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When considering a uniformly moving charged particle, we have the following fields:

$$\vec E = \frac{q(1-\beta^2)}{4\pi\epsilon R_a}\vec R$$ $$\vec B = \frac{1}{c^2}\vec u \times \vec E$$

With $\vec u$ the velocity of the particle.

The Poynting-Vector is $\vec S = \vec E \times \vec H$ which isn't $0$ since the 2 fields are perpendicular, so why do we say that there's no radiation here if there is a non-zero amount of energy being radiated?

Joshua
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1 Answers1

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Hint: Look at the Poynting-vector associated with an electric charge Q and a bar magnet, and then consider $\int\int {\bf E}\times {\bf B} . d{\bf S}$ around a closed surface.

jim
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