The Principle of Least Action and Designing Fiber Optics

Transcription

1 University of Southampton Department of Physics & Astronomy Year 2 Theory Labs The Principle of Least Action an Designing Fiber Optics 1 Purpose of this Moule We will be intereste in esigning fiber optic cables that carry light in particular ways. To change the light path one must carefully choose the refractive inex profile on the cable cross section. How oes the light s path epen on the refractive inex? We will use as our guiing principle Fermat s Principle of Least Time: light propagates between two points so as to minimize its travel time. We will therefore be intereste in the mathematics of fining paths that extremize some quantity. This is known as Calculus of Variation 2 Calculus of Variation Consier a set of curves qs between two points q 1, an q 2,s 2 in the s,q plane we will only consier curves where the trajectory is single value at each value of s. q q 2 q 1 s 2 S Imagine we are intereste in one curve that minimizes the quantity 1

2 S[qs] = s 2 Lq, q,s L is just a number at each point on a given curve etermine by the values of q an s at that point an the graient q = q/. The integral sums these numbers along the line. If the curve that minimizes S is qs we can write the other curves as eviations from it subject to the bounary conitions qs= qs+δ qs δq =δqs 2 =0 The value of S for these curves varies from the value for qs by δs = S[ q+ δq] S[ q] Since qs is the minimum though δs = 0 to lowest orer in δq. Let s calculate S[ q + δq] to orer δq S[ q + δq]= s 2 L q + δq, q + δ q,s s 2 s L q, q,s+δ q 1 q + δq q +... S[ q]+ s 2 s δ q 1 q + δq q Integrating the first term by parts u = / q, v/ = δ q etc s2 δ q q =[δq/ q]s 2 s1 s 2 δq q The first term vanishes since δq vanishes at the en of the path. Thus S[ q + δq] S[ q]= s 2 δq q q This is only zero at orer δq if 2

3 q q = 0 which is the all important Euler Lagrange equation. Q1 Make sure you unerstan this erivation. More generally we coul imagine having more coorinates q q i it s sensible to work with the smallest number of inepenent or generalize coorinates. Show that the Euler Lagrange equations then become a set - one for each q i : q i q i = 0 3 Example 1: Minimum Surface The classic problem, for which Lagrange evelope his Calculus of Variation formalism, is this: Consier two points in the x,y plane. Connect them by a line of your choice an then rotate about the y-axis into the thir imension so the line sweeps out the surface of some volume see the figure. What line shoul you have chosen to minimize the surface area of that volume? y y x z x A general infinitissimal length of the line will contribute the surface area of a cyliner of raius x an height x 2 + y 2 A = 2πx x 2 + y 2 1 3

4 Thus for a particular choice of line yx A = x2 x 1 2πx 1 + y 2 x 2 x The curve that minimizes the surface satisfies the Euler lagrange equation x y y = 0 3 where a prime inicates a erivative with respect to x an L = 2πx 1 + y 2 4 x Q2 Note that L is inepenent of y an hence show that the extremizing curve satisfies the ifferential equation x y x y x = constant,c 1 5 Q3 Hence show that y = C 1 x 2 C1 2 x 6 1/2 Q4 Make the substitution x = C 1 coshφ an fin the solution. What etermines the constants in the solution? 4

5 4 Example 2: Fiber Optics When we look for the path followe by a light ray, the time of flight, accoring to Fermat s principle, is the object that plays the role of the action S above. In a circular cross section fiber optic cable it is sensible to use cylinerical coorinates r,θ,z. If light travels along an infinitessimal path escribe by the general changes r,θ,z then the length of the path is given by l 2 = r 2 + r 2 θ 2 + z 2 7 Q5 Convince yourself we can therefore write the time of flight as ct = finish start L z 8 where L = nr,θ,z r 2 + r 2 θ c is the spee of light in vacuum an n = c/v is the inex of refraction of the material the fibre is mae of, in which the light travels at spee v. Note we have let v vary through the material. In this three imensional case the path that minimizes the time of travel is given by the solution of the two Euler Lagrange equations z r r = 0, z θ θ = 0 10 We will restrict ourselves to problems where the inex of refraction epen only on the raial position r. Q6 Show that the secon equation then implies that nrr 2 θ r 2 + r 2 θ = C 11 5

6 Q7 What can we say about a meriional ray that initially has θ = 0? Note that L is also inepenent of z so / z = 0. Now L z = z + r r + r r + θ θ + θ θ 12 Q8 Using the z inepenence of L an the two Euler Lagrange equations substitute for / r an / θ show that [ L r z r θ ] θ = 0 13 an hence that in the case of the fibre optic nr r 2 + r 2 θ = D 14 Q9 What is the fate of a ray that enters the fibre normal to the surface? We can now solve these two equations for particular nr an fin the path light takes in the fibre. Conversely we can iscover the form of nr neee to make the light follow some esire path. We finish with an example of each. Q10 Semi-Circular Rays Show that if nr= n λr 15 then meriional rays with θ = constant form a connecte series of circular segments in the r z plane. 6

7 Q11 Helix Fin the form of the cylinerically symmetric inex of refraction for which the light rays are helices with r = r 0. If you have a time to spare you coul try to set up a maple program that will solve the ifferential equations numerically an then plot the light s path for any particular choice of n. This worksheet has been prepare from a super book calle Perfect Form by Lemons. The iea of a Principle of Least Action can also be applie to Newtonian particle mechanics proviing an alternative formulation to Newton s Laws. Lemons explores the parallels between the optical an mechanics versions in etail. You ll learn more next year about the mechanics formalism! 7