Mathematical Physics Vol 1

Chapter 3. Examples

70

Solution Given that A has a constant magnitude, it follows that A · A = const . Then d d t ( A · A )= A · d A d t + d A d t · A = 2 A · d A d t = 0. From that, it further follows that A and d A d t are orthogonal if d A d t ̸ = 0.

Problem27

d A d t

d A d t , where A is a scalar function A = | A | .

Prove that A ·

= A ·

Solution Given that A · A = A 2 , it follows that d d t

d d t ( A 2 ) , and we thus obtain

( A · A )=

d A d t

d A

d A d t

d d t

( A · A )= A ·

A = 2 A ·

d t ·

+

.

On the other hand, given that

d d t

d A d t

( A 2 )= 2 A

from the two last relations it follows that

d A d t

d A d t

A ·

= A ·

.

Problem28 If F is a function of x , y , z , t where x , y , z are functions of t prove that d F d t = ∂ F ∂ t + ∂ F ∂ x d x d t + ∂ F ∂ y d y d t + ∂ F ∂ z d z d t .

Proof Assume that

F = F 1 ( x , y , z , t ) i + F 2 ( x , y , z , t ) j + F 3 ( x , y , z , t ) k .