Free Packet 2D
V(x, y) = 0
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, and here it is 0. Nothing pushes the particle about, nothing stands on the floor, and no ground is tinted, since there is no ground V puts out of reach.
The packet starts as a Gaussian at (−4, 0), of width σ = 1, carrying Momentum k₀ (along x) 3. Momentum is speed here, so at Speed 1 the dome slides along x at 3 units a second. It passes x = 0 at t ≈ 1.33 and reaches the wall at x = 8 near t = 4, where it reflects. The energy readout is its own, k₀²/2 + 1/2σ² = 4.5 + 0.5 = 5, with one width term for each direction of the plane.
How it spreads
Building a hump that compact takes a range of momenta, Δk = 1/2σ = 0.5 in each direction. Each of them travels at its own speed, so the packet flattens as it goes: σ(t) = √(1 + t²/4). That is 1.12 at t = 1, 1.41 at 2 and 1.80 at 3. Note that it widens across the direction of travel exactly as fast as along it, though it only moves one way. The spread comes from the momentum the packet has in every direction, not from where it is going. The norm readout holds at 1.0000: the sheet gets lower as it gets wider, and the volume under it never changes.
Where T settles
With nothing in the way T simply climbs. It reads 0.18 at t = 1, then 0.65 at 1.5 and 0.92 at 2. By t = 3 it is all but 1, and it stays there until the wall sends the dome back.
Try
- Halve Packet width to 0.5: a sharper start costs a wider range of momenta, and it spreads four times as fast.
- Set Momentum k₀ (along x) to 0: a still dome that flattens where it stands, spreading at the same rate in every direction.
- Put something in its way, 8 · (r < 1), and the free packet becomes a scattering problem.
- Press Top to watch the footprint grow into a circle wider than the packet started.