Mathematical Physics Vol 1

Chapter 4. Field theory

128

Solution Let us observe the flow of the fluid.

Figure 4.25: Flow of the fluid.

Let us calculate the difference between the flow of fluid entering the elemental volume d V = d x d y d z and the fluid leaving that volume, in the same time interval. The amount of fluid entering this volume, in the direction of y -axis is v y d t d x d z . Let us assume that changes of velocity v y in directions of x and z are negligible, i.e. ∂ v y ∂ x = 0 i ∂ v y ∂ z = 0, due to small dimensions of the observed volume. The amount of fluid that flows out, under the same assumption is

( v y + d v y ) d t d x d z ,

and thus the difference is

( v y + d v y ) d t d x d z − v y d t d x d z = d v y d t d x d z .

This difference can be represented as follows d v y d t d x d z = ∂ v y ∂ y d y + ∂ v y ∂ x d x + ∂ v y ∂ z

d z d t d x d z =

∂ v y ∂ y

d t d V .

It can analogously be shown that changes in the directions of x and z are

d v x d t d y d z , d v z d t d x d y , respectively. The total change is equal to the sum of these changes, i.e.

∂ v y ∂ y

∂ v z ∂ z

∂ v x ∂ x

d ( d V )=

d V d t +

d V d t +

d V d t =

=( div v ) d V d t ⇒

d ( d V ) d t d V

div v =

.

Thus, the divergence of the fluid velocity v represents the change in fluid volume in unit time per unit volume of fluid space.