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?
Fluid Dynamics – Ideal Isotropic Fluid and Stress Tensor Analysis
fluid dynamicsstress-energy-momentum-tensorstress-strainsymmetryterminology
Related Solutions
$T^{ij}$ is nothing more or less than the flow of $i$-momentum across surfaces of constant $j$.1 As a result, the force exerted across a surface $S$ with unit normal one-form components $n_j$ has components $$ F_{(S)}^i = \int\limits_S T^{ij} n_j \,\mathrm{d}^{d-1}x $$ in $d$ dimensions.
The argument for symmetry is not that the cube is static. The argument is that the cube cannot have infinite angular acceleration as its size shrinks. That is, because we are dealing with a continuous fluid, it should be well behaved as we take our region to be arbitrarily small.
Consider a two-dimensional example of a square covering the area $x_0-L/2 < x < x_0+L/2$, $y_0-L/2 < y < y_0+L/2$. The left surface $x = x_0-L/2$ has $n^\mathrm{left}_j = \delta_j^x$ (sign chosen to correspond to flow of momentum into the square), and so the force on our square due to interactions across the left face has components $F_\mathrm{left}^i \approx L T_\mathrm{left}^{ix}$, where $T_\mathrm{left}^{ix}$ is $T^{ix}$ evaluated at the midpoint $(x_0-L/2,y_0)$. On the opposite surface, $n^\mathrm{right}_j = -\delta_j^x$, so $F_\mathrm{right}^i \approx -L T_\mathrm{right}^{ix}$. Similarly, $F_\mathrm{bottom}^i \approx L T_\mathrm{bottom}^{iy}$ and $F_\mathrm{top}^i \approx -L T_\mathrm{top}^{iy}$.
Torque is a $(d-2)$-form: $\tau = {}^*(\tilde{r} \wedge \tilde{F})$, with $\tilde{r}$ and $\tilde{F}$ the one-forms corresponding to displacement $\vec{r}$ and force $\vec{F}$. In 2D, $\tau = \epsilon_{ij} (r^i F^j - r^j F^i)$. If force components $F_\mathrm{left}^i$ are applied at $(x_0-L/2,y_0)$, then $r_\mathrm{left}^i = -(L/2) \delta^i_x$ and $\tau_\mathrm{left} \approx -(1/2) L^2 T^{yx}$. You can also check $\tau_\mathrm{right} \approx -(1/2) L^2 T^{yx}$ and $\tau_\mathrm{bottom} \approx \tau_\mathrm{top} \approx (1/2) L^2 T^{xy}$.
As we shrink the square down, the midpoints at which we evaluate $T^{ij}$ approach one another and we find the total torque is $\tau \approx L^2 (T_\mathrm{center}^{xy} - T_\mathrm{center}^{yx})$. However, the moment of inertia for a square of surface density $\sigma$ is $\sigma L^4/6$. Thus angular acceleration is $$ \alpha = \lim_{L\to0} \frac{6(T^{xy}-T^{yx})}{\sigma L^2} $$ at any point in the fluid. Thus we must have $T^{xy} = T^{yx}$, in order to avoid $\alpha \to \infty$. Note that this argument holds in higher dimensions, in more general settings than fluids, and for more general geometries/spacetimes.2
The argument does not however hold in the linear acceleration case. For example, the net $x$-force will have terms like $L (T_\mathrm{left}^{xx} - T_\mathrm{right}^{xx})$ and $L (T_\mathrm{bottom}^{xy} - T_\mathrm{top}^{xy})$. Even though mass is $\sigma L^2$, which would seem to imply linear acceleration goes as $1/L$, the fact is the pairs of stress tensor components naturally cancel as they are evaluated at the same point ($T_\mathrm{left}, T_\mathrm{right}, T_\mathrm{bottom}, T_\mathrm{top} \to T_\mathrm{center}$). No constraints are imposed from this consideration.
1I avoid saying "in the $j$-direction, since we are really interested in surfaces, and these are characterized by one-forms, not vectors. This is more apparent in non-Cartesian (better still non-diagonal) coordinate systems.
2This symmetry always holds, for any material or field, as long as momentum is conserved. You occasionally see reference to the antisymmetric part of the stress tensor, but this comes from splitting the physics into separate domains, and pretending that momentum is lost when going from one to another (e.g. torques can transfer angular momentum from bulk flow into particle spins, and we choose to treat the latter as some momentum-conservation-violating sink as far as the continuum-modeled fluid is concerned).
For a newtonian fluid, you can write the total stress tensor $\sigma_{ij}$ as $$\sigma_{ij} = -p \delta_{ij} + \tau_{ij}$$ with $$\tau_{ij} = \mu \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right).$$
$-p \delta_{ij}$ is the classic pressure term and $\tau_{ij}$ is the shear stress tensor with $\mu$ the shear viscosity.
When considering a perfect fluid, the shear stress tensor is considered null which lead to take a viscosity equals to 0. You can write the same formula for the thermal component in the energy equation which will give a thermal conductivity also equals to 0.
It is possible to write different behaviour law for the total stress tensor but the main idea will be the same: by neglecting the shear stress, you neglect the viscosity.
These hypothesis leads to consider Euler equations instead of Navier-Stokes equations.
One last thing :
I've found that lots of people (mostly italians, I have to say) define an ideal fluid as an incompressible one.
Be careful! The fluid is never incompressible. The flow may be, depending on Mach number and continuity equation. All fluids are compressible at some point.
Best 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.