#### A.1.2. Volume Changes

Consider a given set of fluid particles occupying a volume d V x ( t )= d 3 x( t ) .  According to the definition

x( t )=x( z,t )            where      z=x( 0 )=x( z,0 )                             (A.1)

we have, at a given time t,

d x i ( t )= j x i ( z,t ) z j d z j   = j x i ( z,t ) z j d x i ( 0 )

so that

d V x ( t )= d 3 x( t )=J d 3 z=Jd V z =Jd V x ( 0 )                                      (A.5)

where J is the jacobian given by

J= ( x 1 , x 2 , x 3 ) ( z 1 , z 2 , z 3 )   =det| x i z j |   =det| x 1 z 1 x 1 z 2 x 1 z 3 x 2 z 1 x 2 z 2 x 2 z 3 x 3 z 1 x 3 z 2 x 3 z 3 |                         (A.8)

Note that J is time dependent through x(t).

Now, as derived in A.1.3,

A δ i j = k a ik A jk

A t = ij A a ij a ij t = ij A ij a ij t

where A is the determinant of the matrix A={ a ij }  and A ij  the cofactor of a ij .  Hence,

J t = ij J ij t ( x i z j )   = ij J ij ( v i z j )   = ijk J ij ( v i x k )( x k z j )

= ik J δ k i ( v i x k )   = i J ( v i x i )

=J x v                                   (A.12)

where we've used

J t ( J t ) z = DJ Dt

v( x,t ) Dx Dt = ( x( z,t ) t ) z

( v i [ x( z,t ),t ] z j ) t = k ( v i ( x,t ) x k ) t ( x k ( z,t ) z j ) t                      (A.11)

Now, d V x ( t )  is a constant for an incompressible fluid.  This means J=1  so that J t =0  and hence x v=0 .