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Levi Civita (Permutation) Symbol Proof

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the following:
    [itex]\varepsilon_{ijk}=
    \left| \begin{array}{ccc}
    \delta_{1i} & \delta_{1j} & \delta_{1k} \\
    \delta_{2i} & \delta_{2j} & \delta_{2k} \\
    \delta_{3i} & \delta_{3j} & \delta_{3k}
    \end{array} \right|[/itex]

    2. Relevant equations
    From my textbook:
    [itex]\hat{e}_3 = \hat{e}_1 \times \hat{e}_2, \quad \hat{e}_1 = \hat{e}_2 \times \hat{e}_3, \quad \ldots \quad \varepsilon_{ijk} \hat{e}_k = \hat{e}_i \times \hat{e}_j \\
    \delta_{ij} = \hat{e}_i \cdot \hat{e}_j [/itex]

    From a website:
    [itex] \varepsilon_{ijk} = (\hat{e}_i \times \hat{e}_j)\cdot\hat{e}_k [/itex]

    3. The attempt at a solution
    I don't even know where to start. My textbook says I should be able to prove the determinant proof using those two relations they provide; however, I have not been able to prove anything.

    It seems as though every continuum mechanics book I've ever seen likes to say "it's easy to show the determinant proof." Apparently it's so easy that no book feels the need to show the derivation. Am I missing any relations? Can someone give me hints or "suggestions" to get me going in the right direction?

    Thanks.
     
    Last edited: Oct 4, 2012
  2. jcsd
  3. Oct 3, 2012 #2

    jedishrfu

    Staff: Mentor

    start with the first formula and sub the second one in

    eijk = (ei x ej) . (ei x ej) dij

    next notice that (ei x ej) is dotted with itself meaning ???
     
  4. Oct 3, 2012 #3
    jedishrfu,

    Thank you very much for your quick response. My wife and I have been looking at your post and can't seem to understand how you got there. Do I need the third equation? I found it on a website but my textbook informs me that I only need the two other equations... When I rearrange those, I seem to get something different than that listed on the website...

    [itex]\varepsilon_{ijk} \hat{e}_k = \hat{e}_i \times \hat{e}_j [/itex]
    [itex]\varepsilon_{ijk} \hat{e}_k \cdot \hat{e}_k = (\hat{e}_i \times \hat{e}_j) \cdot \hat{e}_k [/itex]
    [itex]\varepsilon_{ijk} \delta{kk} = (\hat{e}_i \times \hat{e}_j) \cdot \hat{e}_k [/itex]
    [itex]3\varepsilon_{ijk} = (\hat{e}_i \times \hat{e}_j) \cdot \hat{e}_k [/itex]

    which is not equal to what the website says...

    The only thing that we've gotten when trying to use your hint is the following:

    [itex]\delta_{ij} = \hat{e}_i \cdot \hat{e}_j [/itex]
    [itex]\phantom{\delta_{ij}} = \left( \frac{\hat{e}_j \times \hat{e}_k}{\varepsilon_{ijk}} \right) \cdot \left( \frac{\hat{e}_k \times \hat{e}_i}{\varepsilon_{ijk}} \right) [/itex]

    which can be further manipulated, but it doesn't seem to give anything?

    Furthermore, even if we could get it to the state you mention, I am still slightly baffled. I understand that with something dotted with itself you have the norm squared, but I can't figure out if there is another relation or how to use that one.

    Sorry, I'm sure this is straight forward for you but I've always had a hard time with this damn permutation symbol. I'm used it for countless other identity proofs, but I've never been able to prove the determinant identity of itself.

    Thanks Again.
     
  5. Oct 3, 2012 #4

    gabbagabbahey

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    Gold Member

    Do you mean [itex]\varepsilon_{ijk}= \begin{vmatrix} \delta_{1i} & \delta_{1j} & \delta_{1k} \\ \delta_{2i} & \delta_{2j} & \delta_{2k} \\ \delta_{3i} & \delta_{3j} & \delta_{3k} \end{vmatrix}[/itex]?

    I have no idea what jedishrfu did (or was trying to do) there either, so don't feel too bad.

    Whenever you are doing index gymnastics, at each step, you should check 2 things:

    (1) Does any index occur more than twice in a single term?
    (2) Does each term have the same free indices?

    When taking the dot product of [itex]\varepsilon_{ijk} \hat{e}_k[/itex] (which has an implied summation over [itex]k[/itex]) with [itex]\hat{e}_k[/itex], you need to use a different dummy index for the summation:

    [tex] (\hat{e}_i \times \hat{e}_j) \cdot \hat{e}_k = \varepsilon_{ijm} \hat{e}_m \cdot \hat{e}_k = \varepsilon_{ijm}\delta_{km} = \varepsilon_{ijk}[/tex]
     
  6. Oct 4, 2012 #5

    jedishrfu

    Staff: Mentor

    sorry for he confusion but in your first post I saw a formula relating ek to eixej delta ij and didn't realize the delta ij was part of the 3rd equation. now though it shows it on the 3rd line. did you edit the post?
     
  7. Oct 4, 2012 #6
    Yes, that is absolutely right. It is the determinant of the matrix, not just the matrix itself.

    That is a very clear explanation. I sometimes forget about the "occurance rule." I had the revelation while sleeping that what I did was wrong (mainly, that I can't sum one term ([itex]\delta_{kk}[/itex]) because it was multiplied by [itex]\varepsilon_{ijk}[/itex]). My plan was then to use the kronecker delta as an index switcher [itex]k\rightarrow{k}[/itex], which would leave me with that final formula (although my methodology was incorrect).

    Sorry about that. I did edit the post because I think I had typed an error in one of the supplied equations. You just replied too fast (you probably never hear that!).

    Thanks guys, I'll keep trying.
     
    Last edited: Oct 4, 2012
  8. Oct 4, 2012 #7

    gabbagabbahey

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    If you are allowed to use the well-known fact that the determinant of a 3 x 3 matrix can be written as [itex]\det(\mathbf{A}) = \varepsilon_{lmn} A_{1l} A_{2m} A_{3n}[/itex] (see here and the reference cited therein), then it should be fairly easy to show.

    If not, you will probably just have to expand the determinant using whatever methods you are allowed to use and compare the result to [itex]\varepsilon_{ijk}[/itex].
     
  9. Oct 7, 2012 #8
    Thanks gabbagabbahey. We are allowed the use the [itex]\det(\mathbf{A}) = \varepsilon_{lmn} A_{1l} A_{2m} A_{3n}[/itex] (it was another proof I did).

    The problem I am running into is starting the problem. The way the proofs are expected to be is is working from the left side until it equals the right. This method eliminates the proof by expanding the determinant. I have been trying for days now but can't seem to solve it using the two relations the book lays out (and explicitly says "can be easily proved using these relations"). Does anyone know how to start the proof with the equations I defined earlier in the thread?

    Thanks Again.
     
  10. Oct 9, 2012 #9

    gabbagabbahey

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    Are you sure about this? What is the difference between proving a=b and proving b=a?

    Just use the matrix [itex]\mathbf{A} = \begin{pmatrix} \delta_{1i} & \delta_{1j} & \delta_{1k} \\ \delta_{2i} & \delta_{2j} & \delta_{2k} \\ \delta_{3i} & \delta_{3j} & \delta_{3k} \end{pmatrix} [/itex] and calculate the determinant using the above formula. What is [itex]A_{1l}[/itex]? What is [itex]A_{2m}[/itex]?...
     
  11. Oct 10, 2012 #10
    [itex]\mathrm{det}\begin{pmatrix} \delta_{1i} & \delta_{1j} & \delta_{1k} \\ \delta_{2i} & \delta_{2j} & \delta_{2k} \\ \delta_{3i} & \delta_{3j} & \delta_{3k} \end{pmatrix} = \varepsilon_{ijk} [/itex]
    [itex]\phantom{\mathbf{A} = \begin{pmatrix} \delta_{1i} & \delta_{1j} & \delta_{1k} \\ \delta_{2i} & \delta_{2j} & \delta_{2k} \\ \delta_{3i} & \delta_{3j} & \delta_{3k} \end{pmatrix}} = \mathrm{det}(\delta) [/itex]
    [itex]\phantom{\mathbf{A} = \begin{pmatrix} \delta_{1i} & \delta_{1j} & \delta_{1k} \\ \delta_{2i} & \delta_{2j} & \delta_{2k} \\ \delta_{3i} & \delta_{3j} & \delta_{3k} \end{pmatrix}} = \varepsilon_{lmn}\delta_{il}\delta_{jm}\delta_{kn} [/itex]
    [itex]\phantom{\mathbf{A} = \begin{pmatrix} \delta_{1i} & \delta_{1j} & \delta_{1k} \\ \delta_{2i} & \delta_{2j} & \delta_{2k} \\ \delta_{3i} & \delta_{3j} & \delta_{3k} \end{pmatrix}} = \varepsilon_{ijk} [/itex]

    It's much easier when I stop trying to solve for the other side. I could have done this from the beginning but I was really trying to manipulate the LHS. Thanks for your help.
     
  12. Oct 14, 2012 #11

    gabbagabbahey

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    This isn't quite correct. [itex]A_{1l}[/itex] represents the [itex]l[/itex]th component along the 1st row of the matrix [itex]\mathbf{A}[/itex]. Using [itex]\mathbf{A}=\begin{pmatrix} \delta_{1i} & \delta_{1j} & \delta_{1k} \\ \delta_{2i} & \delta_{2j} & \delta_{2k} \\ \delta_{3i} & \delta_{3j} & \delta_{3k} \end{pmatrix}[/itex], you do not have [itex]A_{1l} = \delta_{il}[/itex]. You do however know that [itex]A_{l1}=\delta_{li}[/itex] (the [itex]l[/itex]th component along the 1st column), so you want to be sure to expand the determinant along the columns of your matrix instead of the rows. Of course [itex]\delta_{il}=\delta_{li}[/itex], so technically there is nothing incorrect in your equations, but your reasoning is not clear. When you expand the determinant along each column you have [itex]\det(\mathbf{A}) = \varepsilon_{lmn} A_{l1} A_{m2} A_{n3}[/itex].
     
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