Changes

<math> g( x_i, t_i, x_f, t_f)\,</math>

<math> g( x_i, t_i, x_f, t_f)\,</math>

Because the propogator is actually in terms of the <i>differences</i> between the x and t values, we will write the difference between the x-vectors as <math>\Delta x</math> and the difference between the times as <math>\Delta t</math>.

Because the propogator is actually in terms of the <i>differences</i> between the x and t values, we will write the difference between the x-vectors as <math>\Delta \mathbf{x}</math> and the difference between the times as <math>\Delta t</math>.

The integral is therefore

The integral is therefore

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

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

We know that  

We know that  

However, this is for a uniform, sourceless wave. Like most generalizations, this is an unrealistic situation in the real world. What we need is a function that can generate a brief pulse. This sounds like a delta function.

However, this is for a uniform, sourceless wave. Like most generalizations, this is an unrealistic situation in the real world. What we need is a function that can generate a brief pulse. This sounds like a delta function.

<math>\frac{d^2g}{dt^2} - \frac{c^2d^2g}{dx^2} = \delta(\Delta x)\delta(\Delta t)</math>

<math>\frac{d^2g}{dt^2} - \frac{c^2d^2g}{dx^2} = \delta(\Delta \mathbf{x})\delta(\Delta t)</math>

This is not an easy equation to solve without using a Fourier transform. Therefore, we'll do just that, with G as the transformed function.

This is not an easy equation to solve without using a Fourier transform. Therefore, we'll do just that, with G as the transformed function.

<math>G = \frac{1}{4\pi^2} \int{g(\Delta x,\Delta t)e^{-ik\Delta x}e^{i\omega \Delta t}d^3x dt}</math>

<math>G = \frac{1}{4\pi^2} \int{g(\Delta x,\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}d^3x dt}</math>

Therefore,

Therefore,

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

<math>g = \frac{1}{4\pi^2} \int{G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta x}e^{i\omega \Delta t}d^3k d\omega}</math>

We will then need to plug this function into the earlier equation. To make this easier, we know that the four-dimensional delta function on the right-hand side can be simplified.

We will then need to plug this function into the earlier equation. To make this easier, we know that the four-dimensional delta function on the right-hand side can be simplified.

<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^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}d^3k d\omega}</math>

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.

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.

<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>

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

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

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