An ideal fluid is the one which cannot support any shearing stress. It also doesn't have viscosity. My question is what does it mean by a fluid to be isotropic? Is an ideal fluid necessarily isotropic and homogeneous? What can we say about the stress tensor of an isotropic fluid?
Answer
Isotropy and homogeneity are different. The former is a consequence of invariance under rotations while the latter comes from invariance under translations. The stress tensor of an isotropic fluid then must be invariant under any orthogonal transformation, and this implies that it is a multiple of the "identity" tensor. More precisely, assume matrix notation for order 2 tensor and let $\sigma$ be the stress tensor of the fluid. If $O\in O(3)$ is any orthogonal transformation, then $$O^T\sigma(x) O = \sigma(x),\qquad\forall x\in\mathbb R^3,$$ or, which is the same, $[\sigma(x), O] = 0$. This is possible only if $\sigma$ is a multiple, at every point, of the identity $3\times 3$ matrix $I$. Therefore there must exists a scalar function $p:\mathbb R^3\to\mathbb R$ such that $$\sigma(x) = p(x) I.$$ In general $p$ need not satisfy any extra assumptions (physically it is just representing a pressure field, so perhaps you might want positivity as well) and if $t\in\mathbb R^3$ is any translation, then $p(x)\neq p(x+t)$ in general. But if the fluid is also homogeneous, then $\sigma(x)=\sigma(x+t)$ for any $t\in\mathbb R^3$, and therefore $p$ must be invariant under translations, so that you get a constant value $p_0$ for which $$\sigma(x) = p_0I,\qquad\forall x\in\mathbb R^3.$$ So isotropy and homogeneity together imply that the stress tensor is just a pressure, which is constant everywhere in space.
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