How to make sense of the Green's function of the 4D wave equation?
For ease of reference in this post equations are numbered as in ref. 1.
The expression given is surprisingly useless for actual calculations. But it seems to be the best we can do with the usual functional notation to express the actual, quite well-defined, distribution. Below I'll try to make it more understandable.
Let's start from the way $(36)$ was derived. The authors in ref. 1 derived it by integrating the Green's function for (5+1)-dimensional wave equation,
$$G_5=\frac1{8\pi^2c^2}\left(\frac{\delta(\tau)}{r^3}+\frac{\delta'(\tau)}{cr^2}\right),\tag{32}$$
where $\tau=t-r/c$, along the line of uniformly distributed sources in 5-dimensional space, using the integral
$$G_{n-1}(r,t)=2\int_r^\infty s(s^2-r^2)^{-1/2}G_n(s,t)ds,\tag{25}$$
where $r=r_{n-1}$ is the radial coordinate in $(n-1)$-dimensional space.
Remember that a Green's function for a wave equation is the impulse response of the equation, i.e. the wave that appears after the action of the unit impulse of infinitesimal size and duration, $f(r,t)=\delta(r)\delta(t)$. Let's replace this impulse with one that is finite at least in one variable, e.g. time. This means that our force function will now be $f(r,t)=\delta(r)F(t)$, where $F$ can be defined as
$$F(t)=\frac{(\eta(t+w)-\eta(t))(w+t)+(\eta(t)-\eta(t-w))(w-t)}{w^2},$$
which is a triangular bump of unit area, with width (duration) $2w$. The choice of triangular shape, rather than a rectangular one, is to make sure we don't get Dirac deltas when differentiating it once.
Then, following equation $(34)$, we'll have the displacement response of the (5+1)-dimensional equation, given by
$$\phi_5(r,t)=\frac1{8\pi^2c^2}\left(\frac{F(\tau)}{r^3}+\frac{F'(\tau)}{cr^2}\right).\tag{34}$$
Now, to find the displacement response $\phi_4(r,t)$ of the (4+1)-dimensional equation, we can use $\phi_5$ instead of $G_5$ in $(25)$. We'll get
$$\phi_4(r,t)= \frac1{4c^3\pi^2r^2w^2} \begin{cases} \sqrt{c^2(t+w)^2-r^2} & \text{if }\,ct\le r<c(t+w),\\ \sqrt{c^2(t+w)^2-r^2}-2\sqrt{c^2t^2-r^2} & \text{if }\,c(t-w)<r<ct,\\ \sqrt{c^2(t+w)^2-r^2}-2\sqrt{c^2t^2-r^2}+\sqrt{c^2(t-w)^2-r^2} & \text{if }\,r\le c(t-w),\\ 0 & \text{otherwise.} \end{cases}$$
Here's a sample of $\phi_4(r,t)$ for $c=1,$ $t=10,$ $w=0.011:$
What happens in the limit of $w\to0$? By cases in the above expression:
- The first case (blue line in the figure above) corresponds to the leading edge of the force function bump, it's located outside of the light cone of the Green's function $G_4$. As $w\to0$, the area under its curve grows unboundedly, tending to $+\infty$.
- The second case (orange) corresponds to the trailing edge of the bump. A zero inside the domain of this case splits the function into a positive and negative parts. The integral of this function times $r^3$ diverges to $-\infty$.
- The third case (green) corresponds to the wake after the force function bump ends. It's negative in the whole its domain, and the integral of it times $r^3$ diverges to $-\infty$. The term itself in the limit of $w\to0$ becomes, for $r<ct$, exactly the second term of $(36)$.
Together, however, the integral $\int_0^\infty r^3\phi_4(r,t)\,\mathrm{d}r$ for $t>w$ remains finite, equal to $\frac t{2\pi^2},$ regardless of the value of $w.$
Conclusions:
- The Green's function does exist and is a well-defined distribution
- The equation $(36)$ formally does make sense
- We can do calculations using $\phi_4$ instead of the $G_4$ from $(36)$, taking the limit $w\to0$ at appropriate times.
References:
1: H. Soodak, M. S. Tiersten, Wakes and waves in N dimensions, Am. J. Phys. 61, 395 (1993)