Consider a possibly moving fluid with constant density ρ and let P denote the pressure in this fluid. If B
is a part of this fluid the force exerted on B by the rest of the fluid is ∫_{∂B}− PndA where
n is the outer normal from B. Assume this is the only force which matters so for example
there is no viscosity in the fluid. Thus the Cauchy stress in rectangular coordinates should
be

( )
− P 0 0
T = |( 0 − P 0 |) .
0 0 − P

Then divT = −∇P. Also suppose the only body force is from gravity, a force of the form −ρgk, so from
the balance of momentum

ρ˙v = − ρgk − ∇P (x). (17.15)

(17.15)

Now in all this, the coordinates are the spacial coordinates, and it is assumed they are rectangular. Thus
x =

(x,y,z)

^{T} and v is the velocity while

˙v

is the total derivative of v =

(v1,v2,v3)

^{T} given by v_{t} + v_{i}v_{,i}.
Take the dot product of both sides of (17.15) with v. This yields

d dz d
(ρ∕2)--|v|2 = − ρg- − --P (x).
dt dt dt

Therefore,

d ( ρ|v|2 )
dt --2--+ ρgz + P (x) = 0,

so there is a constant C^{′} such that

ρ|v-|2- ′
2 + ρgz + P (x) = C

For convenience define γ to be the weight density of this fluid. Thus γ = ρg. Divide by γ.
Then

2
|v|-+ z + P-(x)= C.
2g γ

This is Bernoulli’s^{2}
principle. Note how, if you keep the height the same, then if you raise

|v|

, it follows the pressure
drops.

This is often used to explain the lift of an airplane wing. The top surface is curved, which forces the
air to go faster over the top of the wing, causing a drop in pressure which creates lift. It is
also used to explain the concept of a venturi tube in which the air loses pressure due to being
pinched which causes it to flow faster. In many of these applications, the assumptions used
in which ρ is constant, and there is no other contribution to the traction force on ∂B than
pressure, so in particular, there is no viscosity, are not correct. However, it is hoped that the
effects of these deviations from the ideal situation are small enough that the conclusions are
still roughly true. You can see how using balance of momentum can be used to consider more
difficult situations. For example, you might have a body force which is more involved than
gravity.