1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    \ g_{11}&\ g_{21}&\ g_{31}\\
    g_{12}&\ g_{22}&\ g_{32}\\
    g_{13}&\ g_{23}&\ g_{33}

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Homework Helper
    Gold Member

    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:

    =\ +\ 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  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


    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

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

    = \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


    \ 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

    \ 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

    \ g_{il} &\ g_{im} & \ g_{in}\\
    \ g_{jl} &\ g_{jm} & \ g_{jn}\\
    \ g_{kl} &\ g_{km} & \ g_{kn}


  12. Jul 15, 2009 #11
    Can the moderators please fix the Latex?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook