A Simple Asymmetric Resonant Cavity

by Greg Egan

This web page analyses an extremely simple example of an asymmetric resonant cavity: a tapered quadrilateral in two dimensions. The analysis demonstrates that the net force on the cavity walls is zero, providing a simple counterexample to the absurd claim by the British engineer Roger Shawyer [2] that closing off the ends of a tapered waveguide will produce a cavity that can experience a net force from internal radiation pressure (a claim, sadly, much hyped by the magazine New Scientist [1]).

Readers interested in a more realistic, three-dimensional example should consult this detailed analysis of a truncated spherical cone. The same page includes a proof that the net force is zero for a cavity of any shape whatsoever.

In this simplified analysis, we will treat the field in the cavity as a scalar field: that is, a real number associated with each point in space and time, rather than a vector. This doesn’t prevent waves in the field from carrying momentum and exerting a force (a sound wave in air is a scalar pressure wave, but it still transports momentum). We will require that the field have zero amplitude on the walls of our cavity (just as, for example, a resonant membrane such as a drum skin has zero displacement at its rim).

Rather than starting with a particular cavity, we will consider the standing waves that we can construct by a simple process of adding together a finite number of “plane waves” of equal wavelength, moving in different directions. Plane waves in three dimensions have a succession of parallel planes as their wavefronts (that is, their surfaces of constant phase); these planes are all orthogonal to the direction in which the wave is travelling. In two dimensions, we use the term “plane waves” to describe waves whose wavefronts are a succession of parallel lines, which are again orthogonal to the wave’s direction.

The general form for a plane wave of peak amplitude A, with a wavelength of 2π/k, moving in a direction that makes an angle of θ with the x-axis, and whose phase at x=y=t=0 is φ, is:

wθ,φ(x,y,t) = A sin(k [cos(θ) x + sin(θ) yc t] + φ)

where c is the speed of light in a vacuum.

Suppose we combine six plane waves of equal amplitude and wavelength, whose directions of travel are arranged symmetrically in a six-pointed star. We will start with θ=0 for the first wave, and then increase θ by π/3 for each successive plane wave in our list. Also, we will alternate between phases of 0 and π for successive waves. A phase of π simply changes the sign of the whole wave compared to a phase of 0, so we have:

w1(x,y,t) = A sin(k [xc t])
w2(x,y,t) = –A sin(k [cos(π/3) x + sin(π/3) yc t])
w3(x,y,t) = A sin(k [cos(2π/3) x + sin(2π/3) yc t])
w4(x,y,t) = –A sin(k [–xc t])
w5(x,y,t) = A sin(k [cos(4π/3) x + sin(4π/3) yc t])
w6(x,y,t) = –A sin(k [cos(5π/3) x + sin(5π/3) yc t])

If we add up all of these waves, the result is:

w(x,y,t) = –8A cos(ωt) sin(kx/2) sin[(k/4)(x + y √3)] sin[(k/4)(xy √3)]

where ω = kc is the angular frequency of the waves. Now, this is clearly a standing wave, because it has a fixed spatial pattern (the part independent of t) multiplied by a simple harmonic function in t, cos(ωt). The spatial pattern divides up the xy plane into fixed regions of opposite phase (see the diagram, below left), separated by lines where the amplitude is permanently zero. The lines of zero amplitude are of the form:

x = 2πn/k
x ± y √3 = 4πn/k

where n is an integer; the sloped lines make an angle of ±π/6 with the x axis. We can construct a cavity which will have this standing wave as one of its resonant modes simply by locating the cavity’s walls on lines of zero amplitude: for example, the quadrilateral ABCD on the diagram below. The standing wave then meets the boundary condition of zero amplitude on the cavity walls. In the table (below right) we list the effect of each of the four walls on each of the six travelling waves from which the standing wave was built:

Standing wave built from 6 travelling waves

Each wall reflects various travelling waves, converting them into others in the set of six.

w1 w4, i = 0  
w2w1, i = π/3w3, i = π/3  
w3w6, i = 0  w2, i = π/3
w4w5, i = π/3 w3, i = π/3w1, i = 0
w5  w2, i = 0w6, i = π/3
w6 w5, i = π/3w1, i = π/3 

The angle i quoted for each reflection is the angle of incidence between the direction of the incoming wave and the normal to the wall. The force per unit length exerted on each wall by reflecting a wave will be proportional to the cosine of the angle of incidence, and directed perpendicular to the wall. Since all of our component travelling waves have equal amplitude, and each wall reflects one wave head-on and two at an angle of π/3, the force per unit length (i.e. the pressure, in our two-dimensional model) on all of the walls will be equal. (This simplified analysis neglects the variation in pressure due to the pattern of the field along the wall. However, each wall contains an integral number of half-cycles of the standing wave — the sides of the triangles, in the diagram — and the field varies in exactly the same way along each wall, so a more detailed analysis will confirm that the average pressure over each half-cycle, and hence over each wall, will be equal.)

Since the average pressure on each wall is equal, the net force on this closed figure must sum to zero; it’s a simple exercise in trigonometry to confirm this for the quadrilateral shown.

The details of this analysis don’t change if we alter the size of the quadrilateral, so long as we maintain the boundary conditions. In other words, we can build the cavity from as many equilateral triangles along each side as we like. However, to analyse a cavity with a different taper angle would require a more complicated method, since it’s only possible to build up the standing wave from a finite number of plane waves when the taper angle is π/3.

Finally, we note that this simplified model with a scalar field in two dimensions actually gives us some of the modes of an electromagnetic wave in a three-dimensional cavity. If we extrude the two-dimensional quadrilateral into a prism, producing a triangular prism with one edge lopped off, and if we take the scalar field to be the magnitude of an electric field always perpendicular to the xy plane, then that field will satisfy the boundary conditions for a cavity with conducting walls. The net force on the walls will be zero in the x and y directions for the reasons we’ve given, and zero along the z-axis by symmetry.

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