Scattering Disc
V(x, y) = 8(r < 1)
Open in the app The dials and keys named below are the app's.
What you see
The formula is the potential over the plane. r is fed to it as hypot(x, y), the distance from the origin, so r < 1 is a disc of radius 1 there and the 8 is how high it stands. The packet's energy is k₀²/2 + 1/2σ² = 4.5 + 0.78 = 5.28, well under that, so the floor is tinted right across the disc.
The packet comes in from (−4, 0.6), aimed to pass above the centre of the disc but well inside its edge. It carries Packet width 0.8 and Momentum k₀ (along x) 3, and strikes about 1.3 seconds in at Speed 1.
The scattering
The wavelength is 2π/k₀ ≈ 2.09 and the disc is 2 across, so the obstacle is about one wavelength wide. Nothing here behaves like a ball off a post: the wave wraps the disc. What gets past is thrown up and away from the side it struck. On an arc of radius 6 the density is heaviest between 35° and 50° above the axis, and nearly nothing below it. Straight behind the disc there is a shadow, which diffraction fills back in as the wave travels on. That angular spread, measured far from the target, is what a scattering experiment reports as a cross section.
Where T settles
Once the collision is over, near t = 3.5, the readouts hold at about 0.65 and 0.35. A third of the packet has been thrown back the way it came. The norm readout stays at 1.0000, so whatever leaves one side is accounted for on the other.
Try
- Aim it dead centre, Packet centre y 0. The pattern turns symmetric, and just behind the disc the axis is brighter than the shadow either side of it: the bright point Poisson derived to ridicule the wave theory of light, and Arago then found.
- Packet centre y 2 misses the disc altogether: T goes to 0.90 and the shadow is a dent in one edge of the sheet.
- Make it hard, 30 · (r < 1): less leaks through the disc and T falls to about 0.57.
- Give it something bigger to get round, 8 · (r < 2), and the shadow behind it widens.