Michael Scalora U.S. Army Research, Development, and Engineering Center

1. OPTICS BY THE NUMBERS L’Ottica Attraverso i Numeri Michael Scalora U.S. Army Research, Development, and Engineering Center Redstone Arsenal, Alabama, 35898-5000 & Universita' di Roma "La Sapienza" Dipartimento di Energetica Rome, April-May 2004

2. SVEAT: The Slowly Varying Enevelope Approximation In Time, and The Ability To Inlcude Reflections To All Orders In The BPM Algorithm From the SVEAT to a Vector BPM: Negative Refraction

4. Assuming steady state conditions…

5. Wave front does not distort: Plane Wave propagation Diffraction is very important

6. This equation is of the form: Where: Using the split-step BPM algorithm

8. Apply SVEAT, i.e., SVEA in time only:drop higher temporal derivatives. This assumption means that pulse duration must remain always much longer than the optical cycle at all times. In all kinds of problems, if a pulse is as long as the optical cycle it means trouble for any approximation. So this is a very good approximation almost always.

10. This equation is first order in time. This suggests writing equation in following form:

11. Now, adopting the usual kind of scaling: And choosing

12. This equation is of the form: Which we can ALMOST easily recognized and compare to:

13. N.B.: the differential operator includes ALL longitudinal and spatial derivatives, which means all boundary conditions are left Intact. Integrating this equation must is therefore equivalent to Including longitudinal and transverse reflections to all orders.

14. This is a nasty operator equation, which has this formal solution: The exponential is the product of two non-commuting operators

15. Why is it important to include the index in the denominator? Because that factor accounts for the correct group velocity. Here is how I solve the problem:

16. Add zero: Group terms as follows: And recognize… Solution is accurate up to first order in time

17. Algorithm: Which is solution of Then algbreically manipulate solution to find Work very well in all cases except metals. Special considerations must be made in that case.

18. n=1 n=1.42

19. n=1 n=1.42 Red: without the 1/n2 factor in the operator

20. Example: Assume a PBG structure with cross sectional area as small as 1 mm2, and a Thickness L~10 mirons. The volume is therefore of order V~10-11 m3. I will further assume that the structure is not solidly anchored to the earth, i.e., it is free to move. The incident pulse can be tuned anywhere in the pass band or band gap. The rate of momentum transfer depends on tuning. m E,B before

21. after m E,B E,B Pm

22. momentum density The total momentum at time t is given by: In terms of the Poynting vector The momentum stored inside the object is the difference between the initial total momentum and the instantaneous momentum stored in the field, namely:

23. T~0 Tuned inside the gap... 1

24. Ppbg gr cm/sec 2P0 P0 -P0 Pfield time (l/c=0.33x10-14 sec.) ...Mirror like interaction 2

25. Electromagnetic momentum as a function ot position At different time shots 3

26. At band edge resonance: pulse is almost completely transmitted With some reflections. 4

27. field P(t) (gr cm/sec) pbg t (sec.) 5

28. P(z,t) z Electromagnetic momentum as a function ot position At different time shots 6

29. deceleration stage acceleration stage Velocity (cm/sec) <a> ~ 1011 m/sec2 Time (sec.)

30. Input Pulse Plane Wave: means no diffraction, even though beam width is finite. I.e., each ray travels straight down.

31. Same as slide 1

32. l/4 Z (longitudinal) l/4 x The structure: Cross section of each column is nearly square.

33. In this case the square cross section does not cause ray bending due to simple refraction

34. Same as slide 4.

35. The structure: Cross section of each column is nearly circular. The discretization causes slight imperfections, which can be improved by reducing the integration step. The diameters of each column is close to the l/4 condition, but not sure.

36. Another view: 12 columns long, or ~ 4 microns.

38. Transmitted portion

40. Input

42. Output Output Output Input Gaussian

43. air n=2 - i 0.02 gain Reflections appear to be suppressed

44. air n=2 - i 0.01 gain

45. air n=2

46. air n=2 + i 0.01 loss

47. air n=2 + i 0.02 loss

48. The symbols and the lines indicate the location and direction of motion of the baricenter of the wave packet. n=2+i0.01 n=2 n=2-i0.01

49. 62 nm of Ag air air

50. A discontinuity in m gives a fundamental problem: An infinite derivative for sudden chnages in m However, the H field is continuous across interfaces, just as E is continuous. Symmetrize The equations of motion.