M3/4A16 Assessed Coursework 1 Darryl Holm Due in class Thursday November 6, 2008 #1 Eikonal equation from Fermat s principle

Transcription

1 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 1 M3/416 ssessed Coursework 1 Darryl Holm Due in class Thursday November 6, 2008 #1 Eikonal equation from Fermat s principle #1a Prove that the 3D eikonal equation d nr) dr ) = ṙ 2 n preserves ṙ = 1, where ṙ = dr/. 1) Expanding the 3D eikonal equation yiel ) n ṙ ṙ + nr) r = ṙ 2 n Rearranging yiel r = ṙ ṙ 1 ) n. n Consequently, ṙ r = 0 and the magnitude ṙ is preserved. Evolution under the eikonal equation ten to align ṙ with n/. This is the continuum material version of Snell s Law. #1b Compute the Euler-Lagrange equations when Hamilton s principle is written in the form B B ) dr i 1/2 0 = δ = δ g ijrs)) drj, with 2 = dr i g ij dr j for the Riemannian metric g ij rs)) in 3D with arclength parameter s. Show that these equations may be expressed as, r c + Γ c bers))ṙ b ṙ e = 0 with ṙ b = drb in which the quantities Γ c be r) are defined by Γ c ber) = 1 2 gca [ gae r) b + g abr) e b, c, e {1, 2, 3}, 2) g ] ber), a and g ca is the inverse of the metric, so that g ca g ab = δ c b. Does the eikonal equation emerge when g ab = n 2 r)δ ab? Prove it.

2 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 2 B 0 = δ = = = B B B ) dr i 1/2 B g ijrs)) drj =: δ ) ṙ 2 1/2 ) 1 dr i 2 ṙ δ g ijrs)) drj 1 1 g ij ṙ 2 k ṙi ṙ j d ) ) g kj ṙ j δr k 1 ṙ 1 g ij 2 k ṙi ṙ j g kl r l + g kl m ṙl ṙ m )) δr k, from which equation 2) emerges after rearranging. However, inspection shows that this is not the eikonal equation 1) when g ab = n 2 r)δ ab. #1c Prove that equation 2) preserves ṙ 2 = ṙ i g ij r)ṙ j. What does this tell us about the last part of question #1b? Take the s-derivative of ṙ 2 and follow the path of the variational derivation of the equation. Now substitute from above 1 d 2 ṙ 2 = 1 2 g ij,kṙ k ṙ i ṙ j + ṙ i g ij r j ṙ k g kl r l = 1 2 ṙk g lm,k ṙ l ṙ m ṙ k g kl,m ṙ l ṙ m and rearrange to prove the point that ṙ 2 = ṙ i g ij r)ṙ j is preserved. This tells us that substituting g ab = n 2 r)δ ab into the geodesic equation 2) will not recover the eikonal equation 1). Equations 1) and 2) are different. This is clear, for example, because their conservation laws differ. The eikonal equation 1) preserves the Euclidean condition ṙ = 1, not ṙ = 1, which is preserved by 2). #1d Fourth year students i) Compute the quantities Γ c be r) for g ij = n 2 r)δ ij when n = nr) with r 2 := r a δ ab r b. ii) Write the eikonal equation 1) when the index of refraction nr) de-

3 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 3 pen only on r. iii) Show that the eikonal equation conserves the vector L = r nr)ṙ when index of refraction n depen only on the spherical radius r = r. i) For g ij r) = δ ij n 2 r) with r 2 := r a δ ab r b, the geodesic equation 2) becomes r c = Γ c bers))ṙ b ṙ e = 1 [ gae r) 2 gca + g abr) g ] ber) ṙ b ṙ e b e a = 1 n [δ ceδ ab r a + δ cbδ ] ed r d r a δ ac δ eb ṙ e ṙ b nr or, equivalently, in Euclidean vector form r = 1 [ n ṙ ṙ r ) ] + r ṙ)ṙ nr }{{}}{{} Eikonal Extra ii) In contrast the eikonal equation 1) may be written as nr) r = ṙ ṙ n ) = ṙ ṙ n ) + ṙ 2 n which implies for nr) that iii) Consequently ) nr)ṙ = ṙ 2 n L = with n = dn r dr r ) r nr)ṙ = ṙ 2 r n The RHS vanishes when n = nr) by the previous equation. #2 Hamiltonian formulations ccording to the text the eikonal equation also follows from Fermat s principle in the form B 1 0 = δs = δ 2 n2 rτ)) dr dτ dr B dτ dτ = δ Lr, ṙ) dτ, 3)

4 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 4 with new arclength parameter dτ = n. You may retain the dot notation for d/dτ.) Use this version of Fermat s principle to write the Hamiltonian formulations of the solutions of question #1. s usual, 3rd year students do parts a,b,c, while 4th year and MSc students do parts a,b,c,d.) #2a The fibre derivative of the Lagrangian in 3) above is p = L dr/dτ) = n2 r) dr dτ This defines the canonical momentum and yiel the Legendre transformation for the Hamiltonian Hr, p) = p dr dτ The canonical Hamilton equations are Lr, dr/dτ) = p 2 2n 2 r) dr dτ = H p = 1 n p and dp 2 dτ = H = p 2 n n 3 Substituting the momentum-velocity relation 4) into the momentum equation yiel d n 2 dr ) = n 2 dr 2 n 5) dτ dτ dτ Hence, using dτ = nr) so that d/ = nr)d/dτ one fin ) d dr = n 2 d2 r n dτ + 2 dr ) dr dτ dτ By equation 5) = n dr dr dτ dτ n ) = 1 dr dr n n ) This shows that the geodesic Euler-Lagrange equation for the Lagrangian 3) does recover the eikonal equation in canonical Hamiltonian form. #2b The eikonal equation does emerge when g ab = n 2 r)δ ab for this Lagrangian. #2c Preservation of the Hamiltonian ensures preservation of n 2 dr/dτ 2 = dr/ 2, just as for the eikonal equation, since dτ = nr). 4)

5 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 5 This conservation law bodes well for these equations to recover the eikonal equation. #2d Fourth year students Once we have recovered the eikonal equation, parts i) and ii) follow as before, except now we must change the independent variable by using dτ = n. For iii), it remains to compute the Hamiltonian vector field for L = r p Consequently L = { }, L = r + p p and the RHS vanishes when n = nr). { } ) p 2 L, H = r n n 3 r) #3 R 3 reduction for axisymmetric, translation invariant optical media #3a Compute by chain rule that df dt = {F, H} = F S2 H = F X k ɛ klm S 2 X l H X m, for X 1 = q 2 0, X 2 = p 2 0, X 3 = p q. Done in notes. That S 2 = p q 0 is evident. #3b Show that the Poisson bracket {F, H} = F S 2 H, with definition S 2 = X 1 X 2 X 2 3 satisfies the Jacobi identity. This Poisson bracket may be written equivalently as {F, H} = X k c k ij F X i H X j

6 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 6 From the table {X i, X j } = {, } X 1 X 2 X 3 X 1 0 4X 3 2X 1 X 2 4X 3 0 2X 2 X 3 2X 1 2X 2 0 one identifies c k ij We also have {X i, X j } = c k ijx k. {X l, {X i, X j }} = c k ij{x l, X k } = c k ijc m lkx m. Hence, the Jacobi identity is satisfied as a consequence of {X l, {X i, X j }} + {X i, {X j, X l }} + {X j, {X l, X i }} = c k ij{x l, X k } + c k jl{x i, X k } + c k li{x j, X k } ) = c k ijc m lk + c k jlc m ik + c k lic m jk X m = 0, This is the condition required for the Jacobi identity to hold in terms of the structure constants. Remark. This calculation provides an independent proof of the Jacobi identity for the R 3 bracket in the case of quadratic distinguished functions. #3c Consider the Hamiltonian with the linear combinations H = ay 1 + by 2 + cy 3 6) Y 1 = 1 2 X 1 + X 2 ), Y 2 = 1 2 X 2 X 1 ), Y 3 = X 3, and constant values of a, b, c). Compute the canonical dynamics generated by Hamiltonian 6) on level sets of S 2 > 0 and S 2 = 0. This amounts to computing the intersections of the planes H = ay 1 + by 2 + cy 3 = constant with the hyperboloi of revolution about the Y 1 -axis, S 2 = Y 2 1 Y 2 2 Y 2 3 = constant.

7 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 7 One may solve this problem graphically, algebraically by choosing special cases for the orientations of the family of planes, or most generally by restricting the equations to a level set of either S 2 = constant, or H = constant. Restricting to S 2 = constant hyperboloi of revolution. Each of the family of hyperboloi of revolution S 2 = constant comprises a layer in the hyperbolic onion preserved by axisymmetric ray optics. We use hyperbolic polar coordinates on these layers in analogy to spherical coordinates, Y 1 = S cosh u, Y 2 = S sinh u cos ψ, Y 3 = S sinh u sin ψ. The R 3 bracket thereby transforms into hyperbolic coordinates as {F, H} dy 1 dy 2 dy 3 = {F, H} hyperb S 2 ds dψ d cosh u. Note that the oriented quantity S 2 d cosh u dψ = S 2 dψ d cosh u, is the area element on the hyperboloid corresponding to the constant S 2. On a constant level surface of S 2 the function {F, H} hyperb only depen on cosh u, ψ) so the Poisson bracket for optical motion on any particular hyperboloid is then {F, H} d 3 Y = S 2 ds df dh = S 2 ds {F, H} hyperb dψ d cosh u F = S 2 H ds ψ cosh u H ) F dψ d cosh u. cosh u ψ Being a constant of the motion, the value of S 2 may be absorbed by a choice of units for any given initial condition and the Poisson bracket for the optical motion thereby becomes canonical on each hyperboloid, dψ dz = {ψ, H} hyperb = H cosh u, d cosh u dz = {cosh u, H} hyperb = H ψ. In the Cartesian variables Y 1, Y 2, Y 3 ) R 3, one has cosh u = Y 1 /S and ψ = tan 1 Y 3 /Y 2 ). The Hamiltonian H = ay 1 + by 2 + cy 3 becomes H = as cosh u + bs sinh u cos ψ + cs sinh u sin ψ,

8 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 8 with sinh u = cosh 2 u 1 and sin ψ = 1 cos 2 ψ so that and 1 dψ S dz = 1 H S cosh u 1 d cosh u = 1 H S dz S ψ sinh u cosh u = coth u, = a + b coth u cos ψ + c coth u sin ψ, = b sinh u sin ψ + c sinh u cos ψ. Restricting to the conical surface S 2 = 0 To restrict to the conical surface S 2 = Y1 2 Y2 2 Y3 2 = 0 one chooses coordinates Y 1 = Z, Y 2 = Z cos ψ, Y 3 = Z sin ψ The Poisson bracket for the optical motion thereby becomes canonical on the cone, dψ dz = {ψ, H} cone = H Z, dz dz = {Z, H} cone = H ψ. and 1 dψ S dz = 1 H S Z 1 dz S dz = 1 H S ψ = a + b cos ψ + c sin ψ, = bz sin ψ + cz cos ψ. The equations on the hyperboloi and the cone are a bit complicated. It turns out that a very simple solution is possible on the level sets of the Hamiltonian planes. Restricting to H = constant planes. One may also restrict the equations to a planar level set of H = constant. This tactic is very insightful and particularly simple, because only linear equations arise.

9 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 9 The latter approach summons the linear transformation with constant coefficients, dy 1 dy 2 = B 1 B 2 B 3 dh dx dy 3 C 1 C 2 C 3 dy One fin the constant Jacobian from d 3 Y = B C) dh dx dy. Then one transforms to level sets of H by writing df dz = {F, H}d 3 Y = dc df dh = dh df dc = B C)dH {F, C} xy dx dy Thus, on one of the planes in the family of level sets of H, one fin dx dz dy dz = C y = C x Because C is quadratic and the transformation from Y 1, Y 2, Y 3 ) to H, x, y) is linear, these canonical equations on any of the planar level sets of H are linear. That is, ) [ ] ) ) dx/dz α β x = dy/dz ᾱ β + y γ γ with constants α, β, ᾱ, β, γ, γ). These are linear equations. Direct solution in canonical variables. For the Hamiltonian one has or H = ay 1 + by 2 + cy 3 = a b 2 q 2 + a + b 2 p 2 + c p q q = H = a + b)p + cq p ṗ = H = b a)q cp q q ṗ ) ) ) c a + b q = b a c p

10 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 10 s expected, this is a linear symplectic transformation and the matrix is given by ) c a + b = a b b a c 2 m 1 + a + b 2 m 2 + c m 3. #3d Fourth year students Derive the formula for reconstructing the angle canonically conjugate to S for the canonical dynamics generated by the planar Hamiltonian 6). The volume elements corresponding to the Poisson brackets are d 3 Y =: dy 1 dy 2 dy 3 = d S3 3 d cosh u dψ On a level set of S = p φ this implies canonical variables cosh u, ψ) with symplectic form, d cosh u dψ, and since S = p φ, φ) are also canonically conjugate, one has dp j dq j = ds dφ + d cosh u dψ. One recalls Stokes Theorem on phase space dp j dq j = p j dq j, where the boundary of the phase space area is taken around a loop on a closed orbit. On an invariant hyperboloid S this loop integral becomes ) p dq := p j dq j = Sdφ + cosh u dψ. Thus we may compute the total phase change around a closed periodic orbit on the level set of hyperboloid S from Sdφ = S φ = cosh u dψ + p dq 7) }{{}}{{} Geometric φ Dynamic φ Evidently, one may denote the total change in phase as the sum φ = φ geom + φ dyn,

11 D. D. Holm November 6, 2008 M3/416 Geom Mech Part 1 11 by identifying the corresponding terms in the previous formula. By Stokes theorem, one sees that the geometric phase associated with a periodic motion on a particular hyperboloid is given by the hyperbolic solid angle enclosed by the orbit. Thus, the name: geometric phase.