of indicies on the delta funtion as you have done: i.e. \delta_{ij} = \delta_{ji}[\tex]
 
Last edited:
ya, that's a property of the delta function.
 
again, thanks a lot
 

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