Polarization of a transverse plane wave: linear, circular and elliptical traces, and handedness

Let $k\in {\mathbb{R}}^3\setminus \{0\}$ and $\omega >0$, let $e,b\in {\mathbb{C}}^3$, and let $(E,B)$ be the plane-wave pair of Real plane-wave fields encoded by complex amplitudes on the whole-space region $\mathrm{\Omega }={\mathbb{R}}^3$, $I=\mathbb{R}$, with phase $\theta (x,t)=k\cdot x-\omega t$. Suppose the amplitude conditions of Dispersion, transversality and the two polarization conditions for plane-wave solutions hold, so that $k\cdot e=0$ and $b=k\times e\mathrm{/}\omega$. By that theorem, if the pair is nonzero then it also satisfies the dispersion relation $\mathrm{\mid }k\mathrm{\mid }=\omega \mathrm{/}c$, while the zero pair is permitted for every $k\mathrm{\ne }0$ and $\omega >0$, both on and off this dispersion relation. This item classifies the polarization of that class of transverse plane waves; it adds no field-theoretic content.

Write $e=a+id$ with the real vectors $a=\mathrm{Re}e$ and $d=\mathrm{Im}e$ of Real and imaginary parts, complex conjugation, and modulus, applied componentwise in ${\mathbb{C}}^3$; since $k$ is real and $k\cdot e=0$, the real and imaginary parts of this complex equation give $⟨k,a⟩=⟨k,d⟩=0$. Put $\widehat{k}=k\mathrm{/}\mathrm{\mid }k\mathrm{\mid }$, and let $⟨\cdot ,\cdot ⟩$, $\mathrm{\mid }\cdot \mathrm{\mid }$ and $\times$ be the inner product, norm and cross product of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ and The cross product in ${\mathbb{R}}^3$. Then:

(a) [Trace] At every fixed $x$ one has $E(x,t)=a\cos \theta -d\sin \theta$ with $\theta =\theta (x,t)$. The curve $\gamma :\mathbb{R}\to {\mathbb{R}}^3$, $\gamma (\theta ):=a\cos \theta -d\sin \theta$, lies in the transverse plane $k^{\perp }=\{v\in {\mathbb{R}}^3:⟨k,v⟩=0\}$; it is the image of the unit circle $S^1=\{(u_1,u_2)\in {\mathbb{R}}^2:u_1^2+u_2^2=1\}$ under the $\mathbb{R}$-linear map $(u_1,u_2)\mapsto u_{1}a-u_{2}d$; it is $2\pi$-periodic in the phase parameter $\theta$ and satisfies $\gamma (\theta +\pi )=-\gamma (\theta )$, while the map $t\mapsto E(x,t)$ has time period $T=2\pi \mathrm{/}\omega$. The curve $\gamma$ is called the polarization trace at $x$; it does not depend on $x$, so the classification below is the same at every point.

(b) [Classification] Exactly one of the following three cases holds.

(i) Linear polarization: ${\dim }_{\mathbb{R}}\mathrm{span}\{a,d\}\le 1$. When $\dim \mathrm{span}\{a,d\}=1$, the trace is a nontrivial closed segment in that span, traversed twice per period; when $a=d=0$, the trace is the single point $\{0\}$ and $(E,B)$ is the zero wave.

(ii) Circular polarization: $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }d\mathrm{\mid }>0$ and $⟨a,d⟩=0$. The trace is the circle of radius $\mathrm{\mid }a\mathrm{\mid }$ in $k^{\perp }$, and $\mathrm{\mid }E(x,t)\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }$ for all $(x,t)\in {\mathbb{R}}^3\times \mathbb{R}$.

(iii) Elliptical (nondegenerate, noncircular) polarization: $\dim \mathrm{span}\{a,d\}=2$ and not (ii). Here a nondegenerate ellipse centred at $0$ in a real plane $P$ means the set $L(S^1)$ for a real-linear isomorphism $L:{\mathbb{R}}^2\to P$. The trace is such an ellipse in $\mathrm{span}\{a,d\}$, and the origin lies in the relative interior of the enclosed region $L(\{u\in {\mathbb{R}}^2:\mathrm{\mid }u\mathrm{\mid }\le 1\})$ in that plane. If in addition $⟨a,d⟩=0$, the parametrization is along principal axes and the semiaxis lengths are $\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }d\mathrm{\mid }$; the general principal-axis computation is not asserted by this item.

(c) [Handedness] For $a,d\in k^{\perp }$, the cross product $a\times d$ is orthogonal to $a$ and to $d$ (The cross product is bilinear, alternating, and orthogonal to both factors), hence parallel to $k$; define the helicity sign $H:=⟨a\times d,\widehat{k}⟩\in \mathbb{R}$, whose sign is taken in the order of Order on the reals. One has $H=0$ exactly in the linear case (i), and $H\mathrm{\ne }0$ otherwise. This library calls the polarization right-handed when $H<0$ and left-handed when $H>0$. Geometrically, $H<0$ means that at fixed $x$ the electric vector rotates clockwise as seen by an observer facing the incoming wave (looking along $-\widehat{k}$), and $H>0$ means counterclockwise; here the screen orientation is fixed by any positive orthonormal triad $(u,v,\widehat{k})$, with $u$ to the right and $v$ up for that observer. Equivalently, $⟨E(x,t)\times {\mathrm{\partial }}_{t}E(x,t),\widehat{k}⟩=\omega H$ for every $(x,t)$. The names "right" and "left" handedness are conventional and both conventions occur in the literature, so later statements in this library use only the sign of $H$.

(d) [Status] This is a classification definition. It introduces the trace, the three cases and the sign $H$, and asserts no field equation, no propagation, energy or empirical content. The magnetic amplitude enters only through the two amplitude conditions $k\cdot e=0$ and $b=k\times e\mathrm{/}\omega$ recalled above; the definition covers the zero wave, both on and off the dispersion relation, and no non-plane or non-transverse field is classified.

Well-definedness and conventions

The real vectors $a,d$ are well defined by the componentwise real and imaginary parts of $e$; because $k$ is real, taking real and imaginary parts of $k\cdot e={\sum }_j{k}_j(a_j+i{d}_j)$ gives $⟨k,a⟩=0$ and $⟨k,d⟩=0$, so $a,d\in k^{\perp }$ and $\gamma$ takes values in $k^{\perp }$. The trace formula in (a) is Euler's formula $\exp (i\theta )=\cos \theta +i\sin \theta$ of Euler's formula: $\exp (i\theta )=\cos \theta +i\sin \theta$ for every real $\theta$ together with the field arithmetic on the right-hand side of $E=\mathrm{Re}((a+id)(\cos \theta +i\sin \theta ))=\mathrm{Re}((a\cos \theta -d\sin \theta )+i(a\sin \theta +d\cos \theta ))$ and the real-linearity of $\mathrm{Re}$: the real part is $a\cos \theta -d\sin \theta$, while the imaginary part is not a field. Since $\theta (x,t)=k\cdot x-\omega t$ and $\omega >0$, at fixed $x$ the map $t\mapsto \theta$ is a bijection onto $\mathbb{R}$; hence the set traced by $E(x,t)$ as $t$ varies is the full image $\gamma (\mathbb{R})$, independently of $x$, and $\gamma$ is $2\pi$-periodic with $\gamma (\theta +\pi )=-\gamma (\theta )$ because $\cos$ and $\sin$ shift by a sign under $\theta \mapsto \theta +\pi$. The time period is $T=2\pi \mathrm{/}\omega$, the least positive $s$ with $\theta (x,t+s)=\theta (x,t)-2\pi$ for all $(x,t)$; it is distinct from the phase-parameter period $2\pi$ of $\gamma$.

For the classification, "exactly one of the three cases holds" is checked case by case: if $\dim \mathrm{span}\{a,d\}\le 1$ then (i); otherwise $\dim =2$, and either $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }d\mathrm{\mid }>0$ with $⟨a,d⟩=0$, which is (ii), or not, which is (iii). In case (i) with $\dim =1$, write $d=\lambda a$ with $a\mathrm{\ne }0$ (or symmetrically $a=\mu d$); then $\gamma (\theta )=(\cos \theta -\lambda \sin \theta )a$, whose image is the segment $\{sa:\mathrm{\mid }s\mathrm{\mid }\le \sqrt{1+{\lambda }^2}\}$ in $\mathbb{R}a$, traversed once in each direction between successive extrema of the scalar factor (each such phase interval has length $\pi$). In case (ii), expanding with the bilinearity of the inner product (The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$) gives $\mathrm{\mid }{u}_{1}a-u_{2}d{\mathrm{\mid }}^2=u_1^2\mathrm{\mid }a{\mathrm{\mid }}^2+u_2^2\mathrm{\mid }d{\mathrm{\mid }}^2-2u_1{u}_2⟨a,d⟩=\mathrm{\mid }a{\mathrm{\mid }}^2$ whenever $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }d\mathrm{\mid }$, $⟨a,d⟩=0$ and $u_1^2+u_2^2=1$, so the image of the unit circle under $(u_1,u_2)\mapsto u_{1}a-u_{2}d$ is contained in the circle of radius $\mathrm{\mid }a\mathrm{\mid }$ in the plane $\mathrm{span}\{a,d\}$; conversely, for every $v$ in that plane with $\mathrm{\mid }v\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }$ the coordinates $u_1=⟨v,a⟩\mathrm{/}\mathrm{\mid }a{\mathrm{\mid }}^2$ and $u_2=-⟨v,d⟩\mathrm{/}\mathrm{\mid }d{\mathrm{\mid }}^2$ satisfy $u_1^2+u_2^2=1$ and $u_{1}a-u_{2}d=v$, because $a\mathrm{/}\mathrm{\mid }a\mathrm{\mid }$ and $d\mathrm{/}\mathrm{\mid }d\mathrm{\mid }$ are orthonormal and $v=⟨v,a\mathrm{/}\mathrm{\mid }a\mathrm{\mid }⟩(a\mathrm{/}\mathrm{\mid }a\mathrm{\mid })+⟨v,d\mathrm{/}\mathrm{\mid }d\mathrm{\mid }⟩(d\mathrm{/}\mathrm{\mid }d\mathrm{\mid })$; hence the trace is exactly that circle. In case (iii) the vectors $a,d$ are linearly independent, so $T:(u_1,u_2)\mapsto u_{1}a-u_{2}d$ is a linear isomorphism of ${\mathbb{R}}^2$ onto the two-dimensional plane $\mathrm{span}\{a,d\}$; the image of the closed unit disk under $T$ is a compact convex subset with nonempty interior in that plane, the image of the open disk is its nonempty interior, and the image of the unit circle is its boundary, so the trace is a nondegenerate ellipse centred at $0=T(0)$ with the origin in the relative interior of that enclosed region. The trace itself is its boundary and has empty interior in the plane. If moreover $⟨a,d⟩=0$, then along the principal axes $a,d$ the parametrization $u_{1}a-u_{2}d$ has the stated axis lengths because the two summands are orthogonal: $\mathrm{\mid }{u}_{1}a-u_{2}d{\mathrm{\mid }}^2=u_1^2\mathrm{\mid }a{\mathrm{\mid }}^2+u_2^2\mathrm{\mid }d{\mathrm{\mid }}^2$; no computation of principal axes for general $a,d$ is used or claimed.

For (c), the cross product $a\times d$ is orthogonal to both factors and alternating (The cross product is bilinear, alternating, and orthogonal to both factors), and the real Gram identity The squared cross-product norm is the Gram determinant of two vectors says $\mathrm{\mid }a\times d{\mathrm{\mid }}^2>0$ exactly when $a,d$ are linearly independent. Thus $a\times d=0$ holds exactly when $a$ and $d$ are linearly dependent, that is, in case (i); then $H=0$. Otherwise $\dim \mathrm{span}\{a,d\}=2$, and since $\mathrm{span}\{a,d\}\subseteq k^{\perp }$ are both two-dimensional, they are equal; being orthogonal to $a$ and $d$, the vector $a\times d$ is orthogonal to $k^{\perp }$, whose orthogonal complement in ${\mathbb{R}}^3$ is the line $\mathbb{R}k$, so $a\times d=\mu k$ for a real $\mu$ and $H=\mu \mathrm{\mid }k\mathrm{\mid }\mathrm{\ne }0$. For the rotation identity, differentiating $E=a\cos \theta -d\sin \theta$ along $t$ with ${\mathrm{\partial }}_t\theta =-\omega$ gives ${\mathrm{\partial }}_{t}E=\omega (a\sin \theta +d\cos \theta )$, so $E\times {\mathrm{\partial }}_{t}E=\omega (a\cos \theta -d\sin \theta )\times (a\sin \theta +d\cos \theta )=\omega (({\cos }^2\theta +{\sin }^2\theta )a\times d+\cos \theta \sin \theta (a\times a-d\times d))=\omega a\times d$ by bilinearity and alternation, and taking the inner product with $\widehat{k}$ gives the displayed identity. As a concrete check of the convention, take $\widehat{k}=e_3$ with the positive orthonormal screen triad $(e_1,e_2,e_3)$, so that $e_1$ points right and $e_2$ up for the observer looking along $-e_3$; then $e=e_1-i{e}_2$ has $a=e_1$, $d=-e_2$, $E(0,t)=\cos (\omega t)e_1-\sin (\omega t)e_2$ (clockwise), and $H=⟨e_1\times (-e_2),e_3⟩=-1<0$, consistently with the right-handed naming. This fixes the sign convention of the cited treatment: the source writes $E_0=\alpha -i\beta$ and calls the polarization right-handed if and only if $\beta =\widehat{k}\times \alpha$, which under $a=\alpha$, $d=-\beta$ is $H=-\mathrm{\mid }\alpha {\mathrm{\mid }}^2<0$ in the circular case.

Finally, all amplitudes $e,b$ and vectors $k,a,d$ are constant and all quantities above are computed from them by the operations of the cited items; no limit, no integral and no field equation is used, and the definition is at the level of classification only.