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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?

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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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  • $\begingroup$ @ Phoenix87- Thanks for the illuminating reply. But my second question was that whether an ideal fluid is necessarily isotropic and homogeneous? $\endgroup$ – SRS Feb 5 '15 at 6:02
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    $\begingroup$ with no viscosity there is not viscous stress tensor, hence no off-diagonal components, so you remain with something of the form $\sigma(x) = pI$, which is an isotropic tensor. For the homogeneity some extra conditions are probably needed. $\endgroup$ – Phoenix87 Feb 9 '15 at 13:55
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The stress tensor in an ideal fluid is indeed isotropic, for it can be written as $\sigma_{ij} = -p\delta_{ij}$, $p$ being the hydrostatic pressure. In a real fluid in motion, however, the stress tensor has a "non-isotropic" part, which is also called the "deviatoric" part. It is proportional to the rate of strain tensor. In Newtonian fluids, the constant of proportionality is the coefficient of viscosity. Thus, the stress tensor in a Newtonian fluid is \begin{equation} \sigma_{ij} = -p\delta_{ij} + \mu\left(\frac{\partial v_j}{\partial x_i} + \frac{\partial v_i}{\partial x_j} \right) \end{equation} Note that, if $\vec{v} = 0$ then $\sigma_{ij} = -p\delta_{ij}$. A real fluid, which is isotropic at rest, ceases to be so in motion because of viscosity.

Isotropy, in this and other contexts, is being same in all directions. Homogeneity, on the other hand, is being same at all positions. An ideal fluid need not be homogeneous if it has a variable density. For example, when Lord Rayleigh analysed the instability of a heavy fluid resting on a lighter one, he assumed the fluids to be ideal (and hence isotropic) but they were not homogeneous.

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  • $\begingroup$ What is $\mu$ in this expression? It looks like setting $\mu=0$ gives the expression for ideal fluid back. @AmeyJoshi $\endgroup$ – SRS Dec 17 '19 at 20:59
  • $\begingroup$ $\mu$ is the dynamic viscosity of the fluid. $\endgroup$ – Amey Joshi Dec 18 '19 at 7:57
  • $\begingroup$ So $\sigma_{ij}=-p\delta_{ij}$ is true for Newtonian inviscous fluid, not necessarily ideal fluid (which is also incompressible)? $\endgroup$ – SRS Dec 18 '19 at 12:15
  • $\begingroup$ There is no such thing as Newtonian inviscid fluid. Newtonian fluids have viscosity. They are called Newtonian because they obey Newton's law of viscosity. Ideal fluids are inviscid. They have $\mu = 0$. $\endgroup$ – Amey Joshi Dec 18 '19 at 13:50
  • $\begingroup$ Yes. My bad. So the off-diagonal parts of $\sigma_{ij}$ of a fluid are solely due to viscosity? @AmeyJoshi $\endgroup$ – SRS Jan 2 at 6:10
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As I understand it an isotropic fluid is a fluid whose properties are not dependent on the direction along which they are measured. In a (so-called) Newtonian medium that is isotropic the stress tenssor is related to the strain via \begin{eqnarray} \varepsilon_{v}(p, t) &=& \phi_{v}E_{v}(p, t) \\ \varepsilon_{s}(p, t) &=& \phi_{s}E_{v}(p, t) \end{eqnarray} where $E_{v}$ and $E_{s}$ are (respectively) the scalar isotropic and the zero-trace parts of the strain tensor and $\phi_{s}$ and $\phi_{v}$ are real constants.

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