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I´m reading about tensors. And it´s a bit hard to understand it. If I have understood correctly, a tensor is a linear map that takes one or more vector arguments and return a real value.

For example if you have a velocity vector v. And you want to find the speed in a direction of the direction vector d (with length 1). Than you can write speed = v•d. And than we can think of v as a linear map v[] that takes in a vectorargument and returns a speed. (v[d] = speed.) And v[] will then be a first order tensor. Is this correct thinking?

But how is the exact definition for a tensor for exempel of order n.

Mathstudent
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  • Tensors can mean several things. What material are you using to read about tensors? Answers can be confusing or useless if they are in a different context than what you are studying. – Joonas Ilmavirta Apr 24 '17 at 20:29
  • In my book they use the linear algebra approach I think. I´m using the book "Continuum Mechanics and Thermodynamics, from fundamental concepts to governing equations" – Mathstudent Apr 24 '17 at 20:35
  • Thanks. I hope someone who knows that book can give an answer; I can't tell by the title what the treatment there is like. (There are several things to decide when studying tensors: Do we make a difference between spaces and their duals? Over which rings or fields do we study vector spaces or modules? Are the spaces finite dimensional? Do we study tensors of arbitrary rank or type? Do we restrict ourselves to vector spaces, or do we allow vector bundles over manifolds like in differential geometry? Your book's answers are probably no, real, yes, yes, yes.) – Joonas Ilmavirta Apr 24 '17 at 20:44
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    @JoonasIlmavirta - actually, given that this is a physics-based development, the answers to your questions are probably yes, $\Bbb R$ and $\Bbb C$, no, yes, and the "tensors" are actually tensor fields over manifolds, but always expressed in terms of coordinate patches, so the manifold structure is hidden. – Paul Sinclair Apr 24 '17 at 23:50
  • See my answer – K.defaoite Sep 06 '21 at 00:40

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The tensor algebra of a vector space is the set of all objects that can be constructed using the tensor product of that vector space and its dual space. The vectors in question are the ones from physics and its dual space is the set of row vectors that represent covectors. A covector is a linear map that takes one vector and produces a scalar. It’s components are covariant once so it’s a (0,1)-tensor. The tensor product of n covectors is a linear function of n vectors instead of one, and they’re (0,n)-tensors. For example, the metric tensor takes in two vectors and outputs their dot product. It takes in two vectors and is thus a (0,2)-tensor. Vectors can be thought of as functions that take in covectors and output scalars as V can be identified with V**. The tensor product of n vectors is a linear function of n covectors. These tensors are (n,0)- tensors. An arbitrary (m,n)-tensor can be constructed with a mixed product of vectors and covectors that takes in m covectors and n vectors.

The way to compute the results of applying a tensor to some entries is to apply the first basic element of the tensor to the first entry, the second to the second, and so on, and then perform a summation over what you have left.

The components of a tensor are the tensor being applied to the basis elements.

So a tensor with components $T^{i}_{j,k}$=$T(\epsilon ^{i},e_{j},e_{k})$ can be expressed as follows:

$T^{i}_{j,k}(e_{i}\otimes\epsilon^{j}\otimes\epsilon ^{k})(\epsilon^{a},e_{b},e_{c})$= $T^{i}_{j,k}(e_{i}(\epsilon^{a})\epsilon^{j}(e_{b}) \epsilon^{k}(e_{c}))$=$T^{i}_{j,k}(\delta_{i}^{a}\delta^{j}_{b}\delta^{k}_{c})$

Recalling the krownecker delta index rule, we have it equal to $T^{a}_{b,c}$ after a summation over the repeated indices.

Bible Bot
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