这是一份liverpool利物浦大学PHYS370的成功案例

where the group velocity
$$
\mathbf{v}{g}=\nabla{\mathbf{k}{0} \mid} \omega\left(k{0}\right)=\frac{\mathbf{k}{0}}{\omega\left(\mathbf{k}{0}\right)} .
$$
The last term in the exponent can be neglected if
$$
\frac{q^{2}}{\omega}\left(t-t_{0}\right) \ll 1 \quad \text { or } \quad \frac{L \cdot(\Delta k)^{2}}{k_{0}} \ll 1,
$$

PHYS370 COURSE NOTES ：

$$
h\left(\mathbf{x}-\mathbf{x}{0}-\mathbf{v}{g}\left(t-t_{0}\right)\right)=h\left(\rho, z-z_{0}-v_{g}\left(t-t_{0}\right)\right)
$$
and
$$
\begin{aligned}
\frac{d \mathbf{W}}{d A} &=\frac{\omega_{0} \mathbf{k}{0}}{2 \pi} \int d t\left[h\left(\boldsymbol{\rho}, z-z{0}-v_{g}\left(t-t_{0}\right)\right)\right]^{2} \
&=\frac{\omega_{0}^{2} \hat{k}{0}}{2 \pi} \int{-\infty}^{\infty} d z h^{2}(\boldsymbol{\rho}, z)
\end{aligned}
$$
where
$$
\boldsymbol{\rho}=\hat{\mathbf{e}}{x} x+\hat{\mathbf{e}}{y} y
$$
With $\rho$ located at the target transverse coordinate, that is, $\rho=0$, we have
$$
\frac{d \mathbf{W}}{d A}=\frac{\omega_{0}^{2} \hat{\mathbf{k}}{0}}{2 \pi} \int{-\infty}^{\infty} d z h^{2}(\mathbf{0}, z)
$$