Fig. 4. Force density from a normal incident wave on a lossless slab(d= 80nm)as a function of the relative permittivity Er. The free space wavelength is no= 640nm and u, 1, Ei= 1. Shown are the forces calculated from the distributed Lorentz force(circles)and the Maxwell stress tensor (line ). The background media (region 0 and region 2)is fre integration for the Maxwell stress tensor is the same as that shown in Fig. I except that there is only one boundary at z=0. Again we allow the integration to just enclose the boundary such that Sz-0. The force per unit area on the half space medium is given by F=Re{2.1(x=0-)-2·7(=0+)} The contributions on the two sides of the interface are ,个(=0)=2[2E1(=0)P+(H1(=0)-(=0) +[-0H(2=0)H(z=0-)],(19a) 1=0)=2121(2=0+)P+((=0)P-H(=0) +[-1H(x=0)H(z=0+],(19b) and the force tangential to the boundary is seen to be F1=Re-H(z=0-)H(=0-)+1H2(z=0+)H(z=0+) Upon substitution of the fields on both sides of the boundary, the tangential force is simplified Fr=ERe [(1+Rhs)(1-Rhs)-Po1l Thsl- where Rhs and Ths are the half-space reflection and transmission coefficients, respectively and Pol is given by [8] HokI unko #8324·$15.00USD Received 29 July 2005, revised 12 October 2005; accepted 1 November 2005 (C)2005OSA 14 November 2005/ Vol 13. No 23/ OPTICS EXPRESS 92860 2 4 6 8 10 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 ε r F z (pN/m 2 ) Lorentz Stress Tensor Fig. 4. Force density from a normal incident wave on a lossless slab (d = 80nm) as a function of the relative permittivity εr. The free space wavelength is λ0 = 640nm and µr = 1, Ei = 1. Shown are the forces calculated from the distributed Lorentz force (circles) and the Maxwell stress tensor (line). The background media (region 0 and region 2) is free space. integration for the Maxwell stress tensor is the same as that shown in Fig. 1 except that there is only one boundary at z = 0. Again we allow the integration to just enclose the boundary such that δz → 0. The force per unit area on the half space medium is given by F¯ = 1 2 Re{zˆ·T ¯¯(z = 0 −)−zˆ·T ¯¯(z = 0 +)}. (18) The contributions on the two sides of the interface are zˆ·T ¯¯(z = 0 −) = zˆ h ε0 2 |Ey(z = 0 −)| 2 + µ0 2 ￾ |Hx(z = 0 −)| 2 −|Hz(z = 0 −)| 2
i +xˆ −µ0Hz(z = 0 −)H ∗ x (z = 0 −) , (19a) zˆ·T ¯¯(z = 0 +) = zˆ h ε1 2 |Ey(z = 0 +)| 2 + µ1 2 ￾ |Hx(z = 0 +)| 2 −|Hz(z = 0 +)| 2
i +xˆ −µ1Hz(z = 0 +)H ∗ x (z = 0 +) , (19b) and the force tangential to the boundary is seen to be Fx = 1 2 Re − µ0Hz(z = 0 −)H ∗ x (z = 0 −) + µ1Hz(z = 0 +)H ∗ x (z = 0 +)  . (20) Upon substitution of the fields on both sides of the boundary, the tangential force is simplified to Fx = 1 2 E 2 i kx ω2 Rek0z µ0 ∗ (1+Rhs) (1−Rhs) ∗ − p ∗ 01|Ths| 2  (21) where Rhs and Ths are the half-space reflection and transmission coefficients, respectively and p01 is given by [8] p01 = µ0k1z µ1k0z . (22) (C) 2005 OSA 14 November 2005 / Vol. 13, No. 23 / OPTICS EXPRESS 9286 #8324 - $15.00 USD Received 29 July 2005; revised 12 October 2005; accepted 1 November 2005