- #1
ZetaOfThree
Gold Member
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In the Feynman Lectures on Physics chapter 28, Feynman explains the radiation equation $$\vec{E}=\frac{-q}{4\pi\epsilon_0 c^2}\,
\frac{d^2\hat{e}_{r'}}{dt^2}$$
The fact that the transverse component varies as ##\frac{1}{r}## seems fairly obvious to me since what matters is just the angle through which the charge moves as seen from the distant observer. However, I'm not sure how to show what he claims for the radial component. Can someone help me see clearly why this is true?
\frac{d^2\hat{e}_{r'}}{dt^2}$$
The unit vector ##\hat{e}_{r'}## is pointed toward the apparent position of the charge. Of course, the end of ##\hat{e}_{r'}## goes on a slight curve, so that its acceleration has two components. One is the transverse piece, because the end of it goes up and down, and the other is a radial piece because it stays on a sphere. It is easy to demonstrate that the latter is much smaller and varies as the inverse square of ##r## when ##r## is very great. This is easy to see, for when we imagine that we move a given source farther and farther away, then the wigglings of ##\hat{e}_{r'}## look smaller and smaller, inversely as the distance, but the radial component of acceleration is varying much more rapidly than inversely as the distance.
The fact that the transverse component varies as ##\frac{1}{r}## seems fairly obvious to me since what matters is just the angle through which the charge moves as seen from the distant observer. However, I'm not sure how to show what he claims for the radial component. Can someone help me see clearly why this is true?