Point a beam at a slab and ask: what are the odds it hits something? Give every target an effective area — the cross-section sigma — and the odds become a number. The beam does not fall linearly; it casts an exponential shadow, I = I0 · exp(−n sigma x), dropping to 1/e over one mean free path 1/(n sigma). Down the center the beam goes in, the slab attenuates, the transmission comes out. The blue team builds it; the red team tries to break it.
source Rutherford, E. (1911) The Scattering of alpha and beta Particles by Matter and the Structure of the Atom, Philosophical Magazine, Series 6, 21, 669–688 (no stable open DOI landing — hosted text: math.ubc.ca/~cass/rutherford). Rendered, not quoted.
One target of cross-section sigma (units of area) in a beam. Pack n targets per m³ and a thin layer dx removes a fixed fraction of what reaches it:
dI/I = − n sigma dx. Integrate and the beam is exponential, never linear: I(x) = I0 e−n sigma x. The macroscopic cross-section Sigma = n sigma is the removal rate per metre; its reciprocal 1/(n sigma) is the mean free path.
Live, for the current settings:
An effective target area — the cross-section. I = I0 e−n sigma x with mean free path 1/(n sigma) is the interaction probability behind every neutron in the-nuclear-fission and every reaction rate in the-nuclear-fusion.
Fission asks: will this neutron find a nucleus before it leaks out? That is n sigma x. Fusion asks: how often do two nuclei meet and tunnel? That is flux · n sigma. Same sigma, two questions. Each sphere is the next one's rate constant.
The blue team's live check: recompute I at one mean free path and confirm it is I0/e, and that the intensity never goes negative. If red swaps in the linear law, this badge is where it shows.
Feed the engine four numbers: the incident intensity I0 (normalised to 1), the target number density n (targets per m³), the cross-section sigma (in barns, 1 barn = 1e−28 m² AMBER), and the slab thickness x. Nothing else is needed — area times density times length is the whole story.
A barn is the nuclear scale: roughly the geometric cross-section of a heavy nucleus. Reaction cross-sections range from millibarns to thousands of barns, so sigma is a probability written as an area, not a literal size.
Each slice removes the same fraction, so the curve is exponential. Drag any control — every number is computed live from I0 e−n sigma x, never looked up.
Curve: transmission I/I0 vs depth x. The dashed line marks 1/e at one mean free path.
What the machine proves: a beam through a slab is attenuated exponentially, I = I0 e−n sigma x; the mean free path is 1/(n sigma), where transmission is exactly 1/e ≈ 36.8%; doubling n or sigma halves that path; and the total reaction rate R = flux·n·sigma·V scales linearly with sigma. A numeric RK4 integral of dI/dx = −n sigma I matches the analytic exp to 1e−6.
The blue team's witness (left) recomputes the 1/e point live; the red team (right) tries to make the beam lie.
Worse, one number throws away what Rutherford actually derived: the differential cross-section d(sigma)/d(Omega), the angular pattern. And exp(−n sigma x) assumes a thin pencil beam, single scattering, no build-up — a fat beam in a thick shield needs a build-up factor the clean law omits.
"The cross-section is the physical size of the nucleus." Cut. It is an effective area encoding probability; it can be thousands of barns at a resonance or far below the geometric disk off it.
"Attenuation is linear — twice the slab lets half as much through the removed part." Cut. Each layer removes a fixed fraction, not a fixed amount. The law is exponential — exactly the tamper below.
"After one mean free path the beam is stopped." Kept, corrected. One mfp is an e-folding: 1−1/e ≈ 63% has interacted, but 37% still passes. It is a decay length, not a wall.
The red team's move: swap the exponential for the "obvious" linear attenuation I = I0(1 − n sigma x). It looks fine for thin slabs — then the beam goes negative past one mfp and the 1/e point is wrong.
Switch to I0(1−n sigma x) and the witness (window 7) recomputes I at one mfp, finds 0 instead of I0/e (and negative intensity for thick slabs), and turns red. Nothing is faked; the attack is real and it is caught.