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I think there should be two ways of writing the equation for the impulsive spherical wave from an impulsive point source at the origin, say $\delta(t) \delta(r)$:

$$(4\pi ct)^{-1} \delta(r-ct) \tag{1}$$

$$(4\pi ct)^{-1} \delta(t-r/c) \tag{2}$$

But these are not the same. In (1) the Delta Function has dimension 1/[L] and in (2) the Delta Function has dimension 1/[T].

Also

$$(4\pi ct)^{-1} \delta(r-ct) = (4\pi ct)^{-1} \delta(c(r/c-t)) =(4\pi ct)^{-1} (1/c)\delta((t-r/c))$$

so the two equations differ by (1/c).

My question is which is right and why?

Why would one be preferred over the other?

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The wave from an impulsive point source is the Green's function $G(\vec{r},t)$ for the inhomogenous wave equation. In other words, if we wish to solve the equation $$ - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} + \nabla^2 \psi = \rho(\vec{r},t), $$ we can do so by solving the related equation $$ - \frac{1}{c^2} \frac{\partial^2 G}{\partial t^2} + \nabla^2 G = \delta^3(\vec{r}) \delta(t). \tag{A} $$ in a distributional sense. By convention, the Green's function is usually taken to be the solution to this equation (in a distributional sense) with a delta function of unit weight on the right-hand side. This allows us to then find the solution of a more general inhomogeneous equation via integration: $$ \psi(\vec{r},t) = \iiiint d^3\vec{r}' dt' G(\vec{r} - \vec{r}',t-t') \rho(\vec{r}',t'). \tag{B} $$ (This can be proven by simply applying the wave operator to the right-hand side and bringing the derivatives inside the integral.) If we had used a different weight for the delta functions on the right-hand side, then this equation would require extra pre-factors to compensate.

So which one is right? Dimensionally, the right-hand side of (A) has dimensions of $[\text{length}]^{-3} [\text{time}]^{-1}$, and so by this logic the dimensions of $G$ should be $[\text{length}]^{-1} [\text{time}]^{-1}$. In your question, version (2) has the right units, and so it would be the correct Green's function for this equation.

Of course, if you defined your inhomogeneous wave equation differently---for example, something like $$ - \frac{\partial^2 \psi}{\partial t^2} + c^2 \nabla^2 \psi = \rho(\vec{r},t) $$ then the appropriate Green's function would be the solution to this equation with the right-hand side replaced by a unit-weight delta-function. Via the same dimensional arguments as above, the correct Green's function (so that the solution is still given by eq. (B)) would be $$ G(\vec{r},t) = \frac{1}{4\pi c^3 t} \delta(t-r/c). $$

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  • $\begingroup$ Super answer--thanks. I am surprised that the Green's fn depends on the algebraic arrangement of the wave equation. $\endgroup$ – user45664 May 29 at 23:39
  • $\begingroup$ Very roughly speaking, the Green’s function acts as the “inverse” of a differential operator. From that perspective, it’s not entirely surprising that multiplying the operator by a constant changes the Green’s function, just as multiplying a matrix by a constant changes that inverse. $\endgroup$ – Michael Seifert May 30 at 2:25
  • $\begingroup$ I see--like in $TT^{-1}=\delta$ $\endgroup$ – user45664 May 30 at 18:17

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