Tensors for Physics

396 Appendix: Exercises...

Here one has∇νvμ=εμκνwκ, and consequently

∇·v= 0 ,(∇×v)λ=ελνμεμκνwκ= 2 wλ, ∇νvμ = 0.

7.2 Test Solutions of the Wave Equation(p.102)
Proof that both the ansatz(7.63)and the plane wave(7.64)obey the wave equation.
Furthermore, show that theE-field is perpendicular to the wave vector, and that the
B-field is perpendicular to both.
(i)Ansatz(7.63). From (7.63), i.e. from

Eμ=E(μ^0 )f(ξ ), ξ=k̂νrν−ct,

follows

∇νEμ=(∇νξ)E(μ^0 )f(ξ )′=k̂νE(μ^0 )f(ξ )′,

where the prime indicates the derivative with respect toξ. Clearly,∇νEν=0 implies
̂kνE(ν^0 )=0, theE-field is perpendicular to its direction of propagation. The second

spatial derivative of the field yields

∇ν∇νEμ≡ΔEμ=Eμ(^0 )f(ξ )′′.

Similarly, the time derivative of the ansatz for the theE-field is given by

∂

∂t

Eμ=

(

∂

∂t

ξ

)

E(μ^0 )f(ξ )′=−cE(μ^0 )f(ξ )′,

and the second time derivative is

∂^2

∂t^2

Eμ=c^2 E(μ^0 )f(ξ )′′.

Thus the wave equation

ΔE−

1

c^2

∂^2

∂t^2

E= 0 ,

cf. (7.60), is obeyed.
For theB-field, the ansatz

Bμ=Bμ(^0 )f(ξ ), ξ=k̂νrν−ct,