Relationship between x and y components of E

  • #1

Homework Statement



Before diving into the quantum-mechanical superposition principle, let’s get some practice with superposition in classical physics. Consider an electromagnetic wave propagating in the z-direction, which is a superposition of two linearly polarized waves. The electric field vector in the wave is
E = Ex + Ey, where Ex = a cos(kz − ωt), Ey = b cos(kz − ωt + δ). (1) The parameter δ is a real number between −π/2 and π/2, and indicates by how much the two components are out of phase. Look at the behavior of the electric field at some fixed value of z, say z = 0 for simplicity.

a) [2pt] Describe what the electric fields Ex and Ey are doing as a function of time.

b) [4pt] Show that there is a simple relation between Ex and Ey which does not involve t. Namely, you should find the following: Ex2/a2 + Ey2[/SUP]/ b2 − 2ExEycos δ/ab = constant. (2) Express the constant in the right-hand side of (2) in terms of the phase shift δ.

I am trying to do b[/B]

as I found in a) that

Ex = acos(ωt) and Ey = bcos(ωt - δ)



Homework Equations



E = Ex + Ey

The Attempt at a Solution



Ex = acos(ωt)
Ey = bcos(ωt)

Ex/a = cos(ωt)
Ey/b = cos(ωt - δ)

I assume I can use Euler formula and say e = cosΘ + isinΘ

So I get

Ex/a = ei(ωt)
Ey/b = ei(ωt - δ) = eiωt / eiδ


So

Ey eiδ/b = eiωt

I assume I set them equal to each other but I don't get the terms that I want for the LHS and the RHS becomes 0.
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
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Homework Statement



Before diving into the quantum-mechanical superposition principle, let’s get some practice with superposition in classical physics. Consider an electromagnetic wave propagating in the z-direction, which is a superposition of two linearly polarized waves. The electric field vector in the wave is
E = Ex + Ey, where Ex = a cos(kz − ωt), Ey = b cos(kz − ωt + δ). (1) The parameter δ is a real number between −π/2 and π/2, and indicates by how much the two components are out of phase. Look at the behavior of the electric field at some fixed value of z, say z = 0 for simplicity.

a) [2pt] Describe what the electric fields Ex and Ey are doing as a function of time.

b) [4pt] Show that there is a simple relation between Ex and Ey which does not involve t. Namely, you should find the following: Ex2/a2 + Ey2[/SUP]/ b2 − 2ExEycos δ/ab = constant. (2) Express the constant in the right-hand side of (2) in terms of the phase shift δ.

I am trying to do b[/B]

as I found in a) that

Ex = acos(ωt) and Ey = bcos(ωt - δ)



Homework Equations



E = Ex + Ey

The Attempt at a Solution



Ex = acos(ωt)
Ey = bcos(ωt)

Ex/a = cos(ωt)
Ey/b = cos(ωt - δ)

I assume I can use Euler formula and say e = cosΘ + isinΘ

So I get

Ex/a = ei(ωt)
Ey/b = ei(ωt - δ) = eiωt / eiδ


So

Ey eiδ/b = eiωt

I assume I set them equal to each other but I don't get the terms that I want for the LHS and the RHS becomes 0.

You have to eliminate the time t. Use the addition law for cosine and write both cos(ωt) and sin(ωt) in terms of Ex, Ey and δ. Then use sin2(ωt)+cos2(ωt)=1.
 
  • #3
You have to eliminate the time t. Use the addition law for cosine and write both cos(ωt) and sin(ωt) in terms of Ex, Ey and δ. Then use sin2(ωt)+cos2(ωt)=1.


I cannot rewrite the sinδ terms in terms of Ex and Ey and δ unless I am supposed to use

cosx = sin(x+π/2)
 
  • #5
don't use bold fonts.

First:do not use bold fonts unless you really want to highlight something. As for Addition Laws of Trigonometry see http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

Yes, sorry about that.

What I get is

cos2(ωt)cos2(δ) + sin2(ωt)sin2(δ) + 2cos(ωt)cos(δ)sin(ωt)sin(δ)

Which I can sub to get

(Ex2/a2)cos2(δ) + sin(ωt)sinδ + (Ex/a)sin(2δ)sin(ωt)

However I am unsure how to simplify further
 
  • #6
ehild
Homework Helper
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Yes, sorry about that.

What I get is

cos2(ωt)cos2(δ) + sin2(ωt)sin2(δ) + 2cos(ω)cos(δ)sin(ωt)sin(δ)

Which I can sub to get

(Ex2/a2)cos2(δ) + sin(ωt)sinδ + (2Ex/a)cos(δ)sin(ωt)sin(δ)

However I am unsure how to simplify further
What do you get if you expand cos(ωt-δ) in Ey/b = cos(ωt-δ)?
 
  • #7
cos(ωt)cos(δ) + sin(ωt)sin(δ)

I then squared the entire term. Is that step incorrect?
 
  • #8
ehild
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cos(ωt)cos(δ) + sin(ωt)sin(δ)

I then squared the entire term. Is that step incorrect?
Why did you square the entire term?

You have Ey/b=cos(ωt)cos(δ) + sin(ωt)sin(δ) and Ex/a=cos(ωt). Substitute Ex/a for cos(ωt) in the expression for Ey/b and isolate sin(ωt).
 
  • #9
Ah I squared it because I thought I would simplify after.

Okay then I would get

(Ey/b - Excos(δ)/a)/sinδ = sin(ωt)


Then I use sin2(ωt) + cos2(ωt) = 1?
 
  • #10
ehild
Homework Helper
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Ah I squared it because I thought I would simplify after.

Okay then I would get

(Ey/b - Excos(δ)/a)/sinδ = sin(ωt)


Then I use sin2(ωt) + cos2(ωt) = 1?

Yes.
 
  • #11
Yes.

I get

1/sinδ[Ey2/b2 + Ex2cos2δ/a2-2ExEycosδ/ab] = (1-Ex2/a)

However I am unsure of what to do with the
Ex2cos2δ/a2 on the LHS
 
Last edited:
  • #13
ehild
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