# Does the intensity of dipole radiation fall as $1/r^2$ or $1/r^3$?

I have seen this derivation:

• I want to estimate what is the intensity of the electrical field as function of $r$ the distance from the radiated source ?

• I think it is can modled as pointed source which implies a spherical wave whose intensity goes like $1/r^2$. But I think I remember from school it should be fall like $1/r^3$.

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second link in google for "dipole radiation". Hope this helps. – Robert Filter Mar 8 '13 at 22:02
@RobertFilter didn't see the connection i was looking for in your link – 0x90 Mar 9 '13 at 6:28
please look for the farfield of the dipole radiation which is what you are asking for. – Robert Filter Mar 9 '13 at 18:47

Actually, if you ask about radial dependace of time-avaraged Poynting flux vector, then it depends on where the observer is (far field or near field). If you assume that the observer is far away (more than $\frac{2D^2}{\lambda}$, where $D$ - is the maximum size of your source and $\lambda$ - is the walength (I see you use monocromatic case)) than the intensity should degrade as $\frac{1}{r^2}$ (in case there is no absorbtion in medium, if yes, then it should be a little faster, for example $\frac{1}{r^{2+\alpha}}, \alpha>0$). In the near field region electric and magnetic field components have several terms: $\frac{1}{r},\frac{1}{r^2},\frac{1}{r^3}$, so the Poynting vector should have several components too and they all degrade as $\frac{1}{r^5}$ (and that's even without absorbtion).

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Don't confuse amplitude with intensity. The amplitude of the near field of a dipole falls off as $1/r^3$. The far field however consists of photons. Since the energy content of a photon usually doesn't change by itself (ignoring relativity) the intensity of the far field can only fall of fas $1/r^2$ which means the amplitude falls off as $1/r$. That is because a sphere with x times the radius but the same number of photons distributed over it's surface will have $1/x^2$ of the photon density.

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