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Math Help - Levi-Civita Component Count

  1. #1
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    Cool Levi-Civita Component Count

    Hi, I'm trying to show that a third order system that is skew-symmetric in all of its indices has 21 components equal to zero. I know this is the Levi-Civita pseudotensor. This is what I have so far:

    To find values equal to 0 in the skew-symmetric system, Aijk, the same component must be found to have two equivalent index notations differing in sign. This occurs when at least two of the indices are equal, that is the same component is described when the indices are switched and the sign of the component also changes. There are 3c2 = 3 ways to choose indices to equate. For each of the indices there are 3 choices, thus there are 3x3=9 components described in a third order system when two indices are equal. Multiplying by the number of ways to equate indices we have 3x9=27 components. However, there has been overlap in our counting since each equated index notation will describe the component at Aijk. This occurs three times in each of our counts of components with equated indices, ie. for i=j=k=1,2,3. Since we counted components three different ways, these particular components were counted three times instead of once so we must subtract 2x3=6 from our total count of 27 to get 21.

    Is this right? Is there a short way to say the same thing? Please help me learn to write better proofs.

    Thanks,
    Ultros
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ultros88 View Post
    Hi, I'm trying to show that a third order system that is skew-symmetric in all of its indices has 21 components equal to zero. I know this is the Levi-Civita pseudotensor. This is what I have so far:

    To find values equal to 0 in the skew-symmetric system, Aijk, the same component must be found to have two equivalent index notations differing in sign. This occurs when at least two of the indices are equal, that is the same component is described when the indices are switched and the sign of the component also changes. There are 3c2 = 3 ways to choose indices to equate. For each of the indices there are 3 choices, thus there are 3x3=9 components described in a third order system when two indices are equal. Multiplying by the number of ways to equate indices we have 3x9=27 components. However, there has been overlap in our counting since each equated index notation will describe the component at Aijk. This occurs three times in each of our counts of components with equated indices, ie. for i=j=k=1,2,3. Since we counted components three different ways, these particular components were counted three times instead of once so we must subtract 2x3=6 from our total count of 27 to get 21.

    Is this right? Is there a short way to say the same thing? Please help me learn to write better proofs.

    Thanks,
    Ultros
    That is right, there are 9 ways that the first two indices may be equal, and 9 ways the second two may be equal, and nime ways that the first and last amy be equal, but we have counted the three cases with all the indices equal three times, so the total number of ways is 3*9-2*3=21.

    RonL
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