Harmonic Plane Wave in Homogeneous Medium - pt. 1

ECE 5368/6358               han q le - copyrighted
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1. Introduction and background

See LectSet 1

2. Monochromatic traveling wave: basic

2.1 Vector representation

2.1.1 Vector relationship

Electric field is a vector:            E={(Subscript[E, x]),(Subscript[E, y]),(Subscript[E, z])}                     (2.1.1a)
same with magnetic:                 H={(Subscript[H, x]),(Subscript[H, y]),(Subscript[H, z])}                    (2.1.1b)
For plane wave, each component has the same phase function:
                                                      (e^(i (k . r -ω t)))                      (2.1.2a)
where:         k = ((ω )/c)(Sqrt[μ ϵ]) = (Subscript[k, 0])(Sqrt[μ ϵ]) =((2π)/λ)(Sqrt[μ ϵ])                  (2.1.2b)
Let's check their relationship with Maxwell's equation
                                                  ∇×E+(μ/c)((∂ H)/(∂ t)) = 0             (2.1.3)

Calculate ∇ × E

∇×(Overscript[E, ->])  =  (({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {(∂_x), (∂_y), (∂_z)}, {(Subscript[E, x]), (Subscript[E, y]), (Subscript[E, z])}})) = (({{(OverHat[x]), (OverHat[y]), (OverHat[z])}, {(∂_x), (∂_y), (∂_z)}, {(Subscript[E, x])(e^(i k . r )), (Subscript[E, y])(e^(i k . r )), (Subscript[E, z])(e^(i k . r ))}})) (e^(-i ω t))
                =i (({{(OverHat[x]), (OverHat[y]), (OverHat[z])}}))(({{(Subscript[k, y]) (Subscript[E, z])-(Subscript[k, z]) (Subscript[E, y])}, {(Subscript[k, z]) (Subscript[E, x])-(Subscript[k, x]) (Subscript[E, z])}, {(Subscript[k, x]) (Subscript[E, y])-(Subscript[k, y]) (Subscript[E, x])}})) (e^(i (k . r -ω t)))= i k ×(Overscript[E, ->])     (2.1.4)

This is a basic relation for a harmonic plane wave 
                                                ∇×(Overscript[E, ->])  = i k ×(Overscript[E, ->])                (2.1.5)

k=. ; x=. ; y=. ; z=. ;
Curl[{(Subscript[F, x])(E^(I  ((Subscript[k, 1]) x+(Subscript[k, 2]) y +(Subscript[k, 3]) z ))), (Subscript[F, y])(E^(I ((Subscript[k, 1]) x+(Subscript[k, 2]) y +(Subscript[k, 3]) z ) )), (Subscript[F, z])(E^(I  ((Subscript[k, 1]) x+(Subscript[k, 2]) y +(Subscript[k, 3]) z ))) },{x,y,z}]

FullSimplify[%]

{(Subscript[k, 1]) ,(Subscript[k, 2]) , (Subscript[k, 3]) } \[Cross] {(Subscript[F, x]) , (Subscript[F, y]) , (Subscript[F, z])}

Hence, we obtain the relation in (2.1.5)

Calculate (μ/c)((∂ H)/(∂ t))

(μ/c)((∂ H)/(∂ t))=-i  (μ/c) ω H

Apply Maxwell equation for the two terms, express H as function of E

∇×(Overscript[E, ->]) = i k ×(Overscript[E, ->]) = - (μ/c)((∂ H)/(∂ t)) =i  (μ/c) ω H 
                                              k ×(Overscript[E, ->]) =μ (Subscript[k, 0]) H 
                                             (Subscript[k, 0])(Sqrt[μ ϵ])  (OverHat[k]) ×(Overscript[E, ->]) =μ (Subscript[k, 0]) H

(Sqrt[ (ϵ/μ)])  (OverHat[k]) ×(Overscript[E, ->]) = H

Do the same for Maxwell equation:   ∇×H-(ϵ/c)((∂ E)/(∂ t)) =0

∇×H =i k × H =(ϵ/c)((∂ E)/(∂ t)) =-i (ϵ/c) ω E   
                                      (Subscript[k, 0])(Sqrt[μ ϵ])  (OverHat[k]) × H =-ϵ (Subscript[k, 0]) E

(Sqrt[(μ/ϵ) ])  (OverHat[k]) × H =- E

What can you conclude about the angles between the three vectors: k, E, and H?

What is the relationship between magnitude of E and H?

From above:                    (Sqrt[ (ϵ/μ)])   |E|=|H|

Calculate the Poynting vector for a plane wave

S=(c/(4 π))E×H 
                                             =(c/(4 π))E×((Sqrt[ (ϵ/μ)])  (OverHat[k]) ×(Overscript[E, ->]))=(c/(4 π))(Sqrt[ (ϵ/μ)]) (OverHat[k]) (((|E|))^2)    
For harmonic wave, the time-average of ((Cos[ωt])^2) or ((Sin[ωt])^2) is (1/2), hence:
                                         ⟨(Overscript[S, →])⟩=(c/(8 π))(Sqrt[(ϵ/μ)])(((|E|))^2)

Note: we must use real quantities in calculating S, but use the conjugate of H if using the fields in complex form for time-averaged. Example: let (OverVector[E])=(OverHat[x]) A Cos[k z-ω t]
                                    (OverVector[H])=(OverHat[y]) (Sqrt[ (ϵ/μ)])A Cos[k z-ω t] 
Then,     (OverVector[S])=(OverHat[x]) × (OverHat[y]) (c/(4 π))(Sqrt[ (ϵ/μ)])(A^2) ((Cos[k z-ω t])^2)=(OverHat[z])  (c/(4 π))(Sqrt[ (ϵ/μ)])(A^2) ((Cos[k z-ω t])^2) 
This is “instantaneous intensity,” but really not meaningful. What is physically meaningful is the time-averaged intensity, which is what we measure, and what is relevant in optical systems:
                                       I=|⟨(OverVector[S])⟩|=(c/(8 π))(Sqrt[ (ϵ/μ)]) (A^2)

2.1.2 Linear poplarization.

The orientation of the E field vector is generally called “polarization” of the light. However, polarization is also a more general concept than a specific quantity. It is about the description of the properties of the vectorial nature of the E and H field. A question is: why we pick the orientation of E field as the preferred description for polarization but not H field? There are 2 reasons: 1- it’s just a convention because E field is more familiar, and 2- H field is really an axial vector, not a true vector as we will see. Hence, using E field orientation as the polarization is more intuitive to understand in terms of vector orientation.

If the orientation of the E field is always in a plane, we define that to be linear polarization.

Because k, E, and H are mutually orthogonal, we can choose coordinate: z= along k, E along x and H along y  for the case of linear polarization. We will examine polarization in depth in later chapter.

Demo Apps

Is linear polarization the only possible polarization for plane wave?

2.1.3 Circular poplarization.

We know that k, E, and H are mutually orthogonal, and we choose above each vector to be always in a plane. But there is nothing forcing it that way.

Is this wave possible?

2.2 Single wave traveling

If we deal with ONLY one component of the vector, we can treat it as a scalar field.

And we can have a 2D wave: move the k-vector angle.