Vectors and Tensors R. Shankar Subramanian - Noppa

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that is, b× a points opposite to a× b. We can write a × b = e a b (1) 1 1 e a b (2) 2 2 e a b (3) 3 3 There is a compact representation of a determinant that helps us write a× b = ε ab ijk i j (Note that k is a free index. The actual symbol chosen for it is not important; what matters is that the right side has one free index, making it a vector) ε ijk is called the permutation symbol ε = 0 if any two of the indices are the same ijk =+ 1if i, j, k form an even permutation of 1, 2, 3 [example: 1,2,3] =−1if i, j, k We can assign a geometric interpretation to a and b is θ , then a⋅ b = a b cos θ form an odd permutation of 1, 2, 3 [example: 2, 1, 3] a⋅ b and a× b. If the angle between the two vectors and the length of a× b is ab sinθ . You may also recognize absinθ as the area of the parallelogram formed by a and b as two adjacent sides. Given this, it is straightforward to see that a⋅ b× c = ε ab c ijk i j k is the volume of the parallelepiped with sides ab , ,andc Second Order Tensors . This is called the triple scalar product. Note that we did not define vector division. The closest we come is in the definition of second-order tensors! 4
Imagine Instead, we write a b =Τ a =Τ⋅ b A tensor (unless explicitly stated otherwise we’ll only be talking about “second-order” and shall therefore omit saying it every time) “operates” on a vector to yield another vector. It is very useful to think of tensors as operators as you’ll see later. Note the “dot” product above. Using ideas from vectors, we can see how the above equation may be written in index notation. a i =Τijb j It is important to note that b ⋅Τ would be biΤ ij and would be different from Τ⋅ b in general. The two underbars in Τ now take on a clear significance; we are referring to a doubly subscripted entity. We can think of a tensor as a sum of components in the same way as a vector. For this, we use the following result. e ⋅Τ⋅ e =Τ Scalar () i ( j) ij We’re not using index notation here Thus, to get Τ 23 we would find e( ) ⋅Τ⋅e 2 ( 3 ). We can then think of T as a sum. Τ=Τ e e +Τ e e +Τ e e () 1 () 1 () 1 ( 2) () 1 ( 3) 11 12 13 +Τ e e +Τ e e +Τ e e ( 2) () 1 ( 2) ( 2) ( 2) ( 3) 21 22 23 +Τ e e +Τ e e +Τ e e ( 3) () 1 ( 3) ( 2) ( 3) ( 3) 31 32 33 What are the quantities e () e 1 ( 2) and others like them? They are called dyads. They are a basis set for representing tensors. Each is a tensor that only has one component in this basis set. Note that e e ≠ e e . () i ( j) ( j) ( i) You can see that tensors and matrices have a lot in common! 5
Page 1 and 2: 2005 Vectors and Tensors R. Shankar
Page 3: Vector Operations The entity v has
Page 7 and 8: Let A or A ij be the tensor. ↓ ab
Page 9 and 10: and has three roots λ1, λ2, and
Page 11 and 12: Divergence Theorem If the volume V
Page 13: integral on the left side becomes t

Vectors and Tensors R. Shankar Subramanian - Noppa

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