Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question from someone just starting to study tensors (sorry if it's silly):

So I understand (maybe?) that tensors are basically about coordinate transformations (and things that are invariant under said transformations), and that in writing out a tensor, we use notation that represents functions that perform the coordinate transformations.

But when I see a tensor of, say, rank 3 or 4 written out, it looks like we're jumping from one coordinate system to another to another, before we arrive from the original coordinate system to the final one. If you're only really starting at one coordinate system, and ending at another, why can't you treat each tensor as just a single coordinate transformation, i.e. a function that takes you directly from the original coordinate system to the final one?

share|improve this question
    
I'm not sure what you mean. You don't need to think about tensors in terms of changing coordinates at all. –  Qiaochu Yuan Aug 13 '12 at 2:12
1  
One place where higher order tensors occur in analysis is Stokes' theorem. Generally, high rank tensors are just general multilinear maps, so they are quite natural... –  tomasz Aug 13 '12 at 2:13
1  
(just expanding on Qiaochu Yuan's comment) $A = A_mdx^m$ or $A = \bar{A}_j d\bar{x}^j$ the one-form $A$ could either be written in the $(x^{\mu})$ or in the $({\bar{x}}^{j})$ coordinate charts. The object $A$ is itself coordinate-free. The coordinate transformations of the differentials then force particular transformations on the components of $A$ in barred or unbarred coordinate charts. It's not that $A$ is a coordinate change, rather, it's components change. $A$ in contrast is what it is no matter how you observe it. –  James S. Cook Aug 13 '12 at 2:28
    
In representation theory, new representations can be created from old ones by forming tensor powers of the original representation, and often such powers higher than the second are important. –  KCd Aug 13 '12 at 6:13

2 Answers 2

Tensors of rank 2, that change coordinates, are only one particular type of tensors. Not all tensors do that!

In particular:

  • Tensors of rank 0 are scalars;
  • Tensors of rank 1 are vectors or differential forms;
  • Tensors of rank 2 are quadratic forms, coordinate transformations, or any other linear object that can be represented by a single matrix;
  • Tensors of higher rank are linear maps of higher rank. They are not changes of coordinates!

By the way, in the physics literature, it is often said that the metric tensor is a change of coordinates (between covariant and contravariant). This is mathematically incorrect. Covariant and contravariant vectors are mathematically distinct objects!

share|improve this answer

To add to what is written above, an early and well known application of higher order tensors was in the mechanics of deformable bodies (more particularly, in the linearized theory of elasticity), due to Valdemar Voigt (circa 1898)

For instance, $\sigma$, a stress tensor and $\epsilon$, a strain tensor, are related by the the tensor equation

$$ \mathbf{\sigma} = \mathbf{C \epsilon} $$

Here $\sigma$ and $\epsilon$ are second order tensors and $C$ is a fourth-order tensor.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.