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Is this tensor identity correct?

  1. Jul 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I do not know if the following is correct;if it is,I will be able to save some calculation while doing a problem.Can you please let me know if it is true:

    [tex]\epsilon_{ijk}\*\epsilon_{lmn} =

    \left(\begin{array}{ccc}\ g_\ {11}&\ g_\ {21}&\ g_\ {31}\\ g_\ {12}&\ g_\ {22}&\ g_\ {32}\\ g_\ {13}&\ g_\ {23}&\ g_\ {33}\end{array}\right)[/tex]

    2. Relevant equations

    definition of [tex]\epsilon_\ {ijk}[/tex]

    definition of the "metric" tensor [tex]\ g_{ij}[/tex]

    3. The attempt at a solution

    let me see first if the latex output has come OK.
     
    Last edited by a moderator: Jul 11, 2009
  2. jcsd
  3. Jul 11, 2009 #2
    I am sorry.Last time there was a problem and the Latex output failed.


    I am to prove [tex]\ det[g_{ij}]=\ V^2[/tex] where [tex]\[g_{ij}][/tex] is metric tensor and [tex]\ V=\epsilon_i\cdot\epsilon_j\times\epsilon_k[/tex]

    Now,I was wondering if the identity

    [tex]\epsilon_{ijk}\epsilon_{lmn}=
    \begin{vmatrix}
    \ g_{11}&\ g_{21}&\ g_{31}\\
    g_{12}&\ g_{22}&\ g_{32}\\
    g_{13}&\ g_{23}&\ g_{33}
    \end{vmatrix}[/tex]

    is correct---in which case the problem is solved easily.Note that the antisymmetric tensors are both co-variant.

    Surely,the best way to know whether it is correct or not is to prove it...but somehow I am not getting it.If it is correct can anyone give me some hint?
     
  4. Jul 11, 2009 #3

    HallsofIvy

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    I fixed your Latex. You had {cc} where you wanted {ccc} for three columns. Also you had "\{" where you only wanted "{".

    You give as a "relevant equation" "the definition of [itex]\epsilon_{ijk}[/itex]. Okay, what is that definition?
     
  5. Jul 11, 2009 #4
    We know [tex]\ g_{ij}=\ e_i\cdot\ e_j[/tex]

    Now, [tex]\ V^2=\ V\ V=\epsilon_{ijk}\epsilon_{lmn}[/tex]

    My idea was if we use the defiiniton of [tex]\epsilon_{ijk}=\ e_i\cdot\ e_j\times\ e_k[/tex]

    then after some vector manipulation we would get [tex]\ g_{ij}=\ e_i\cdot\ e_j[/tex],but that did not help much...

    I believe the equation is correct for it reduces to a known result when one of the epsilons are covariant and the other is contravariant.
     
  6. Jul 11, 2009 #5

    gabbagabbahey

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    Your definition of [itex]V[/itex] makes very little sense to me. Are [itex]\epsilon_i[/itex] orthogonal unit vectors sush that [itex]\epsilon_i\epsilon^j=\delta_{ij}[/itex]? How are you defining the dot and cross products between covariant vectors?

    I can tell you just from glancing at your proposed identity that there is no way it is correct. On the LHS of your "identity" you have a 6th rank covariant tensor (Assuming the definition you are using for [itex]\epsilon_{ijk}[/itex] makes it a tensor and not a tensor density----there are at least two different definitions for [itex]\epsilon_{ijk}[/itex], so you need to tell us which one your course/textbook uses!) while on the RHS you have the determinant of a matrix; which is a scalar....how can a 6th rank tensor possibly be equal to a scalar?!

    No, assuming [itex]V[/itex] is a vector (again, your definition is not at all clear to me), then [itex]V^2=V_{a}V^{a}[/itex] is a scalar, why do you think it is equal to [itex]\epsilon_{ijk}\epsilon_{lmn}[/itex]???
     
    Last edited: Jul 11, 2009
  7. Jul 11, 2009 #6
    Let me tell you step by step:

    A.Neither [tex]\ V[/tex] is a vector and [tex]\epsilon_i[/tex] are orthogonal unit vectors. My proposition was V is the volume of the parallelopiped whose sides are three three linearly independent vectors: [tex]\ e_i,\ e_j,\ e_k[/tex].In the first post there was a typo and I meant the unit vectors are [tex]\ e_i[/tex] and not [tex]\epsilon_i[/tex].And I do not propose [itex]\epsilon_i\epsilon^j=\delta_{ij}[/itex]

    Regarding definiing dot and cross products between covariant vectors:

    With this do you agree: [tex]\epsilon_{ijk}=(\ e_j\times\ e_k)\cdot\ e_i[/tex]?

    Now,we also know:

    [tex]\epsilon_{ijk}
    =\ +\ V ,[\ i,\ j,\ k \ is\ an\ even\ permutation\ of (\ 1,\ 2,\ 3)]\\
    =\ -\ V ,[\ i,\ j,\ k \ is\ an\ odd\ permutation\ of (\ 1,\ 2,\ 3)]\\
    =0 ,[\ two\ or\ more\ indices\ are\ equal].[/tex]

    Hence, for orthonormal system, V=1 as [tex]\epsilon_{ijk}=\ 1[/tex] for even permutation.

    Yes,I can clearly see it's wrrong...I am new to tensors.

    However, then why the definition that [itex]\epsilon_{ijk}=\pm\ V[/itex] etc. for even or odd permutations of 1,2,3 is correct---[itex]\epsilon_{ijk}[/itex] is also a rank three tensor...Is not it?
     
    Last edited: Jul 11, 2009
  8. Jul 11, 2009 #7

    robphy

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  9. Jul 11, 2009 #8
    OK...I got it...[tex]\epsilon_{ijk}=\ +\ V[/tex] means ijk-th component of epsilon antisymmatric matrix is equal to V.

    So,can we write ...[tex]\epsilon_{ijk}\epsilon_{lmn}=\ V\ V=\ V^2[/tex] to mean that as we multiply ijk-th element of [tex]\epsilon[/tex] tensor and lmn-th element of [tex]\epsilon[/tex] tensor, we get the squared volume of the parallelopiped spanned by the vectors [tex]\ e_i,\ e_j,\ e_k[/tex]?
     
  10. Jul 11, 2009 #9
    Let us refer to the link provided by robphy.Refer to the section:1.1 and 1.2---relation to Kronecker Delta and generalisation to n dimensions.

    I think there is a problem with covariant and contravariant scheme written out there.[tex]\epsilon_{ijk}\epsilon^{lmn}[/tex] and not [tex]\epsilon_{ijk}\epsilon_{lmn}[/tex] is equal to the det [Kronecker delta] as long as we are not using orthonormal basis.by [Kronecker delta] I mean the matrix written there.

    I thought [tex]\epsilon_{ijk}\epsilon_{lmn}[/tex] is equal to det[g_ij] as

    [tex]\ g_{ij}=\ e_i\cdot\ e_j[/tex] and

    [tex]\ g_{ij}=\ e_i\cdot\ e^j=\delta_i^j[/tex]

    Could I convey my thought?
     
  11. Jul 15, 2009 #10
    Let [tex]\ e_i,\ e_j,\ e_k[/tex] form a basis,not necessarily orthogonal.Then,
    [tex]\ e_i\cdot(\ e_j\times\ e_k)[/tex] is the volume |V| spanned by the bases.
    In the deleted post I intended to prove
    [tex]\ e_i\cdot(\ e_j\times\ e_k)=\epsilon_{ijk}[/tex]

    Thus,it follows that for a non-orthogonal basis,
    [tex]\ e_i\cdot(\ e_j\times\ e_k)=\epsilon_{ijk}=+V[/tex]
    So,for even permutation of ijk,ijk-th element of [tex]\epsilon[/tex] tensor=+V

    [tex]\ e_i\cdot(\ e_j\times\ e_k)=\epsilon_{ijk}=-V[/tex]
    So,for odd permutation of ijk,ijk-th element of [tex]\epsilon[/tex] tensor=-V

    etc.

    However,if the basis is orthonormal,we would have |V|=1.

    Thus,[tex]\ V^2=\epsilon_{ijk}\epsilon_{lmn}=[\ e_i\cdot(\ e_j\times\ e_k)][\ e_l\cdot(\ e_m\times\ e_n)][/tex] for even permutation of {i,j,k} and {l,m,n} where {i,j,k},{l,m,n}={1,2,3}

    Let [tex]\ e_i=i,\ e_j=j,\ e_k=k,\ e_l=l,\ e_m=m,\ e_n=n[/tex]

    I was failing to evaluate the following determinant:

    [tex]\begin{vmatrix}\ i_1&\ i_2&\ i_3\\
    j_1&\ j_2&\ j_3\\
    k_1&\ k_2&\ k_3
    \end{vmatrix}

    \begin{vmatrix}\ l_1&\ l_2&\ l_3\\
    m_1&\ m_2&\ m_3\\
    n_1&\ n_2&\ n_3
    \end{vmatrix}

    = \begin{vmatrix}\ i_1\ l_1\ +\ i_2\ l_2\ + \ i_3\ l_3 &\ i_1\ m_1\ + \ i_2\ m_2\ + \ i_3\ m_3 & \ i_1\ n_1\ +\ i_2\ n_2\ + \ i_3\ n_3\\

    \ j_1\ l_1\ +\ j_2\ l_2\ + \ j_3\ l_3 &\ j_1\ m_1\ + \ j_2\ m_2\ + \ j_3\ m_3 &\ j_1\ n_1\ +\ j_2\ n_2\ + \ j_3\ n_3\\

    \ k_1\ l_1\ +\ k_2\ l_2\ + \ k_3\ l_3 &\ k_1\ m_1\ + \ k_2\ m_2\ + \ k_3\ m_3 & \ k_1\ n_1\ +\ k_2\ n_2\ + \ k_3\ n_3

    \end{vmatrix}

    =\begin{vmatrix}
    \ i\cdot\ l &\ i\cdot\ m & \ i\cdot\ n\\
    \ j\cdot\ l &\ j\cdot\ m & \ j\cdot\ n\\
    \ k\cdot\ l &\ k\cdot\ m & \ k\cdot\ n
    \end{vmatrix}

    =\begin{vmatrix}
    \ e_i\cdot\ e_l &\ e_i\cdot\ e_m & \ e_i\cdot\ e_n\\
    \ e_j\cdot\ e_l &\ e_j\cdot\ e_m & \ e_j\cdot\ e_n\\
    \ e_k\cdot\ e_l &\ e_k\cdot\ e_m & \ e_k\cdot\ e_n
    \end{vmatrix}

    =\begin{vmatrix}
    \ g_{il} &\ g_{im} & \ g_{in}\\
    \ g_{jl} &\ g_{jm} & \ g_{jn}\\
    \ g_{kl} &\ g_{km} & \ g_{kn}
    \end{vmatrix}

    =det[g]

    [/tex]
     
  12. Jul 15, 2009 #11
    Can the moderators please fix the Latex?
     
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