The mass of fluid at a point is not well defined,
since a point has zero volume. Yet to apply Newton's second law, we need the
mass of the object which is accelerating. For us this is the mass of the bag of
water. To advance towards our goal of writing Newton's second law in terms of
fields, we use the density,
, of the
fluid. Since density can be defined as the mass of our bag divided by the
volume of the bag, we can safely consider the limit of the density as the
volume shrinks to a point. The ratio can stay constant, even though both the
mass and the volume go to zero.
We write the second law for our bag of water:
where
is the mass
of the bag. Dividing both sides by the volume of the bag,
, we
have
where
is the net
force on a unit volume of fluid at the location of interest. Note that
is a field
quantity.
THE STANDARD CONTRIBUTIONS TO THE TOTAL FORCE
We consider the force of gravity on a unit volume,
and the force on a unit volume by its neighbors. Other forces are possible, and
will be included for the present as
.
The force of gravity on the unit volume is simply
where
is the local
gravitational field strength.
THE PRESSURE GRADIENT FORCE
The force on our unit volume by its neighbors will be expressed in terms of the pressure in the fluid, and its variations. The pressure is the force that a fluid exerts on a unit area which it touches. The force associated with pressure is always perpendicular to the area that it touches.
 |
| Cube of water under pressure, |
with  |
We choose our unit volume to be a cube with a simple relation to the x,y,z coordinate axes, as shown. We will calculate the net force due to all the neighbors, in the x direction, and then extend the result to the other two directions. Since pressure exerts a force normal to the surface, only the left and right faces of the cube experience a force with non-zero x component. The net force due to pressure has x component
Here,
is an
average pressure, averaged over the left surface of the cube.
is a similar
average on the right.) Note that the force upon the right face has negative x
direction.
For our cube, the area on the left is the same as the area on the right, so the area factors out of the expression for the force, leaving the pressure difference to be calculated. We calculate the difference by expanding the pressure on the right in terms of the pressure on the left, using a Taylor's series:
(The derivatives are evaluated on the left face.) We
require that the distance
be so small
that we may drop all but the first derivative term. We can calculate the
pressure difference
and determine
the net x component of the force due to pressure variation:
where
is the area
of each face of the cube, and
is the volume
of the cube.
This means that the x component of the pressure-caused force on a unit volume of fluid is
We are now prepared to write the x component of Newton's second law for a unit volume of fluid:
Note well that the y and z components of velocity
have an influence on this equation for the x component of Newton's law: They
appear (along with the x component of velocity) in the scalar term
.
The y and z components of the second law are similar. We may combine the three component equations into a vector form, term by term:
The acceleration term on the left becomes, as we have seen:
The gravitational term becomes
and similarly the "other" force term becomes
.
The pressure term is less familiar in appearance, and more interesting. The vector representation is
is called "the gradient of P." P is a scalar, but the
gradient of P is a vector. The direction of this vector is in the direction of
maximum rate of change of P. (The gradient of P points to larger P, so the
negative, 
points to lower P.) The magnitude of this vector is the derivative along the
direction of greatest change.
Now we can write the vector form of Newton's second law for fluids, in terms of fields:
This is called Euler's equation." To keep with tradition we divide both sides by the density and have the standard form of Euler's equation:
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