Difference between pressure and stress tensor What is the difference between hydrostatic pressure and stress tensor? 
 A: Consider a continuous body $C$ and a closed subset $V$ with boundary $\partial V$ completely included in $C$ (notice that $\partial V \subset V$). 
If $p \in \partial V$, the external part of $C$, namely $C \setminus V$ acts on $p$ (actually on a neighbourhood of $p$ in $\partial V$) by means of a superficial force
 $$d\vec{f} = \vec{s}(p, \vec{n}) dS(p)\:,$$
 where $\vec{n}$ is the outgoing unit normal vector at $p$ and $dS(p)$ the infinitesimal surface around $p$ included in the larger surface made of the boundary $\partial V$. So $dS(p)$ is normal to $\vec{n}$.
The total force acting on $V$ due to the part of $C$ outside $V$ is:
$$\vec{F}_V = \int_{+\partial V} \vec{s}(p, \vec{n}) dS(p)\quad (1)$$ 
The map $ C \times \mathbb{S}^2 \ni (p, \vec{n}) \mapsto \vec{s}(p, \vec{n})$ is called stress function. 
Notice that a point $p \in C$ has several stress vectors applied on it, in view of the various choices of $\vec{n}$. 
Moreover $ \vec{s}(p, \vec{n})$ is not necessarily parallel to $\vec{n}$, it could have components normal to $\vec{n}$ (so parallel to $dS(p)$) describing friction forces.
A fundamental theorem due to Cauchy establishes  that, under some quite general hypotheses on the continuous body, the map $\mathbb{S}^2 \ni  \vec {n} \mapsto \vec{s}(p, \vec{n})$ is the restriction of a linear map. As a consequence it is the action of a tensor field (I use Einstein's convention on indices below)
$$s^a(p, \vec{n}) = \sigma(p)^{ab} n_b$$
where $n_a$ are the covariant components of $\vec{n}$ and $s^a$ the contravariant components of $\vec{s}$ (actually using orthonormal frames there is no distinction between these two types of components).  
The Cauchy stress-tensor $\sigma$ is always symmetric: $\sigma(p)^{ab}= \sigma(p)^{ba}$ (it can be proved by imposing the standard relation between momenta of forces and angular momentum).
The identity (1) can be re-written taking advantage of divergence theorem:
$$F^a_V = \int_{V} \sigma^{ab}, _b d^3x$$
where $,_b$ denotes the derivative with respect to $x^b$. This is the starting point to write down in local form "F=ma" for continuous bodies.
A very special case is that of an   isotropic $\sigma$:
$$\sigma(p)^{ab} = -P(p)\delta^{ab}\qquad (3)$$ 
where $P(p)\geq 0$ defines a scalar field. Here all the stress vectors  are always normal to the boundary of $V$ no matter how you fix that portion of continuous body, and the total force  is purely  compressive. The scalar $P$ is the pressure. Non-viscous fluids have the stress tensor of the form (3). Even viscous fluids have that form if the fluid is in an equilibrium configuration.
The presence of viscous forces adds  some non isotropic (i.e. non-diagonal) terms to the RHS of (3).  
A: The hydrostatic pressure is one-third of the trace of the stress tensor. In other words, the pressure is the average of the normal stresses on a given volume. 
If you are using equations that include both pressure and a stress tensor, care must be taken to make sure that the diagonal components of the stress tensor don't also include the pressure terms.
The off-diagonal terms of the stress tensor are the shear stresses on the element. So the differences between the stress tensor and the pressure are:


*

*Pressure is a scalar, the stress tensor is a tensor

*Pressure is only based on the normal stress on the element, the stress tensor includes both normal and shear terms

