Material Derivative

The time rate of change of a quantity such as temperature or velocity of a material particle is known as a material derivative, and is denoted by a superimposed dot or by $D/Dt$.

(i)

When referential or Lagrangian description of a quantity is used, i.e.,

$$\theta = \theta (X_1,\ X_2,\ X_3,\ t)$$

then

$$\dot\theta = \frac{D\theta}{Dt} = \frac{\partial \theta}{\partial t}\bigg\vert _{X_i - \mbox{fixed}}.$$ (3.5.1)

(ii)

When spatial or Eulerian description of a quantity is used, i.e.,

$$\theta = \theta(x_1,\ x_2,\ x_3,\ t),$$

where

$$x_i = x_i(X_1,\ X_2,\ X_3,\ t),$$

then

$$\dot\theta = \frac{D\theta}{Dt} = \frac{\partial\theta}{\partial ... ...x_j}\hspace{.3in}\frac{\partial x_j}{\partial t}\bigg\vert _{X_i-\mbox{fixed}}.$$ (3.5.2)

In rectangular Cartesian coordinates

$$v_j = \frac{\partial x_j}{\partial t}\bigg\vert _{X_i - \mbox{fixed}}$$ (3.5.3)

is the $j$th component of the velocity of a material particle. Therefore, in rectangular Cartesian coordinates,

$$\dot\theta = \frac{\partial\theta}{\partial t} + \frac{\partial\theta}{\partial x_j} v_j.$$ (3.5.4)

Note that in (3.5.4) $\theta$ is given in the spatial description.

Example: Given the motion

$$x_i = X_i (1 + t),\ t\ge 0,$$

find the spatial description of the velocity field.

Solution

$$x_i =\ X_i (1 + t)$$

$$v_i =\ \dot x_i = X_i = \frac{x_i}{1 + t}.$$

Example: Given

$$\theta = 2(x^2_1 + x^2_2)\ \ x_i = X_i (1 + t).$$

Find, at $t = 1$, the rate of change of temperature of the material particle which in the reference configuration was at $(1,1,1)$.

Solution

$$\hspace*{.7in} \theta =\ 2 (x^2_1 + x^2_2)$$

$$=\ 2 [X^2_1 (1 + t)^2 + X^2_2(1 + t)^2]$$

$$\dot\theta =\ 2[2X^2_1(1 + t) + 2X^2_2(1 + t)]$$

$$\therefore \dot\theta\ \ t = 1\ \ (1,1,1) = 16.$$

$$(ii)\hspace*{.7in} \dot\theta =\ \frac{\partial\theta}{\partial t} + \frac{\partial\theta}{\partial x_j}\ \frac{\partial x_j}{\partial t}$$

$$=\ 0 + 4x_1\frac{x_1}{1 + t} + 4x_2\frac{x_2}{1 + t}.$$

At $t = 1$, for the material particle $(1,1,1)$

$$x_i =\ 2(\delta_{i1} + \delta_{i2} + \delta_{i3})$$

$$\therefore \hspace*{.3in} \dot\theta =\ \frac{4(2)^2}{1 + 1} + \frac{4(2)^2}{1 + 1} = 16.$$

Exercise: The motion of a continuous medium is defined by the equations

$$x_1 =\ \frac{1}{2}(X_1 + X_2)e^t + \frac{1}{2}(X_1 - X_2) e^{-t},\ x_2 = \frac{1}{2}(X_1 + X_2)e^t-\frac{1}{2} (X_1 - X_2)e^{-t},$$

$$x_3 =\ X_3,\hspace{2in} 0 \le t < .$$

(a)

Express the velocity components in terms of referential coordinates and time.

(b)

Express the velocity components in terms of spatial coordinates and time.

Exercise: Given the motion of a body to be $x_i = (X_1 + ktX_2)\delta_{i1} + X_2\delta_{i2} + X_3\delta_{i3}$, and the temperature field by $\theta = x_1 + x_2$, find $\dot\theta$ for the particle which currently is at the place $(1,1,1)$.