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Contracted Epsilon Identity

  • Thread starter BOAS
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  • #1
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Hi,

I am confused about how I arrive at the contracted epsilon identity. [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

1. Homework Statement


Show that [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

Homework Equations




The Attempt at a Solution


[/B]
From the relation between the Levi-civita symbol and the Kronecker delta, I compute [itex]\epsilon_{ijk} \epsilon_{imn}[/itex] by finding the determinant of the following matrix.

[itex]\epsilon_{ijk} \epsilon_{imn} = det \left[ \begin{array}{cccc} \delta_{ii} & \delta_{im} & \delta_{in} \\ \delta_{ji} & \delta_{jm} & \delta_{jn} \\ \delta_{ki} & \delta_{km} & \delta_{kn} \end{array} \right][/itex] which yields

[itex]\epsilon_{ijk} \epsilon_{imn} = \delta_{ii} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) - \delta_{im} (\delta_{ji} \delta_{kn} - \delta_{jn} \delta_{ki}) + \delta_{in} (\delta_{ji} \delta_{km} - \delta_{jm} \delta_{ki})[/itex]


I am confused about how to progress.

Thanks for any help you can give.
 

Answers and Replies

  • #2
Geofleur
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For starters, what does ## \delta_{ii} ## equal? Note that you need to sum over repeated indices.
 
  • #3
555
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For starters, what does ## \delta_{ii} ## equal? Note that you need to sum over repeated indices.
[tex]\delta_{ii} = 3[/tex]

This seems to be the only repeated indice.
 
  • #4
RUber
Homework Helper
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Hi,

I am confused about how I arrive at the contracted epsilon identity. [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

1. Homework Statement


Show that [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

3. The Attempt at a Solution

[itex]\epsilon_{ijk} \epsilon_{imn} = \delta_{ii} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) - \delta_{im} (\delta_{ji} \delta_{kn} - \delta_{jn} \delta_{ki}) + \delta_{in} (\delta_{ji} \delta_{km} - \delta_{jm} \delta_{ki})[/itex]


I am confused about how to progress.

Thanks for any help you can give.
Note that your first term in the expansion: ##\delta_{ii} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km})## looks a lot like the final result you are looking for.
Then, the challenge should be to show that ##- \delta_{im} (\delta_{ji} \delta_{kn} - \delta_{jn} \delta_{ki}) + \delta_{in} (\delta_{ji} \delta_{km} - \delta_{jm} \delta_{ki}) = 0## in all cases. To do this, think about what must be true for any term to not be zero, and show that it implies another opposite term must also not be zero.
 
  • #5
RUber
Homework Helper
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[tex]\delta_{ii} = 3[/tex]

This seems to be the only repeated indice.
You seem to be using a different definition of the Kronecker delta? Usually the only possible outcomes are 0 or 1.
 
  • #6
555
19
You seem to be using a different definition of the Kronecker delta? Usually the only possible outcomes are 0 or 1.
I thought the idea was that [itex]\delta_{ii}[/itex] implied the summation of [itex]\delta_{11}, \delta_{22}, and \delta_{33}[/itex], which are each respectively equal to 1.
 
  • #7
Geofleur
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That's right, ## \delta_{ii} = \delta_{11} + \delta_{22} + \delta_{33} = 3 ##. Now eliminate the ## \delta ##'s in front of the other two terms and see what happens.
 
  • #8
RUber
Homework Helper
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I see. I was thinking one term at a time, rather than the sum over the terms.
In that case, I get the same result as Geofleur.
 
  • #9
555
19
That's right, ## \delta_{ii} = \delta_{11} + \delta_{22} + \delta_{33} = 3 ##. Now eliminate the ## \delta ##'s in front of the other two terms and see what happens.
I am unsure of how to evaluate these deltas.

[itex]\delta_{im} = 0[/itex] unless [itex]i = m[/itex] so do they both just vanish?
 
  • #10
Geofleur
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They do vanish unless ## i = m ##. In that case, the ##\delta##'s in front become ones, and the ## i ##'s inside the parenthesized terms become ## m ##'s.
 
  • #11
Geofleur
Science Advisor
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Note that the ##i##'s are repeated and thus being summed over.
 
  • #12
555
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They do vanish unless ## i = m ##. In that case, the ##\delta##'s in front become ones, and the ## i ##'s inside the parenthesized terms become ## m ##'s.
ahh, I see.

so I have that [itex]\epsilon_{ijk} \epsilon_{imn} = 3 \delta_{jm} \delta_{kn} - 3 \delta_{jn} \delta_{km} - \delta_{jm} \delta_{kn} + \delta_{jn} \delta_{km} + \delta_{jn} \delta_{km} - \delta_{jm} \delta_{kn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/itex]

Thank you very much for your help!
 
  • #13
Ray Vickson
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Hi,

I am confused about how I arrive at the contracted epsilon identity. [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

1. Homework Statement


Show that [tex]\epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}[/tex]

Homework Equations




The Attempt at a Solution


[/B]

Thanks for any help you can give.
The factor ##\epsilon_{ijk}## vanishes unless ##ijk## is a permutation of ##123##, so for any pair ##j \neq k## the required ##i## is uniquely determined. Then, for that ##i##, ##\epsilon_{imn}## vanishes unless ##mn## is a permutation of ##jk##. Thus, for a nonzero term on the left, we need either ##j = m## and ##k = n## (in which case the left-hand-side is ##(\pm1 )^2 = +1##, or ##j = n## and ##k = m## (in which case the left-hand-side is ##(-1)(+1) = -1##). That is, the nonzero values of the left-hand-side are the same as the nonzero values of ##\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}##. The same is true of the zero values, so the two sides must be equal.
 

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