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 $ct$.

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  • $\begingroup$ 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 $\int_{-\infty}^{\infty} e^{i \omega t} \delta (t-r/c) \, dt = e^{i \omega r/c} = e^{i k_0 r}$ ? $\endgroup$ – xyz Mar 4 '14 at 4:14
  • 6
    $\begingroup$ @James: The problem is $f(t)=\delta(r-ct) \neq \delta(t-r/c)$. Rather, $f(t)=\delta(r-ct) = c^{-1}\delta(t-r/c)$. $\endgroup$ – DumpsterDoofus Mar 4 '14 at 4:22

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