3.2 Motion of a Continuum
63
Dotting this expression with the velocity yields
v · ∇T = e
−αt
x 1 + 2x 1 e
−αt
+ 2x 2 x 3
.
Inserting these expressions into the above equation for dT /dt provides the desired
result.
Equation (3.26) also can be used to differentiate vectors and higher-order tensors.
For example, the acceleration field for a continuum is given by
a =
d
dt
v(R, t) =
∂v
∂t
a =
d
dt
v(r, t) =
∂v
∂t
+ v · ∇v
(3.29)
for the material and spatial descriptions, respectively.
Example 3.4 The velocity field is given in the spatial form v(r, t). Determine the
Cartesian components of the acceleration field a(r, t).
Solution
Substituting a = a i e i , v = v i e i , and ∇ = e i ∂/∂x i into Eq. (3.29) 2 yields
a i e i =
∂
∂t
(v i e i ) + (v k e k ) ·
e i
∂
∂x i
(v j e j )
= e i
∂v i
∂t
+ e j v k δ ki
∂v j
∂x i
= e i
∂v i
∂t
+ e j v i
∂v j
∂x i
= e i
∂v i
∂t
+ v j
∂v i
∂x j
.
The resulting acceleration components agree with those computed using a i =
dv i /dt and Eq. (3.28).
In Eqs. (3.25) and (3.27), the first term represents the local rate of change of φ
at a fixed position r, and the second term gives the convective rate of change as the
particle travels through space. To better understand the physical meaning of these
terms, we let φ represent temperature and consider two special cases for a particle
moving along the x-axis (Fig. 3.3). In the first case, the temperature field is spatially
uniform at all times (∇φ = ∂φ/∂x 1 = 0), but it changes in time (∂φ/dt = 0);