Antisymmetric and symmetric tensors. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: Antisymmetric and symmetric tensors. 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. However, the connection is not a tensor? S = 0, i.e. A (or . 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.) Using 1.2.8 and 1.10.11, the norm of a second order tensor A, denoted by . 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 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 A), is 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. 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 . * 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. (NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. 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 …. 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. Antisymmetric and symmetric tensors I think your teacher means Frobenius product.In the context of tensor analysis (e.g. 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. 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: 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. Similarly, just as the dot product is zero for orthogonal vectors, when the double contraction of two tensors . There is also the case of an anti-symmetric tensor that is only anti-symmetric in specified pairs of indices. I agree with the symmetry described of both objects. If a tensor changes sign under exchange of each pair of its indices, then the tensor is completely (or totally) antisymmetric. There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. 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) Obviously if something is equivalent to negative itself, it is zero, so for any repeated index value, the element is zero. the product of a symmetric tensor times an antisym-metric one is equal to zero. 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. = δ ilδ jm −δ imδ jl imδ jl a symmetric tensor an. 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