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. 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 *The proof that the product of two tensors of rank 2, one symmetric and one antisymmetric is zero is simple. If a tensor changes sign under exchange of each pair of its indices, then the tensor is completely (or totally) antisymmetric. A (or . S = 0, i.e. Similarly, just as the dot product is zero for orthogonal vectors, when the double contraction of two tensors . 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. Show that [tex]\epsilon_{ijk}a_{ij} = 0[/tex] for all k if and only if [tex]a_{ij}[/tex] is symmetric. 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. 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. 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 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.) Antisymmetric and symmetric 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 …. (NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. Antisymmetric and symmetric tensors 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 . This makes many vector identities easy to prove. 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. A), is I think your teacher means Frobenius product.In the context of tensor analysis (e.g. I agree with the symmetry described of both objects. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: However, the connection is not a tensor? 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