Is this tensor identity correct?

  • Thread starter neelakash
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  • #1
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Homework Statement



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]

Homework Equations



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

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

The Attempt at a Solution



let me see first if the latex output has come OK.
 
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Answers and Replies

  • #2
511
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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?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
965
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?
 
  • #4
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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.
 
  • #5
gabbagabbahey
Homework Helper
Gold Member
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7
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]

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?

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]

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?!

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

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

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:
  • #6
511
1
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.

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?!

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:
  • #8
511
1
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]?
 
  • #9
511
1
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?
 
  • #10
511
1
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]
 
  • #11
511
1
Can the moderators please fix the Latex?
 

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