Mathematical Physics Vol 1

Chapter 4. Field theory

122

where ∆ s is the distance between points P and Q . b) Give a physical interpretation of the value lim ∆ s → 0 ∆ T ∆ s = d T d s . c) Prove that d T d s = ∇ T · d r d s .

Solution a) ∆ T represents the temperature increment in the transition from point P topoint Q , ∆ s the distance between these two points, whereas their quotient ∆ T ∆ s represents the mean value of temperature change per unit length in the direction of PQ . b) As the increment is equal to ∆ T = ∂ T ∂ x ∆ x + ∂ T ∂ y ∆ y + ∂ T ∂ z ∆ z + higher order in finitesimal values ∆ x , ∆ y , ∆ z , it follows that

∆ s → 0

∆ z ∆ s

∆ T ∆ s

∂ T ∂ x

∆ x ∆ s

∂ T ∂ y

∆ y ∆ s

∂ T ∂ z

lim ∆ s → 0

= lim

+

+

and consequently

∂ T ∂ x

∂ T ∂ y

∂ T ∂ z

d T d s

d x d s

d y d s

d z d s

=

+

+

.

d T d s represents the temperature change in point P , in the direction of PQ .

Thus,

c)

∂ T ∂ x

∂ T ∂ y ∂ T ∂ y

∂ T ∂ z

d T d s

d x d s

d y d s

d z d s

=

+

+

=

=

k ·

k =

∂ T ∂ x

∂ T ∂ z

d x d s

d y d s

d z d s

i +

j +

i +

j +

d r d s

= ∇ T ·

.

Problem38 Find the increment of the function φ = x 2 yz + 4 xz 2 in point A ( 1 , − 2 , − 1 ) , in the direction of vector v = 2 i − j − 2 k .

Solution As the gradient in an arbitrary point is

∇ φ = ∇ ( x 2 yz + 4 xz 2 )=( 2 xyz + 4 z 2 ) i + x 2 z j +( x 2 y + 8 xz ) k ,

it follows that the gradient in point A ( 1 , − 2 , − 1 ) is ∇ φ = 8 i − j − 10 k .