# Fluids Lecture 01

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### 1 Hydrostatics

Hydrostatics is the interaction of short and long range forces in a fluid.

Long-range forces act on all elements ofa fluid. Gravity is the most important example of a long-range force. Electromagnetic forces may be important if the fluid is charged or conductive and carrying a current. By definition, long-range forces vary slowly over distance, so we often use the fact that they are constant over small (differential) distances.

Short-range forces are more complicated. Short-range forces tend to interact with a fluid as a function of the surface area of some volume element. The effects of short-range forces are then propagated through the fluid by the stress tensor $\Sigma$ . Typically, there are symmetries between a surface force and the stress tensor. An obvious one is that if the surface force changes sign, so should the stress tensor. To find others, we’re going to have to invoke some formalism.

Let us say that $\sigma _{ij}$ is the component of force in the $i$ direction acting on a surface area with its normal in the $j$ direction. Thus $\Sigma$ is given by:

$\Sigma ={\begin{pmatrix}\sigma _{21}&\sigma _{22}&\dots \\\vdots &&\\\end{pmatrix}}\,\!$ The diagonal components of $\Sigma$ are the normal stresses, and the others are the more interesting tangential stresses. Linear algebra tells us that one may always choose principal axes such that $\Sigma$ is diagonal. We also know that the trace of $\Sigma$ is invariant.

Fluids are defined as being unable to withstand any tendency by applied forces to deform its shape without changing its volume. For example, if $\Sigma$ is given by:

$\Sigma ={\begin{pmatrix}{1 \over 3}&0&0\\0&{1 \over 3}&0\\0&0&{1 \over 3}\\\end{pmatrix}}\,\!$ Then it is isotropically compressive, and a fluid is able to withstand this stress without deformation. However, this compression results in a hydrostatic pressure. Generally, for a fluid at rest, the hydrostatic pressure is given by:

$p=-{1 \over 3}tr(\Sigma )\,\!$ Note that if the diagonal elements of $\Sigma$ are not identical, then the fluid cannot be at rest (it is deforming).

For a fluid in hydrostatic equilibrium, the sum of the surface forces must be zero, giving us that the force per volume is:

$F=\div p\,\!$ ### 2 Dynamics of Ideal Fluids (Dry Water)

${\frac {\partial \rho }{\partial t}}+\nabla \cdot (\rho u)=0\,\!$ 