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: 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 . Obviously if something is equivalent to negative itself, it is zero, so for any repeated index value, the element is zero. Antisymmetric and symmetric tensors. However, the connection is not a tensor? 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 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. A), is 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.) There is also the case of an anti-symmetric tensor that is only anti-symmetric in specified pairs of indices. 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. *The proof that the product of two tensors of rank 2, one symmetric and one antisymmetric is zero is simple. This makes many vector identities easy to prove. (NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. the product of a symmetric tensor times an antisym-metric one is equal to zero. * I have in some calculation that **My book says because** is symmetric and is antisymmetric. 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. I agree with the symmetry described of both objects. 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 There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. 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. Show that [tex]\epsilon_{ijk}a_{ij} = 0[/tex] for all k if and only if [tex]a_{ij}[/tex] is symmetric. Antisymmetric and symmetric tensors 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) Using 1.2.8 and 1.10.11, the norm of a second order tensor A, denoted by . and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: 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 …. If a tensor changes sign under exchange of each pair of its indices, then the tensor is completely (or totally) antisymmetric. S = 0, i.e. 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. Antisymmetric and symmetric tensors. 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