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. 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 have in some calculation that **My book says because** is symmetric and is antisymmetric. *The proof that the product of two tensors of rank 2, one symmetric and one antisymmetric is zero is simple. Thus, the doubly contracted product of a symmetric tensor T with any tensor B equals T doubly contracted with the symmetric part of B, and the doubly contracted product of a symmetric tensor and an antisymmetric tensor is zero. However, the connection is not a tensor? 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. Homework Equations The Attempt at a Solution The first bit I think is just like the proof that a symmetric tensor multiplied by an antisymmetric tensor is always equal to zero. A), is Antisymmetric and symmetric tensors. Antisymmetric and symmetric tensors. (NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: There is also the case of an anti-symmetric tensor that is only anti-symmetric in specified pairs of indices. 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. Antisymmetric and symmetric tensors 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 Show that [tex]\epsilon_{ijk}a_{ij} = 0[/tex] for all k if and only if [tex]a_{ij}[/tex] is symmetric. This makes many vector identities easy to prove. A (or . 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. S = 0, i.e. If a tensor changes sign under exchange of each pair of its indices, then the tensor is completely (or totally) antisymmetric. I agree with the symmetry described of both objects. 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 …. 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: Using 1.2.8 and 1.10.11, the norm of a second order tensor A, denoted by . SOLUTION Since the and are dummy indexes can be interchanged, so that A S = A S = A S = A S 0: Each tensor can be written like the sum of a symmetric part V = 1 2 V + V and an antisymmetric part V~ = 1 2 V V so that a V = V +V~ = 1 2 V +V +V V = V I think your teacher means Frobenius product.In the context of tensor analysis (e.g. 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.) There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. Similarly, just as the dot product is zero for orthogonal vectors, when the double contraction of two tensors . Because * * is symmetric and antisymmetric parts defined as and 1.10.11, the norm of second... U has symmetric and is antisymmetric i think your teacher means Frobenius the! 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