Tensors and ranks

[shounakbhatta]

Main Question or Discussion Point

Hello All,

I am quiet new to this subject.I am unable to get over tensors and ranks.

(1) Does tensor order has to do something with the rank of the matrix?

(2) What doe tensor order 2,3.....means?

If I have a 2 by 2 matrix, does that means it is a 2nd.order tensor and 3 by 3 matrix means a third order tensor and so on.......

Does tensor have relations with matrix?

Please excuse me if my question is too mundane?

Thanks,

-- Shounak

 

[haushofer]

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.

 

[CJ2116]

Starting with your second point first:

(2) What doe tensor order 2,3.....means?

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!

 

[Mark M]

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.

 

[Chestermiller]

Check out PFs Math & Science Learning Materials, Calculus and Beyond Learning Materials, "Introduction to Tensor Calculus and Continuum Mechanics" thread. It has an excellent PDF document on tensor analysis.

Chet

 

[pervect]

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.

 

[HallsofIvy]

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.

 

[shounakbhatta]

Hello All,

Thanks for responding. Actually, it was ok with me till 2nd.order tensor, but higher one like 3rd.or 4th.order tension, I was unable to visualize also.

Well, then dimension is related to the order of a tensor. For a 3d array, the tensor is 3rd.order, for a 2d.array it is 2nd.order tensor right?

But one thing is confusing me. For a stress tensor it is a 2nd.order tensor, represented by a cube.
For a stress energy tensor, the matrix is:

[t00,t01,t02,t03
t10,t11,t12,t13
t20,t21,t22,t23
t30,t31,t32,t33]

What would this tensor called? Tensor of order 4?
Please if you can explain me.

Regards,

-- Shounak

 

[haushofer]

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.

 

[shounakbhatta]

Hello,

May be I am making a wrong question.

What I am trying to guess is a 3x3 matrix, say a stress, is a second order tensor.
Similarly, for a 4x4 matrix, I mean to say, is it a third order tensor? As I have mentioned earlier, a stress-energy tensor, a 4 by 4 matrix.....

As you have told stress-energy tensor is a 2nd.order tensor.

I want to understand, that a matrix, if it is 3 by 3, 4 by 4 or 5 by 5.....the tensor is also related to it?

What I understand is indices mean row,col.

Please consider it if my question is irrelevant.

-- Shounak

 

[Chestermiller]

Maybe this will help. Sometimes tensors are expressed as the summation of components times the unit vectors in the coordinate directions. In the expression for a vector, for example, there is one unit vector in each term of the summation, and the components each have one index, corresponding to the particular direction. For a second order tensor (e.g., stress), each term of the summation has a component with two indices attached to it, and two unit vectors "multiplying" the components. The two unit vectors are placed in juxtaposition to one another, and the combination is called a dyad. The second order tensor fulfills its function in life only when it is dotted with another vector. The way this is done is to dot the vector with the unit vector in each dyad that is adjacent to the vector. This has the effect of mapping the vector into a new vector. In the case of the stress tensor, for example, when you dot the stress tensor with a unit normal vector to a plane, you get the traction vector acting on the plane.

The dimensionality of the space has nothing to do with order of the tensor. The order of the tensor is determined by the number of unit vectors that are included within each term of the summation, and the corresponding number of indices on the components of the tensor. In the case of a third order tensor, for example, there are three unit vectors in each term, and three indices on each component. In 3D space, there would be 27 terms in the summation.

 

[shounakbhatta]

Ok, thanks. Slowly getting the concept.

For stress-energy, a second order tensor, bcoz.it has two indices t01......

That means for a third order tensor, it would look like t001,t002 and so on.....?

Am I right?

Please correct me if I am wrong as I am a slow learner.

-- Shounak

 

[Chestermiller]

Yes. That is right (not the slow learner part).

 

[shounakbhatta]

Thank you very much for all the help.

I can get hold of the concept now.

-- Shounak