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.

  • 2
    $\begingroup$ A fluid can only support the bulk modulus $${\rm d}V = -K V {\rm d} P$$ $\endgroup$ Commented Jul 12, 2014 at 0:18
  • $\begingroup$ If the normal stress term ultimately relates to the bulk modulus and volumetric strain, can you provide some physical insight into how this actually manifests itself? $\endgroup$
    – Bryson S.
    Commented Jul 12, 2014 at 0:25
  • $\begingroup$ I guess if you apply equal pressures on all sides of a cube there is going to be a change in volume only and no change in shape. There will be no "flow", just an overall scaling. $\endgroup$ Commented Jul 12, 2014 at 1:16
  • $\begingroup$ Is this anything more than the everyday concept of suction? $\endgroup$
    – Phil Frost
    Commented Jul 12, 2014 at 1:38
  • $\begingroup$ @Phil Frost Suction is just relative differences in pressure. What I am talking about effectively amounts to negative pressure (tension). $\endgroup$
    – Bryson S.
    Commented Jul 12, 2014 at 2:31

2 Answers 2


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.

  • $\begingroup$ Thanks, dmckee for merging the two replies, I did not know how to do it myself. $\endgroup$ Commented Jul 12, 2014 at 17:47
  • $\begingroup$ One last comment to finish my answer: a tensile normal stress component may exist, but that does not imply tensile pressure. You'd require a cohesive mechanism to develop tensile, hydrostatic pressure and a gas has no such mechanism (gases do have weak London-type forces, but they are very short range). Also, the additional caveat is that we're assuming the fluid is well described as a continuum. That's not always true, especially for rarefied gases. $\endgroup$ Commented Jul 12, 2014 at 18:42
  • $\begingroup$ This is great stuff, but are you saying that the presence of tensile normal stress is just illusory? Or more to the point, can a fluid (specifically a gas) support tensile forces? How? $\endgroup$
    – Bryson S.
    Commented Jul 12, 2014 at 20:04

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.

  • $\begingroup$ Is there any known mechanism whereby this negative pressure could be supported in a gas? $\endgroup$
    – Bryson S.
    Commented Jul 12, 2014 at 20:00
  • $\begingroup$ @Bryson S.: I am not sure if negative pressure is possible for gases, but if it is, the mechanism should be the same: molecular attraction. $\endgroup$
    – akhmeteli
    Commented Jul 12, 2014 at 22:59
  • $\begingroup$ But aren't the atoms/molecules spaced to far apart for these forces to have any discernible effect? $\endgroup$
    – Bryson S.
    Commented Jul 12, 2014 at 23:36
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    $\begingroup$ @Bryson S.: There is no such thing as "too far" in physics, there is only ""too far compared to something". In this case, I guess, you should compare the potential energy of attraction with the kinetic energy. Again, it is not obvious that negative absolute pressure is possible in gases, it is not obvious that it's impossible. $\endgroup$
    – akhmeteli
    Commented Jul 12, 2014 at 23:46

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