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The radial radiant-energy density of a Gaussian beam gives energy per increment of radius, rather than per unit surface area. For local radiant exposure $H(R)$ and encircled radiant energy of a Gaussian beam $Q(R)$,Its units are $\mathrm{J}{\mathrm{m}}^{-1}$, while $H$ has units $\mathrm{J}{\mathrm{m}}^{-2}$. The primary Gaussian profile uses $w$ as the irradiance $e^{-2}$ radius.^[ref]

For an original narrow-annulus comparison, take $U=1\mu \mathrm{J}$ and $w=0.80\mathrm{mm}$. The central disk from zero to $0.020\mathrm{mm}$ receives about $1.249\mathrm{nJ}$. The annulus from $0.390$ to $0.410\mathrm{mm}$ receives about $30.32\mathrm{nJ}$. Both span the same radial width, but the outer annulus has much more area. Equal radial width is not equal receiving area.

The shell density is maximal at $R=w/2$ and vanishes at $R=0$, despite the local exposure being largest on axis. Dividing $g$ by $U$ gives a probability density function over nonnegative radius: it is a Rayleigh distribution with scale $w/2$. This is an energy-weighted mathematical representation, not an additional photon-counting assumption. Differences between successive centred-aperture energy measurements estimate shell energies; dividing by their radial separation approximates $g$ when that separation is sufficiently small.

The radiant exposure of a Gaussian beam is its time-integrated transverse irradiance. At a fixed plane normal to a circular Gaussian beam, suppose the Gaussian beam radius $w$ is unchanged during a pulse and denotes the radius where irradiance is $e^{-2}$ of the on-axis value. With total laser pulse energy $U$,The primary beam account specifies this radius convention and the corresponding irradiance normalization; the radiant exposure parent derives its time integral.^[ref]

An original aperture example asks how much of that energy a finite receiver collects. Integrating over a centred disk of radius $R$ gives the encircled radiant energy of a Gaussian beam:For stipulated $U=2\mu \mathrm{J}$, $w=0.80\mathrm{mm}$ and $R=0.40\mathrm{mm}$, the receiver collects only $39.35$percent, or $0.7869\mu \mathrm{J}$. Its spatially averaged radiant exposure is about $1.566\mathrm{J}{\mathrm{m}}^{-2}$, compared with on-axis $H(0)\approx 1.989\mathrm{J}{\mathrm{m}}^{-2}$. Dividing the total $U$ by this receiver's area would count energy that missed it.

To collect a chosen fraction $0<f<1$ under this ideal model, the centred aperture needs $R=w\sqrt{-\log (1-f)/2}$. For95percent, $R\approx 1.224w$. A possible experimental check measures integrated received energy through several known apertures and compares their ratios. Aperture centring, detector linearity and any profile change during the pulse affect that comparison. A time-varying radius must remain inside the time integral rather than being replaced by an unspecified average radius.