Is This Tensor Identity Valid?

[neelakash]

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:

\epsilon_{ijk}\*\epsilon_{lmn} =<br /> <br /> \left(\begin{array}{ccc}\ g_\ {11}&amp;\ g_\ {21}&amp;\ g_\ {31}\\ g_\ {12}&amp;\ g_\ {22}&amp;\ g_\ {32}\\ g_\ {13}&amp;\ g_\ {23}&amp;\ g_\ {33}\end{array}\right)

Homework Equations

definition of \epsilon_\ {ijk}

definition of the "metric" tensor \ g_{ij}

The Attempt at a Solution

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

 

[neelakash]

I am sorry.Last time there was a problem and the Latex output failed.


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

Now,I was wondering if the identity

\epsilon_{ijk}\epsilon_{lmn}=<br /> \begin{vmatrix}<br /> \ g_{11}&amp;\ g_{21}&amp;\ g_{31}\\<br /> g_{12}&amp;\ g_{22}&amp;\ g_{32}\\<br /> g_{13}&amp;\ g_{23}&amp;\ g_{33}<br /> \end{vmatrix}

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?

 

[HallsofIvy]

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 \epsilon_{ijk}. Okay, what is that definition?

 

[neelakash]

We know \ g_{ij}=\ e_i\cdot\ e_j

Now, \ V^2=\ V\ V=\epsilon_{ijk}\epsilon_{lmn}

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

then after some vector manipulation we would get \ g_{ij}=\ e_i\cdot\ e_j,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.

 

[gabbagabbahey]

I am sorry.Last time there was a problem and the Latex output failed.I am to prove \ det[g_{ij}]=\ V^2 where \[g_{ij}] is metric tensor and \ V=\epsilon_i\cdot\epsilon_j\times\epsilon_k

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

Now,I was wondering if the identity

\epsilon_{ijk}\epsilon_{lmn}=<br /> \begin{vmatrix}<br /> \ g_{11}&amp;\ g_{21}&amp;\ g_{31}\\<br /> g_{12}&amp;\ g_{22}&amp;\ g_{32}\\<br /> g_{13}&amp;\ g_{23}&amp;\ g_{33}<br /> \end{vmatrix}

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 \epsilon_{ijk} makes it a tensor and not a tensor density----there are at least two different definitions for \epsilon_{ijk}, 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 \ g_{ij}=\ e_i\cdot\ e_j

Now, \ V^2=\ V\ V=\epsilon_{ijk}\epsilon_{lmn}

No, assuming V is a vector (again, your definition is not at all clear to me), then V^2=V_{a}V^{a} is a scalar, why do you think it is equal to \epsilon_{ijk}\epsilon_{lmn}?

 

[neelakash]

Let me tell you step by step:

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

Regarding definiing dot and cross products between covariant vectors:

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

Now,we also know:

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

Hence, for orthonormal system, V=1 as \epsilon_{ijk}=\ 1 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 \epsilon_{ijk} makes it a tensor and not a tensor density----there are at least two different definitions for \epsilon_{ijk}, 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 \epsilon_{ijk}=\pm\ V etc. for even or odd permutations of 1,2,3 is correct---\epsilon_{ijk} is also a rank three tensor...Is not it?

 

[robphy]

http://en.wikipedia.org/wiki/Levi-Civita_symbol may be helpful

 

[neelakash]

OK...I got it...\epsilon_{ijk}=\ +\ V means ijk-th component of epsilon antisymmatric matrix is equal to V.

So,can we write ...\epsilon_{ijk}\epsilon_{lmn}=\ V\ V=\ V^2 to mean that as we multiply ijk-th element of \epsilon tensor and lmn-th element of \epsilon tensor, we get the squared volume of the parallelopiped spanned by the vectors \ e_i,\ e_j,\ e_k?

 

[neelakash]

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.\epsilon_{ijk}\epsilon^{lmn} and not \epsilon_{ijk}\epsilon_{lmn} 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 \epsilon_{ijk}\epsilon_{lmn} is equal to det[g_ij] as

\ g_{ij}=\ e_i\cdot\ e_j and

\ g_{ij}=\ e_i\cdot\ e^j=\delta_i^j

Could I convey my thought?

 

[neelakash]

Let \ e_i,\ e_j,\ e_k form a basis,not necessarily orthogonal.Then,
\ e_i\cdot(\ e_j\times\ e_k) is the volume |V| spanned by the bases.
In the deleted post I intended to prove
\ e_i\cdot(\ e_j\times\ e_k)=\epsilon_{ijk}

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

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

etc.

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

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

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

I was failing to evaluate the following determinant:

\begin{vmatrix}\ i_1&amp;\ i_2&amp;\ i_3\\<br /> j_1&amp;\ j_2&amp;\ j_3\\<br /> k_1&amp;\ k_2&amp;\ k_3<br /> \end{vmatrix}<br /> <br /> \begin{vmatrix}\ l_1&amp;\ l_2&amp;\ l_3\\<br /> m_1&amp;\ m_2&amp;\ m_3\\<br /> n_1&amp;\ n_2&amp;\ n_3<br /> \end{vmatrix}<br /> <br /> = \begin{vmatrix}\ i_1\ l_1\ +\ i_2\ l_2\ + \ i_3\ l_3 &amp;\ i_1\ m_1\ + \ i_2\ m_2\ + \ i_3\ m_3 &amp; \ i_1\ n_1\ +\ i_2\ n_2\ + \ i_3\ n_3\\<br /> <br /> \ j_1\ l_1\ +\ j_2\ l_2\ + \ j_3\ l_3 &amp;\ j_1\ m_1\ + \ j_2\ m_2\ + \ j_3\ m_3 &amp;\ j_1\ n_1\ +\ j_2\ n_2\ + \ j_3\ n_3\\<br /> <br /> \ k_1\ l_1\ +\ k_2\ l_2\ + \ k_3\ l_3 &amp;\ k_1\ m_1\ + \ k_2\ m_2\ + \ k_3\ m_3 &amp; \ k_1\ n_1\ +\ k_2\ n_2\ + \ k_3\ n_3<br /> <br /> \end{vmatrix}<br /> <br /> =\begin{vmatrix}<br /> \ i\cdot\ l &amp;\ i\cdot\ m &amp; \ i\cdot\ n\\<br /> \ j\cdot\ l &amp;\ j\cdot\ m &amp; \ j\cdot\ n\\<br /> \ k\cdot\ l &amp;\ k\cdot\ m &amp; \ k\cdot\ n<br /> \end{vmatrix}<br /> <br /> =\begin{vmatrix}<br /> \ e_i\cdot\ e_l &amp;\ e_i\cdot\ e_m &amp; \ e_i\cdot\ e_n\\<br /> \ e_j\cdot\ e_l &amp;\ e_j\cdot\ e_m &amp; \ e_j\cdot\ e_n\\<br /> \ e_k\cdot\ e_l &amp;\ e_k\cdot\ e_m &amp; \ e_k\cdot\ e_n<br /> \end{vmatrix}<br /> <br /> =\begin{vmatrix}<br /> \ g_{il} &amp;\ g_{im} &amp; \ g_{in}\\<br /> \ g_{jl} &amp;\ g_{jm} &amp; \ g_{jn}\\<br /> \ g_{kl} &amp;\ g_{km} &amp; \ g_{kn}<br /> \end{vmatrix}<br /> <br /> =det[g]<br /> <br />

 

[neelakash]

Can the moderators please fix the Latex?