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. (27.9)
(27.9)
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
∂y (t,x)
Fij(x) =--i-----.
∂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 27.4.1ρ_{0}
(x)
= ρ
(t,y (t,x))
det
(F)
and in any reasonable physicalmotion 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 27.9. 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 Vt = ρ(t,y)dV = ρ(t,y(t,x ))det(F )dV
Vt ∫ V0
= mass of V0 = ρ0(x )dV.
V0
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
∫
ρ(t,y)v (t,y)dV
Vt
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 27.4.1 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
∫
∂Vt s(t,y,n)dA
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,
∫ [∂v- -∂v ∂yi] ∫ ∫
V ρ (t,y) ∂t + ∂yi ∂t dV (y ) = V b(t,y)dV (y)+ ∂B eiTij(t,y)njdA,
t t t
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) +e ∂-(Tij(t,y))
∂t ∂yi ∂t i ∂yj
≡ b (t,y) +div(T )
where here divT is a vector whose i^{th} component is given by
∂Tij-
(divT)i = ∂yj .
The term
∂v
∂t
+
-∂v-
∂yi
∂yi
∂t
, 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
-ys(s,t)×-yt(s,t)-
n = |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 27.10 is the sum of integrals over small
pieces of the form
(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 27.12 equals
∫ ∂xα ∂xβ ∂ya ∂yb
Tij (x (s,t))-∂s-∂t-εjab∂x-∂x--dsdt
D α β
∫ ∂xα ∂xβ ∂ya ∂yb
= Tij(x (s,t))∂s--∂t-εcabδjc∂xα∂x-βdsdt
D
◜-=◞δ◟jc-◝
∫ ∂x α∂xβ ∂yc ∂xp∂ya ∂yb
= Tij(x(s,t))-∂s--∂t-εcab∂xp-∂yj∂xα-∂xβdsdt
D