What is the difference between hydrostatic pressure and stress tensor?
[Physics] Difference between pressure and stress tensor
fluid-staticspressurestress-energy-momentum-tensorthermodynamics
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Pressure is defined as force per unit area applied to an object in a direction perpendicular to the surface. And naturally pressure can cause stress inside an object. Whereas stress is the property of the body under load and is related to the internal forces. It is defined as a reaction produced by the molecules of the body under some action which may produce some deformation. The intensity of these additional forces produced per unit area is known as stress (pretty picture from wikipedia):
EDIT PER COMMENTS
Overburden Pressure or lithostatic pressure is a case where the gravity force of the object's own mass creates pressure and results in stress on the soil or rock column. This stress increases as the mass (or depth) increases. This type of stress is uniform because the gravity force is uniform.
http://commons.wvc.edu/rdawes/G101OCL/Basics/earthquakes.html
Included in lithostatic pressure are the weight of the atmosphere and, if beneath an ocean or lake, the weight of the column of water above that point in the earth. However, compared to the pressure caused by the weight of rocks above, the amount of pressure due to the weight of water and air above a rock is negligible, except at the earth's surface. The only way for lithostatic pressure on a rock to change is for the rock's depth within the earth to change.
Since this is a uniform force applied throughout the substance due to mostly to the substance itself, the terms pressure and stress are somewhat interchangeable because pressure can be viewed as both an external and internal force.
For a case where they are not equal, just look that the image of the ruler. If pressure is applied at the far end (top of image) it creates unequal stress inside the ruler, especially where the internal stress is high at the corners.
Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an $x$-directed force pushing along an interface of constant $y$; this would be $\sigma_{xy}$. If we assemble all such combinations $\sigma_{ij}$, the collection of them is the stress tensor.
Pressure is part of the stress tensor. The diagonal elements form the pressure. For example, $\sigma_{xx}$ measures how much $x$-force pushes in the $x$-direction. Think of your hand pressing against the wall, i.e. applying pressure.
Given that pressure is one type of stress, we should have a name for the other type (the off-diagonal elements of the tensor), and we do: shear. Both pressure and shear can be internal or external -- actually, I'm not sure I can think of a real distinction between internal and external.
A gas in a box has a pressure (and in fact $\sigma_{xx} = \sigma_{yy} = \sigma_{zz}$, as is often the case), and I suppose this could be called "internal." But you could squeeze the box, applying more pressure from an external source.
Perhaps when people say "pressure is internal" they mean the following. $\sigma$ has some nice properties, including being symmetric and diagonalizable. Diagonalizability means we can transform our coordinates such that all shear vanishes, at least at a point. But we cannot get rid of all pressure by coordinate transformations. In fact, the trace $\sigma_{xx} + \sigma_{yy} + \sigma_{zz}$ is invariant under such transformations, and so we often define the scalar $p$ as $1/3$ this sum, even when the three components are different.
1Now the word "tensor" has a very precise meaning in linear algebra and differential geometry and tensors are very beautiful things when fully understood. But here I'll just use it as a synonym for "matrix."
Best Answer
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).