= Gaussian pulse energy and peak power
{c}
{title2=$U=P_{\rm pk}\tau\sqrt{\pi/(4\ln2)}$}
For an isolated <Gaussian pulse>, the relation between <laser pulse energy> $U$ and <peak power> $P_{\rm pk}$ depends on the pulse-width convention. Here $P(t)$ is the total incident <radiant power>, the background is zero, and $\tau>0$ denotes the power <full width at half maximum>. Integrating $P$ over the entire pulse defines its positive energy $U>0$.
Pulse <energy>, <peak power> and <average power> resolve different information. Consider a <Gaussian pulse> with
$$
P(t)=P_{\rm pk}\exp\!\left[-4\ln2\left(\frac t\tau\right)^2\right].
$$
The chosen <pulse duration> $\tau$ is the <full width at half maximum> of <radiant power>: $P(\pm\tau/2)=P_{\rm pk}/2$. It refers to <radiant power>, rather than the <wave amplitude> of the <electric field>. The <Gaussian integral> gives the relation between <Gaussian pulse energy and peak power>:
$$
U=P_{\rm pk}\tau\sqrt{\frac{\pi}{4\ln2}},\qquad
\boxed{P_{\rm pk}=\sqrt{\frac{4\ln2}{\pi}}\frac U\tau
\simeq0.939437\,\frac U\tau.}
$$
The <pulse energy shape factor> is $\kappa=U/(P_{\rm pk}\tau)$. It equals $1.064467$ for this power profile; a rectangular pulse of duration $\tau$ has $\kappa=1$. Knowing only a width and an integrated <energy> therefore requires a pulse-shape assumption before inferring <peak power>.
An equivalent form makes the normalization transparent. Write the pulse centre as $t_0$, and let $\sigma$ denote the <standard deviation> of its <energy-normalized pulse profile>:
$$
P(t)=\frac{U}{\sqrt{2\pi}\sigma}
\exp\!\left[-\frac{(t-t_0)^2}{2\sigma^2}\right],
\qquad \tau=2\sqrt{2\ln2}\,\sigma.
$$
The ratio $p(t)=P(t)/U$ is a <probability density function> in the mathematical sense: it is nonnegative, has units $\mathrm{s^{-1}}$, and integrates to one. Here it weights the pulse's <energy> over time. Its <normal distribution> profile has <mean> $t_0$ and <variance> $\sigma^2$. This gives $P_{\rm pk}=U/(\sqrt{2\pi}\sigma)$ without a separate area estimate.
For a stipulated $\sigma=2\,\mathrm{ns}$ and $U=1\,\mathrm{nJ}$, the power <full width at half maximum> is $4.70964\,\mathrm{ns}$ and \b[the peak power is $0.19947\,\mathrm W$]. Confusing the <standard deviation> with the <full width at half maximum> would insert the wrong duration into the conversion.
The <Gaussian pulse field-envelope width> is a second distinct convention. If a fixed spatial field profile gives $P(t)=K|A(t)|^2$ for a constant $K>0$, the <wave amplitude> envelope obeys
$$
|A(t)|=|A(t_0)|e^{-(t-t_0)^2/(4\sigma^2)}.
$$
Consequently its amplitude <full width at half maximum> is $\tau_E=\sqrt2\,\tau$. Inserting $\tau_E$ into the formula for a power-width input underestimates the true <peak power> by a factor $1/\sqrt2$. State which measured waveform defines the width.
An electrical <photodetector> trace can also be broadened by its <impulse response>. The optical pulse-width inference therefore needs a characterized detector response.<Calibration Service for Low-level Pulsed-Laser Radiometers at 1.06 µm: Pulse Energy and Peak Power by Rodney W. Leonhardt>[§5.2, printed p. 13: detector convolution and bandwidth-limited pulse reproduction]{ref}
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