In fluid mechanics it is assumed, that the normal components of the stress tensor are all the same and identical to the pressure p: $\sigma_{xx}= \sigma_{yy}=\sigma_{zz} = p$

Where does this come from? In solid materials it is not the case in general, but why in fluids? In all my texts on fluid mechanics this assumption is not justified. What particular property of the state "fluid" is responsible for the assertion

$\sigma_{xy}= p \delta_{i,j}+\tau_{i,j}$ where $\tau$ is the viscous stress tensor.


3 Answers 3


Even for a fluid, the diagonal components of the stress tensor are not each equal to pressure (or minus the pressure, depending on the sign convention used for the stress tensor) unless the fluid is not deforming (static equilibrium) or the fluid is inviscid or the diagonal components of the viscous stresses are zero. Otherwise, the trace of the stress tensor is equal to 3 times the pressure (assuming incompressible flow), and the diagonal components of the viscous stresses contribute to the diagonal components of the overall stress tensor (as your final equation shows).


The formula $\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-p$ is true only for a stationary fluid. For a fluid in motion the three normal stresses will be different from each other, and then pressure is defined to be their average: $-p\equiv(\sigma_{xx}+\sigma_{yy}+\sigma_{zz})/3$.

For a stationary fluid, $\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-p$ is a consequence of the fact that a fluid cannot remain at rest when acted upon by shear stress, no matter how small its magnitude (in fact this is taken to be the definition of simple fluids like air and water; solids on the other hand deform to finite extent and come to rest). This definition implies that in a stationary fluid there can be no shear stresses; every area element within the fluid must experience a purely normal stress. Then it is simple enough to prove that this normal stress must be independent of the orientation of the area element at a given point.



Basically, the $\sigma_{i,j}$ is defined as the tensor for the surface forces on the fluid, and those will be the pressure and the friction. The tensor $\sigma_{i,j}$ reads: the surface force in the direction i over the face with normal j of the element of fluid. Then, since the pressure acts perpendicularly to each face of the fluid, it will be the diagonal elements and since the friction is normal to the faces of the element of fluid it will be in the off-diagnal elements.

Let see it a tiny bit more formally

First comes the derivation of the dynamics equation for an element of a fluid (which is Newton's second law).

This is,

$\int \rho \frac{D v_i}{Dt}dv = \int f_i dv + \int \sigma_{i,j} n_j ds$

where on the r.h.s. we have two terms because the first is for volume forces and the second for surface forces.

Using Gauss theorem: $\int \sigma_{i,j} n_j ds = \int \partial_j \sigma_{i,j} dv$


$\rho \frac{D v_i}{Dt} = f_i + \partial_j \sigma_{i,j}$

We have volume forces (like gravity) and surface forces, like pressure (which is normal to each face of the fluid), and -in viscous fluids, friction forces (that are parallel to each face of the fluid element).

Friction will be not null only when you have relative velocity between the fluid element and another surface, which can be a wall or also another element of fluid.

For the volume forces they act on the center of each element as a vector. For the surface forces, you have to define the surface on which the force is acting and its direction. Then, whereas in volume forces we needed one vector to define it, for a surface force we need 2 vectors: one that defines the surface where the force is acting (typically a unit vector normal to the surface) and another to define the direction in which the force is acting.

Typically, instead of two vectors one two dimensional tensor is used: $\sigma_{i,j}$. (whereas for a volume force we had a 'one dimensional tensor' $f_{i}$)

And this is basically the answer again:

Then, $\sigma_{i,j}$ can be defined as: "The force in the direction $i$ on the surface with normal $j$".

From that, can be seen that the pressure will be the diagonal elements, because the pressure acts on all the sides of a volume element in a perpendicular direction.

But for the friction is the opposite, there'll be no elements of friction in the diagonal, but rather in the non-diagonal elements, because friction is parallel to the surfaces.

Viscous fluids case.

When deriving $\tau_{i,j}$ you wan it to be null for 1)translation and 2)rotation.

Because of 1) is that $\tau_{i,j}$ cannot depend on the velocity of the fluid. But what restriction brings 2)?

Let's see what's the velocity field for the rotation case:

$\bar{v} = \bar{\omega} \times \bar{r}$

$v_i = \epsilon_{ijk} \omega_j r_k$

This means that you have to construct $\tau_{i,j}$ in a symmetric way (then, that vector field will vanish in $\tau_{i,j}$)

Then, $\tau_{i,j}$ can look like:

$\alpha(\partial_{i}v_{j} + \partial_{j}v_{i})$

And there's still another symmetric combination of the space derivatives of the velocities:

$\beta \partial_l v_{l} \delta_{i,j}$


$\tau_{i,j} = \alpha(\partial_{i}v_{j} + \partial_{j}v_{i})$ + $\beta \partial_l v_{l} \delta_{i,j}$

Hence, the diagonal components vanish and you recover the case without friction.

$\tau_{i,j}$ can of course depend on higher order derivatives and in those cases the fluid is called non-newtonian.

I hope this was your question. Tell me otherwise ;)


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