Antisymmetric and symmetric tensors I think your teacher means Frobenius product.In the context of tensor analysis (e.g. Show that [tex]\epsilon_{ijk}a_{ij} = 0[/tex] for all k if and only if [tex]a_{ij}[/tex] is symmetric. A), is widely used in mechanics, think about $\int \boldsymbol{\sigma}:\boldsymbol{\epsilon}\,\mathrm{d}\Omega$, if you know the weak form of elastostatics), it is a natural inner product for 2nd order tensors, whose coordinates can be represented in matrices. S = 0, i.e. Antisymmetric and symmetric tensors. Using 1.2.8 and 1.10.11, the norm of a second order tensor A, denoted by . If a tensor changes sign under exchange of each pair of its indices, then the tensor is completely (or totally) antisymmetric. A and B is zero, one says that the tensors are orthogonal, A :B =tr(ATB)=0, A,B orthogonal (1.10.13) 1.10.4 The Norm of a Tensor . Antisymmetric Tensor By definition, A µν = −A νµ,so A νµ = L ν αL µ βA αβ = −L ν αL µ βA βα = −L µ βL ν αA βα = −A µν (3) So, antisymmetry is also preserved under Lorentz transformations. There is also the case of an anti-symmetric tensor that is only anti-symmetric in specified pairs of indices. *The proof that the product of two tensors of rank 2, one symmetric and one antisymmetric is zero is simple. the product of a symmetric tensor times an antisym-metric one is equal to zero. Obviously if something is equivalent to negative itself, it is zero, so for any repeated index value, the element is zero. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. I see that if it is symmetric, the second relation is 0, and if antisymmetric, the first first relation is zero, so that you recover the same tensor) I agree with the symmetry described of both objects. * I have in some calculation that **My book says because** is symmetric and is antisymmetric. Thanks Evgeny, I used Tr(AB T) = Tr(A T B) Tr(A T B)=Tr(AB) and Tr(AB T)=Tr(A(-B))=-Tr(AB) So Tr(AB)=-Tr(AB), therefore Tr(AB)=0 But if it can be done along the lines I tried with indexes, I'd really like to see that - I am looking for opportunities to practice Indexing A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components …. This makes many vector identities easy to prove. Antisymmetric and symmetric tensors. There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: Similarly, just as the dot product is zero for orthogonal vectors, when the double contraction of two tensors . A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components [math]U_{ijk\dots}[/math] and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: The alternating tensor can be used to write down the vector equation z = x × y in suffix notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 −x 3y 2, as required.) 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