In ch 30-7 of the Lectures, Feynman explains that the field of a plane of oscillating charges at a point P is proportional to the velocity of the charges, considered at the appropriate retarded time (retarded by the vertical distance from the point P).(adsbygoogle = window.adsbygoogle || []).push({});

Feynman derives this formula only for large distance but says that it also works for any other distance.

This implies that directly at the plane of the charges, the electrical field is in phase with the (negative) velocity of the charges (and the same is true at a distance of one, two, three... wavelengths).

What is puzzling me is that this does not seem to allow for correct reflection of light.

If I have an incident em wave, this wave accelerates the charges. The acceleration is in phase with the wave (if the charges are sufficiently free as in a metal), so the vleocity of the charges is shifted by 90° in phase (when the acceleration is maximal, the velocity is zero - at the turning point of the oscillation). If the electrical field is in phase with the velocity, then the electrical field created by the plane of charges is phase-shifted by 90°, not by 180° as it should be to cancel the incoming wave.

Probably I'm making a stupid mistake somewhere, just cannot figure out where.

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# Light Reflection and Feynmans plane of oscillating charges

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