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I am dealing with a Reynolds stress tensor, which arises in Navier-Stokes equations and is symmetric. By using an Ultrasound based instrument, I can read the velocity components of the fluid along three non orthogonal axis named 1-2-3. I can transform the measured velocities along these three axis in a Cartesian three orthogonal axis system x-y-z by writing the transformation matrix A and hence:

$\begin{Bmatrix} u\\ v\\ w \end{Bmatrix}= \mathbf{A}\cdot \begin{Bmatrix} u_1\\ u_2\\ u_3 \end{Bmatrix} $

where $\begin{Bmatrix} u\\ v\\ w \end{Bmatrix}$ is the velocity vector in a Cartesian orthogonal coordinate system, $u_1,u_2$ and $u_3$ are the three measured velocities along the three non orthogonal axes, $\mathbf{A}$ is the transformation matrix. The tensor transformation is:

$\mathbf{Q}^*=\mathbf{A}\cdot\mathbf{Q}\cdot\mathbf{A}^T$.

where $\mathbf{Q}$ is the Reynolds stress tensor as computed in the 1-2-3 instrument coordinate system:

$\mathbf{Q}=\begin{bmatrix} u_1^{\prime}u_1^{\prime}&u_1^{\prime}u_2^{\prime} &u_1^{\prime}u_3^{\prime} \\ u_2^{\prime}u_1^{\prime}&u_2^{\prime}u_2^{\prime} &u_2^{\prime}u_3^{\prime} \\ u_3^{\prime}u_1^{\prime}&u_3^{\prime}u_2^{\prime} &u_3^{\prime}u_3^{\prime} \end{bmatrix}$

and $\mathbf{Q}^*$ is the Reynolds stress tensor in the x-y-z coordinate system:

$\mathbf{Q}=\begin{bmatrix} u^{\prime}u^{\prime}&u^{\prime}v^{\prime} &u^{\prime}w^{\prime} \\ v^{\prime}u^{\prime}&v^{\prime}v^{\prime} &v^{\prime}w^{\prime} \\ w^{\prime}u^{\prime}&w^{\prime}v^{\prime} &w^{\prime}w^{\prime} \end{bmatrix}$

The prime indicates that that the variable is the fluctuating part and the overline indicates the mean value, e.g. $u=\overline{u}+u^{\prime}$.

One half of the trace of the tensor $\mathbf{Q}^*$ is the turbulent kinetic energy of the fluid (per unit mass) and should be invariant. Indeed it is invariant as long as the transformation matrix $\mathbf{A}$ is orthogonal, i.e. $\mathbf{A}^{-1}=\mathbf{A}^T$ but it is not for general transformation matrix. In fact the trace of the tensor $\mathbf{Q}$ is different from the trace of the tensor $\mathbf{Q}^*$ if $\mathbf{}$ is the non-orthogonal transformation matrix between the two coordinate system 1-2-3 and x-y-z.

I wonder if the invariants of a tensor are really invariant (i.e. assume the same value) only for specific class of transformations (e.g. for similarity transformations).

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    Are you sure your tensor transformation is correct? Assuming your tensor is $(1, 1)$, it should be $Q \mapsto AQA^{-1}$, and $A^T = A^{-1}$ only for orthogonal matrices :)2011-08-15
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    @Alexei: I believe the transformation is correct. If you write $\vec{v}$ to be the velocity vector, the tensor $\mathbf{Q} = \vec{v}\otimes \vec{v}$ or $\vec{v}~\vec{v}^T$. In other words, $\mathbf{Q}$ is a type (0,2) or (2,0) tensor, so the coordinate transformation law is correct. (What you wrote would be the transformation law for a linear transformation, a type (1,1) tensor.)2011-08-15
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    @Willie: Well, yeah, I agree with you. But stress tensor canonically *is* a (1,1)-tensor, it maps direction to the force acting in it (or is force a covector? I dunno). If it's (0,2) tensor, though, then you first have to raise an index with a metric, just like you wrote.2011-08-15
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    @Alexei: mmm, I don't think I completely agree. The stress-energy tensors for classical field theories is, like you say, canonically a (1,1) tensor; but the [Cauchy stress tensor](http://en.wikipedia.org/wiki/Cauchy_stress_tensor) is defined to be (0,2), and in using it one usually consider its transpose (which is a (2,0) tensor) acting on directions *through dot product*. We in the modern age have the benefit of hindsight that stress tensors *should* be (1,1). But I don't think Reynolds or Cauchy or Biot had that opportunity.2011-08-15

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