# Derivation of Green's Function for Wave Equation

In the textbook Modern Methods in Analytical Acoustics (Crighton-1992, Amazon link to 2013 edition) the following relates the 3D Green's function in the time-domain to the frequency domain $g(x-y)$: \begin{align} g\left(\mathbf{x-y}\right)&=-\frac{c^2}{4\pi cr}\int_{-\infty}^\infty\delta\left(r-ct\right)e^{i\omega t}\,dt,\qquad r=\left|\mathbf{x-y}\right|\\ &=-\frac{1}{4\pi r}e^{ik_0t},\tag{2.158} \end{align}

I cannot see how the integration has eliminated the variable c. To me the integration should leave $e^{i k_0 r}$ where $k_0=\omega/c$ and then the answer should be $$g(\mathbf{x-y}) = -\frac{c}{4 \pi r} e^{i k_0 r}$$ which I know to be incorrect by a factor of c.

Is the text wrong? And if so, then how do I derive the correct expression for $g(\mathbf{x-y})$?

Note: the $-c^2$ factor is used to relate the Green's function $G(x,y,t) = \frac{\delta(r-ct)}{4 \pi c r}$ for the equation $$(\frac{\partial^2}{\partial t^2} - c^2 \nabla^2) G(x,y,t) = \delta |x-y|$$ to the reduced wave equation: $$( \nabla^2 + k_0^2) G(x,y,\omega) = -\frac{1}{c^2} \delta |x-y|$$

In Mathematica:

Refine[-(c^2/(4 π c r)) Integrate[
DiracDelta[r - c t] Exp[I ω t], {t, -∞, ∞}],
Assumptions -> {r ∈ Reals, c > 0}]


The output is

$$-\frac{e^{\frac{i r \omega }{c}}}{4 \pi r}=-\frac{e^{irk_0}}{4 \pi r}$$ The extra factor of $$c$$ is eliminated since the $$\delta$$ function has an argument of $$ct$$.

• I am still puzzled by the result. My logic is: $\delta(r-ct) = \delta(t-r/c)$ and therefore by the translation property of the delta function $\int_{-\infty}^{\infty} e^{i \omega t} \delta (t-r/c) \, dt = e^{i \omega r/c} = e^{i k_0 r}$ ? – xyz Mar 4 '14 at 4:14
• @James: The problem is $f(t)=\delta(r-ct) \neq \delta(t-r/c)$. Rather, $f(t)=\delta(r-ct) = c^{-1}\delta(t-r/c)$. – DumpsterDoofus Mar 4 '14 at 4:22