# How to make sense of the Green's function of the 4+1D wave equation?

In the paper "Wakes and waves in N dimensions" by Harry Soodak and Martin S. Tiersten, equation $$(36)$$ gives the Green's function for the 4+1D wave equation in the following form:

$$G_4(r,t)=\frac1{4\pi^2c^3}\left(\frac{\delta(t-r/c)}{r(t^2-r^2/c^2)^{1/2}}-\frac{\eta(t-r/c)}{c(t^2-r^2/c^2)^{3/2}}\right),\tag{36}$$

where $$r=\sqrt{x^2+y^2+z^2+w^2}$$ is distance from origin, $$c$$ is wave propagation speed, $$\delta(\cdot)$$ is the Dirac delta, and $$\eta(\cdot)$$ is the Heaviside step function.

Trying to understand its meaning, I've stumbled upon the fact that the first term contains Dirac delta with singularity at $$r=ct$$, multiplied by another function, which is singular (has algebraic branch point) at the very same point. This means that not only is the multiplier infinite at the delta's singularity, but it doesn't even have a well-defined complex phase at this point.

How should this expression be interpreted? Does it even make sense? If not, does Green's function not exist at all in this case?

• Right: the "Dirac delta" singularity is along a whole cone, not just a point. This was a critical component of Dirac's contribution. Not just "Dirac delta at a point", which Heaviside and others had used some decades earlier, in lower-dimensional physical problems (telegraph cables and such). Commented Oct 15, 2020 at 22:46
• I think his (highly justified) problem here is the fact that it is not clear what the definition of $\delta_0(x)/\sqrt{x}$ since there is no general definition in the theory of distribution of the product of two singular distributions. Commented Oct 15, 2020 at 23:10

1. The retarded Green's function$$^1$$
\begin{align} G_{\rm ret}(\vec{r},t) ~=~&i\theta(t)G_C(\vec{r},t) ~\stackrel{(B)}{=}~2\theta(t) \bar{G}(\vec{r},t) ~\stackrel{(C)}{=}~2\theta(t)~{\rm Re}\,G_F(\vec{r},t),\tag{A}\cr iG_C(\vec{r},t)~=~&G_{\rm ret}(\vec{r},t)-G_{\rm adv}(\vec{r},t) ~=~2~{\rm sgn}(t)\bar{G}(\vec{r},t),\tag{B}\cr \bar{G}(\vec{r},t) ~=~&\frac{G_{\rm ret}(\vec{r},t)+G_{\rm adv}(\vec{r},t)}{2} ~=~{\rm Re}\,G_F(\vec{r},t),\tag{C}\cr G_F(\vec{r},t) ~=~&\lim_{\epsilon\searrow 0^+}iG_E(\vec{r},(i+\epsilon)t),\tag{D}\cr {\rm Re}\,G_F(\vec{r},t)~\stackrel{(D)}{=}~&-\lim_{\epsilon\searrow 0^+}{\rm Im}\,G_E(\vec{r},(i+\epsilon)t)\tag{E} \cr ~\stackrel{(H)}{=}~&\lim_{\epsilon\searrow 0^+} \left\{ \begin{array}{lcl} \frac{1}{2}{\rm Im}\sqrt{-t^2+i\epsilon} &{\rm for}& d= 1,\cr \frac{1}{4\pi}{\rm Im} \ln(r^2-t^2+i\epsilon) &{\rm for}& d= 2,\cr -\frac{1}{(d-2){\rm Vol}(\mathbb{S}^{d-1})}{\rm Im} \frac{1}{(r^2-t^2+i\epsilon)^{d/2-1}} &{\rm for}& d\geq 3, \end{array}\right.\tag{F} \end{align} for the d'Alembertian $$(\partial_t^2-\vec{\nabla}^2)G(\vec{r},t) ~=~\delta^{d-1}(\vec{r})\delta(t) \tag{G}$$ in $$d$$-dimensional Minkowski spacetime can be derived as a generalized function by analytic continuation/Wick rotation of the Euclidean Green's function $$G_E(\vec{r})~=~\left\{ \begin{array}{lcccl} -\frac{\sqrt{r^2+0^+}}{2}&=& -\frac{r}{2} &{\rm for}& d=1, \cr -\frac{\ln(r^2+0^+)}{4\pi}&=&-\frac{\ln(r)}{2\pi} &{\rm for}& d=2, \cr \frac{(r^2+0^+)^{1-d/2}}{(d-2){\rm Vol}(\mathbb{S}^{d-1})}&=&\frac{r^{2-d}}{(d-2){\rm Vol}(\mathbb{S}^{d-1})} &{\rm for}& d\geq 3,\end{array}\right. \tag{H}$$ for the Laplacian $$-\vec{\nabla}^2 G_E(\vec{r})~=~\delta^d(\vec{r}) \tag{I}$$ in $$d$$ dimensions.

2. OP's expression (36) is supposed to be the retarded Green's function $$G_{\rm ret}(\vec{r},t)$$ in 4+1D. It is a linear combination of a singular function times a distribution. OP is correct that eq. (36) is not well-defined in distribution theory. This is why we propose to use analytic continuation/Wick rotation (D) instead.

3. If one rewrites the retarded Green's functions $$G_d(r^2,t)$$ with a square radial argument $$r^2$$ (and regularization $$\epsilon>0$$) then one may check that the formulas (F) satisfy the recurrence relations (25) and (29) of Ref. 1 in the following form $$G_{d+2}(r^2,t)~=~-\frac{1}{\pi} \frac{\partial G_d(r^2,t)}{\partial (r^2)}\tag{29'}$$ and$$^2$$ $$G_{d-1}(r^2,t)~=~2\int_{\mathbb{R}_+}\!\mathrm{d}s~G_d(s^2+r^2,t). \tag{25'}$$ Ref. 1 argues that dimensional reduction from $$d$$ to $$d\!-\!p$$ dimensions corresponds to inserting a $$(d\!-\!p)$$-dimensional delta function source with a $$p$$-dimensional transversal support in $$d$$ dimensions. If $$p=1$$, this yields eq. (25'). If $$p=2$$, one gets $$G_{d-2}(r^2,t)~=~\int_{\mathbb{R}_+}\!2\pi s\mathrm{d}s~G_d(s^2+r^2,t) ~=~\pi\int_{r^2}^{\infty}\!\mathrm{d}(s^2)~G_d(s^2,t). \tag{28'}$$ Differentiation of eq. (28') yields eq. (29').

4. More calculations: \begin{align}{\rm Re}\,&G_F(\vec{r},t) ~\stackrel{(E)}{=}~ -{\rm Im}\,G_E(\vec{r},(i+\epsilon)t)\cr ~\stackrel{(F)}{=}~&\left\{ \begin{array}{lclcl} \frac{1}{2}{\rm Im}\sqrt{-t^2+i\epsilon} &=& \frac{|t|}{2} &{\rm for}& d=1, \cr \frac{1}{4\pi}{\rm Im}\ln(r^2-t^2+i\epsilon) &=& \frac{1}{4}\theta(t^2-r^2) &{\rm for}& d=2, \cr -\frac{1}{4\pi}{\rm Im}\frac{1}{\sqrt{r^2-t^2+i\epsilon}} &=&-\frac{1}{4\pi}{\rm Im}\sqrt{\frac{r^2-t^2-i\epsilon}{(r^2-t^2)^2+\epsilon^2}}\cr &=&\frac{1}{4\pi}\sqrt{\frac{\sqrt{(r^2-t^2)^2+\epsilon^2}-(r^2-t^2)}{2((r^2-t^2)^2+\epsilon^2)}}\cr &=&\frac{\theta(t^2-r^2)}{4\pi\sqrt{|t^2-r^2|}}+\text{sing. terms} &{\rm for}& d=3, \cr -\frac{1}{4\pi^2}{\rm Im}\frac{1}{r^2-t^2+i\epsilon} &=&-\frac{1}{4\pi^2}{\rm Im}\frac{r^2-t^2-i\epsilon}{(r^2-t^2)^2+\epsilon^2}\cr &=&\frac{1}{4\pi^2}\frac{\epsilon}{(r^2-t^2)^2+\epsilon^2}\cr &=&\frac{1}{4\pi}\delta(t^2-r^2) &{\rm for}& d=4, \cr -\frac{1}{8\pi^2}{\rm Im}\frac{1}{(r^2-t^2+i\epsilon)^{3/2}} &=&-\frac{1}{8\pi^2}{\rm Im}\sqrt{\frac{(r^2-t^2-i\epsilon)^3}{((r^2-t^2)^2+\epsilon^2)^3}} \cr &=&\frac{1}{8\pi^2}{\rm sgn}(r^2-t^2-\epsilon^2/3)\cr &\times&\sqrt{\frac{\sqrt{((r^2-t^2)^2+\epsilon^2)^3}-(r^2-t^2)^3+3\epsilon^2(r^2-t^2)}{2((r^2-t^2)^2+\epsilon^2)^3}} \cr &=&-\frac{\theta(t^2-r^2)}{8\pi^2|t^2-r^2|^{3/2}}+\text{sing. terms} &{\rm for}& d=5. \cr \end{array}\right.\end{align} \tag{J} Here the singular terms have support on the light-cone $$\{(\vec{r},t)\in\mathbb{R}^d | r^2=t^2\}$$. Therefore \begin{align} G_{\rm ret}(\vec{r},t) ~\stackrel{(A)}{=}~& 2\theta(t)~{\rm Re}\,G_F(\vec{r},t)\cr ~\stackrel{(J)}{=}~&\left\{ \begin{array}{lcl} t^+=\max(t,0) &{\rm for}& d=1, \cr \frac{1}{2}\theta(t-r) &{\rm for}& d=2, \cr \frac{\theta(t-r)}{2\pi\sqrt{t^2-r^2}} +\text{sing. terms}&{\rm for}& d=3, \cr \frac{\delta(t-r)}{4\pi r} &{\rm for}& d= 4, \cr -\frac{\theta(t-r)}{4\pi^2(t^2-r^2)^{3/2}}+\text{sing. terms} &{\rm for}& d=5.\end{array}\right.\end{align} \tag{K} Eq. (K) makes contact to the ill-defined expressions (35) & (36) of Ref. 1.

References:

1. H. Soodak & M.S. Tiersten, Wakes and waves in $$N$$ dimensions, Am. J. Phys. 61 (1993) 395.

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$$^1$$ In this answer, we put the speed of light $$c=1$$. The limit symbol $$\lim_{\epsilon\searrow 0^+}$$ is implicitly implied from now on. Here $${\rm Vol}(\mathbb{S}^{d-1}) ~=~ 2\frac{\pi^{d/2}}{\Gamma(\frac{d}{2})}\tag{L}$$ is the volume of the $$(d\!-\!1)$$-dimensional sphere $$\mathbb{S}^{d-1}$$. So the constant in eqs. (F) & (H) is $$(d\!-\!2){\rm Vol}(\mathbb{S}^{d-1}) ~=~ 4\frac{\pi^{d/2}}{\Gamma(\frac{d}{2}-1)}.\tag{M}$$

$$^2$$ Concretely, eq. (25') was checked for low $$d$$ (high enough that the integral (25') is convergent). Now use the recurrence relation (29') to extend eq. (25') to arbitrary high $$d$$. Eq. (29') does not hold if we remove the regularization $$\epsilon>0$$.

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

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:

1. 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 $$+\infty$$.
2. 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 $$-\infty$$.
3. 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 $$r, exactly the second term of $$(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)

This is a good question, and my answer here is not a complete answer, but here already my point of view. Mathematically speaking, there is no general definition in the theory of distribution of the product of two singular distributions. Here it seems your problem is the definition of the distribution $$\frac{\delta_0(t-rc)}{r\sqrt{t^2-r^2c^2}}$$, which can be written $$\frac{\delta_0(t-rc)}{r\sqrt{t^2-r^2c^2}} = \frac{1}{r\sqrt{t+rc}} \frac{\delta_0(t-rc)}{\sqrt{t-rc}},$$ and so the more precise problem is to find the definition of $$\frac{\delta_0(x)}{\sqrt{|x|}}$$. The meaning could be the solution of $$f$$ of the equation $$|x|^{1/2}\,f(x) = \delta_0.$$ However this equation has no solutions. Indeed, by homogeneity, the solution should be the finite part distribution $$f(x) =$$ fp$$(|x|^{-3/2})$$) defined as a distribution by $$\langle f,\varphi\rangle = \langle\mathrm{fp}(|x|^{-3/2}),\varphi\rangle = ∫\frac{\varphi(x)-\varphi(0)}{|x|^{3/2}}\,\mathrm d x.$$ However, $$|x|^{1/2}\,f(x)$$ does not have meaning either in this case (restricted to tests functions with $$\varphi(0)=0$$, we would have $$f(x)\,|x|^{1/2} = \frac{1}{|x|}$$ which would still be different from $$\delta_0$$).

I think this paper is just very formal and not rigorous at all mathematically. The best way to understand what should be a multiple of the true result would be to go over all the computations in a mathematically rigorous way.