I have a transformation matrix $K$ which transfers elastic constitutive matrix $C$ between two coordinate systems. According to textbooks such as T.C.T. Ting's "Anisotropic Elasticity", the elastic matrix in the new coordinate system $C^*$ should be


My question is: If $C$ is isotropic, then $C^*=C$ no matter what $K$ is. But it seems that we cannot obtain this result from the above equation. For example, if we let $C$ be an identity matrix, then $C^* \neq I$ because $K$ and $K^T$ are generally not orthogonal.

Am I wrong?

P.S. relevant question here

  • $\begingroup$ I think there is a misunderstanding here... Forgot about the elastic tensor, which is really hard to visualize it. Just think about a two-dimensional vector. $A = (A_{x},A_{y})$. If you transform your system of coordinate for example move it and rotate it somehow, you will get a new vector $A^{'}=(A^{'}_{x},A^{'}_{y})$ but obviously $A$ is not equal to $A^{'}$. So what? The thing here is that if you transform the vector by rotation and moving, the invariant of the vector will be the same. In this case, the invariant is the Euclidean norm of the 2D vector $|A| = |A^{'}|$. $\endgroup$ Commented Sep 18, 2019 at 15:16
  • $\begingroup$ In your case for a fourth order elastic tensor $\mathbf{C}$, when you transform it linearly, the invariants of this tensor, which are related to its eigenvalues, will remain the same and not necessarily the $\mathbf{C}$ and $\mathbf{C}^{*}$ one by one for their members. $\endgroup$ Commented Sep 18, 2019 at 15:18
  • $\begingroup$ In order to find out what will remain constant in case of a fourth order tensor $\mathbf{C}$, see this discussion here: imechanica.org/node/6771 $\endgroup$ Commented Sep 18, 2019 at 15:46
  • $\begingroup$ Thanks for your answer. It seems that the question could be turned into: Does an isotropic fourth-order tensor invariant under a coordinate system transfer? I think the answer should be YES. $\endgroup$
    – hillyuan
    Commented Sep 19, 2019 at 0:54
  • $\begingroup$ I think the question is somehow meaningless if you want to word it in that way. It's obvious that any nth-order tensor members will be changed under linear transformations. But, the only thing is that invariants of any nth-order tensor will remain constant under linear transformations, which again is obvious... $\endgroup$ Commented Sep 19, 2019 at 1:45

1 Answer 1


If you are looking for validity of the equation for elastic tensor linear transformation, here is a quick and easy explanation:

From Hooke's law, we know that: $\sigma = \mathbf{C} \cdot \varepsilon$, where $\sigma$ is the stress tensor, $\varepsilon$ is the strain tensor, and $\mathbf{C}$ is the elastic tensor. Let's say you want to transform the coordinate linearly as:

$$\sigma^{'} = K_{\sigma} \sigma$$

$$\varepsilon^{'} = K_{\varepsilon} \varepsilon$$

Where $\sigma^{'}$ and $\varepsilon^{'}$ are stress and strain tensors in the new coordinate and $K_{\sigma}$ and $K_{\varepsilon}$ are the transformation matrices for stress and strain tensor respectively. We know that transforming the coordinates linearly should not have any effect on Hooke's law:

$$\sigma^{'} = \mathbf{C}^{*} \cdot \varepsilon^{'}$$


$$K_{\sigma} \sigma = \mathbf{C}^{*} \cdot K_{\varepsilon} \varepsilon$$


$$\sigma = K_{\sigma}^{-1} \mathbf{C}^{*} \cdot K_{\varepsilon} \varepsilon$$

But we had:

$$\sigma = \mathbf{C} \cdot \varepsilon$$


$$\mathbf{C} = K_{\sigma}^{-1} \mathbf{C}^{*} K_{\varepsilon}$$


$$\mathbf{C}^{*} = K_{\sigma} \mathbf{C} K_{\varepsilon}^{-1}$$

But $K_{\sigma}$ and $K_{\varepsilon}$ are connected together based on Reuter's matrices:

$$K_{\varepsilon} = R K_{\sigma} R^{-1}$$

Where $R$ and $R^{-1}$ are Reuter's matrices:

$$R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

$$R^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{bmatrix}$$

So finally, you would have:

$$\mathbf{C}^{*} = K_{\sigma} \mathbf{C} R K^{-1}_{\sigma} R^{-1}$$

Check the validity of the derivation from Gerdeen's book at eq. 8.65.

But if you are looking to understand what will remain constant after linear transformation, I refer you to these blogs and Wikipedia page to find out tensor invariants are directly related to their eigenvalues: https://imechanica.org/node/6771 and https://en.wikipedia.org/wiki/Invariants_of_tensors

  • $\begingroup$ If you are using Voigt notation, K should be different for stress and strain. $\endgroup$
    – nicoguaro
    Commented Sep 19, 2019 at 19:00
  • 1
    $\begingroup$ @nicoguaro Done! $\endgroup$ Commented Sep 19, 2019 at 19:43
  • $\begingroup$ Thanks, @AloneProgrammer. What I want to confirm is if $C^*=C$ when C is an isotropic elastic matrix and R arbitrary. Maybe the last equation shows this? $\endgroup$
    – hillyuan
    Commented Sep 26, 2019 at 1:47
  • $\begingroup$ @hillyuan What do you mean arbitrary $R$? $R$ is not arbitrary at all... $\endgroup$ Commented Sep 26, 2019 at 1:54
  • $\begingroup$ @AloneProgrammer Sorry! $R$ is a constant matrix! What I supposed to say is $C$ under and arbitrary rotation. $\endgroup$
    – hillyuan
    Commented Sep 26, 2019 at 2:32

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