# Causality of the wave equation

1. May 21, 2015

### Coffee_

Consider the wave equation $\nabla^{2} f - \frac{1}{c^{2}} \frac{\partial ^{2} f}{\partial t^{2}} = \delta(r) \delta(t)$ where there is no wave before $t=0$

The solution will be something up to a constants like $f=\frac{\delta(r-ct)}{r}$.

So we have a dirac delta function that spreads out as a circular wave.

Let's now consider a time $t=t'>0$ and an $r=r'$ where we find such a peak. If I now use the same reasoning as in the beginning on this peak, I find that this peak also should make these circular waves in space. This is Huygens prinicple it seems to me.

Is it correct to say that yes, this indeed does happen. Any position at any time where we find such a peak will act as a source for new circular peak-waves again. They just happen to cancel out in the backwards direction and so we find the forward moving evoltuion.

QUESTION:

1) Are there any fatal flaws in my reasoning above?

2) If not so much, then how does it come that we do find the evolution by solving the differential equation without having to consider these ''new'' sources. Where in the solution is this implicitly hidden? Where is it hidden that the backwards new sources all cance

Last edited: May 21, 2015
2. May 22, 2015

### theodoros.mihos

Huygens principle is a trick, talks about frontwaves. Frontwaves are something like a surface on which the wave have the same phase, and ignore backwaves which position been closer to source. I think that assumed an infinity wave with well known ω and k than can represented by the solution for wave equation for 2D in this case:
$$y(r,\theta,t) = \frac{A}{r}\sin(\omega{t}-\mathbf{k}\cdot\mathbf{r})$$
But in this case we have a wave pulse on the form:
$$y(r,\theta,t)=\frac{A}{r}\delta(r-ct)$$
I believe this is a much different function.
https://en.wikipedia.org/wiki/Huygens–Fresnel_principle

Last edited: May 22, 2015
3. May 22, 2015

### Coffee_

I think I shouldn't have mentioned Huygens at all. My question could be summed up as:

When solving the wave equation analytically given certain initial and boundary condition we find a unique solution for how the oscillations will evolve in time and space. At a time $t>0$ so after the oscillations started evolving we pick a random point where $y$ is not $0$. Will this excitation act as another source that will disperse according to the wave equation? If yes, why don't we have to even think about this when solving the wave equation analytically? Somewhere in the solution we must have implicitly dealt with this fact.

4. May 22, 2015

### theodoros.mihos

5. May 22, 2015

### Delta²

Well i believe the answer is that this excitation acts "partially" as another source. It is not exactly the same as having a source at that point because the source injects energy into the medium, while a simple excitation of the medium just allows the energy to pass from the previous points of the medium to the next points, it serves as the mechanism of the self propagation of the wave.