How does the green function for the wave equation in three dimensions preserve the ordering of noises between a speaker and a listener I was provided with the following equation in class for the Green's function of a three dimensional wave equation:

However, I am confused as to how this form of the Greens function preserves the ordering of noises between a speaker and a listener. Any explanation would be much appreciated.
 A: This Green’s function is for a single sound pulse emitted from $r=0$ at $t=0$. It doesn’t explicitly describe a sequence of sounds.
However, from it you can see that the sound arrives at distance $r$ at a time $r/c$ after it is emitted. So noises emitted later are going to arrive later.
For a better understanding of this, consider the more general Green’s function $G(\mathbf{r},t;\mathbf{r}’,t’)$ for a pulse emitted from $\mathbf{r}’$ at $t’$, and think about convolving it with a sound source that emits sound over time.
A: Implicit in this solution is the fact that the origin of the coordinate system is located at the speaker, at the moment in time in which they emit a pulse of sound. 
In other words, the speaker is located at the origin, so $ \mathbf{r}_{speaker} = \langle 0, 0, 0 \rangle $, and they emit a pulse of sound at $ t = 0 $. 
I assume by ''ordering of noises'' you mean the order in time. 
It's best illustrated with an example. Consider a listener located at $ \mathbf{r}_{listener} = \langle 10, 0, 0 \rangle $. 
For simplicity, set $ y $ and $ z $ to zero, to only look at sound on the x-axis. Then $ \mathbf{r} = \langle x, 0, 0 \rangle $.
When $ |\mathbf{r}| - ct = |x| - ct = 0 $, a pulse will exist at the listener's location (remember that the $ \delta $ function is zero when its argument is zero). Plugging in $ x = 10 $, you find that at $ t = 10 / c$, a pulse will arrive at the listener's location. 
Also implicit in this solution is the fact that time is restricted to taking only non-negative values ($ t > 0 $).
For any given location at some distance $ |{\mathbf{r}}|$, the pulse will pass that location at some $ t > 0 $ that satisfies $ t = |{\mathbf{r}}| / c $. Which is a later time than $ t = 0 $.
In this sense, the temporal ordering between the speaker and listener is preserved because for a listener at any location distinct from the speaker's location will hear the sound at some time later than the time of speaking.
