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I have studied in my Physics course that one of the drawbacks of Rutherford's atomic model was that when an electron will revolve around the nucleus, it is undergoing acceleration and so it should radiate energy and consequentially fall into the nucleus.

Similarly when a charged particle is projected in the plane perpendicular to a uniform magnetic field it executes uniform circular motion withradius $r=mv/qB$.

My question is why isn't the charged particle radiating energy here? Even in this case the charged particle is accelerating, just as it was in Rutherford's model of the atom. So shouldn't the radius decrease in this case also?

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3 Answers 3

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A charged particle circulating in a magnetic field does radiate energy, and it is called synchrotron radiation. All circular particle accelerators have energy losses due to this radiation.

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    $\begingroup$ It's only synchrotron radiation if the particle velocity is relativistic. Otherwise it's cyclotron radiation. The emission dynamics are quite different for the two cases, hence the nitpick. :) $\endgroup$
    – Kitchi
    Commented Nov 8, 2012 at 13:37
  • $\begingroup$ so does that mean that the radius will keep on decreasing with time? $\endgroup$
    – awareeye
    Commented Nov 8, 2012 at 15:03
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    $\begingroup$ Yes. The radius decreases, that is why accelerators have to feed pulses to beams to keep them at the same momentum/ radius $\endgroup$
    – anna v
    Commented Nov 8, 2012 at 15:08
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    $\begingroup$ @Kitchi both cyclotron and Synchrotron are at relativistic energies. When one talks of particles one is at fractions of c. It is at the ultra high energies, when the harmonics of the cyclotron radiation become so dense it becomes a continuum. astro.umd.edu/~miller/teaching/astr601/lecture16.pdf $\endgroup$
    – anna v
    Commented Nov 8, 2012 at 15:11
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    $\begingroup$ @annav - Synchrotron radiation is only for relativistic motion, because then you will see relativistic beaming of the radiation in the direction of motion. Cyclotron radiation is in the non-relativistic limit, and as long as $\lambda$ >> size of the radiating system, we can approximate a dipole radiation field. These two links explain that in much more detail. (pdf warning). $\endgroup$
    – Kitchi
    Commented Nov 8, 2012 at 16:17
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I only add that there is even a natural (not artificial) source of synchrotron radiation: Crab Nebula (remnant of supernova observed in 1054). The reason why we see it (even in small telescopes) is thus quite diferrent from other celestial sources (where we observe light from hot stars or acretion discs, or excited gas).

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This is a really good question and it puzzled me for many years.

Physics is composed of many different theories, and as history has progressed, physicists have refined them or created new ones. My fav example is Newton's theory of gravity, which is great at explaining gravity for most applications; however, Einstein's are much more accurate and a more thorough explanation.

Electrons orbiting around a nucleus is true for electrodynamics. The question you are posing was a huge issue (like you mentioned above) for many years. It was a big reason why people thought the E&M theory was not complete.

The reason it does not radiate is explained by quantum mechanics. In QM particles are waves and particles. The electron is a waved function. The wave function surrounding a central potential will have bound states. These bound states DO NOT MOVE; the wave function will not change when in a bound state. Because the bound states dont move, the electric field from the electron does not move, so nothing is radiated. This is a subtle and very sweet concept.

Side note: a central potential is something with only radial dependence, and is the kind of potential a nucleus makes.

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  • $\begingroup$ What do you mean by "The electron is a waved function"? $\endgroup$
    – Gonenc
    Commented Jul 20, 2015 at 15:51
  • $\begingroup$ As the answers from 3 years ago say, a charged particle does radiate when in circular motion around a uniform magnetic field. So there is no "reason it does not radiate" when it actually does radiate. $\endgroup$
    – Jim
    Commented Jul 20, 2015 at 16:27

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