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Volume and Surface Forces

Generally speaking, fluids are acted upon by two distinct types of force. The first type is long range in nature--that is, such that it decreases relatively slowly with increasing distance between interacting elements--and is capable of completely penetrating into the interior of a fluid. Gravity is an obvious example of a long-range force. One consequence of the relatively slow variation of long-range forces with position is that they act equally on all of the fluid contained within a sufficiently small volume element. In this situation, the net force acting on the element becomes directly proportional to its volume. For this reason, long-range forces are often called volume forces. In the following, we shall write the total volume force acting at time $ t$ on the fluid contained within a small volume element of magnitude $ d V$ , centered on a fixed point whose position vector is $ {\bf r}$ , as

$\displaystyle {\bf F}({\bf r},t)\,d V.$ (1.1)

The second type of force is short range in nature, and is most conveniently modeled as momentum transport within the fluid. Such transport is generally due to a combination of the mutual forces exerted by contiguous molecules, and momentum fluxes caused by relative molecular motion. Suppose that $ \pi$ $ _x({\bf r}, t)$ is the net flux density of $ x$ -directed fluid momentum due to short-range forces at position $ {\bf r}$ and time $ t$ . In other words, suppose that, at position $ {\bf r}$ and time $ t$ , as a direct consequence of short-range forces, $ x$ -momentum is flowing at the rate of $ \vert$$ \pi$ $ _x\vert$ newton-seconds per meter squared per second in the direction of vector $ \pi$ $ _x$ . Consider an infinitesimal plane surface element, $ d{\bf S} = {\bf n}\,dS$ , located at point $ {\bf r}$ . Here, $ dS$ is the area of the element, and $ {\bf n}$ its unit normal. (See Section A.7.) The fluid which lies on that side of the element toward which $ {\bf n}$ points is said to lie on its positive side, and vice versa. The net flux of $ x$ -momentum across the element (in the direction of $ {\bf n}$ ) is $ \pi$ $ _x \cdot d{\bf S}$ newtons, which implies (from Newton's second law of motion) that the fluid on the positive side of the surface element experiences a force $ \pi$ $ _x \cdot d{\bf S}$ in the $ x$ -direction due to short-range interaction with the fluid on the negative side. According to Newton's third law of motion, the fluid on the negative side of the surface experiences a force $ -$$ \pi$ $ _x \cdot d{\bf S}$ in the $ x$ -direction due to interaction with the fluid on the positive side. Short-range forces are often called surface forces, because they are directly proportional to the area of the surface element across which they act. Let $ \pi$ $ _y({\bf r},t)$ and $ \pi$ $ _z({\bf r},t)$ be the net flux density of $ y$ - and $ z$ -momentum, respectively, at position $ {\bf r}$ and time $ t$ . By a straightforward extension of previous argument, the net surface force exerted by the fluid on the positive side of some planar surface element, $ d{\bf S}$ , on the fluid on its negative side is

$\displaystyle {\bf f} = (-$$\displaystyle \mbox{\boldmath$\pi$}$$\displaystyle _x\cdot d{\bf S},\, -$$\displaystyle \mbox{\boldmath$\pi$}$$\displaystyle _y\cdot d{\bf S},\, -$$\displaystyle \mbox{\boldmath$\pi$}$$\displaystyle _z\cdot d{\bf S}).$ (1.2)

In tensor notation (see Appendix B), the previous equation can be written

$\displaystyle f_i = \sigma_{ij}\,dS_j,$ (1.3)

where $ \sigma_{11}= -($$ \pi$ $ _x)_x$ , $ \sigma_{12} = -($$ \pi$ $ _x)_y$ , $ \sigma_{21}=-($$ \pi$ $ _y)_x$ , et cetera. (Note that, because the subscript $ j$ is repeated in the previous equation, it is assumed to be summed from $ 1$ to $ 3$ . Hence, $ \sigma_{ij}\,dS_j$ is shorthand for $ \sum_{j=1,3}\sigma_{ij}\,dS_j$ .) Here, the $ \sigma_{ij}({\bf r}, t)$ are termed the local stresses in the fluid at position $ {\bf r}$ and time $ t$ , and have units of force per unit area. Moreover, the $ \sigma_{ij}$ are the components of a second-order tensor (see Appendix B) known as the stress tensor. [This follows because the $ f_i$ are the components of a first-order tensor (as all forces are proper vectors), and the $ dS_i$ are the components of an arbitrary first-order tensor (as surface elements are also proper vectors--see Section A.7--and Equation (1.3) holds for surface elements whose normals point in any direction), so application of the quotient rule (see Section B.3) to Equation (1.3) reveals that the $ \sigma_{ij}$ transform under rotation of the coordinate axes as the components of a second-order tensor.] We can interpret $ \sigma_{ij}({\bf r}, t)$ as the $ i$ -component of the force per unit area exerted, at position $ {\bf r}$ and time $ t$ , across a plane surface element normal to the $ j$ -direction. The three diagonal components of $ \sigma_{ij}$ are termed normal stresses, because each of them gives the normal component of the force per unit area acting across a plane surface element parallel to one of the Cartesian coordinate planes. The six non-diagonal components are termed shear stresses, because they drive shearing motion in which parallel layers of fluid slide relative to one another.


next up previous
Next: General Properties of Stress Up: Mathematical Models of Fluid Previous: What is a Fluid?
Richard Fitzpatrick 2016-03-31