# Tensors of order 3

I'm wondering what a tensor of order 3 looks like, and what it's purposes are. I've seen them written down before, but they look like matrices; I'm probably not understanding the concept well. How is it different from a matrix?

[Revised question, based on the comments below.]

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"It looks flat"... do you mean something like a matrix, but with rows, columns, and a third dimension, representing a three dimensional array of coefficients? If so, the name you are looking for is tensor of order 3 (row- and column-vectors are tensors of order 1, and matrices are tensors of order 2). – Niel de Beaudrap Aug 29 '11 at 2:00
Yes. Tensor of order... Hmm... That's what I'm looking for, thanks! – alexy13 Aug 29 '11 at 2:10
To answer your question, now that we know you're interested in tensors: we don't usually write them down the way we would matrices. In fact, most places we use tensors, we try to avoid even writing out matrices if we can help it! –– Anyone want to write an answer that describes how equations with tensors are written, e.g. Einstein notation or the like? – Niel de Beaudrap Aug 29 '11 at 2:15
@alexy13: I've revised your question to better match what you were asking about, hope you don't mind. – Niel de Beaudrap Aug 29 '11 at 2:43
The question seems quite different with the changes, at least the answers doesn't seem to match anymore - maybe better put it into a new question? – Johannes L Aug 29 '11 at 13:03

Let's make this clear:

To rotate any object in a linear room of $n$ dimentions, you need a matrix(=a second-order tensor) with $n$ rows and $n$ columns. Such a rotation matrix $A$ is a special case of a base transformation, you recognize it by it's determinant: $det(A)=+1$ (I'm not so super sure here.)

You don't use third-order tensors to rotate anything. So what's a $n$th- order tensor? There is some more abstract defintion, but for now, just imagine it as a matrix with $n$, instead of 2, indices: $T=a_{a b c d ...}$ So a third-order tensor has three indices. You can imagine it like a many matrix that is not a rectangle on the paper, but a cuboid in the room.

The matrix is a special case of a second-order ("two-dimensional") tensor. The common interpretation of tensors is as multilinear functionals. When you apply a matrix on two vectors (collapsing multiplication), you get a scalar. This function is linear in both vectors: $c= \sum \limits_{ij} a_{ij} b_i d_j$.

You can do the same with a third-order (or "three-dimensional") tensor:$c= \sum \limits_{ijk} a_{ij} b_i d_j e_k$. This is now linear in all three vectors. Or you do such: $c_{ij}= \sum \limits_{k} a_{ijk} e_k$, thus getting a second-order tensor, which we usually call a matrix.

The order of a tensor doesn't refer to the amount of entrys.

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I'm not sure he really means to talk about rotations, and he's already clearified that he doesn't mean the number of entries. It would probably be more helpful to him if your answer just focused on tensors (perhaps with an example which includes the special case of a tensor of order 3, while making it clear that it generalizes to higher orders) and not on misconceptions that he might have based on the original phrasing of the question. – Niel de Beaudrap Aug 29 '11 at 15:30
Well, since the first answer ist still referring to rotation matrices, I was afraid that this may confuse readers. But yes, my answer isn't exactly clear and short. – Konstantin Schubert Aug 29 '11 at 15:53

In my opinion, the best way to understand a rotation matrix is the Rodrigues's formula (as J.M. mentioned), or called axis-angle representation. Given a rotation axis and a rotation angle, then you can write out the rotation matrix.

For a 2D rotation matrix, it can still be interpreted by a axis-angle representation. But the rotation axis of a 2D rotation matrix is always the z-axis!!! As a comparison, the rotation axis of a 3D matrix can be arbitrary vector. You can get details from wiki.

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Just so that no-one downvotes this answer: this was an appropriate answer for a reasonable interpretation to the question as it originally appeared. – Niel de Beaudrap Aug 29 '11 at 10:51
Shiyu maybe should reword his/her answer for the changed question. +1 for good answer before the question was changed. – alexy13 Aug 29 '11 at 12:02