Introduction
No continuum mechanics course can claim to be complete without a discussion of
material derivatives.
The material derivative computes the time rate of change of any quantity such as
temperature or velocity (which gives acceleration) for a portion of a material
moving with a velocity, \({\bf v}\). If the material is a fluid, then the movement
is simply the flow field.
The sketch to the right shows a fluid flowing through a converging nozzle.
Clearly any particle of fluid speeds up as it flows along the decreasing
cross-section path.
But if you were to focus on a single \((x,y)\) point
in space, you would not notice this acceleration because the fluid
velocity at that point appears constant.
The problem with this interpretation is that you are not following the same
particle to see it speeding up over time.
The material derivative effectively corrects for this confusing effect to
give a true rate of change of a quantity.
There are in fact many other names for the material derivative. They include
total derivative, convective derivative, substantial derivative,
substantive derivative, and still others.
Calculation of the Material Derivative
There are many symbolic representations of the material derivative. The most popular is
\({D() \over Dt}\), although \({d() \over dt}\) may also be used. The quantity
being differentiated goes inside the parentheses.
Both representations are shorthand for
\[
{D() \over Dt} = {\partial () \over \partial t} + {\bf v} \cdot {\partial () \over \partial {\bf x}}
\]
Note that the last derivative is with respect to \({\bf x}\), not \({\bf X}\). This
makes the equation ideal for Eulerian based applications such as fluid mechanics. It is
arrived at by applying the chain rule to Eulerian quantities that are dependent on
time and position. For example, the material derivative of
temperature, \(T\), is
\[
\begin{eqnarray}
{d \over dt}T(t,x,y,z) & = &
{\partial \, T \over \partial t} +
\left( {\partial \, T \over \partial x} \right) \left( {\partial x \over \partial t} \right) +
\left( {\partial \, T \over \partial y} \right) \left( {\partial y \over \partial t} \right) +
\left( {\partial \, T \over \partial z} \right) \left( {\partial z \over \partial t} \right)
\\
\\
& = &
{\partial \, T \over \partial t} +
v_x {\partial \, T \over \partial x} +
v_y {\partial \, T \over \partial y} +
v_z {\partial \, T \over \partial z}
\\
\\
& = &
{\partial \, T \over \partial t} +
{\bf v} \cdot \nabla T
\\
\\
& = &
{D T \over D t}
\end{eqnarray}
\]
This is written in tensor notation as
\[
{D T \over D t} = T_{,t} + v_i T_{,i}
\]
A second example: the material derivative of velocity gives acceleration.
\[
\begin{eqnarray}
{\bf a} =
{d \over dt}{\bf v}(t,x,y,z) & = &
{\partial {\bf v} \over \partial t} +
v_x {\partial {\bf v} \over \partial x} +
v_y {\partial {\bf v} \over \partial y} +
v_z {\partial {\bf v} \over \partial z}
\\
\\
& = &
{\partial {\bf v} \over \partial t} +
{\bf v} \cdot \nabla {\bf v}
\\
\\
& = &
{D {\bf v} \over D t}
\end{eqnarray}
\]
This is written in tensor notation as
\[
a_i = {D {\bf v} \over D t} = v_{i,t} + v_k v_{i,k}
\]
Note that \(i\) is the free index here. So the above equation for the
material derivative of velocity is actually three equations, one for each
component.
\[
\begin{eqnarray}
a_x & = &
{\partial v_x \over \partial t} +
v_x {\partial v_x \over \partial x} +
v_y {\partial v_x \over \partial y} +
v_z {\partial v_x \over \partial z} \\
\\
a_y & = &
{\partial v_y \over \partial t} +
v_x {\partial v_y \over \partial x} +
v_y {\partial v_y \over \partial y} +
v_z {\partial v_y \over \partial z} \\
\\
a_z & = &
{\partial v_z \over \partial t} +
v_x {\partial v_z \over \partial x} +
v_y {\partial v_z \over \partial y} +
v_z {\partial v_z \over \partial z}
\end{eqnarray}
\]
Material Derivative Particulars
Note how important it is to write \((v_k v_{i,k})\) not \((v_i v_{i,k})\).
This dictates which components are automatically summed over the three dimensions.
Also, it helps to apply a rigorous mathematical interpretation
to each partial derivative in order to minimize any confusion. For example,
\({\partial {\bf v} \over \partial t}\) implies that \(x, y,\) and \(z\) are
held constant. So this term captures the transient changes in the flow field
at each fixed point in space. If all the fluid is always flowing past a point
at the same velocity, then this term is zero. On the other hand, the
\({\bf v} \cdot {\partial {\bf v} \over \partial {\bf x}}\) term implies
that the calculations are done at a 'snapshot' in time.
Material Derivative Example
This example involves a ball being thrown straight up into the air. The usual description
of its position, \(y\), at any time, \(t\), is
\[
y = Y + v_o t - {1 \over 2} g \, t^2
\]
and its velocity is
\[
v = {dy \over dt} = v_o - g \, t
\]
and its acceleration is
\[
a = {dv \over dt} = -g
\]
So not surprisingly (at all), its acceleration is constant and equal to the acceleration
of gravity. This was easy because the position at any instant is an explicit
function of time and its initial position, \(Y\).
But what if instead of position as a function of time, the velocity field was given
as a function of position? This is typical of fluid flows. The velocity field
could then be written as
\[
v_y = \pm \sqrt{v_o^{\,2} - 2 g (y - y_o)}
\]
where the positive root applies to the ball rising, and the negative one represents it
falling. It is not readily apparent at all in this case that the velocity equation,
as a function of position, actually represents a constant acceleration equal to gravity.
But it does. To see this, apply the material derivative.
First, note that this problem is 1-D, so things simplify to
\[
a = {D v_y \over D t} = {\partial v_y \over \partial t} + v_y {\partial v_y \over \partial y}
\]
\({\partial v_y \over \partial t}\) is zero because the velocity, as described here,
is not transient. It does not change with time at any given location. \(v_y\) is given
above. And \({\partial v_y \over \partial y}\) is
\[
{\partial v_y \over \partial y} = {\mp g \over \sqrt{v_o^{\,2} - 2 g (y - y_o)}}
\]
And multiplying \(v_y {\partial v_y \over \partial y}\) out gives
\[
\begin{eqnarray}
v_y {\partial v_y \over \partial y} & = & \pm \sqrt{v_o^{\,2} - 2 g (y - y_o)}
\left( {\mp g \over \sqrt{v_o^{\,2} - 2 g (y - y_o)}} \right) \\
\\
& = & -g
\end{eqnarray}
\]
So this does indeed give the exact same constant acceleration of gravity as before.
