Tensors and ranks

No. The rank of a tensor indicates the number of indices. Only rank two tensors can be represented as a matrix. The rank of this matrix on its turn tells you something about the number of linear independent columns of this matrix. As such it is clear that these two notions of rank don't have anything to do with each other.

 

Starting with your second point first:

In order to answer this, I think that you need to be careful and make the distinction between covariant and contravariant components.

For example, a tensor of rank (1,0) is what you would normally consider to be a vector. It has one contravariant component.

A tensor of rank (0,1) is something called a dual vector (see either Schutz' GR book or MTW for an excellent geometric picture of what exactly these look like!) This has one covariant component.

The idea of constructing a matrix out of these comes from taking outer products of them.

Suppose for example that we had a vector (usually represented as a column)
$$v^\alpha e_{\alpha} = \left(\begin{array}{c}a\\b\\c\\d\end{array}\right)$$

and a dual vector
$$u_{\beta}\omega^{\beta}=(h,i,j,k)$$

We can construct a new (1,1) tensor by taking their tensor product:

$$v^{\alpha} e_{\alpha}\otimes u_{\beta}\omega^{\beta}=\left(\begin{array}{c}a\\b\\c\\d\end{array}\right)\otimes (h,i,j,k)=\left(\begin{array}{ccc}ah & ai & aj & ak\\bh & bi & bj & bk\\ch & ci & cj&ck\\dh & di & dj & dk\end{array}\right)$$

So, if you have a (2,1) tensor this is going to be a 3D "cube" and so on.

I hope this was somewhat helpful and sorry if I'm being sloppy!

 

A tensor's order (or rank or degree) is the number of free indices the tensor has. So, a tensor ##T^{\mu \nu}## has an order of two.

So, no. A matrix is an order two tensor, since it has two indices (row and column). The rank of a matrix is the number of linearly independent row or column vectors (row rank and column rank respectively) in the matrix.

 

Lets just go through a very short list of tensors that have familiar equivalents.

The total rank of a tensor is the sum of all of its two subranks.

A rank 0 tensor (total rank) is a scalar, a number.

A rank 1 tensor is a vector.

A rank 2 tensor is a matrix.

The subdivision of ranks has a definite meaning, but I'm not going to attempt to explain it in this post.

 

In a given coordinate system, a second order tensor can be represented by a matrix. If the coordinate system has dimension 2, the matrix is 2 by 2, if dimension 3, 3 by 3, etc.

In a given coordinate system, a third order tensor would have to be represented by a three dimensional array. If the coordinate system has dimension 2, that would be "2 by 2 by 2", in three dimensions "3 by 3 by 3"- you could think of it as consisting of 3 3 by 3 matrices, one in front of the other.

"covariant" and "contravarient" is a whole different matter. If you like, you could think of a "covariant vector" as a "row" matrix, one row and n columns, and a "contravariant vector" as a "column" matrix, one column and n rows, so that the "act" on one another by matrix multiplication.

 

A cube would represent a third order tensor, having three free indices. The stress tensor is of rank two, hence can be represented by a matrix. You just wrote it down. I don't understand how you can think of this as order four; in that case every matrix element should have two free indices, being matrices by themselves.