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The formula for point wave source is

$A(r)=\frac{a_0 \sin(\omega t + \phi)}{r^2}$

and has a singularity in zero.

So, if I integrate this function over any finite volume, I will get infinity.

So, what is the function which should be integrated over a volume, to get total formula of the wave? emitted by non-point source of coherent elements?

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No, because a finite volume has the volume element going like $r^2 drd\Omega$ that removes your singularity. – Jon Feb 5 '12 at 20:52
Why do you have $r^2$ in numerator? – Suzan Cioc Feb 6 '12 at 7:28
This is the just the volume element in 3d. It takes the $r^2$ factor. Please, could you make more explicit your problem? – Jon Feb 6 '12 at 9:25
Probably I don't understand it myself yet :> – Suzan Cioc Feb 6 '12 at 18:05
If you want, I can put this as an answer so we can clarify. – Jon Feb 6 '12 at 18:15
up vote 1 down vote accepted

Let us consider a spherical wave going like

$$\phi(r,t)=a_0e^{ikr\cos(\theta)-i\omega t+i\psi}\frac{1}{r}$$

typical of a radiation field. In your case, one neglect the spatial dependence and finally gets, taking the real part,

$$\phi(r,t)=a_0\cos({\omega t+\psi})\frac{1}{r}.$$

As you know, you have to take the square of it to get energy density and then integrate on the volume. So,

$$W=\int_V d^3x\phi^2(r,t)$$

and we assume that the volume is finite and spherical with radius $R$. But now, in spherical coordinates, we have $d^3x=r^2drd\Omega$ and you can immediately integrate on the solid angle obtaining $4\pi$. This will give

$$W=4\pi\int_0^R r^2dra_0^2\frac{1}{r^2}\cos^2({\omega t+\psi})=4\pi Ra_0^2\cos^2({\omega t+\psi})$$

where you see that the volume element compensates your singular behavior giving a finite energy as one physically should expect. If you have in mind a lamp you will notice that the radiated energy does not burn you out.

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