This example is a little more substantial than the above. It concerns the balance of momentum for a
continuum. To see a full description of all the physics involved, you should consult a book on continuum
mechanics. The situation is of a material in three dimensions and it deforms and moves about in three
dimensions. This means this material is not a rigid body. Let B_{0} denote an open set identifying a chunk of
this material at time t = 0 and let B_{t} be an open set which identifies the same chunk of material at time
t > 0.
Let y
(t,x )
=
(y1(t,x ),y2 (t,x),y3(t,x))
denote the position with respect to Cartesian
coordinates at time t of the point whose position at time t = 0 is x =
(x1,x2,x3)
. The coordinates x
are sometimes called the reference coordinates and sometimes the material coordinates and
sometimes the Lagrangian coordinates. The coordinates y are called the Eulerian coordinates
or sometimes the spacial coordinates and the function
(t,x)
→ y
(t,x)
is called the motion.
Thus
y(0,x) = x. (17.9)
(17.9)
The derivative,
D2y (t,x) ≡ Dxy (t,x )
is called the deformation gradient. Recall the notation means you fix t and consider the function
x → y
(t,x)
, taking its derivative. Since it is a linear transformation, it is represented by the usual matrix,
whose ij^{th} entry is given by
F (x) = ∂yi(t,x).
ij ∂xj
Let ρ
(t,y)
denote the density of the material at time t at the point y and let ρ_{0}
(x)
denote the density of
the material at the point x. Thus ρ_{0}
(x)
= ρ
(0,x)
= ρ
(0,y(0,x))
. The first task is to consider
the relationship between ρ
(t,y)
and ρ_{0}
(x)
. The following picture is useful to illustrate the
ideas.
PICT
Lemma 17.4.1ρ_{0}
(x)
= ρ
(t,y(t,x))
det
(F)
and in any reasonable physical motion det
(F )
> 0.
Proof:Let V_{0} represent a small chunk of material at t = 0 and let V_{t} represent the same chunk of
material at time t. I will be a little sloppy and refer to V_{0} as the small chunk of material at time t = 0 and
V_{t} as the chunk of material at time t rather than an open set representing the chunk of material. Then by
the change of variables formula for multiple integrals,
∫ ∫
dV = |det(F)|dV.
Vt V0
If det
(F )
= 0 for some t the above formula shows that the chunk of material went from positive volume to
zero volume and this is not physically possible. Therefore, it is impossible that det
(F )
can equal zero.
However, at t = 0, F = I, the identity because of (17.9). Therefore, det
(F )
= 1 at t = 0 and if it is
assumed t → det
(F)
is continuous it follows by the intermediate value theorem that det
(F )
> 0 for all
t.
Of course it is not known for sure that this function is continuous but the above shows why it is at least
reasonable to expect det
(F)
> 0.
Now using the change of variables formula
∫ ∫
mass of V = ρ(t,y)dV = ρ(t,y(t,x ))det(F) dV
t Vt V0
∫
= mass of V0 = V ρ0 (x )dV.
0
Since V_{0} is arbitrary, it follows
ρ0(x) = ρ (t,y (t,x))det(F )
as claimed. Note this shows that det
(F )
is a magnification factor for the density.
Now consider a small chunk of material, V_{t} at time t which corresponds to V_{0} at time t = 0. The total
linear momentum of this material at time t is
∫
Vt ρ (t,y)v (t,y)dV
where v is the velocity. By Newton’s second law, the time rate of change of this linear momentum should
equal the total force acting on the chunk of material. In the following derivation, dV
(y)
will indicate the
integration is taking place with respect to the variable, y. By Lemma 17.4.1 and the change of variables
formula for multiple integrals
Having taken the derivative of the total momentum, it is time to consider the total force acting on the
chunk of material.
The force comes from two sources, a body force b and a force which acts on the boundary of the chunk
of material called a traction force. Typically, the body force is something like gravity in which case,
b = −gρk, assuming the Cartesian coordinate system has been chosen in the usual manner. The traction
force is of the form
∫
s(t,y,n )dA
∂Vt
where n is the unit exterior normal. Thus the traction force depends on position, time, and the orientation
of the boundary of V_{t}. Cauchy showed the existence of a linear transformation T
(t,y)
such that
T
(t,y)
n = s
(t,y,n)
. It follows there is a matrix T_{ij}
(t,y)
such that the i^{th} component of s is
given by s_{i}
(t,y,n)
= T_{ij}
(t,y)
n_{j}. Cauchy also showed this matrix is symmetric, T_{ij} = T_{ji}. It
is called the Cauchy stress. Using Newton’s second law to equate the time derivative of the
total linear momentum with the applied forces and using the usual repeated index summation
convention,
the sum taken over repeated indices. Here is where the divergence theorem is used. In the last integral,
the multiplication by n_{j} is exchanged for the j^{th} partial derivative and an integral over V_{t}.
Thus
the sum taken over repeated indices. Since V_{t} was arbitrary, it follows
[ ]
ρ(t,y ) ∂v-+ ∂v-∂yi = b (t,y)+ ei∂-(Tij(t,y-))
∂t ∂yi ∂t ∂yj
≡ b (t,y)+ div(T)
where here divT is a vector whose i^{th} component is given by
(divT) = ∂Tij.
i ∂yj
The term
∂∂vt
+
∂∂yvi
∂y∂ti
, is the total derivative with respect to t of the velocity v. Thus you might see this
written as
ρ˙v = b + div(T).
The above formulation of the balance of momentum involves the spatial coordinates y but people also
like to formulate momentum balance in terms of the material coordinates x. Of course this changes
everything.
The momentum in terms of the material coordinates is
As indicated earlier, this is a physical derivation, so the mathematical questions related to interchange of
limit operations are ignored. This must equal the total applied force. Thus using the repeated index
summation convention,
the first term on the right being the contribution of the body force given per unit volume in the material
coordinates and the last term being the traction force discussed earlier. The task is to write this last
integral as one over ∂V_{0}. For y ∈ ∂V_{t} there is a unit outer normal n. Here y = y
(t,x )
for x ∈ ∂V_{0}. Then
define N to be the unit outer normal to V_{0} at the point x. Near the point y ∈ ∂V_{t} the surface ∂V_{t} is given
parametrically in the form y = y
(s,t)
for
(s,t)
∈ D ⊆ ℝ^{2} and it can be assumed the unit normal to ∂V_{t}
near this point is
n = ys-(s,t)×-yt(s,t)-
|ys (s,t)× yt(s,t)|
with the area element given by
|ys (s,t)× yt(s,t)|
dsdt. This is true for y ∈ P_{t}⊆ ∂V_{t}, a small piece
of ∂V_{t}. Therefore, the last integral in (17.10) is the sum of integrals over small pieces of the
form
Summation over repeated indices is used. Remember y = y
(t,x)
and it is always assumed the mapping
x → y
(t,x)
is one to one and so, since on the surface ∂V_{t} near y, the points are functions of
(s,t)
, it
follows x is also a function of
(s,t)
. Now by the properties of the cross product, this last integral
equals
∫ ( )
T (x (s,t)) ∂xα-∂xβ -∂y-× -∂y- dsdt (17.12)
D ij ∂s ∂t ∂x α ∂x β j
(17.12)
where here x
(s,t)
is the point of ∂V_{0} which corresponds with y
(s,t)
∈ ∂V_{t}. Thus
Tij (x (s,t)) = Tij(y (s,t)).
(Perhaps this is a slight abuse of notation because T_{ij} is defined on ∂V_{t}, not on ∂V_{0}, but it avoids
introducing extra symbols.) Next (17.12) equals
∫
Tij(x (s,t)) ∂xα-∂xβεjab-∂ya∂yb-dsdt
D ∂s ∂t ∂x α∂xβ