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Radiation pressure on a dielectric wedge

Masud Mansuripur, Armis R. Zakharian, and Jerome V. Moloney

Open Access Open Access

Abstract

The force of electromagnetic radiation on a dielectric medium may be derived by a direct application of the Lorentz law of classical electrodynamics. While the light’s electric field acts upon the (induced) bound charges in the medium, its magnetic field exerts a force on the bound currents. We use the example of a wedge-shaped solid dielectric, immersed in a transparent liquid and illuminated at Brewster’s angle, to demonstrate that the linear momentum of the electromagnetic field within dielectrics has neither the Minkowski nor the Abraham form; rather, the correct expression for momentum density has equal contributions from both. The time rate of change of the incident momentum thus expressed is equal to the force exerted on the wedge plus that experienced by the surrounding liquid.

©2005 Optical Society of America

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Figures (4)
Fig. 1.
Fig. 1. Dielectric prism of index n, illuminated with a p-polarized plane wave at Brewster’s angle θB. The prism’s apex angle is twice the refracted angle θ′B, and the beam’s footprint on the prism’s face is a. The H-field magnitude inside the slab is the same as that outside, but the E-field inside is reduced by a factor of n relative to the outside field. The bound charges on the upper and lower surfaces feel the force F s of the E-field. The net force experienced by the prism is along the x-axis, being the sum of the forces on the upper and lower surfaces as well as Fw1 and Fw2 , which are exerted at the beam’s edges (i.e., side-walls) within the dielectric.

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Fig. 2.
Fig. 2. Dielectric prism of index n 2 in a host liquid of index n 1, illuminated with a p-polarized plane wave at Brewster’s angle θB. The apex angle is 2θ′B, and the footprint of the beam on each side of the wedge is a. The liquid experiences a force Fw at the edges of the beam where the (expansive) force on one edge is not compensated by the force on the opposite edge. The net force exerted on the system is along the x-axis and is the sum of the forces F s on the upper and lower solid-liquid interfaces, Fw1 and Fw2 at the beam’s edges within the prism, and Fw at the beam’s edges inside the liquid.

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Fig. 3.
Fig. 3. A small gap at the interface between a solid (dielectric constant=ε 2) and a liquid (dielectric constant=ε 1) helps to explain the existence of separate interfacial charges. If the perpendicular E-field in the gap is denoted by Eg , the continuity of D ⊥ across the gap yields the magnitude of E ⊥ in the two media as Eg/ε 1 and Eg/ε 2, as shown. The surface charge densities are then given by the E ⊥ discontinuity at each interface.

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Fig. 4.
Fig. 4. Gaussian beam (λo=0.65µm, FWHM=6.5 µm at z=0) is incident at θB=56.31° onto a prism of (relative) refractive index n=1.5 and apex angle ϕ=67.38°. Since the beam is p-polarized and incidence is at Brewster’s angle, no significant reflections occur at either the first or the second surface. In (a)–(c) the prism is in free space, whereas in (d) it is in water. The computational cell size in all cases is Δx=Δz=5 nm. (a) Time-snapshot of the H-field, which consists only of the Hy component. (b) Magnitude of the E-field, which consists of Ex and Ez components. (c) Force density distribution inside the prism of index n=1.5 surrounded by free space. (d) Force density distribution inside and outside a prism of index n 2=1.995 immersed in water (n 1=1.33).

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Equations (29)

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F = ρ b E + J b × B ,
F ( edge ) = 1 4 ε o ( ε − 1 ) ∣ E ∣ 2 .
p = 1 4 ( n 2 + 1 ) n ε o ∣ E ∣ 2 ⁄ c .
σ b = ε o E o ( 1 − 1 ⁄ n 2 ) sin θ B .
F ∥ = 1 2 σ b E ∥ = 1 2 ε o E o 2 ( 1 − 1 ⁄ n 2 ) sin θ B cos θ B .
F ∥ x = a ε o E o 2 ( 1 − 1 ⁄ n 2 ) sin 2 θ B cos θ B .
F ⊥ = 1 2 σ b E ⊥ = 1 4 ε o E o 2 ( 1 − 1 ⁄ n 2 ) ( 1 + 1 ⁄ n 2 ) sin 2 θ B .
F ⊥ x = 1 2 a ε o E o 2 ( 1 − 1 ⁄ n 4 ) sin 2 θ B cos θ B .
F w 1 + F w 2 = − 1 2 a ε o E o 2 ( 1 − 1 ⁄ n 2 ) sin θ ′ B .
F x ( total ) = ( a cos θ B ) ε o E o 2 [ ( n 2 − 1 ) ⁄ ( n 2 + 1 ) ] .
σ b = ε o [ 1 − ( n 1 2 ⁄ n 2 2 ) ] E o sin θ B .
F x ( surface ) = a ε o ( 1 − n 1 2 ⁄ n 2 2 ) [ 1 + 1 2 ( 1 + n 1 2 ⁄ n 2 2 ) ] E o 2 sin 2 θ B cos θ B .
F w 1 + F w 2 = − 1 2 a ε o [ n 1 2 − ( n 1 2 ⁄ n 2 2 ) ] E o 2 sin θ ′ B .
F x ( liquid ) = 2 F w sin 2 θ ′ B = 1 2 a sin θ B ( n 1 2 − 1 ) ε o E o 2 sin 2 θ ′ B .
F x ( total ) = 1 2 a cos θ B ( n 1 2 + 1 ) ε o E o 2 cos 2 θ ′ B .
F 1 = 1 4 ε o [ ( 1 ⁄ ε 1 ) 2 − 1 ] ∣ E g ∣ 2 .
F 2 = 1 4 ε o [ 1 − ( 1 ⁄ ε 2 ) 2 ] ∣ E g ∣ 2 .
F o = ± 1 4 σ 1 σ 2 ⁄ ε o = ± 1 4 ε o [ 1 − ( 1 ⁄ ε 1 ) ] [ 1 − ( 1 ⁄ ε 2 ) ] ∣ E g ∣ 2 .
σ 2 = ε o [ 1 − ( 1 ⁄ ε 2 ) ] E g = ε o [ ε 1 − ( ε 1 ⁄ ε 2 ) ] E o sin θ B .
F ∥ x = a σ 2 E ∥ cos θ ′ B = a ε o [ ε 1 − ( ε 1 ⁄ ε 2 ) ] E o 2 sin 2 θ B cos θ B .
F ⊥ x = 1 2 a ε o [ ε 1 2 − ( ε 1 ⁄ ε 2 ) 2 ] E o 2 sin 2 θ B cos θ B .
F x ( solid ) = 1 2 ( a cos θ B ̟ ) ε o E o 2 ε 1 ( ε 1 + 1 ) ( ε 2 − 1 ) ( ε 1 + ε 2 ) .
F x ( liquid ) = − 1 2 ( a cos θ B ̟ ) ε o E o 2 ε 2 ( ε 1 2 − 1 ) ⁄ ( ε 1 + ε 2 ) .
F ̂ ⊥ x = 1 2 a ε o [ ε 1 − ( ε 1 ⁄ ε 2 ) ] [ 1 + ( ε 1 ⁄ ε 2 ) ] E o 2 sin 2 θ B cos θ B .
F ̂ x ( solid ) = ( a cos θ B ̟ ) ε o E o 2 ε 1 ( ε 2 − 1 ) ⁄ ( ε 1 + ε 2 ) .
F ̂ x ( liquid ) = − 1 2 ( a cos θ B ̟ ) ε o E o 2 ( ε 1 − 1 ) .
F ︽ x ( solid ) = 1 2 ( a cos θ B ̟ ) ε o E o 2 ( ε 1 − 1 ) ( ε 2 − ε 1 ) ( ε 2 + ε 1 ) .
S z ( x , z = 0 ) = 1 2 E o H o exp [ − 2 ( x ⁄ x o ) 2 ] .
∫ S z ( x , z = 0 ) d x = π ⁄ 8 E o H o x o = 6.49 × 10 − 3 W ⁄ m .
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