Invariants of a Reynolds stress tensor

Question

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).

Answer 1

The problem is that you are not defining "trace" correctly.

The trace of a rank 2 tensor is only equivalent to the sum of the diagonal entries of its matrix representation if the tensor is type (1,1). (Type (1,1) here means that the tensor represents a linear transformation; or equivalently that the matrix representation transforms like $\mathbf{Q} \to A\mathbf{Q}A^{-1}$ under change of basis/coordinates in the vector space.)

The Reynold stress tensor as you defined it is a type (0,2) (or type (2,0), depending on your convention) tensor, so its transformation law is $\mathbf{Q} \to A\mathbf{Q}A^T$. This type of tensors cannot be "traced"...until you specify what you are tracing with respect to. To be able to trace, you need to first multiply a type (0,2) tensor against a type (2,0) tensor to form a type (1,1) tensor, and then you can perform the trace. The type (2,0) tensor in this case is the symmetric bilinear form that corresponds to the Euclidean inner product.

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In other words, going back to the more physical part of the story: the kinetic energy is an invariant, but you are just computing it wrong. Think about the case where you have just a particle. The kinetic energy of a particle is $\frac12 m \vec{v}\cdot\vec{v}$. In an orthonormal coordinate system

$$ \vec{v}\cdot \vec{v} = (v_x)^2 + (v_y)^2 + (v_z)^2 $$

but in a coordinate system that is not orthonormal, the dot product is not given by just the sum of the squares of the components! (For example, consider spherical or cylindrical coordinates!)