Mathematical Physics Vol 1

4.6 Examples

135

Solution

∇ × V =

∂ ∂ z k × ( xz 3 i − 2 x 2 yz j + 2 yz 4 k )=

∂ ∂ x

∂ ∂ y

i +

j +

i

j

k

∂ ∂ x ∂ ∂ z xz 3 − 2 x 2 yz 2 yz 4 ∂ ∂ y

=

=

=

∂ z

i + +

∂ x

∂ ( − 2 x 2 yz )

∂ ( xz 3 )

∂ ( 2 yz 4 )

∂ ( 2 yz 4 )

j +

∂ y −

∂ z −

∂ y

∂ ( − 2 x 2 yz ) ∂ x

∂ ( xz 3 )

k =

−

=( 2 z 4 + 2 xz ) i +( 3 xz 2 ) j − ( 4 xyz ) k .

For point A ( 1 , − 1 , 1 ) we obtain ∇ × V A

= 3 j + 4 k .

Exercise 64 Observe the expression ∇ · ( V × W ) . a) Prove that

∇ · ( V × W )=( ∇ × V ) · W − V · ( ∇ × W ) . b) Find the value for the expression if W = r and ∇ × V = 0. c) Prove that, if V and W are irrotational (potential) fields, then the field obtained as their vector product is solenoidal (rotational). d) It follows from b) and c) that the vector field of a vector product of a position vector and any irrotational vector field is a solenoidal field. Prove this statement. e) The vector field of a vector product of a position vector and a conservative force is always a solenoidal field. Prove this statement.

Solution a) Given that

∂ ∂ x V x

∂ ∂ y V y

∂ ∂ z

∇ · ( V × W )=

=

V z W x W y W z

(4.159)

∂ ∂ x

∂ ∂ y

∂ ∂ z

( V y W z − V z W y )+

( V z W x − V x W z )+

( V x W y − V y W x ) ,

=

after differentiating and grouping, we obtain W x ∂ V z ∂ y − ∂ V y ∂ z + W y ∂ V x ∂ z − ∂ V z ∂ x

+ W z

∂ V x ∂ y − ∂ y . ∂ W x

∂ V y ∂ x −

(4.160)

− V x

∂ z −

V y

∂ W z ∂ x −

V z

∂ W y

∂ W y

∂ W z

∂ W x

∂ y −

∂ z −

∂ x −