Page 240 - Intro to Tensor Calculus

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Substituting the relations from the equations (2.3.66) into the triple scalar product gives

j
p
k
n
k
i
m
j
0
∆V = e ijk (a + u i ,m a )(b + u b )(c + u c ).
,n
,p
Expanding the triple scalar product and employing the result from Exercise 1.4, problem 34, we find the
simplified result gives us the dilatation
∆V − ∆V r
0
Θ= = u ,r = div (~u). (2.3.67)
∆V
That is, the dilatation is the divergence of the displacement field. If the divergence of the displacement field
is zero, there is no volume change and the material is said to be incompressible. If the divergence of the
displacement field is different from zero, the material is said to be compressible.
Note that the strain e ij is expressible in terms of the displacement field by the relation
1 mn r
e ij = (u i,j + u j,i ), and consequently g e mn = u . (2.3.68)
,r
2
Hence, for an orthogonal system of coordinates the dilatation can be expressed in terms of the strain elements
along the main diagonal.

Conservation of Mass

Consider the material in an arbitrary region R of a continuum. Let % = %(x, y, z, t) denote the density
3
of the material within the region. Assume that the dimension of the density % is gm/cm in the cgs system
of units. We shall assume that the region R is bounded by a closed surface S with exterior unit normal ~n
defined everywhere on the surface. Further, we let ~v = ~v(x, y, z, t) denote a velocity field associated with all
points within the continuum. The velocity field has units of cm/sec in the cgs system of units. Neglecting
sources and sinks, the law of conservation of mass examines all the material entering and leaving a region R.
ZZZ
Enclosed within R is the material mass m where m = %dτ with dimensions of gm in the cgs system of
R
units. Here dτ denotes an element of volume inside the region R. The change of mass with time is obtained
by differentiating the above relation. Differentiating the mass produces the equation
ZZZ
∂m ∂%
= dτ (2.3.69)
∂t R ∂t
and has the dimensions of gm/sec.
Consider also the surface integral
ZZ
I = %~v · ˆndσ (2.3.70)
S
where dσ is an element of surface area on the surface S which encloses R and ˆn is the exterior unit normal
vector to the surface S. The dimensions of the integral I is determined by examining the dimensions of each
term in the integrand of I. We find that
gm cm 2 gm
[I]= · · cm =
cm 3 sec sec
and so the dimension of I is the same as the dimensions for the change of mass within the region R. The
surface integral I is the flux rate of material crossing the surface of R and represents the change of mass