Mathematical Physics Vol 1

Chapter 4. Field theory

132

Solution If we refer to Example 51b on p.129 and introduce the notation φ = U , that is, A = ∇ V , then ∇ · ( U ∇ V )=( ∇ U ) · ( ∇ V )+ U ( ∇ · ∇ V )=( ∇ U ) · ( ∇ V )+ U ∇ 2 V , (4.157) ∇ · ( V ∇ U )= =( ∇ V ) · ( ∇ U )+ V ∇ 2 U . (4.158) The relation (4.158) was obtained from (4.157) by replacing U by V and V by U . Subtracting the left and right sides and using the divergence property (the divergence of the difference is the difference of the divergence), we obtain ∇ · ( U ∇ V ) − ∇ · ( V ∇ U )= = ∇ · ( U ∇ V − V ∇ U )= =( ∇ U ) · ( ∇ V )+ U ∇ 2 V − [( ∇ V ) · ( ∇ U )+ V ∇ 2 U ]= = U ∇ 2 V − V ∇ 2 U .

Problem55 Find the constant a for which the vector field

V =( x + 3 y ) i +( y − 2 z ) j +( x + az ) k ,

is solenoid.

Solution The condition for a vector field to be solenoid is ∇ · V = 0 (see p. 92). Given that ∇ · V = ∂ ( x + 3 y ) ∂ x + ∂ ( y − 2 z ) ∂ y + ∂ ( x + az ) ∂ z = 1 + 1 + a , this condition yields ∇ · V = a + 2 = 0 , that is, a = − 2.