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Posts Tagged ‘complex’

Geometry of general Jones vector (problem 2.8)

Posted by peeterjoot on August 9, 2012

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Another problem from [1].

Problem

The general case is represented by the Jones vector

\begin{aligned}\begin{bmatrix}A \\ B e^{i\Delta}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.1.1)

Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle

\begin{aligned}\frac{1}{{2}} \tan^{-1} \left( \frac{2 A B \cos \Delta }{A^2 - B^2} \right),\end{aligned} \hspace{\stretch{1}}(1.1.2)

with the x axis.

Solution

Prior to attempting the problem as stated, let’s explore the algebra of a parametric representation of an ellipse, rotated at an angle \theta as in figure (1). The equation of the ellipse in the rotated coordinates is

Figure 1: Rotated ellipse

 

\begin{aligned}\begin{bmatrix}x' \\ y'\end{bmatrix}=\begin{bmatrix}a \cos u \\ b \sin u\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.2.3)

which is easily seen to have the required form

\begin{aligned}\left( \frac{x'}{a} \right)^2+\left( \frac{y'}{b} \right)^2 = 1.\end{aligned} \hspace{\stretch{1}}(1.2.4)

We’d like to express x' and y' in the “fixed” frame. Consider figure (2) where our coordinate conventions are illustrated. With

Figure 2: 2d rotation of frame

 

\begin{aligned}\begin{bmatrix}\hat{\mathbf{x}}' \\ \hat{\mathbf{y}}'\end{bmatrix}=\begin{bmatrix}\hat{\mathbf{x}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta} \\ \hat{\mathbf{y}} e^{\hat{\mathbf{x}} \hat{\mathbf{y}} \theta}\end{bmatrix}=\begin{bmatrix}\hat{\mathbf{x}} \cos \theta + \hat{\mathbf{y}} \sin\theta \\ \hat{\mathbf{y}} \cos \theta - \hat{\mathbf{x}} \sin\theta\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.2.5)

and x \hat{\mathbf{x}} + y\hat{\mathbf{y}} = x' \hat{\mathbf{x}} + y' \hat{\mathbf{y}} we find

\begin{aligned}\begin{bmatrix}x' \\ y'\end{bmatrix}=\begin{bmatrix}\cos \theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.2.22)

so that the equation of the ellipse can be stated as

\begin{aligned}\begin{bmatrix}\cos \theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}a \cos u \\ b \sin u\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.2.7)

or

\begin{aligned}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}\cos \theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}a \cos u \\ b \sin u\end{bmatrix}=\begin{bmatrix}a \cos \theta \cos u - b \sin \theta \sin u \\ a \sin \theta \cos u + b \cos \theta \sin u\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.2.8)

Observing that

\begin{aligned}\cos u + \alpha \sin u = \text{Real}\left( (1 + i \alpha) e^{-i u} \right)\end{aligned} \hspace{\stretch{1}}(1.2.9)

we have, with \text{atan2} = \text{atan2}(x, y) a Jones vector representation of our rotated ellipse

\begin{aligned}\begin{bmatrix}x \\ y\end{bmatrix}=\text{Real}\begin{bmatrix}( a \cos \theta - i b \sin\theta ) e^{-iu} \\ ( a \sin \theta + i b \cos\theta ) e^{-iu}\end{bmatrix}=\text{Real}\begin{bmatrix}\sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta } e^{i \text{atan2}(a \cos\theta, -b\sin\theta) - i u} \\ \sqrt{ a^2 \sin^2 \theta + b^2 \cos^2 \theta } e^{i \text{atan2}(a \sin\theta, b\cos\theta) - i u}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.2.10)

Since we can absorb a constant phase factor into our -iu argument, we can write this as

\begin{aligned}\begin{bmatrix}x \\ y\end{bmatrix}=\text{Real}\left(\begin{bmatrix}\sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta } \\ \sqrt{ a^2 \sin^2 \theta + b^2 \cos^2 \theta } e^{i \text{atan2}(a \sin\theta, b\cos\theta) -i \text{atan2}(a \cos\theta, -b\sin\theta)} \end{bmatrix} e^{-i u'}\right).\end{aligned} \hspace{\stretch{1}}(1.2.11)

This has the required form once we make the identifications

\begin{aligned}A = \sqrt{ a^2 \cos^2 \theta + b^2 \sin^2 \theta }\end{aligned} \hspace{\stretch{1}}(1.2.12)

\begin{aligned}B = \sqrt{ a^2 \sin^2 \theta + b^2 \cos^2 \theta } \end{aligned} \hspace{\stretch{1}}(1.2.13)

\begin{aligned}\Delta =\text{atan2}(a \sin\theta, b\cos\theta) - \text{atan2}(a \cos\theta, -b\sin\theta).\end{aligned} \hspace{\stretch{1}}(1.2.14)

What isn’t obvious is that we can do this for any A, B, and \Delta. Portions of this problem I tried in Mathematica starting from the elliptic equation derived in section 8.1.3 of [2]. I’d used Mathematica since on paper I found the rotation angle that eliminated the cross terms to always be 45 degrees, but this turns out to have been because I’d first used a change of variables that scaled the equation. Here’s the whole procedure without any such scaling to arrive at the desired result for this problem. Our starting point is the Jones specified field, again as above I’ve using -iu = i (k z - \omega t)

\begin{aligned}\mathbf{E} = \text{Real}\left( \begin{bmatrix}A \\ B e^{i \Delta}\end{bmatrix}e^{-i u}\right)=\begin{bmatrix}A \cos u \\ B \cos ( \Delta - u )\end{bmatrix}e^{-i u}\end{aligned} \hspace{\stretch{1}}(1.2.15)

We need our cosine angle addition formula

\begin{aligned}\cos( a + b ) = \text{Real} \left( (\cos a + i \sin a)(\cos b + i \sin b)\right) =\cos a \cos b - \sin a \sin b.\end{aligned} \hspace{\stretch{1}}(1.2.16)

Using this and writing \mathbf{E} = (x, y) we have

\begin{aligned}x = A \cos u\end{aligned} \hspace{\stretch{1}}(1.2.17)

\begin{aligned}y = B ( \cos \Delta \cos u + \sin \Delta \sin u ).\end{aligned} \hspace{\stretch{1}}(1.2.18)

Subtracting x \cos \Delta/A from y/B we have

\begin{aligned}\frac{y}{B} - \frac{x}{A} \cos \Delta = \sin \Delta \sin u.\end{aligned} \hspace{\stretch{1}}(1.2.27)

Squaring this and using \sin^2 u = 1 - \cos^2 u, and 1.2.17 we have

\begin{aligned}\left( \frac{y}{B} - \frac{x}{A} \cos \Delta \right)^2 = \sin^2 \Delta \left( 1 - \frac{x^2}{A^2} \right),\end{aligned} \hspace{\stretch{1}}(1.2.27)

which expands and simplifies to

\begin{aligned}\left( \frac{x}{A} \right)^2 +\left( \frac{y}{B} \right)^2 - 2 \left( \frac{x}{A} \right)\left( \frac{y}{B} \right)\cos \Delta = \sin^2 \Delta,\end{aligned} \hspace{\stretch{1}}(1.2.27)

which is an equation of a rotated ellipse as desired. Let’s figure out the angle of rotation required to kill the cross terms. Writing a = 1/A, b = 1/B and rotating our primed coordinate frame by \theta degrees

\begin{aligned}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}\cos \theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}x' \\ y'\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.2.22)

we have

\begin{aligned}\begin{aligned}\sin^2 \Delta &=a^2 (x' \cos \theta - y'\sin\theta)^2+b^2 ( x' \sin\theta + y' \cos\theta)^2 \\ &- 2 a b (x' \cos \theta - y'\sin\theta)( x'\sin\theta + y'\cos\theta)\cos \Delta \\ &=(x')^2 ( a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2 a b \cos \theta \sin \theta \cos \Delta ) \\ &+(y')^2 ( a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2 a b \cos \theta \sin \theta \cos \Delta ) \\ &+ 2 x' y' ( (b^2 -a^2) \cos \theta \sin\theta + a b (\sin^2 \theta - \cos^2 \theta) \cos \Delta ).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.27)

To kill off the cross term we require

\begin{aligned}\begin{aligned}0 &= (b^2 -a^2) \cos \theta \sin\theta + a b (\sin^2 \theta - \cos^2 \theta) \cos \Delta \\ &= \frac{1}{{2}} (b^2 -a^2) \sin (2 \theta) - a b \cos (2 \theta) \cos \Delta,\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.27)

or

\begin{aligned}\tan (2 \theta) = \frac{2 a b \cos \Delta}{b^2 - a^2} = \frac{2 A B \cos \Delta}{A^2 - B^2}.\end{aligned} \hspace{\stretch{1}}(1.2.27)

This yields 1.1.2 as desired. We also end up with expressions for our major and minor axis lengths, which are respectively for \sin \Delta \ne 0

\begin{aligned}\sin\Delta/ \sqrt{ b^2 + (a^2 - b^2) \cos^2 \theta - a b \sin (2 \theta) \cos \Delta }\end{aligned} \hspace{\stretch{1}}(1.2.27)

\begin{aligned}\sin\Delta/\sqrt{ b^2 + (a^2 - b^2)\sin^2 \theta + a b \sin (2 \theta) \cos \Delta },\end{aligned} \hspace{\stretch{1}}(1.2.27)

which completes the task of determining the geometry of the elliptic parameterization we see results from the general Jones vector description.

References

[1] G.R. Fowles. Introduction to modern optics. Dover Pubns, 1989.

[2] E. Hecht. Optics. 1998.

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Complex form of Poynting relationship

Posted by peeterjoot on August 2, 2012

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This is a problem from [1], something that I’d tried back when reading [2] but in a way that involved Geometric Algebra and the covariant representation of the energy momentum tensor. Let’s try this with plain old complex vector algebra instead.

Question: Average Poynting flux for complex 2D fields (problem 2.4)

Given a complex field phasor representation of the form

\begin{aligned}\tilde{\mathbf{E}} = \mathbf{E}_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}\end{aligned} \hspace{\stretch{1}}(1.0.1)

\begin{aligned}\tilde{\mathbf{H}} = \mathbf{H}_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Here we allow the components of \mathbf{E}_0 and \mathbf{H}_0 to be complex. As usual our fields are defined as the real parts of the phasors

\begin{aligned}\mathbf{E} = \text{Real}( \tilde{\mathbf{E}} )\end{aligned} \hspace{\stretch{1}}(1.0.3)

\begin{aligned}\mathbf{H} = \text{Real}( \tilde{\mathbf{H}} ).\end{aligned} \hspace{\stretch{1}}(1.0.4)

Show that the average Poynting vector has the value

\begin{aligned}\left\langle{{ \mathbf{S} }}\right\rangle = \left\langle{{ \mathbf{E} \times \mathbf{H} }}\right\rangle = \frac{1}{{2}} \text{Real}( \mathbf{E}_0 \times \mathbf{H}_0^{*} ).\end{aligned} \hspace{\stretch{1}}(1.0.5)

Answer

While the text works with two dimensional quantities in the x,y plane, I found this problem easier when tackled in three dimensions. Suppose we write the complex phasor components as

\begin{aligned}\mathbf{E}_0 = \sum_k (\mathbf{E}_{kr} + i \mathbf{E}_{ki}) \mathbf{e}_k = \sum_k {\left\lvert{\mathbf{E}_k}\right\rvert} e^{i \phi_k} \mathbf{e}_k\end{aligned} \hspace{\stretch{1}}(1.0.6)

\begin{aligned}\mathbf{H}_0 = \sum_k (\mathbf{H}_{kr} + i \mathbf{H}_{ki}) \mathbf{e}_k = \sum_k {\left\lvert{\mathbf{H}_k}\right\rvert} e^{i \psi_k} \mathbf{e}_k,\end{aligned} \hspace{\stretch{1}}(1.0.7)

and also write \phi_k' = \phi_k + \mathbf{k} \cdot \mathbf{x}, and \psi_k' = \psi_k + \mathbf{k} \cdot \mathbf{x}, then our (real) fields are

\begin{aligned}\mathbf{E} = \sum_k {\left\lvert{\mathbf{E}_k}\right\rvert} \cos(\phi_k' - \omega t) \mathbf{e}_k\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}\mathbf{H} = \sum_k {\left\lvert{\mathbf{H}_k}\right\rvert} \cos(\psi_k' - \omega t) \mathbf{e}_k,\end{aligned} \hspace{\stretch{1}}(1.0.9)

and our Poynting vector before averaging (in these units) is

\begin{aligned}\mathbf{E} \times \mathbf{H} = \sum_{klm} {\left\lvert{\mathbf{E}_k}\right\rvert} {\left\lvert{\mathbf{H}_l}\right\rvert} \cos(\phi_k' - \omega t) \cos(\psi_l' - \omega t) \epsilon_{klm} \mathbf{e}_m.\end{aligned} \hspace{\stretch{1}}(1.0.10)

We are tasked with computing the average of cosines

\begin{aligned}\left\langle{{ \cos(a - \omega t) \cos(b - \omega t) }}\right\rangle=\frac{1}{{T}} \int_0^T \cos(a - \omega t) \cos(b - \omega t) dt=\frac{1}{{\omega T}} \int_0^T \cos(a - \omega t) \cos(b - \omega t) \omega dt=\frac{1}{{2 \pi}} \int_0^{2 \pi}\cos(a - u) \cos(b - u) du=\frac{1}{{4 \pi}} \int_0^{2 \pi}\cos(a + b - 2 u) + \cos(a - b) du=\frac{1}{{2}} \cos(a - b).\end{aligned} \hspace{\stretch{1}}(1.0.11)

So, our average Poynting vector is

\begin{aligned}\left\langle{{\mathbf{E} \times \mathbf{H}}}\right\rangle = \frac{1}{{2}} \sum_{klm} {\left\lvert{\mathbf{E}_k}\right\rvert} {\left\lvert{\mathbf{H}_l}\right\rvert} \cos(\phi_k - \psi_l) \epsilon_{klm} \mathbf{e}_m.\end{aligned} \hspace{\stretch{1}}(1.0.12)

We have only to compare this to the desired expression

\begin{aligned}\frac{1}{{2}} \text{Real}( \mathbf{E}_0 \times \mathbf{H}_0^{*} )= \frac{1}{{2}} \sum_{klm} \text{Real}\left({\left\lvert{\mathbf{E}_k}\right\rvert} e^{i\phi_k}{\left\lvert{\mathbf{H}_l}\right\rvert} e^{-i\psi_l}\right)\epsilon_{klm} \mathbf{e}_m = \frac{1}{{2}} \sum_{klm} {\left\lvert{\mathbf{E}_k}\right\rvert} {\left\lvert{\mathbf{H}_l}\right\rvert} \cos( \phi_k - \psi_l )\epsilon_{klm} \mathbf{e}_m.\end{aligned} \hspace{\stretch{1}}(1.0.13)

This proves the desired result.

References

[1] G.R. Fowles. Introduction to modern optics. Dover Pubns, 1989.

[2] JD Jackson. Classical Electrodynamics Wiley. John Wiley and Sons, 2nd edition, 1975.

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