Mathematical Physics Vol 1

Chapter 4. Field theory

120

c) As in the previous example, we first express 1 r coordinate system, and then compute the gradient

with respect to the Cartesian

! = ∇ h ( x 2 + y 2 + z 2 ) − 1

2 i ⇒

1 r

1 p x 2 + y 2 + z 2

∇ φ = ∇ ∇ φ = i

= ∇

2 + j

2 +

∂ ∂ x

∂ ∂ y

( x 2 + y 2 + z 2 ) − 1

( x 2 + y 2 + z 2 ) − 1

+ k

2 ⇒

∂ ∂ z

( x 2 + y 2 + z 2 ) − 1

∇ φ = i −

2 · 2 x + j −

2 · 2 y +

1 2

1 2

( x 2 + y 2 + z 2 ) − 3

( x 2 + y 2 + z 2 ) − 3

+ k −

2 · 2 z ⇒

1 2

( x 2 + y 2 + z 2 ) − 3

x i + y j + z k ( x 2 + y 2 + z 2 )

r r 3

∇ φ = −

= −

.

3 2

Problem34 Prove that ∇ r n = nr n − 2 r .

Solution

∇ r n = ∇ q ( x 2 + y 2 + z 2 ) n

= ∇ ( x 2 + y 2 + z 2 ) n / 2 ⇒

∇ r n = i h

n 2 ( x 2 + y 2 + z 2 ) ( n / 2 ) − 1 · 2 x i + j h

n 2 ( x 2 + y 2 + z 2 ) ( n / 2 ) − 1 · 2 y i +

+ k h

n 2 ( x 2 + y 2 + z 2 ) ( n / 2 ) − 1 · 2 z i ⇒

∇ r n = n ( x i + y j + z k )( x 2 + y 2 + z 2 ) ( n / 2 ) − 1 ∇ r n = nr n − 2 r .

R Note that in both this and previous example the solution can be obtained in a simpler way. See example 39 on page 123.

Problem35 Prove that grad f is a vector orthogonal to the surface given by the function f ( x , y , z )= c , where c is a constant.