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I write the elastic deformation $E$ for an incompressible material in spherical symmetry (Have a look at here Eq.(5.32) ). Since we are incompressible, $\mathbf{det} \,E=1$, so in spherical coordinates we can write : $$E=\mathbf{diag}(1/\alpha^2,\alpha,\alpha) $$

with $\alpha=r/R$, $r$ being the radius in the current configuration and $R$ in the initial stress-free configuration. We have $1/\alpha^2=dr/dR$.

So what I would expect is that the elastic deformation respects the rotational invariance : $$E.v=w \implies E.(R.v)=R.w$$ where $v,w$ are vectors and $R$ is the rotation matrix.

To simplify I write everything in cartesian coordinates : $E_c=M_{S\rightarrow C}\,E\,M_{C\rightarrow S}$ where $E_c$ is the elastic strain in cartesian coordinates and $M_{i\rightarrow j}$ is the matrix that transforms from a base $i$ to a base $j$.

The rotational invariance thus writes : $E_c=R^{T}E_c\,R$.

I take : $$M_{C\rightarrow S}=\left( \begin{array}{ccc} \sin (\theta ) \cos (\phi ) & \cos (\theta ) \cos (\phi ) & -\sin (\phi ) \\ \sin (\theta ) \sin (\phi ) & \cos (\theta ) \sin (\phi ) & \cos (\phi ) \\ \cos (\theta ) & -\sin (\theta ) & 0 \\ \end{array} \right)$$ $$M_{S\rightarrow C}=\left( \begin{array}{ccc} \sin (\theta ) \cos (\phi ) & \sin (\theta ) \sin (\phi ) & \cos (\theta ) \\ \cos (\theta ) \cos (\phi ) & \cos (\theta ) \sin (\phi ) & -\sin (\theta ) \\ -\sin (\phi ) & \cos (\phi ) & 0 \\ \end{array} \right)$$ I'm taking a rotation around the $z$-axis : $$R=\left( \begin{array}{ccc} \cos (A) & -\sin (A) & 0 \\ \sin (A) & \cos (A) & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ And : $$R^{T}E_c\,R=\tiny{\left( \begin{array}{ccc} \frac{\sin ^2(A+\theta ) \left(a^3 \sin ^2(\phi )+\cos ^2(\phi )\right)}{a^2}+a \cos ^2(A+\theta ) & -\frac{\left(a^3-1\right) \cos ^2(\phi ) \sin (2 (A+\theta ))}{2 a^2} & \frac{\left(a^3-1\right) \sin (\phi ) \cos (\phi ) \sin (A+\theta )}{a^2} \\ -\frac{\left(a^3-1\right) \cos ^2(\phi ) \sin (2 (A+\theta ))}{2 a^2} & \cos ^2(A+\theta ) \left(\frac{\cos ^2(\phi )}{a^2}+a \sin ^2(\phi )\right)+a \sin ^2(A+\theta ) & \frac{\left(a^3-1\right) \sin (\phi ) \cos (\phi ) \cos (A+\theta )}{a^2} \\ \frac{\left(a^3-1\right) \sin (\phi ) \cos (\phi ) \sin (A+\theta )}{a^2} & \frac{\left(a^3-1\right) \sin (\phi ) \cos (\phi ) \cos (A+\theta )}{a^2} & \frac{\sin ^2(\phi )}{a^2}+a \cos ^2(\phi ) \\ \end{array} \right)}\\ \neq E_c$$

The problem is that I don't find that the left-hand-side and the right hand side are equal. Did I make a mistake ?

I write here my code Mathematica :

"matrix to transform from spherical to cartesian coordinates"
MSC = CoordinateTransformData["Spherical" -> "Cartesian", 
   "OrthonormalBasisRotation", {r, \[Theta], \[Phi]}];
" matrix to transform from cartesian to spherical coordinates"
MCS = Inverse[MSC];
"rotation axe z"
R = {{Cos[A], -Sin[A], 0}, {Sin[A], Cos[A], 0}, {0, 0, 1}};
iR = Inverse[R]

RCS = FullSimplify[Dot[MCS, R]]
iRSC = FullSimplify[Dot[iR, MSC]]
F = DiagonalMatrix[{1/a^2, a, a}]
verif = FullSimplify[Dot[iRSC, RCS]]
test = FullSimplify[Dot[iRSC, F, RCS]]
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The displacement field $\vec u$ might be spherically symmetric, but that doesn't mean that the gradient of $\vec u$ (or $\vec u$ itself) needs to be invariant to all rotations. They will only be invariant to rotations around the $r$-axis.

Incompressibility is just a red herring here.

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