Mathematical Physics Vol 1

Chapter 4. Field theory

142

Solution a) Let us start from the definition of the rotor

i

j

k

∂ ∂ x

∂ ∂ y

∂ ∂ z

rot v = ∇ × v =

,

v x

v y

v z

where v x = x + 2 y + az , v y = bx − 3 y − z , v z = 4 x + cy + 2 z . It follows that rot v =( c + 1 ) i +( a − 4 ) j +( b + 2 ) k . According to the condition of this example the vector field is potential, namely rot v =0. This condition is fulfilled if a = 4 , b = 2 , c = − 1, and it follows that v =( x + 2 y + 4 z ) i +( 2 x − 3 y − z ) j +( 4 x − y + 2 z ) k . b) Given that v = ∇ φ = i + j + k ,

∂φ ∂ x

∂φ ∂ y

∂φ ∂ z

it follows, according to a), that ∂φ ∂ x

= x + 2 y + 4 z ,

(4.166)

∂φ ∂ y ∂φ ∂ z

= 2 x − 3 y − z , = 4 x − y + 2 z .

(4.167)

(4.168)

As, in this case, r ot v = rotgrad ϕ = 0, this system of equations is integrable. By integrating the relation (4.166) over x , assuming that y and z are constant, we obtain φ = x 2 2 + 2 xy + 4 xz + f ( y , z ) , (4.169) where f ( y , z ) is a function of y and z . By differentiating the equation (4.169) by y and equating it with the equation (4.167) we obtain

∂φ ∂ y

∂ f ( y , z ) ∂ y

= 2 x +

= 2 x − 3 y − z .

(4.170)

It follows that

∂ f ( y , z ) ∂ y

= − 3 y − z .

(4.171)

By solving the equation (4.171), assuming that z is constant, we obtain

3 y 2

f ( y , z )= −

yz + g ( z ) .

(4.172)

2 −

Substituting the equation (4.172) into the equation (4.169) we obtain the follow ing expression for the required function

x 2 2

3 y 2

φ =

+ 2 xy + 4 xz + −

yz + g ( z )

(4.173)

2 −