## Transformation of a tensor to a new coordinate system

The transformation (rotation) of a tensor into a new coordinate system is a common problem in rock mechanics and in continuum mechanics in general. In this article we will present the necessary equations and an example case. We will use the stress tensor as example.

Consider a rigid body in equilibrium and a coordinate system. The stress state of any internal point of this body is given by the stress tensor (cf. Fig. 1):

$\sigma_{ij}=\left[\begin{array}{ccc}\sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}\end{array}\right]$
(1)

Now consider a second coordinate system that is rotated with respect to the original coordinate system. The stress tensor may be transformed in the new coordinate system according to the tensor transformation rule (cf. Fig. 2):

$\sigma’_{ij}=\alpha_{im}\alpha_{jn}\sigma_{mn}$
(2)

where the summation convention has been used. The quantity:

$\alpha_{ij}=\left[\begin{array}{ccc}\alpha_{11} & \alpha_{12} & \alpha_{13} \\ \alpha_{21} & \alpha_{22} & \alpha_{23} \\ \alpha_{31} & \alpha_{32} & \alpha_{33}\end{array}\right]$
(3)

is the rotation tensor with elements the direction cosines between the axes of the two coordinate systems. It should be noted that this tensor is not symmetric. The angle for each direction cosine is measured from the primed (new) system to the unprimed (old) system (Fig. 2).

### Example

Assume the following stress tensor:

$\sigma_{ij}=\left[\begin{array}{ccc}5 & 3 & 1\\ 3 & 2 & 0\\ 1 & 0 & 4\end{array}\right]$
(4)

Rotate the coordinate system by $$60$$ degrees around axis $$1$$ and transform the stress tensor into this new coordinate system.

Axis $$1$$ in the primed (new) coordinate system coincides with the unprimed (old) axis $$1$$ thus the angle between them is $$0$$. Moreover the angles between primed axis $$1$$ and unprimed axes $$2$$ and $$3$$ are both $$\dfrac{\pi}{2}$$. The angle between the new axis $$2$$ and the old axis $$1$$ is $$\dfrac{\pi}{2}$$. The angle between primed axis $$2$$ and unprimed axis $$2$$ is $$\dfrac{\pi}{3}$$ while the angle between primed axis $$2$$ and unprimed axis $$3$$ is $$\dfrac{\pi}{6}$$. Primed axis $$3$$ is normal to unprimed axis $$1$$ (angle $$\dfrac{\pi}{2}$$ ). The angles between new axis $$3$$ and old axes $$2$$ and $$3$$ are $$\dfrac{5\pi}{6}$$ and $$\dfrac{\pi}{3}$$, respectively. Thus, the rotation tensor takes the form:

$\alpha_{ij}=\left[\begin{array}{ccc}\cos0 & \cos\dfrac{\pi}{2} & \cos\dfrac{\pi}{2} \\ \cos\dfrac{\pi}{2} & \cos\dfrac{\pi}{3} & \cos\dfrac{\pi}{6} \\ \cos\dfrac{\pi}{2} & \cos\dfrac{5\pi}{6} & \cos\dfrac{\pi}{3}\end{array}\right]$
(5)

From the tensor transformation rule Eq. (2) we can calculate each element of the new stress tensor as follows:

$\begin{array}{rl} \sigma’_{11}=&\alpha_{11}^{2}\sigma_{11}+\alpha_{12}^{2}\sigma_{22}+\alpha_{13}^{2}\sigma_{33}\\ &+2\alpha_{11}\alpha_{12}\sigma_{12}+2\alpha_{11}\alpha_{13}\sigma_{13}+2\alpha_{12}\alpha_{13}\sigma_{23}\\ \sigma’_{22}=&\alpha_{21}^{2}\sigma_{11}+\alpha_{22}^{2}\sigma_{22}+\alpha_{23}^{2}\sigma_{33}\\ &+2\alpha_{21}\alpha_{22}\sigma_{12}+2\alpha_{21}\alpha_{23}\sigma_{13}+2\alpha_{22}\alpha_{23}\sigma_{23}\\ \sigma’_{33}=&\alpha_{31}^{2}\sigma_{11}+\alpha_{32}^{2}\sigma_{22}+\alpha_{33}^{2}\sigma_{33}\\ &+2\alpha_{31}\alpha_{32}\sigma_{12}+2\alpha_{31}\alpha_{33}\sigma_{13}+2\alpha_{32}\alpha_{33}\sigma_{23}\\ \sigma’_{12}=&\alpha_{11}\alpha_{21}\sigma_{11}+\alpha_{12}\alpha_{22}\sigma_{22}+\alpha_{13}\alpha_{23}\sigma_{33}\\ &+(\alpha_{11}\alpha_{22}+\alpha_{12}\alpha_{21})\sigma_{12}+(\alpha_{12}\alpha_{23}+\alpha_{13}\alpha_{22})\sigma_{23}\\ &+(\alpha_{11}\alpha_{23}+\alpha_{13}\alpha_{21})\sigma_{13}\\ \sigma’_{23}=&\alpha_{21}\alpha_{31}\sigma_{11}+\alpha_{22}\alpha_{32}\sigma_{22}+\alpha_{23}\alpha_{33}\sigma_{33}\\ &+(\alpha_{21}\alpha_{32}+\alpha_{22}\alpha_{31})\sigma_{12}+(\alpha_{22}\alpha_{33}+\alpha_{23}\alpha_{32})\sigma_{23}\\ &+(\alpha_{21}\alpha_{33}+\alpha_{23}\alpha_{31})\sigma_{13}\\ \sigma’_{13}=&\alpha_{11}\alpha_{31}\sigma_{11}+\alpha_{12}\alpha_{32}\sigma_{22}+\alpha_{13}\alpha_{33}\sigma_{33}\\ &+(\alpha_{11}\alpha_{32}+\alpha_{12}\alpha_{31})\sigma_{12}+(\alpha_{12}\alpha_{33}+\alpha_{13}\alpha_{32})\sigma_{23}\\ &+(\alpha_{11}\alpha_{33}+\alpha_{13}\alpha_{31})\sigma_{13}\end{array}$
(6)

Hence the new stress tensor is (approximately):

$\sigma’_{ij}=\left[\begin{array}{ccc} 5 & 2.4 & -2.1 \\ 2.4 & 3.5 & 0.9 \\ -2.1 & 0.9 & 2.5 \end{array}\right]$
(7)