Skew tracing in a step-index optical fiber using Geometric Algebra

Angeleene S. Ang,1, ∗ Quirino M. Sugon, Jr.,2, 1 and Daniel J. McNamara2, 1 1Ateneo de Manila University, Department of Physics, Loyola Heights, Quezon City, Philippines 1108 2Manila Observatory, Upper Atmosphere Division, Ateneo de Manila University Campus (Dated: April 14, 2015) We used Geometric Algebra to compute the paths of skew rays in a cylindrical, step-index mul- timode optical fiber. To do this, we used the vector addition form for the law of propagation, the exponential of an imaginary vector form for the law of refraction, and the juxtaposed vector product form for the law of reflection. In particular, the exponential forms of the vector rotations enables us to take advantage of the addition or subtraction of exponential arguments of two rotated vectors in the derivation of the ray tracing invariants in cylindrical and spherical coordinates. We showed that the light rays inside the optical fiber trace a polygonal helical path characterized by three invariants that relate successive reflections inside the fiber: the ray path distance, the difference in axial dis- tances, and the difference in the azimuthal angles. We also rederived the known generalized formula for the numerical for skew rays, which simplifies to the standard form for meridional rays.

I. INTRODUCTION To construct a ray tracing algorithm, we need to de- scribe rays mathematically. We can do this in three ways. Optical fibers are waveguides used for optical commu- The first way is by labeling points, lines, and angles, nication as first illustrated by Tyndall in 1870[1]. Single- then use geometry and trigonometry to determine the mode fibers have radii of about 8 microns, while multi- scalar invariants, e.g., the axial path length between re- mode fibers have about 60 microns[2]. The description flections, axial angle of incident and reflected rays, and of light propagation in single-mode fibers require wave the numerical aperture. But the system for naming these , while that in multimode fibers, the ray approxi- geometrical quantities vary from author to author, mak- mation is sufficient.[3] In this paper, we shall only talk ing it difficult to compare results. [3, 11–14] To solve the of multimode fibers using ray or geometric optics, i.e, no problem of the angle naming conventions, we define the diffraction, interference, or polarization. directions of the vectors in terms of the polar angles θ Optical fibers can be classified depending on the func- and the azimuthal angle φ in spherical coordinates, dis- tional dependence of the on the fiber tinguished only by their subscripts, e.g. θσ and φσ for radius: step-index or graded-index (GRIN). Step-index the incident ray σ and φη for the normal vector η. can be multi-step[4] while, graded-index fibers can have The second way is to use vectors for ray tracing. In refractive index functions that are parabolic. Here, we general, we need five vectors: the initial position of the shall focus only on step-index fibers, where the refractive ray r, the final position of the ray r0, the propagation index of the core is constant and light rays are guided vector σ, the reflected vector σ0, and the vector normal through total internal reflection. to the surface η0. The task of ray tracing then is to If the rays lie only on a single plane, the rays are said to relate these variables using a set of equations, such as be meridional; if the rays trace a discrete helix, the rays those given in Klein and Furtak[15], though in a slightly are skew. Most textbooks describe only the meridional different form involving the concavity function cσηk = rays and mention skew rays only in passing. The reason ±1: for this is that the mathematics used to describe skew rays is difficult. [5–7] rk+1 = rk + sk+1 σk+1, (1a) The study of geometric optics ray tracing for skew rays σ0 = σ − 2c η cos β , (1b) is still an important problem for two reasons. The first k k σηk k k is pedagogical: skew ray tracing for laser light in large nk+1σk+1 = nkσk fibers, e.g., HeNe laser in 1 cm diameter glass fiber, can + cσηk(nk+1 cos βk+1 − nk cos βk)ηk, (1c) be used to verify the helical properties of skew rays in a student laboratory. The second is practical: skew rays which corresponds to propagation, reflection, and refrac- tracing equations can serve as the starting point for the tion. But these ray tracing equations are difficult to ap- analysis of electric fields at the fiber walls[8] and can ply if we wish to compute for analytic solutions, such as provide the basis of ray tracing algorithms for optical reflected ray and final position after a certain number of arXiv:1504.02906v1 [physics.optics] 11 Apr 2015 fibers[9], which may even include the Goos-H¨anchen ef- reflections within the fiber. That is why ray tracing algo- fect for reflection[10]. rithms do not use analytical solutions but rather iterative computations using the equations described[9, 10]. The third way is to use vectors in ray tracing but within the framework of the Eikonal method. [16, 17] ∗ [email protected] Here, the position R of the ray satisfies a vector differ- 2 ential equation: II. GEOMETRIC ALGEBRA FOR GEOMETRIC OPTICS d  dR n = ∇n, (2) A. Geometric Algebra ds ds 1. Vectors and Imaginary Numbers where s is the propagation distance and n is the refractive index. This method lends itself readily to the analysis of Geometric Algebra Cl3,0, known as Pauli Algebra,[20] invariants in step-index fibers as shown by Zubia et al. is generated by three spatial unit vectors e1, e2, and e3 [4]. This method may also be applied to graded-index which satisfy the orthonormality axioms [27]: fibers using computational methods[9, 18, 19]. This dif- 2 ferential equation approach may be powerful and lends ej = 1, (5a) itself well to computational methods, but it does not ex- e e = −e e . (5b) ploit the simple vector geometry of rays for step-index j k k j fibers. That is, the square of a vector simply is unity and the In this paper, we shall not use the Eikonal method juxtaposition multiplication product of two distinct unit nor the reflection and refraction equations in Eqs. (1b) vectors anti-commute with each other. and (1c). Instead, we shall construct the ray tracing We can generalize these multiplication rules for two ar- equations using the exponential rotational operators and bitrary vectors a and b in 3D using the Pauli Identity[20, direct vector products in Geometric Algebra: 21] ab = a · b + i(a × b), (6) ab = a · b + i a × b, (3a) where i = e e e . It can be shown that i is an imaginary inθ 1 2 3 e = cos θ + in sin θ, (3b) number that commutes with both scalars and vectors. If we express a as the sum of its perpendicular and which are the Pauli identity[20, 21] and Euler’s Theo- parallel vector components with respect to b, then rem, respectively. For example, the law of reflection and a = a + a . (7) refraction can be expressed as ⊥ k Using the Pauli Identity in Eq. (6), we can show that

σk+1 = −ηkσkηk, (4a) akb = ak · b = bak, (8a) σ = σ eicσηkeσηk(βk−βk+1). (4b) k+1 k a⊥b = a⊥ × b = −ba⊥. (8b) Equation (8) is the 3D generalization of the orthonormal- The direct vector product form of the law of ity axiom in Eq. (5). reflection[21–23] in Eq. (4a) and the exponential rota- tion form of the law of refraction[24] in Eq. (4b) were al- ready known before. These laws and their variants were 2. Rotations used in geometric optics for the derivation of ray tracing equations for spherical and mirrors for different Let θ be a scalar and n be a unit vector. Since (in)2 = ray cases: finite skew, paraxial skew, finite meridional, −1, then we may use Euler’s Theorem: and paraxial meridional[25, 26]. Equations (4a) and (4b) were also used before in the derivation of skew ray tracing einθ = cos θ + in sin θ. (9) equations in optical fibers[14]; however, the treatment in this three-page extended conference abstract is lim- Now, suppose a⊥ and ak are the components of a per- ited and does not include the discussion of the numerical pendicular and parallel to n. Multiplying these from aperture. We shall correct these in this paper. the left of Eq. (9), and using the commutation and anti- commutation rules in Eq. (8), we obtain We shall divide the paper into six sections. Section I is inθ inθ Introduction. In Section II, we shall describe the basics ake = e ak, (10a) of Geometric Algebra and how it can be used for vector a einθ = e−inθa . (10b) rotations in different coordinate systems. We shall use ⊥ ⊥ this algebra to describe the Laws of Propagation, Reflec- Notice that a change in sign of the exponential’s argu- tion, and Refraction in Geometric Optics. In Section III, ment occurs only when the exponential is multiplied to we shall derive the distances and angles of propagation a⊥. of the light rays inside the fiber and compute the fiber’s If we expand the left side of Eq. (10b), we get numerical aperture. In Section IV, we shall summarize inθ the ray tracing invariants. Section V is Conclusions. a⊥e = a⊥(cos θ + in sin θ). (11) 3

e3 z

sk+1σk+1 θ rk+1

cos θ e3

rk e2 φ sin θ e y 1 e ie 3 φ e1 x

FIG. 1: Relations of the Cylindrical Coordinate System FIG. 2: A vector sk+1σk+1 is defined using two vectors as compared with Cartesian and Spherical Polar. with the same origin point, rk and rk+1.

Distributing the terms, and using the Pauli identity in B. Geometric Optics Eq. (6), we obtain 1. Law of Propagation inθ a⊥e = a⊥ cos θ − (a⊥ × n) sin θ, (12)

inθ If a light ray moves from its initial position rk at the since a⊥ · n = 0. Geometrically, a⊥e is the vector a⊥ th k interface by a distance sk+1 in the direction of the rotated counterclockwise about a unit vector n. unit vector σk+1, then the ray’s final position rk+1 at the (k + 1)th interface is [24]

3. Coordinate Systems rk+1 = rk + sk+1 σk+1. (18)

(see Fig. 2). Note that we can express r , r , σ as We claim that the unit radial vector e in spherical k k+1 k r the sum of e and e eie3φ as given in Eq. (16) : coordinates is given by [23] 3 1

ie3φk rk = zke3 + rk sin θke1e , (19a) e = e−ie3φ/2 e eie2θeie3φ/2. (13) r 3 ie3φk+1 rk+1 = zk+1e3 + rk+1 sin θk+1e1e , (19b)

ie3φσk+1 To show this, we note that from Eq. (12), σk+1 = cos θσ(k+1)e3 + sin θσ(k+1)e1e , (19c)

ie2θ e3e = e3 cos θ + e1 sin θ, (14) then Eq. (18) can be separated into simultaneous equa- tions for each component. so that Eq. (13) may be written as rk+1 sin θk+1 cos φk+1 = rk sin θk cos φk+

−ie3φ/2 ie3φ/2 s sin θ cos φ , er = e (e3 cos θ + e1 sin θ) e . (15) k+1 σ(k+1) σ(k+1) (20a)

Distributing the terms and using the relations in Eq. (10), rk+1 sin θk+1 sin φk+1 = rk sin θk sin φk+ we get sk+1 sin θσ(k+1) sin φσ(k+1),

ie3φ (20b) er = e3 cos θ + e1 sin θ e . (16) rk+1 cos θk+1 = rk cos θk + sk+1 cos θσ(k+1). Hence, (20c)

er = e1 sin θ cos φ + e2 sin φ sin θ + e3 cos θ. (17) 2. Law of Reflection Equation (17) is the representation of the unit spherical th radial vector er in rectangular coordinates, as shown in Let σk be the direction of the ray as it hits the k Figure 1. interface with a unit normal vector ηk. The new direction 4

σk+1 after it hits the interface is [23] where σk is the incident vector, σk+1 is the refracted vector, cσηk is the concavity function in Eq. (27), and −iηkπ/2 iηkπ/2 σk+1 = −e σke . (21) eσηk is the axis of rotation σ × η That is, the reflected ray σk+1 is the negative of the k k eσηk = , (31) incident ray σk rotated counterclockwise about ηk by |σk × ηk| an angle π. Expanding the exponentials using the Euler identity Eq. (9), then Eq. (21) reduces to with ηk is the unit vector normal to the surface. Geomet- rically, Eq. (30) says that the refracted ray is the incident ray rotated either clockwise (if the concavity function is σk+1 = −ηkσkηk, (22) negative/interface is convex) or counterclockwise (if the which is a known form for the reflection law in Geometric concavity function is positive/interface is concave) about Algebra. [23] eσηk by the difference of the incident angle and the re- 0 If we rewrite the product ηkσk in Eq. (22) using the fracted angle, β−β . We will use primed variables for the Pauli Identity in Eq. (6) and distribute the rightmost ηk, refracted values. Also, we shall mostly deal with concave we get surfaces inside the fiber, so that we can immediately set cσηk = +1. σk+1 = −(ηk · σk)ηk + i(ηk × σk)ηk. (23)

Applying the Pauli identity again for the second term III. SKEW RAYS IN OPTICAL FIBERS results to A. Refraction at Point Pi σk+1 = −ηk(ηk · σk) + (ηk × σk) × ηk, (24)

Let σi be the direction of the initial ray as it strikes since i(ηk × σk) · ηk = 0. Finally, expanding the triple the flat end of the fiber: cross product on right hand side, Eq. (24) becomes

ie3φσi σi = sin θσi e1e + cos θσi e3, (32) σk+1 = −2ηk(ηk · σk) + σk. (25) which is similar in form to Eq. (19c). If the ray σi strikes Let us define β as the angle of incidence, so that k the fiber at the position ri, then the outward normal vector at the surface is ηk · σk = cσηk cos βk, (26) ηi = −e3. (33) where cσηk is the concavity function, Multiplying Eq. (32) and Eq. (33), and using the ex- σk · ηk cσηk = , (27) ponential identities in Eq. (8), we get |σk · ηk| ie3φσ σiηi = sin θσ ie2e i − cos θσ . (34) whose values are ±1. Thus, Eq. (25) reduces to i i Expanding the left-hand side of Eq. (34) using the Pauli σk+1 = −2ηk cos βk + σk, (28) Identity in Eq. (6), and separating the scalar and vector parts, we obtain which is the same expression for the law of reflection in Klein and Furtak. [15, 24] σi · ηi = − cos θσi , (35a)

ie3φσi σi × ηi = sin θσi e2e . (35b)

3. Law of Refraction Substituting these into the definition of the the concavity function and the rotation axis in Eqs. (27) and (31), we The Law of Refraction is given by arrive at

0 0 n sin β = n sin β , (29) cos θσi cσηi = − = −1, (36a) | cos θσi | 0 where n and n are the indices of refraction at both sides ie3φσ eσηi = e2e i , (36b) of the interface, while β and β0 are the angles of incidence and refraction. since θσi > 0. The refraction law in Eq. (29) may also be reformulated We can find the angle βi between the unit vectors σi in geometric algebra as follows [24]: and ηi using their dot product:

0 icση eση (β−β ) σk+1 = σk e , (30) σi · ηi = cos βi = cos θσi , (37) 5

e3 P2

P1

z0e3 r0

e2

P0 Rη0 0 φη0 βi Pi

βi ηi e1 σi FIG. 4: Vector r can be expressed as the sum of a (a) The incident ray σi strikes the point Pi entrance end of the 0 fiber where the normal vector is ηi. The angle of incidence is rotating vector on the xy plane, R η0, and a vertical 0 βi and the angle of refraction is βi. From point Pi, the light component, z0e3. ray enters the fiber and strikes the cylindrical walls at points P0, P1, and P2. so that Eq. (40) reduces to 0 P1 0 −iβ eσηi σ0 = e3e i . (42) Expanding the exponential in Eq. (42), we get

0 0 ie3φσi 0 P0 0 σ0 = sin βi e1e + cos βi e3. (43) P2

On the other hand, using the expansion of σ0 in Eq. (19c), we have ri

ie3φσ0 0 σ0 = cos θσ0 e3 + sin θσ0 e1e . (44) Pi

σi Thus, equating Eqs. (43) and (44), we get 0 θσ0 = βi, (45a)

φσ0 = φσi . (45b) 0 0 0 (b) Top view of Fig. (3a), with the points P0, P1, and P2 as the projections of the points P0, P1, and P2 on the fiber’s entrance. which is the same relations in [14], once we have changed the subscripts to match their notation. FIG. 3: The propagation of the light ray inside the optical fiber in perspective view and top view B. Propagation from Pi to P0 so that We define the entry point of the light ray when it in- tersects the base of the cylinder as ri: βi = θσi . (38) ie3φi 0 ri = ρi e1e = ρi cos φie1 + ρi sin φie2. (46) Using Snell’s Law, the refracted angle βi is given by  n  β0 = sin−1 sin β , (39) From position ri, the ray moves in the direction of σ0 de- i n0 i fined in Eq. (44), until it strikes the walls of the cylinder where we set n = 1 for air (see Fig. 3). at position r0, so that The law of refraction as given by Eq. (30) for k = 0 is r0 = s0σ0 + ri, (47) 0 i(βi−β )eσηi σ0 = σie i , (40) where s0 is the distance of propagation. If we define the where cση1 = 1. We can express σi as a rotation of e3 radius of the cylinder to be R, then the position r0 can about the axis eση1 by a clockwise angle βi, be written as

−ieσηiβi ie3φη σi = e3e , (41) r0 = R e1e 0 + z0e3, (48) 6

where φη0 is the azimuthal angle and z0 is the distance along the axis, as illustrated in Figure 4. Substituting Eq. (47) into Eq. (48), we get

ie3φη s0σ0 + ri = R e1e 0 + z0e3. (49) r0 Notice that there are three unknowns in this equation: s0+σ0 s0, φη0 , and z0. ηi Let us separate the axial and radial components of Eq. (49) to obtain s0−σ0 σi s0 cos θσ0 e3 = z0e3, (50a) FIG. 5: If we extend s σ backwards, we obtain a ie3φσ ie3φi ie3φη 0+ 0 s0 sin θσ e1e 0 + ρi e1e = R e1e 0 . (50b) 0 vector of length s0+, which will terminate at the fiber’s wall, extended in the −e3 direction. Equation (50a) gives us our first equation relating s0 and z0:

that for our case, s0− is a virtual backward propagation z0 = s0 cos θσ . (51) 0 from the fiber entrance.

Now, squaring both sides of Eq. (50b) and using the Lastly, we solve for φη0 . Separating Eq. (50b) into the exponential identities in Eq. (10), we get components of e1 and e2, we get

2 2 2 2 s0 sin θσ0 cos φσ0 + ρi cos φi = R cos φη0 , (58a) s0 sin θσ0 + ρi + 2s0ρi sin θσ0 cos(φi − φσ0 ) = R . (52)

s0 sin θσ0 sin φσ0 + ρi cos φi = R sin φη0 . (58b)

Solving for s0 sin θσ0 using the quadratic formula results to Dividing Eq. (58a) by Eq. (58b), we obtain q y 2 η0 2 2 tan φη = , (59) s0 sin θσ0 = ± R − ρi sin (φi − φσ0 ) 0 xη0 − ρi cos(φi − φσ ), (53) 0 where after simplifying the terms. x = s sin θ cos φ + ρ cos φ , (60a) Notice that there are two possible solutions: η0 0 σ0 σ0 i i

yη0 = s0 sin θσ0 sin φσ0 + ρi sin φi. (60b) h q 1 2 2 2 s0± = ± R − ρi sin (φi − φσ0 ) We note that s0 sin θσ0 is given in Eq. (53), so that the sin θσ0 i rectangular coordinates xη0 and yη0 of the normal vector − ρi cos(φi − φσ0 ) , (54a) η0 are already defined in terms of the initial parameters. Thus, the cylindrical coordinates of the normal vector η0 The distance s0+ is the distance traveled by the light ray can now be determined, since the radius R of the fiber is inside the fiber from the fiber’s entrance to the cylindrical constant and the azimuthal angle φη0 is given by Eq. (59). interface, while the distance s0− is the distance traveled by the light ray if it propagates opposite to the direction σ0. The difference between these two distances is C. Refraction at P0: Numerical Aperture 2 q 2 2 2 At point P , on the fiber’s cylindrical wall, the incident s0+ − s0− = R − ρi sin (φi − φσ0 ), (55) 0 sin θσ 0 ray is σ0 defined in Eq. (44), and the normal vector is which is a new result (see Fig. 5). If ρ = 0, then we get ie3φη i η0 = e1e 0 . (61) the meridional case: Their product is 2R s − s = . (56) 0+ 0− ie (φ −φ ) ie φ sin θ 3 σ0 η0 3 η0 σ0 σ0η0 = sin θσ0 e − i cos θσ0 e2e , (62)

On the other hand, if ρi = R, where we used the exponential indentities in Eq. (10). Separating the scalar and imaginary vector parts, we get 2R| cos(φi − φσ0 )| s0+ − s0− = , (57) sin θσ0 σ0 · η0 = sin θσ0 cos(φσ0 − φη0 ), (63) σ × η = e cos θ sin φ − e cos θ cos φ which is similar to the form of the wall-to-wall propaga- 0 0 1 σ0 η0 2 σ0 η0 tion distance given by Cozannet and Treheux[28], except + e3 sin θσ0 sin(φσ0 − φη0 ), (64) 7 after removing the imaginary number i in the second 0 equation. If we define ψ as the angle between σ0 and η0, then Eq. (63) yields χ 0 cos ψ = sin θσ0 cos(φσ0 − φη0 ), (65) which is essentially the same as that of Potter[11] and Senior[12]. 0 Since ψ is also the angle of incidence at point P0, then φ by Snell’s Law we have i − φ σ ρ i n0 sin ψ0 = n sin ψ, (66) R FIG. 6: The skew factor χ as a function of the ratios of where ψ is the angle of refraction outside at point P0. the relative radii ρ/R and the azimuthal angular For the ray to be trapped, we set ψ = π/2, so that difference φi − φσi n sin ψ0 = . (67) n0 Substituting Eq. (75) back into Eq. (71) results to Combining Eq. (65) and Eq. (67), p 2 n02 − n2 n 2 2 NA = . (76) 1 = + sin θσ cos (φσ − φη ), (68) s ρ n02 0 0 0 0 sin θ + i cos(φ − φ ) R σ0 R i σ0 then solving for sin θσ0 , we get Combining this with the expression for s in Eq. (54a) p 0 n02 − n2 and using the identity φσ = φσ from Eq. (45b), we sin θ = . (69) 0 i σ0 0 arrive at n | cos(φσ0 − φη0 )| p Now, the numerical aperture NA of an optical fiber is NA = χ n02 − n2, (77) defined as where 0 NA = n sin θσ . (70) 0 1 χ = r . (78) We combine Eqs. (69) and (70) to get ρ2 1 − i sin2(φ − φ ) p R2 i σi n02 − n2 NA = . (71) | cos(φσ0 − φη0 )| Equation (77) forms the expression for the numerical aperture NA for the skew rays in optical fibers. Here Equation (71) is the expression for the numerical aper- we introduce the skew factor χ which is defined in terms ture for skew rays in optical fibers as given by Potter[11, of the initial ray parameters at the fiber entrance: the po- 29] and Senior[12]. sition ri = (ρi, φi, 0) in cylindrical coordinates where the For the meridional approximation, we set φη0 = φσ0 , incident ray σi = (1, φσi , θσi ) in spherical coordinates. so that Eq. (71) reduces to the known standard form for Notice that the skew factor depends on the azimuthal the numerical aperture as given by Klein and Furtak[15] angle φσ of the incident ray, but not on the polar angle and Potter[11]: i θσi . The expression for the numerical aperture NA in terms p 02 2 NA = n − n . (72) of the skew factor χ is a new result. To arrive at the If we expand the denominator in Eq. (71), we get meridional case in Eq. (72), we set the radial distance ρi = 0, to get the skew factor χ = 1. On the other hand, if we set the initial position at the edge of the fiber, i.e. cos(φσ0 − φη0 ) = cos φσ0 cos φη0 + sin φσ0 sin φη0 . (73) ρi = R, we get Since 1 χ = , (79) cos φη0 = xη0 /R, (74a) p 2 cos (φσ0 − φη0 )

sin φη0 = yη0 /R, (74b) which leads to Eq. (71), the numerical aperture for skew then using Eq. (60) together with trigonometric indenti- rays given in the literature. Fig. 6 shows the plot of ties, we obtain the skew factor χ as a function of the normalized radial

s0 ρi distance ρi/R and the difference of the azimuthal angles cos(φσ0 − φη0 ) = sin θσ0 + cos(φi − φσ0 ). (75) φ − φ . R R i σi 8

D. Reflection at P0 where s1 and σ1 is the ray’s propagating distance and direction, respectively. This is shown in Figure 7. If we use the expression of r in Eq. (48) and express r in a At position r0 on the interface, the light ray’s propa- 0 1 similar form, then Eq. (88) becomes gation vector is σ0 and the normal vector to the interface is η0: ie3φη ie3φη s1σ1 + Re1e 0 + z0e3 = Re1e 1 + z1e3. (89) ie φ σ = cos θ e + sin θ e e 3 σ0 , (80a) 0 σ0 3 σ0 1 Separating Eq. (89) into its radial and axial components, ie3φη η0 = e1e 0 . (80b) we get

ie3φσ1 ie3φη0 ie3φη1 To obtain the new propagation vector σ1 after reflection, s1 sin θσ1 e1e + Re1e = Re1e , (90a) we use the law of reflection in Eq. (22): s1 cos θσ1 e3 + z0e3 = z1e3. (90b)

σ1 = −η0σ0η0. (81) Notice that there are three unknowns in Eq. (90): s1,

φη1 , z1. Substituting the expressions for η0 in Eq. (80b), and σ0 To solve for the propagation distance s1, we first square in Eq. (80a) into Eq. (81), yields both sides of Eq. (90a):

ie3(2φη0 −φσ0 ) 2 σ1 = − sin θσ e1e + cos θσ e3, (82) 2 2 2 0 0 s1 sin θσ1 + R + 2s1R sin θσ1 cos(φη0 − φσ1 ) = R . (91) after distributing the terms and simplifying the result. Hence, Eq. (82) is the desired equation for the new propagation 2R cos(φ − φ ) vector σ . η0 σ1 1 s1 = − . (92) Using the definition of σ1 as written in Eq. (19c), sin θσ1

ie3φσ 1 To solve for φη1 , we expand Eq. (90a) into its e1 and σ1 = sin θσ1 e1e + cos θσ1 e3, (83) e2 components: and comparing this with the expression for σ1 in Eq. (83), we obtain [14] R cos φη1 = s1 sin θσ1 cos φσ1 + R cos φη0 , (93a)

R sin φη1 = s1 sin θσ1 sin φσ1 + R sin φη0 . (93b) θσ1 = −θσ0 , (84a) Dividing the two equations and using the expression for φσ1 = 2φη0 − φσ0. (84b) s1 in Eq. (92), we get Notice that Eq. (84a) is not proper as we previously de- sin φ cos 2φ − cos φ sin 2φ fined θ to have values between 0 to π. η0 σ1 η0 σ1 tan φη1 = , (94) In order to avoid having to write the negative sign − cos φη0 cos 2φσ1 − sin φη0 sin 2φσ1 before the θ, we can use the following relation derived after using the double angle identities. Hence, from the Euler identity:

− sin(2φσ1 − φη0 ) ie3(±π) tan φ = . (95) e = cos(±π) + ie3 sin(±π) = −1, (85) η1 − cos(2φσ1 − φη0 ) so that Eq. (82) becomes Note that we retained the negative signs in both the nu- merator and denominator so that we can properly deter- ie3(2φη0 −φσ0±π) σ1 = sin θσ0 e1e + cos θσ0 e3. (86) mine the angle φη1 which is from 0 to 2π. Hence, IV. SKEW RAY INVARIANTS θσ1 = θσ0 , (87a) φσ = 2φη − φσ0 ± π, (87b) 1 0 A. Propagation Distance Between Reflections which is similar to what we have derived in a previous paper[14]. Notice that Eq. (87a) is now in the proper The procedure for computing the propagation of the form of θ, except that the sign of π in Eq. (87b) must be light ray from P1 to P2 is the same as the procedure determined for individual cases. from P0 to P1. Therefore, we can simply change the subscripts to obtain the following relations for the prop-

agation distance s2, the propagation polar angle θσ2 , and E. Propagation from P0 to P1 the propagation azimuthal angle φσ2: 2R cos(φ − φ ) From the propagation law in Eq. (18), we know that η1 σ2 s2 = − , (96a) the final position of the ray depends on its initial position sin θσ2 r : 0 θσ2 = θσ1 , (96b) φ = 2φ − φ ± π. (96c) r1 = s1σ1 + r0, (88) σ2 η1 σ1 9

Taking the difference of Eq. (102) and Eq. (103), and

imposing s2 = s1 in Eq. (100) and θσ2 = θσ1 in Eq. (96b), then η1 ∆z = z2 − z1 = z1 − z0. (104)

s1σ1 Thus, the axial distance ∆z between wall-to-wall propa- r1 gations, defined as the pitch, is constant. If we substitute the expression for z1 in Eq. (102) and s in Eq. (92), we get z1 1

∆z = −2R cos(φη0 − φσ1 ) cot θσ1 , (105) η0 r0 which is similar in form to that given in Eq. (22) of Cozannet and Treheux[28]. Alternatively, we can express ∆z in terms of s1 in Eq. (92) to get

∆z = s1 cos θσ1 , (106)

FIG. 7: The light ray traveled from point r0 to point r1 which is an invariant given in Love and Snyder[3]. Note on the fiber’s walls, at a distance s1, along the direction that this invariant reduces only to the ray half-period if σ1. The normal vectors at the two points are η0 and η1. the rays are meridional.

Notice that the angle θσ remains constant through reflec- C. Change in Azimuthal Angle Between tions. Our aim is to show that the propagation distance Reflections s is also constant between reflections, i.e., s1 = s2. If we substitute Eqs. (96b) and (96c) into Eq. (96a), We know from Eq. (95) that the azimuthal angle φη we get 1 at point P1 is given by

2R cos(−φη1 + φσ1 )   s2 = , (97) −1 − sin(2φσ1 − φη0 ) φη = tan , (107) sin θσ1 1 − cos(2φσ1 − φη0 ) since cos(φ − π) = − cos φ. Expanding Eq. (97) using a so that trigonometric identity, and using the expression for the sines and cosines of φη1 in Eq. (95), φη1 = 2φσ1 − φη0 + kπ, (108) sin φ = − sin(2φ − φ ), (98) η1 σ1 η0 where k is +1, −1, or 0. cos φ = − cos(2φ − φ ), (99) η1 σ1 η0 To find k, we substitute the expression for φη1 in

Eq. (108), for φσ2 in Eq. (96c), and sin(θσ1 ) in Eq. (96b) we obtain to Eq. (96a) to obtain 2R cos(φ − φ ) s = − η0 σ1 . (100) 2 2R cos(2φσ1 − φη0 − 2φη1 + φσ1 + (k ∓ 1)π) sin(θσ1 ) s2 = − . sin θσ1 Comparing this with the expression for s1 in Eq. (92), (109) we arrive at Substituting the expression for φη1 in Eq. (108) again and rearranging the terms, we get s1 = s2. (101)

2R cos(−φσ1 + φη0 + (−k ∓ 1)π) Thus, the propagation distance between two reflections s2 = − . (110) sin θ is constant. σ1

Comparing this the expression for s2 in Eq. (100), we arrive at k = ∓1 , so that Eq. (108) reduces to B. Axial Distance Between Reflections

φη1 = 2φσ1 − φη0 ∓ π. (111) The axial position z1 of point P1 is given in Eq. (90b): Using a similar process, we can show that the form of

z1 = s1 cos θσ1 + z0. (102) φη2 and φη3 are similar to φη1 , Similarly, we can show that for point P , 2 φη2 = 2φσ2 − φη1 ∓ π, (112a)

φη3 = 2φσ3 − φη2 ∓ π, (112b) z2 = s2 cos θσ2 + z1. (103) 10 so that ray path distance, (2) the change in axial distances, and (3) the change in the azimuthal angles. We also showed

φη2 − φη1 = 2(φσ2 − φη1 ) ∓ π, (113a) that the numerical aperture for skew rays we obtained is the same as that of the literature. φη3 − φη2 = 2(φσ3 − φη2 ) ∓ π. (113b) To derive the ray tracing invariants, we did not use Replacing the subscripts of Eq. (96c) to obtain an ex- the reflection and refraction laws in Klein and Furtak[15]. pression for φσ3 , Rather, we used two alternative techniques: (a) the Pauli Identity which expresses the geometric product of two

φσ3 = 2φη3 − φσ2 ± π, (114) vectors as a sum of their dot and imaginary cross prod- ucts and (b) the Euler’s Theorem which expresses the and using the the expressions for φσ3 in Eq. (114) and exponential of an imaginary vector as a sum of the co- φη2 in Eq. (112a), Eq. (113b) becomes sine of the magnitude of the vector and the product of the normalized imaginary vector with the sine of the vec- φ − φ = 2(φ − φ ) ∓ π. (115) η3 η2 σ2 η1 tor’s magnitude. These two theorems allow us to express vector rotations in exponential form, which enables us to Comparing Eq. (113a) with Eq. (115), we arrive at take advantage of the addition or subtraction of exponen- tial arguments of two rotated vectors in the derivation of ∆φ = φη − φη = φη − φη . (116) 3 2 2 1 the ray tracing invariants. Thus, the change in the azimuthal angle ∆φ between Many of the equations for the invariants were already consecutive reflections is a constant. known before, except maybe for the change in the az- So now we have two invariants between consecutive imuthal angle between reflections. Also, we obtained a reflections: the change in azimuthal angle ∆φ and the new expression for the numerical aperture, which allows change in the axial propagation distance ∆z. From this, the point of entry of light to be an arbitrary point in we conclude that the rays inside a fiber trace a polygonal the fiber’s entrance, and not limited to the center or the helix. edge as given by previous authors. Finally, our use of standard notations for the angles θ and φ for cylindri- cal and spherical coordinates, differentiated only by sub- V. CONCLUSIONS scripts would hopefully help in visualizing the complex geometry of skew ray tracing. In this paper, we used Geometric Algebra to compute In future papers, we shall extend our work on skew the paths of skew rays in a cylindrical, step-index optical rays to ray tracing in circular, conical, and toroidal fibers, fiber. To do this, we used the vector addition form for the which may be step-index, multistep-index, or graded in- law of propagation, the exponential of an imaginary vec- dex. tor form for the law of refraction, and the vector product form for the law of reflection. In addition, we used the spherical angles θ and φ to describe the directions of rays ACKNOWLEDGEMENT in space, but expressed in cylindrical coordinates in ex- ponential form. We showed that the light rays inside the Thanks to James Hernandez for checking the read- optical fiber trace a polygonal helical path characterized ability of the manuscript. This work was supported by by three invariants between sucessive reflections: (1) the Manila Observatory and Ateneo de Manila University.

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