Changes

<math>\delta = \frac{1}{4\pi^2} \int{e^{-ik\Delta x}e^{i\omega \Delta t}d^3k d\omega}</math>

<math>\delta = \frac{1}{4\pi^2} \int{e^{-ik\Delta x}e^{i\omega \Delta t}d^3k d\omega}</math>

Now, we have an equation with integrals on both  

Now, we have an equation with integrals on both sides. Since both of these integrals have the same limits and integrands, they must be integrals of equal functions. We can simply drop the integrals.

sides. By definition, these integrands are equal,  

<math> \frac{d^2 \frac{1}{4\pi^2} G(\Delta x,\Delta t)e^{-ik\Delta x}e^{i\omega \Delta t}}{dt^2}-c^2\frac{d^2\frac{1}{4\pi^2} G(\Delta x,\Delta t)e^{-ik\Delta x}e^{i\omega \Delta t}}{dx^2} = \frac{1}{4\pi^2} e^{-ik\Delta x}e^{i\omega \Delta t}</math>

so we can drop the integral signs.

This is a complicated equation, but it can be solved for G. Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

(eqn)

<math>\int{dti} \int{f(x_i,t_i)g(\Delta x, \Delta t) dx_i dy_i}</math>

 

== Color of the Laser ==

== Color of the Laser ==