0.3 The kinematics of fluid motion  (Page 8/9)

Thus the antisymmetric part of the velocity gradient tensor corresponds to rigid body rotation, and, if the motion is a rigid one (composed of a translation plus a rotation), the symmetric part of the velocity gradient tensor will vanish. For this reason the tensor $e_{ij}$ is called the deformation or rate of strain tensor and its vanishing is necessary and sufficient for the motion to be without deformation, that is, rigid.

Physical interpretation of the (rate of) deformation tensor

The (rate of) deformation tensor is what distinguishes fluid motion from rigid body motion. Recall that a rigid body is one in which the relative distance between two points in the body does not change. We show here that the (rate of) deformation tensor describes the rate of change of the relative distance between two particles in a fluid. Also, it describes the rate of change of the angle between three particles in the fluid.

First we will see how the distance between two material points change during the motion. The length of an infinitesimal line segment from $P$ to $Q$ is $ds$ , where

now $P$ and $Q$ are the material particles $\xi$ and $\xi +d\xi$ so that $d{\xi }_j$ and $d{\xi }_k$ do not change during the motion. Also, recall that $D{x}_i/Dt=v_i$ . Thus

by symmetry. However,

Thus

by symmetry, or

Now $d{x}_i/ds$ is the ith component of a unit vector in the direction of the segment $PQ$ , so that this equation says that the rate of change of the length of the segment as a fraction of its length is related to its direction through the deformation tensor.

In particular, if $PQ$ is parallel to the coordinate axis 01 we have $d\mathbf{x}/ds={\mathbf{e}}_{\left(1\right)}$ and

Thus $e_{11}$ is the rate of longitudinal strain of an element parallel to the 01 axis. Similar interpretations apply to $e_{22}$ and $e_{33}$ .

Now let's examine the angle between two line segments during the motion. Consider the segment, $PQ$ and $PR$ where $PQ$ is the segment $\xi +\mathbf{d}\xi$ as before and PR is the segment $\xi +\mathbf{d}{\xi }^{\text{'}}$ . If $\theta$ is the angle between them and $d{s}^{\text{'}}$ is the length of $PR$ , we have from the scalar product,

Differentiating with respect to time we have

since $d{v}_i^{\text{'}}=(\partial v_i/\partial x_j)d{x}_j^{\text{'}}$ . The $i$ and $j$ are dummy suffixes so we may interchange them in the first term on the right, then performing differentiation we have after dividing by $ds$ $d{s}^{\text{'}}$

Now suppose that $d{x}^{\text{'}}$ is parallel to the axis 01 and $dx$ to the axis 02, so that $(d{x}_i^{\text{'}}/d{s}^{\text{'}})\text{and}(d{x}_j/ds)={\delta }_{j2}\text{and}{\theta }_{12}=\pi /2$ . Then

Thus $e_{12}$ is to be interpreted as one-half the rate of decrease of the angle between two segments originally parallel to the 01 and 02 axes respectively. Similar interpretations are appropriate to $e_{23}$ and $e_{31}$ .