Particles scattered out of the beam must be missing from it: behind the target there is a shadow. The shadow is made by interference between the scattered wave and the incident one straight ahead, and so it depends on the forward amplitude. Counting shows that the total cross section is fixed by the imaginary part of the forward amplitude alone: $\sigma=(4\pi/k)\operatorname{Im}f(0)$, the optical theorem. In partial waves it holds wave by wave, $\operatorname{Im}a_{\ell}=|a_{\ell}|^{2}$ for $a_{\ell}=e^{i\delta_{\ell}}\sin\delta_{\ell}$, which puts every elastic amplitude on one circle in the complex plane. It holds too when the target absorbs particles, for the total of scattered and absorbed. A black disc of radius $a$, absorbing every partial wave up to $ka$, absorbs $\pi a^{2}$ and scatters as much again, into a narrow forward peak that fills in its shadow: $\sigma=2\pi a^{2}$.

A pillar in a stream leaves a wake behind it, and the water missing from the wake is exactly the water turned aside. Measure the wake straight downstream and you know how much the pillar deflects in all directions.

The complex plane of a partial-wave amplitude $a=e^{i\delta}\sin\delta$, real part across and imaginary part up. As $\delta$ runs from 0 to $\pi$, $$ goes once round the circle of radius about , on which : the optical theorem wave by wave. The arrow is ; the top, , is the unitarity limit. Absorption pulls inside the circle.

The total cross section, of scattering and of absorption together, is fixed by the forward amplitude:

• Partial wave by partial wave, $\operatorname{Im}a_{\ell}=|a_{\ell}|^{2}$ for : every elastic amplitude lies on the circle of radius about .

In the forward direction every Legendre polynomial is 1, so the forward amplitude is the plain sum of the partial amplitudes. Its imaginary part collects $\sin^{2}\delta_{\ell}$ from each, which is exactly what the cross section collects. With absorption the same comparison goes through for the total, with $1-\operatorname{Re}S_{\ell}$ in place of .

$P_{\ell}(1)=1$.

$\operatorname{Im}(e^{i\delta}\sin\delta)=\sin^{2}\delta$.

The cross section from the phase shifts, compared term by term.

Elastic and absorbed add: $|1-S|^{2}+1-|S|^{2}=2(1-\operatorname{Re}S)$; and .

The total cross section is $4\pi/k$ times the imaginary part of the forward amplitude.

Particles scattered away are missing from the forward beam; that loss comes from the interference of the scattered and incident waves straight ahead.

When the target absorbs, $|S_{\ell}|<1$, and the theorem gives the total of the scattered and the absorbed.

Every elastic partial-wave amplitude $e^{i\delta}\sin\delta$ lies on the circle of radius $\tfrac12$ about , where its imaginary part equals its squared modulus.

Great work! You now know how to:

• derive the optical theorem from the partial waves
• extend it to targets that absorb
• explain the shadow of a black disc