Every atom contains charges and charges emit EM wave when they are accelerated. So if an atom is accelerated, the charges inside will also accelerate. Which means that the charges inside will also emit EM wave.

But I don't think I'm right. Because then everything in earth is accelerating around the sun. And so we should everything in the earth to be emiting some kind of EM wave. So where am I terribly wrong?


1 Answer 1


Because the positive and negative charges are very close together the waves they produce would almost exactly cancel. As a result almost no energy would be detected from them by a detector somewhere else in the universe.

  • $\begingroup$ But the positive and negative have some distance between them. So the waves they will produce have some phase difference. Then why will the phase difference always be destructive and thus cancelling both waves. Can't it happen that the waves have a phase difference of 90 degrees and so they will not cancel each other completely? $\endgroup$ Commented Jun 4, 2020 at 4:38
  • $\begingroup$ The distance between the positive and negative charges would need to be significant compared with the size of the wave. With a period of 1 year the wavelength would be 1 light-year, and the diameter of an atom is not significant. $\endgroup$
    – Peter
    Commented Jun 4, 2020 at 6:28
  • $\begingroup$ If the distance between the charges are negligible compared to the wavelength, then shouldn't the waves be in phase with each other? $\endgroup$ Commented Jun 4, 2020 at 7:11
  • $\begingroup$ No, because their charges are opposite. $\endgroup$
    – Peter
    Commented Jun 4, 2020 at 7:22
  • 1
    $\begingroup$ I am not certain, but I think collisions between molecules move electrons into higher energy level within the molecule's energy levels (which are much more complex than those of a single atom). The electrons can emit energy when they drop back to a lower energy level. I'm not sure how this works for, say neon gas. It might be a good question to ask if it hasn't been already asked. $\endgroup$
    – Peter
    Commented Jun 4, 2020 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.