Radar cross-section

Remarks

The limit exists because the scattered field decays like $1/R$ in the far field. In terms of the far-field vector amplitude $F$, with $E_s(R\hat{x}) \approx F(\hat{x}) e^{ikR}/R$, one has $\sigma(\hat{x}) = 4\pi \abs{F(\hat{x})}^2 / \abs{E_i}^2$. It is often given in decibels relative to one square metre (dBsm).

Example: a perfectly conducting sphere of radius $a \gg \lambda$ has monostatic radar cross-section close to $\pi a^2$. A flat perfectly conducting plate of area $A$, large compared with $\lambda^2$, seen at normal incidence has $\sigma \approx 4\pi A^2 / \lambda^2$. For $A = 1$ m$^2$ and $\lambda = 3$ cm this is about $1.4 \times 10^4$ m$^2$, far larger than the physical area.

Non-example: the quantity $4\pi R^2 \abs{E_s}^2 / \abs{E_i}^2$ at a fixed finite distance $R$ in the near field is not the radar cross-section, since it still depends on $R$.