Levi-Civita Symbol multiplied by itself

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    Levi-civita Symbol
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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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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