= Encircled radiant energy of a Gaussian beam
{title2=$Q(R)=U(1-e^{-2R^2/w^2})$}
The \b[encircled radiant energy of a Gaussian beam] is the <radiant energy> inside a centred circular receiver at a specified transverse plane. For fixed <Gaussian beam radius> $w$ and total <laser pulse energy> $U$,
$$
Q(R)=U\left(1-e^{-2R^2/w^2}\right).
$$
The <radiant exposure of a Gaussian beam> contains the radial integration and aperture-capture plot. Its $w$ is the irradiance $e^{-2}$ radius, as specified in the primary beam account.<Propagation of a Gaussian Beam Through a Thin Lens by Francesco D'Angelo>[Equation1 and radius definition]{ref}
An original annular example distinguishes energy near an axis from energy between two radii. Subtracting two encircled values gives
$$
Q(R_2)-Q(R_1)=U\left(e^{-2R_1^2/w^2}-e^{-2R_2^2/w^2}\right).
$$
Choose $U=5\,\mathrm{\mu J}$, $w=0.50\,\mathrm{mm}$, $R_1=0.25\,\mathrm{mm}$ and $R_2=0.50\,\mathrm{mm}$. The annulus receives $5(e^{-1/2}-e^{-2})\approx2.356\,\mathrm{\mu J}$, or 47.12 percent of the full beam energy. A bright central point does not imply that most energy lies in an arbitrarily small disk around it.
Indeed, the <radial radiant-energy density of a Gaussian beam> is $dQ/dR=2\pi R H(R)$, where $H$ is local <radiant exposure>. For this profile it is largest at $R=w/2$, even though $H$ is largest at zero radius. Increasing a centred aperture and comparing integrated detector readings can test the cumulative prediction. Detector linearity, aperture alignment and the finite collection boundary must be controlled; a displaced aperture is not described by the centred formula.
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