How can a gas support tensile stresses? In working through a rigorous derivation of the compressible Navier-Stokes equations, I find that the momentum flux in the X-direction should be driven not only by the normal pressure gradient $\frac{\partial p}{\partial x}$ and shear stress terms $\frac{\partial(\tau_{yx})}{\partial x}$ and $\frac{\partial(\tau_{zx})}{\partial x}$, but also by the gradient of the normal stress $\frac{\partial(\tau_{xx})}{\partial x}$. It's intuitively clear to me how adjacent lamina moving at different speeds can transfer momentum across their interface, and so the shear stress terms in the momentum equation are readily intelligible. The normal stress term, on the other hand, is far less intuitive because I cannot see how a freely-deforming fluid can support tensile stresses. Positive normal stresses (i.e. compression) are not that hard to understand, but it's proving exceedingly difficult to fully envisage an element of a fluid "pulling on" an adjacent element in a way even remotely analogous to the behavior of a solid under the same conditions. I am also unclear on the difference between "pressure" and "normal stress" in the fluid. How exactly are these terms different? I am interested primarily with gases not liquids.
 A: Let's take your last question first. Let the stress tensor at a point (x,y,z) in the fluid be given as $\sigma$. You can pick a Cartesian basis $\{ e_1, e_2, e_3 \}$ and express the components of the tensor in that basis
$$
\begin{bmatrix}
\sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
\sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\
\sigma_{xz} & \sigma_{yz} & \sigma_{zz} 
\end{bmatrix}
$$
The normal stresses are simply  $\sigma_{xx},\sigma_{yy}$ and $\sigma_{zz}$. It is important to realize that these stresses will have different values in another basis.  
Clearly, you can't attach too much physical significance to things that are basis dependent. However, it is a theorem of continuum mechanics that you can ALWAYS find at least one basis in which the off-diagonal (shear terms) are zero. In this basis, the tensor components are
$$
\begin{bmatrix}
\sigma_{1} & 0 & 0 \\
0 & \sigma_{2} & 0 \\
0 & 0 & \sigma_{3} 
\end{bmatrix}
$$
These numbers have actual physical significance. $\max({\sigma_1, \sigma_2, \sigma_3})$ is the largest principal normal stress at the point. Similarly, $\min({\sigma_1, \sigma_2, \sigma_3})$ is the smallest normal stress at that point. It is not too hard to realize that $\sigma_1, \sigma_2, \sigma_3$ are the eigenvalues of the stress tensor.
On the other hand, the pressure is (-1/3) times the trace of the stress tensor, i.e.
$$
p = -\frac{1}{3} \sigma_{jj}
$$
The trace is an invariant of the stress tensor, so if you take the sum of the diagonals of the stress tensor in any basis you'll get the same value. Mathematically,
$$
Tr([\beta][\sigma_{ij}][\beta]^T) = Tr(\sigma_{ij})
$$
So you see that the pressure and the normal stress are very different entities indeed. In particular, the pressure is ISOTROPIC - it has no preferred direction.

Now consider a state of pure shear in a fluid. To keep matters simple, we'll assume planar flow and ignore out of plane components.
The stress tensor for pure shear in our standard basis looks like this
$$
\begin{bmatrix}
0 & \tau  \\
\tau & 0  \\
\end{bmatrix}
$$
Looks like the normal stresses are zero, right? Not so fast. As this is a symmetric real tensor, you can ALWAYS find another basis in which you have normal stress components!
In fact, if you solve the eigenvalue problem setting
$\det (\sigma-\lambda I )=0$, you get principal normal stresses of $\pm \tau$. 
So, in a coordinate system with basis vectors $e'_1 = \{ (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \}$ and $e'_2 = \{ (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \}$ rather than $\{(1,0), (0,1)\}$, you get a stress tensor from a situation of "pure" shear that looks like this
$$
\begin{bmatrix}
\tau & 0  \\
0  & -\tau  \\
\end{bmatrix}
$$
You can easily verify this by carrying out the change of basis yourself. 
So what looks like "pure shear" in one basis is biaxial normal stresses in another basis. Since the signs are different, you have both tensile and compressive normal stresses in your fluid. 
A: It looks like the question boils down (at least in part) to the following: can a fluid have negative ABSOLUTE pressure? This question has been discussed here several times. My take is: it can (although such state is probably metastable in the best case), because the force between two molecules can be attractive. See, e.g., http://www.youtube.com/watch?v=BickMFHAZR0 , where they discuss how trees taller than 10m can deliver water to their top. I don't know though if a gas, rather than a liquid, can have negative pressure. In cosmology, the so-called Chaplygin gas is considered though. 
