Method of Rotated Reference Frames

Transcription

1 Method of Rotated Reference Frames Vadim Markel in collaboration with George Panasyuk Manabu Machida John Schotland Radiology/Bioengineering UPenn, Philadelphia

2 Outline

3 MOTIVATION (the inverse problems perspective)

4 ρ s S zs α(r) zd z Given a data function φ( ρ, ρ ) s d which is measured for multiple pairs ( ρ, ρ ), find the absorption s d D ρ d coefficient α( r) inside the slab 4

5 Linearized Integral Equation φ( ρ, ρ ) = Γ( ρ, ρ ; r) ( r)d 3 s d s d δα r φ ( ρ, ρ ) s d = I ( ρs, zs; ρd, zd) I0 ( ρs, zs; ρd, zd) I ( ρ, z ; ρ, z ) (measurable data-function) 0 α ( r) = α + δα( r) 0 s s d d Γ ( ρ, ρ ) = G ( ρ, z ; r) G ( r; ρ, z ) s d 0 s s 0 d d (first Born approximation)

6 How To Invert? Numerical SVD approach: φ = Γ δα i ij j j j φ i φ =Γδα Γ δα φ = Γ Γδα * * + * 1 * ( ) = ΓΓ Γ reg φ

7 Analytical SVD approach: Making use of the translational invariance φ ( q, q ) = φ ( ρ, ρ ) e d ρ d ρ i ( q s ρ s+ q d ρ d ) 2 2 s d s d s d qs = q /2 + p, qd = q /2 p; Data function: ψ ( q, p) = φ ( q / 2 + p, q / 2 p) L ψ ( q, p) = g ( q / 2 + p; z) g ( q / 2 p; z) δα ( q; z)dz 0 iρ q 2 δα ( q; ) = δα ( ρ, ) d ρ z dz 2 d q G ( ρ, z ; ρ, z) = g s z e ( 2π ) 0 s s 2 s ( 2π ) d ( q; z) e 2 d q G0 ( ρ, z; ρd, zd) = g ( ; ) 2 d q z e iq ( ρ ρ ) s iq ( ρ ρ) d 7

8 6cm S.D.Konecky, G.Y.Panasyuk, K.Lee, V.Markel, A.G.Yodh and J.C.Schotland Imaging complex structures with diffuse light Optics Express 16(7), (2008)

9 In the case of RTE, we need the plane-wave decomposition of the Greens function, of the form 2 d q G (,; rsr ˆ, sˆ ) = g(;,; qz sˆ z, sˆ ) e ( 2π ) 0 2 iq ( ρ ρ ) and the integral kernel 2 Γ ( q, p; z) = g( q/2 + p; z, zˆ ; z, sˆ ) g( q/2 p; z,; sˆ z, zˆ )ds We then get the integral equations of the form z d ψ( qp, ) = Γ( qp, ; z) δα ( q; z)dz z s s d

10 Motivation Continued (the spectral methods)

11 Solve system of equations : ( z+ W) x = b Find σ ( ) = Im = Im ( + ) 1 z bx b z W b for M different values of z, where z W = x+ iδ is a complex number is an N N matrix x, b are vectors of length N

12 The "naive" method: 1) Choose z 2) Make a matrix A= z+ W 3) Solve Ax = b 4) Goto step (1) Computational complexity: M N 3

13 The spectral method: 1) Diagonalize W (find eigenvectors n and eigenvalues w ) 2) For every z, x n 2 n n b cnδ = ; σ = 2 2 z+ w ( x+ w ) + δ n n n n Computational complexity: N + M N 3 2 However, if the whole vector x is not needed, the complexity may be as low as 3 N M N +

14 Spectral Method for the RTE? 2 sˆ μ ˆ ˆ ˆ ˆ ˆ ˆ t I r s μs A s s I r s s ε r s RTE: ( + )(,) = (, )(, )d + (,) Where is the "spectral variable"? How can we write this equation in the form ( z+ W) I = ε? μ and μ do not qualify... t s We can try to expand I( rs, ˆ) into a a 3D Fourier integral with respect to r and into the basis of ordinary spherical harmonics Y ( θϕ, ) with respect to sˆ... lm...and see if the equation can be cast into the desired form.

15 The Conventional Method of Spherical Harmonics This results in the following system of equations with respect to the vector of the expansion coefficients I( k) ( k - the Fourier variable) : ia k I( k) + ia k I( k) + ia k I( k) + D I( k) = ε ( k) ( x) ( y) ( z) x y z A, A, ( x) ( y) A ( z) are different matrices. ( x) * Alm, l m = sinθcos ϕylm( θϕ, ) Yl m ( θϕ, )sin θdθdϕ, etc.... This rather awe-inspiring set of equations has perhaps only academic interest. K.M.Case, P.F.Zweifel, Linear Transport Theory

16 Rotated Reference Frames The usual sperical harmonics are defined in the laboratory reference frame. Then θ and ϕ are the polar angles of the unit vector sˆ in that frame. THE MAIN IDEA: For each value of the Fourier variable k, use spherical harmonics defined in a reference frame whose z-axis is aligned with the direction of k. We call such frames "rotated". Spherical harmonics defined in the rotaded frame are denoted by Y ( sk ˆ; ˆ ).

17 Rotation of the Laboratory Frame (x,y,z). z y' z' l ˆ ˆ l Y (; ) (,,0) () ˆ lm sk = Dm m ϕk θk Ylm s m = l θ k y Wigner D-functions Euler angles x ϕ θ Spherical functions in the laboratory frame x'

18 ( sˆ + μ ) I = μ AI + ε t (,) ˆ (,) ˆ d ikr 3 I rs = I kse k ikr ε(,) rsˆ = ε( ks,) ˆ e d ( ik sˆ + μ ) I = μ AI + ε t I ( ks, ˆ) = F ( k) Y ( sk ˆ; ˆ ) lm, ε ( ks, ˆ) = E ( k) Y ( sk ˆ; ˆ ) lm, lm lm A(, ss) AY (; skˆ) Y (; skˆ) * ˆˆ = ˆ ˆ l lm lm lm, s s lm lm 3 k

19 lm ik R F ( k) + σ F ( k) = E ( k) lm lm l m l lm lm σ = μ + μ (1 A ) l a s l R = sˆ kˆ Y (; sk ˆ ˆ) Y (; sk ˆ ˆ)ds= l lm * 2 lm lm l m b( m) = [ b( m) b ( m) ] = δ δ + δ mm l l = l 1 l+ 1 l = l+ 1 l m l

20 RTE in the Angular Basis of Rotated Spherical Functions ikr I( k) + D I( k) = E( k) Scalar spectral parameter Block-tridiagonal real symmetric matrix Source term Diagonal matrix Slm, l m = δ ll δmm [ μa + μs(1 Al )] Parameters of the phase function. (For the HG l model, A = g ) l

21 Let D W = SS = S RS 1 1 ikr I( k) + D I( k) = E( k) ikw S I k S E 1 ( + 1) ( ) = ( ) I k S ikw S E ( ) = (1 + ) ( ) k k W ψ = λ ψ μ μ μ I ( k) = μ 1 1 S ψ ψ μ μ S E 1+ ikλ μ ( k)

22 The Spectral Solution ˆ ˆ (,) d σ 1+ ikλ 1 1 lm ψμ ψμ S E( k) Ylm( s; k) ikr 3 I rsˆ = e k lm ε(,) rsˆ = δ( r r) δ( sˆ sˆ ) ikr 0 * ˆ E ˆ lm( k) = e Y ( 3 lm s0; k) (2 π ) l μ μ The integral is not easy but doable. Details in J.Phys.A 39, 115 (2006)

23 ˆ m * ˆ ˆ ˆ ˆ = lm χll l m 0 m= l, l = m G (,; rsr, s) Y (; sr) ( R) Y ( s; Rˆ) m m ( 1) χll ( R) = ( 1) 2π σ σ l l l M n 3 M= l n, λ > 0 λn l φ ( M) φ ( M) l ( M ) l l l + 2 j,0 l l + 2 j,0 R ClMl,,, MClml,,, m kl l + 2 j j= 0 λn( M ) l = min( l, l ) μ = ( Mn, ) lm ψ ( ) Mn = δmm l φn M n n r ŝ ŝ 0 R r 0

24 Some Properties of the Eigenvalues

25 If λ < 1/ μ - continuous spectrum t If λ > 1/ μ - discrete spectrum t Bounds on the diffuse eigenvalue λ =max[ λ (0)] d D= cμλ 2 2 β1 + β2 λd β1 + β2 bl (0) 1 βl = ; β1 = ; σσ 3 μ [ μ + (1 A ) μ ] l l 1 a a 1 s β 2 = 2 15[ μ + (1 A) μ ][ μ + (1 A ) μ ] a 2 d a 1 s a 2 s n Bound on the gap: Δλ= λ λ d next largest Δλ max[ β + β β β,0]

26 Density, Current, and the Fick s Law 1 1 ( ) (, ˆ)d ; ( ) (, ˆ) ˆd c c 1 0 u( r) = 2l+ 1 χ ˆ ˆ 0l( r) Pl( s0 r) c 2 2 u r = I r s s J r = I r s s s l= 0 rˆ 0 Jr ( ) = 2l+ 1 χ ˆ ˆ 1l( r) Pl( s0 r) 3 l= 0 r λd If κ = 1, then Δ λλ J next largest = D u, where D= cμλ a 2 d

27 Infinite Space, Point Uni-Directional (Sharply-Peaked) Source (a) forward and backward propagation Angular dependence of the specific intensity for forward (a) and backward (b) μ μ -5 propagation obtained at lmax = 21, g = 0.98 and a/ s = The distance to the source z is assumed to be positive for forward propagation and negative for backward propagation.

28 Infinite Space, Point Uni-Directional (Sharply-Peaked) Source (b) off-axis propagation ŝ 0 z α ŝ Two cases: a) sˆ in the xy plane b) sˆ in the yz plane y ŝ β x

29 Angular distribution of specific intensity for off-axis propagation (relatively small absorption) -5 Parameters: g = 0.98 and μa/ μs = 6 10 (a), (b), = 0.03 (c), (d).

30 Angular distribution of specific intensity for off-axis propagation (relatively large absorption) Parameters: g = 0.98 and μ / μ = 0.2. a s

31 Plane-Wave Decomposition ˆ ˆ (,) d σ 1+ ikλ 1 1 lm ψμ ψμ S E( k) Ylm( s; k) ikr 3 I rsˆ = e k 2 dq ( 0 ) * ˆ ˆ iq ρ ρ lm ˆ ˆ ˆ lm lm 0 l m 0 lm, l m G (,; rsr, s) = e Y (;) szg (;, qz z ) Y ( s;) zˆ 2π ( ) dk g q z z ˆ ( ) ˆ 0 = e D q+ zk K q + k D q+ zk 2π z ikz ( z z0 ) 2 2 ( ;, ) ( ) ( ) Kw ( ) = lm μ S 1 l ψ μ ψ 1+ iwλ S μ z z z 1 μ μ iϕ J iθ J μ ; D ( kˆ ) = e cos θ = k / q + k k z k z e k y 2 2 z

32 i( m m ) ϕq l l lm e g ( q; z, z ) = sgn( z z ) d [ iτ ( qλ )] lm Q ( q) = q + 1/ λ μ 0 0 σσ l l 2 2 μ 2 cos[ τ( )] 1, sin[ τ( )] l+ l + m+ m l [ ] mm1 m = l m = l μλ, > Qμ ( q) z z0 e l 1ψ μ ψ 2 μ 2 mm τ λ 2 μ λμqμ( q) lm l m d [ i ( q )] i x = + x i x = ix α D ( αβγ,, ) = e d ( β) e l im l im γ mm mm μ μ

33 Plane waves: Evanescent Waves kr I = e F sˆ k k = k = knˆ k 2 k( ), 1/ λμ, nˆ - real unit vector F ( ˆ) = lm Y ( ˆ; ˆ k s ψ μ lm s k) lm Evanescent waves: k = q± zˆ + q zˆ = k k = i q 1/ λμ, 0, 1/ λμ iq k =± q + 1/ λ z z 2 2 n

34 kˆ r exp( imϕ k ) l Iˆ = exp Y ( ˆ; ˆ lm ) dmm ( θ ) l φn ( M) Mn sz k k λmn lm σ l (plane wave modes) exp( imϕ ) I = exp i Q ( q) z Y ( sz ˆ; ˆ) ( 1) d ( θ ) l φ ( M) [ q ρ ] ( ± ) q l+ m l qmn Mn lm m, M k n lm σ l (evanescent modes) G V 0 2 dq ( ) ( ) (,; ˆ 0, ˆ ˆ ˆ 0) 2 I ± rsr s = qmn(,) rs V Mn I q qmn( r0, s0) (2 π ) qmn = Q Mn 1 ( q) λ Mn 2 Mn

35 Half-space problem Z>0 Scattering medium Propagationd escribed by the RTE z<0 Nonscattering medium d 2 q I = A ( q) I (2 ) 2 Mn π Mn ( + ) qmn

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39 CONCLUSIONS The method of rotated reference frames takes advantage of all symmetries of the RTE (symmetry with respect to rotations and inversions of the reference frame). The angular and spatial dependence of the obtained solutions is expressed in terms of analytical functions. The analytical part of the solution is of considerable mathematical complexity. This is traded for relative simplicity of the numerical part. We believe that we have reduced the numerical part of the computations to the absolute minimum allowed by the mathematical structure of the RTE.

40 Publications: 1. V.A.Markel, Modified spherical harmonics method for solving the radiative transport equation, Letter to the Editor, Waves in Random Media 14(1), L13-L19 (2004). 2. G.Y.Panasyuk, J.C.Schotland, and V.A.Markel, Radiative transport equation in rotated reference frames, Journal of Physics A, 39(1), (2006). 3. J.C.Schotland and V.A.Markel, Fourier-Laplace structure of the inverse scattering problem for the radiative transport equation, Inverse Problems and Imaging 1(1), (2007). Available on the web at