Relationship between x and y components of E

[jasonchiang97]

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: E_x²/a² + E_y²[/SUP]/ b² − 2E_xE_ycos 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

E_x = acos(ωt) and E_y = bcos(ωt - δ)

Homework Equations

E = E_x + E_y

The Attempt at a Solution

E_x = acos(ωt)
E_y = bcos(ωt)

E_x/a = cos(ωt)
E_y/b = cos(ωt - δ)

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

So I get

E_x/a = e^i(ωt)
E_y/b = e^{i(ωt - δ)} = e^iωt / e^iδ

So

E_y e^iδ/b = e^iωt

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.

 

[ehild]

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: E_x²/a² + E_y²[/SUP]/ b² − 2E_xE_ycos 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

E_x = acos(ωt) and E_y = bcos(ωt - δ)

Homework Equations

E = E_x + E_y

The Attempt at a Solution

E_x = acos(ωt)
E_y = bcos(ωt)

E_x/a = cos(ωt)
E_y/b = cos(ωt - δ)

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

So I get

E_x/a = e^i(ωt)
E_y/b = e^{i(ωt - δ)} = e^iωt / e^iδ

So

E_y e^iδ/b = e^iωt

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.

 

[jasonchiang97]

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 E_x and E_y and δ unless I am supposed to use

cosx = sin(x+π/2)

 

[ehild]

I cannot rewrite the sinδ terms in terms of E_x and E_y and δ unless I am supposed to use

cosx = sin(x+π/2)

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

 

[jasonchiang97]

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

(E_x²/a²)cos²(δ) + sin(ωt)sinδ + (E_x/a)sin(2δ)sin(ωt)

However I am unsure how to simplify further

 

[ehild]

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

(E_x²/a²)cos²(δ) + sin(ωt)sinδ + (2E_x/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)?

 

[jasonchiang97]

cos(ωt)cos(δ) + sin(ωt)sin(δ)

I then squared the entire term. Is that step incorrect?

 

[ehild]

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).

 

[jasonchiang97]

Ah I squared it because I thought I would simplify after.

Okay then I would get

(E_y/b - E_xcos(δ)/a)/sinδ = sin(ωt)Then I use sin²(ωt) + cos²(ωt) = 1?

 

[ehild]

Ah I squared it because I thought I would simplify after.

Okay then I would get

(E_y/b - E_xcos(δ)/a)/sinδ = sin(ωt)Then I use sin²(ωt) + cos²(ωt) = 1?

Yes.

 

[jasonchiang97]

Yes.

I get

1/sinδ[E_y²/b² + E_x²cos²δ/a²-2E_xE_ycosδ/ab] = (1-E_x²/a)

However I am unsure of what to do with the
E_x²cos²δ/a² on the LHS

 

[jasonchiang97]

Thanks for your help! I figured it out

 

[ehild]

Thanks for your help! I figured it out

You are welcome.