Question #276690

Let E:=E

x

i

^

+E

y

j

^

+E

z

k

^

 and H:=H

x

i

^

+H

y

j

^

+H

z

k

^

 be two vectors assumed to have continuous partial derivatives (of second order at least) with

respect to position and time. Suppose further that E

 and H

 satisfy the equations:


∇⋅E=0,∇⋅H=0,∇×E=−1

c


H

∂t


,∇×H=1

c


E

∂t


prove that E

 and H

 satisfy the equation


2

E

i

=1

c

2


2

E

i

∂t

2


 and ∇

2

H

i

=1

c

2


2

H

i

∂t

2


Here, i=x,y

 or z

.

Hint: Use the fact that∇×(∇×V)=∇(∇⋅V)−∇

2

V.





1
Expert's answer
2021-12-09T15:33:51-0500

×(×E)=1c(×H)t=1c2E2t2∇×(∇×E)=-\frac{1}{c}\frac{\partial (∇×H)}{\partial t}=-\frac{1}{c^2}\frac{\partial E^2}{\partial t^2}


×(×E)=(E)2E=2E∇×(∇×E)=∇(∇⋅E)−∇^2E=−∇^2E


2E=1c2E2t2∇^2E=\frac{1}{c^2}\frac{\partial E^2}{\partial t^2}


2Ei=1c2Ei2t2∇^2E_i=\frac{1}{c^2}\frac{\partial E_i^2}{\partial t^2}



×(×H)=1c(×E)t=1c2H2t2∇×(∇×H)=\frac{1}{c}\frac{\partial (∇×E)}{\partial t}=-\frac{1}{c^2}\frac{\partial H^2}{\partial t^2}


×(×H)=(H)2H=2H∇×(∇×H)=∇(∇⋅H)−∇^2H=−∇^2H


2H=1c2H2t2∇^2H=\frac{1}{c^2}\frac{\partial H^2}{\partial t^2}


2Hi=1c2Hi2t2∇^2H_i=\frac{1}{c^2}\frac{\partial H_i^2}{\partial t^2}


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