Mathematical Physics Vol 1

4.6 Examples

139

follows that

i j k ω 1 ω 2 ω 3 x y z

∇ × v = ∇ × ( ω × r )= ∇ ×

=

= ∇ × [( ω 2 z − ω 3 y ) i +( ω 3 x − ω 1 z ) j +( ω 1 y − ω 2 x ) k ]=

i

j

k

∂ ∂ x ∂ ∂ z ω 2 z − ω 3 y ω 3 x − ω 1 z ω 1 y − ω 2 x = = 2 ( ω 1 i + ω 2 j + ω 3 k )= 2 ω . ∂ ∂ y

=

Exercise 68

Let

∂ B ∂ t

∇ · E = 0 , ∇ · B = 0 ,

∇ × E = −

,

(4.163)

∂ E ∂ t

1 c 2

∇ × B =

.

Prove that E and B satisfy the equation ∇ 2 u = a 2 ∂ 2 u ∂ t 2 electric field, B is the magnetic induction, and a a constant.

, where E is the strength of the

Solution Observe the expression ∇ × ( ∇ × E ) , which can be expressed in the form (see Example 66, p.137) ∇ × ( ∇ × E )= − ∇ 2 E + ∇ ( ∇ · E ) . Using the initial assumption (4.163) ( ∇ · E = 0 ∧ ∇ × E = − ∂ B ∂ t ) we obtain ∇ × − ∂ B ∂ t = − ∇ 2 E , that is − ∂ ∂ t ( ∇ × B )= − ∇ 2 E . If we now use the condition ∇ × B = 1 c 2 ∂ E ∂ t , we finally obtain for the field E 1 c 2 ∂ E ∂ t = ∇ 2 E . (4.164) Similarly, for the field B we obtain 1 c 2 ∂ 2 B ∂ t 2 = ∇ 2 B . (4.165)