I don't understand how to obtain second equation with first part in the equation ∇×→A0e−j→k⋅→r=−j→k×→A0e−j→k⋅→r.
Answer
∇≡[∂x1∂x1∂x1∂x2∂x1∂x3],A=[A1A2A3],k⋅x=k1x1+k2x2+k3x3
∇×(Ae−ik⋅x)=[e1e2e3∂x1∂x1∂x1∂x2∂x1∂x3A1e−ik⋅xA2e−ik⋅xA3e−ik⋅x]=[A3∂e−ik⋅x∂x2−A2∂e−ik⋅x∂x3A1∂e−ik⋅x∂x3−A3∂e−ik⋅x∂x1A2∂e−ik⋅x∂x1−A1∂e−ik⋅x∂x2]=−ie−ik⋅x[k2A3−k3A2k3A1−k1A3k1A2−k2A1]=−ie−ik⋅x[e1e2e3k1k2k3A1A2A3]⏟k×A
More generally :
If ψ(x1,x2,x3) and A(x1,x2,x3) are scalar and vector functions respectively of the coordinates in R3 then ∇×(ψA)=∇ψ×A+ψ∇×A121212
Proof of identity (04): ∇×(ψA)=[e1e2e3∂x1∂x1∂x1∂x2∂x1∂x3ψA1ψA2ψA3]=[∂(ψA3)∂x2−∂(ψA2)∂x3∂(ψA1)∂x3−∂(ψA3)∂x1∂(ψA2)∂x1−∂(ψA1)∂x2]=[A3∂ψ∂x2+ψ∂A3∂x2−A2∂ψ∂x3−ψ∂A2∂x3A1∂ψ∂x3+ψ∂A1∂x3−A3∂ψ∂x1−ψ∂A3∂x1A2∂ψ∂x1+ψ∂A2∂x1−A1∂ψ∂x2−ψ∂A1∂x2]=[A3∂ψ∂x2−A2∂ψ∂x3A1∂ψ∂x3−A3∂ψ∂x1A2∂ψ∂x1−A1∂ψ∂x2]+[ψ∂A3∂x2−ψ∂A2∂x3ψ∂A1∂x3−ψ∂A3∂x1ψ∂A2∂x1−ψ∂A1∂x2]=[e1e2e3∂ψ∂x1∂ψ∂x2∂ψ∂x3A1A2A3]⏟∇ψ×A+ψ[e1e2e3∂x1∂x1∂x1∂x2∂x1∂x3A1A2A3]⏟∇×A=∇ψ×A+ψ∇×A
No comments:
Post a Comment