# Rotating tensors (different from vectors?)

1. Sep 9, 2010

### koab1mjr

Hi all

I am taking my first grad level class on stress and elasticity and ran into a bit of a wall. We are dealing with a 3x3 stress tensor which describes the state of stress of a given point. Now various text book problems have questions where given one state describe the state if the element were roateated in some fashion. Now we have just been deriving some formulas and use of the mohr's circle for this but I was thinking "hey why don't i just rotate the tensor". I asked the professor and was told that was perfectly fine if you know how to do it. So since I know linear algebra (undergrad level) and how to deal with vectors and tensors being a glorified vector no sweat.

So i am working on some HW and I try it out. What I find is my answer is incorrect because my matrix is not symetric so obviously something is wrong. Now my approach is wrong as I am told by my professor today but he did not give me much direction as too how to fix so why i am here

What I have been doing is treating the tensor like a matrix created by putting together vectors as rows or columns. I take each of the vectors apply my rotation matrix to them and voila a rotated tensor. This is of course wrong but I am looking for help as too my flaw and more so how I could go about learning how to rotate tensors. I feel it should not be much different than a transofmration of a vector but I could be wrong

any help is much appreciated.

2. Sep 10, 2010

### Fredrik

Staff Emeritus
To rotate a vector: x'=Rx, where R is an orthogonal matrix. By definition of matrix multiplication, this can be written as $x'_i=R_{ij}x_j$. (Indices that appear twice are summed over). To rotate a tensor with two indices:

$$T_{ij}'=R_{ik}R_{jm}T_{km}=R_{ik}(R^T)_{mj}T_{km}=R_{ik}T_{km}(R^T)_{mj}=(RTR^T)_{ij}$$

So T'=RTRT.

Last edited: Sep 10, 2010
3. Sep 10, 2010

### D H

Staff Emeritus
A bit nit picky, but you aren't really rotating the vector (or matrix). You are transforming it. It is still the same vector (matrix) underneath. What is being rotated is the frame of reference. The vector (matrix) is being transformed to this rotated frame. Compare this with the vector corresponding to the second hand on an analog clock. In fifteen seconds that vector rotates by 90 degrees -- and in this case it is the vector that is rotating, not the frame of reference.

Another nit-picky point: Fredrik's result applies to type (0,2) tensors such as a stress tensors and moment of inertia tensor. The Hessian of some scalar function transforms differently.

4. Sep 10, 2010

### koab1mjr

OK so just so I got this i take my transformation operator and I calc the transpose then conclude with t' = AtA'.... With A being the matrix. I will give it a shot.... I will keep you posted. Thanks!

5. Sep 11, 2010

### qbert

depends on your point of view.
see http://http://en.wikipedia.org/wiki/Active_and_passive_transformation" [Broken]

Last edited by a moderator: May 4, 2017
6. Jun 20, 2012

### sacsac13

Hello everybody,
i read carefully what you wrote. Right now i am wondering if it is possible to rotate a tensor of order 4. How does this work? I have no idea if this is possible but it would be interessting.
bye sacsac
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