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4. THE EQUATION OF MOTION
Newton’s law for a fluid element is

dV
F ,
dt
(4.1)
where  is the mass density (mass per unit volume), V is the velocity of the fluid
element, and F is the force per unit volume acting on the element. The latter is
composed of two types: volumetric forces and surface forces.
Volumetric forces act throughout the volume of the fluid element. They can be
thought of as acting at the centroid. Examples are:
1. Gravity: Fg  g , where g is the gravitational acceleration. If the force is a
central force, then Fg   , where  is the gravitational potential.
2. Electromagnetic forces: Since the fluid can conduct electricity, it can have a
current density J    n q V (with the sum being over all species of ions
and electrons, and n , q , and V the number density, electric charge, and
velocity of species  ), and, in principle, a net electric charge per unit
volume, q . The electromagnetic force (per unit volume) are then the
electric force, Fq  q E ( E is the electric field), and the Lorentz force,
FL  J  B ( B is the magnetic field).
Surface forces are more complicated. Consider the forces acting on a surface S . We
assume the convention that the material in front of S exerts a force on the material
behind S that is given by
F  S P ,
(4.2)
Fi  Si Pij .
(4.3)
or
We consider three orientations for S , along each of the three coordinate directions.
If S  ê1 ,
F  P11ê1  P12 ê 2  P13ê 3 ,
(4.4)
which is a vector. Similarly, if S  ê 2 , then
F  P21ê1  P22 ê 2  P23ê 3 ,
(4.5)
and if S  ê 3 , then
F  P31ê1  P32 ê 2  P33ê 3 .
(4.6)
It therefore takes 9 numbers to define the force on the surface S . These are the
components of the stress tensor, Pij . (We will not prove that P is a tensor, but it is!)
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The total surface force acting on a fluid element is the sum of the forces on its faces.
We want the total force acting on the volume. Since F has been defined as the as the
force exerted by the material in front of S acting on the material behind S , all of the
material within the element is behind the faces, and the net force is the negative of the
surface forces, i.e.,
F  —
 dS  P ,
(4.7)
S
or, by Gauss’ theorem,
F     PdV .
(4.8)
V
As V  0 , we obtain the net force per unit volume as
f   P .
(4.9)
This is the volumetric equivalent of the surface forces.
considering only surface forces is then

dV
  P .
dt
The equation of motion
(4.10)
We will now prove that the stress tensor is symmetric, i.e., Pij  Pji . Consider the
angular momentum per unit volume of fluid,
L  r  V .
(4.11)
Using Equation (4.10), the total (i.e., Lagrangian) time rate of change of L for a fluid
with fixed volume V0 is
L   r 
V0
dV
dV    r   PdV ,
dt
V0
(4.12)
or, in Cartesian tensor notation
Li    ijk rj l Plk dV .
(4.13)
V0
Since l rj   lj , we can write
 
rj l Plk   l rj Plk  Pjk ,
(4.14)
so that
 
Li     ijk  l rj Plk  Pjk  dV ,
V0


   i  ijk rj Plk dV    ijk Plk dV ,
V0
V0
 —
 dSlijk rj Plk   ijk Plk dV ,
S0
(4.15)
V0
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where we have used Gauss’ theorem, and S0 is the surface bounding V0 . We recognize
the first term on the right hand side of Equation (4.15) as the total external torque applied
to the surface of the volume. The remaining term is the rate of change of internal angular
momentum of the fluid. In the absence of applied torque, we require Li  0 , or

P dV  0 ,
(4.16)
ijk lk
V0
which can be written as
 2 
1
ijk

Pjk   ikj Pkj dV 
V0
 2  P
1
ijk
jk

 Pkj dV  0 .
(4.17)
V0
(The first expression comes from interchanging the dummy indices j and k ; the second
follows from the properties of  ijk .) Since Equation (4.15) must hold for an arbitrary
volume, we have Pjk  Pkj , which is the desired result.
The symmetry of the stress tensor is a very general result; it can be considered a
general principle of physics. It is independent of the properties of the medium, which can
be solid, liquid, or gas (or even plasma). It is required to prevent the internal angular
momentum of the system from increasing without bound.
The stress tensor (indeed, any tensor) can always be decomposed as
P  pI   ,
(4.18)
Pij  p ij  ij .
(4.19)
or
The first term on the right hand side is called the scalar pressure. The second term is
called the viscous stress tensor.
Including all the volumetric and equivalent volumetric forces, Equation (4.1)
becomes

dV
 q E  J  B  p     ,
dt
(4.20)
which is the Lagrangian form of the equation of motion. The Eulerian form is
 V

 V  V   q E  J  B  p     .
 t


(4.21)
As always, they are equivalent.
We now compute WV  V  FV , the work done on a volume element by the viscous
force FV     :


WV  Vi  j  ji   j Vi  ji   ji  jVi ,
or
WV      V    : V .
(4.22)
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Then

V0
WV dV       V dV    : VdV ,
V0
V0
—
 dS    V    : VdV .
S0
(4.23)
V0
The first term on the right hand side is the work done on the surface S ; the second term
is the work done throughout the volume. In the absence of the surface term, kinetic
energy is lost from the fluid if  :V  0 ; it must show up as internal (thermal) energy.
We therefore identify the volumetric viscous heating rate as
QV   : V .
(4.24)
We remark that the equation of motion has introduced 6 new dependent variables;
i.e., the 6 independent components of the stress tensor..This is a further example of the
closure problem first mentioned in Section 3.
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