Question #277014

Let E:=Exi+Eyj+Ezk  and H:=Hxi+Hyj+Hzk 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

∇^2Ei =1/c^2. ∂^2Ei/∂t^2  and ∇^2Hi=1/c^2. ∂^2Hi/∂t^2

Here, i=x,y  or z.

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



1
Expert's answer
2021-12-09T14:57:12-0500

×(×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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