# Second Rank Tensors [duplicate]

I'm a little confused, for the twentieth time, on what tensors are. I thought they were a generalization of matrices-but then they have special transformation rules. I'm looking for a concise definition of what a tensor is. In Griffiths, he says that an $n$ rank tensor transforms with $n$ components of a rotation matrix, why is this? I'll attach what I'm referencing.

• Possible duplicates: physics.stackexchange.com/q/20437/2451 , physics.stackexchange.com/q/32011/2451 and links therein. – Qmechanic Mar 8 '14 at 23:17
• Have you attempted an internet and/or physics.SE search? There is an enormous amount already written about this. It is important to note that the notion of "tensor" being used depends on the context. The type of tensor you're referring to is a tensor w.r.t the rotation group, but one can define tensors w.r.t other groups. Mathematically, a notion that is central to many of the formulations is that of a "mutlilinear transformation" on a vector space. – joshphysics Mar 8 '14 at 23:38
• I have been searching for a while now... I get confused between all the different explanations I find, I wish I could find a bare bones description of them mathematically. A multilinear transformation sounds good, but then I don't really get this. Why are there two copies of R being applied? Whenever I've seen a map on a matrix, it's only taken one copy. So why do second rank tensors take two? – user24082 Mar 8 '14 at 23:44
• @Anthony en.wikipedia.org/wiki/Tensor#As_multilinear_maps The number of copies of $R$ is the number of "slots" in the multilinear map; this comes from the fact that when you transform the components of the tensor from one basis to another, you need to transform each index, and there are $n$ such indices on a tensor of rank $n$. – joshphysics Mar 9 '14 at 0:14
• Why do you need to transform each index? In things like matrices in R^2 we only apply a rotation matrix once, don't we? I mean, I would things that since a matrix is just an amalgamation of vectors in R^2.... – user24082 Mar 9 '14 at 2:16

Succintly put, all rank-$2$ tensors may be represented as matrices w.r.t. to a particular basis choice. All matrices may be interpreted as rank-$2$ tensors provided you've fixed a basis. The transformation law is then just a consequence of basis independence!