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: $$ ( \nabla^2 + k_0^2) G(x,y,\omega) = -\frac{1}{c^2} \delta |x-y| $$


1 Answer 1


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 $ct$.

  • $\begingroup$ 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, 2014 at 4:14
  • 6
    $\begingroup$ @James: The problem is $f(t)=\delta(r-ct) \neq \delta(t-r/c)$. Rather, $f(t)=\delta(r-ct) = c^{-1}\delta(t-r/c)$. $\endgroup$ Mar 4, 2014 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.