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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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    Related: Does a nonzero Poynting vector mean that there is propagation of energy? – Alfred Centauri Jun 19 '16 at 12:10
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    The fields are tied to the particle, there is propagation in the sense that the fields follow the particle ; to have radiation the fields should be propagating independently of a charge source, by the definition of radiation. – anna v Jun 19 '16 at 12:43
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    I understand that, but that doesn't explain the behaviour of the Poynting-vector or am I not seeing it? – Joshua Jun 19 '16 at 13:00
  • Well, technically any wave is EM radiation. Usually though, when talking about the Poynting vector people use it to determine the power being transmitted in the wave $P = \frac{1}{2} \lvert \vec P \rvert $. Since power is energy per unit time, I guess you could multiply it by the transmission length (in seconds) to get the total energy $U_{total} = P t_{transmission}$ – M Barbosa Jun 19 '16 at 13:46

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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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