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Here's my problem.

Intensity of radiation follows an inverse square law with distance.

https://byjus.com/physics/inverse-square-law/

Electromagnetic fields can be derived from the Lienard Wiechert equations:

$E(r,t)=Kq(1-n\cdot\beta)^{-3}(\frac{n-\beta}{\gamma^2|r-r_s|^2}+\frac{n\times((n-\beta)\times\dot\beta)}{|r-r_s|})$

This equation shows the electric force from a source charge on a target charge at distance $|r-r_s|$. The basic force appears to vary with distance by $ \frac{1}{|r-r_s|^2}$ while the radiation force varies with distance by $\frac{1}{|r-r_s|}$.

What am I missing?

Edit: The baseline electric FORCE declines by the inverse square. The radiation force declines inverse linear, but intensity=energy declines by the square.

Thank you Jensen Paull and Dale for clearing that up for me.

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2 Answers 2

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You appear to be missing the fact that the $E$ field and the intensity are different quantities. There is no requirement that both quantities behave the same as a function of $r$.

In particular, in the far field the intensity is proportional to the square of the $E$ field, so you would expect that if the intensity decays as $1/r^2$ then the $E$ field would decay as $1/r$, which is precisely the relationship you found.

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Intensity, which is also Power/area. Is derived from the poynting vector.

Which is proportional to ExB

B also follows 1/r and thus the cross product falls off like 1/r^2

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  • $\begingroup$ Thank you! So the formula for E is basicly for point charges? To actually figure the force you'd care about the cross-section of the charge, compared to the surface area of a sphere that radius? $\endgroup$
    – J Thomas
    Dec 22, 2021 at 17:23
  • $\begingroup$ The formula you have wrote are for point charges. intensity is defined as the energy passing through a surface per second per unit area. You don't take a cross section of a charge. That doesn't make sense. You define some surface and then compute the surface flux integral of the poynting vector to find out how much energy is passing through that surface per second. Intensity is a way of measuring energy not force. Force is E * charge. $\endgroup$
    – jensen paull
    Dec 22, 2021 at 17:32
  • $\begingroup$ What you wrote makes sense, so radiation would fall off by the inverse square. So applying the same reasoning, i would expect the electric force from a charge that is not accelerated to fall off at a higher power, and magnetism also. When I google this I get various less-technical explanations that say radiation is inverse-square and so is electric charge both the same. hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html#isqe Also I find sites talking about intensity at a point. Dammit? $\endgroup$
    – J Thomas
    Dec 22, 2021 at 19:48
  • $\begingroup$ A stationary charge does not radiate since the poynting vector is zero. It requires both electric and magnetic fields to be present. Intensity in the context of radiation is POWER/AREA or Energy/(second*(area)) which is basically FLOWING energy. If there is no energy flow then there is no poynting vector. The site you linked describes 2 different things. The first one is that the MAGNITUDE of the E field falls off like 1/r^2 and the second shows INTENSITY of radiation. The sites that describe "intensity" at a point. Are either A- talking about the flow of energy passing some surface, such that $\endgroup$
    – jensen paull
    Dec 22, 2021 at 19:55
  • $\begingroup$ The poynting vector in that surface is constant. Or B they are describing the ENERGY of the E field at a specific point in space. ENERGY, E field magnitude and INTENSITYS are 3 different things $\endgroup$
    – jensen paull
    Dec 22, 2021 at 19:56

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