4
$\begingroup$

Is accelerating charges the only way to produce EM radiation?

What about gamma radiation emitted from the nucleus of an atom? Does that count as accelerating charges? How?

Are there other ways of producing EM radiation? Or are these the only two ways?

What about collision of particles? What about emission of a photon as an electron falls to a lower n state?

Would this be a correct way to describe it (noting this is for introductory level, so a grasp of the basics would be handy without confusing with the specifics):

All forms of EM radiation are produced by accelerating charged particles. The only exception is gamma radiation, which is produced from the nucleus of radioactive atoms.

However, note that some physicists would argue that gamma radiation released from the nucleus of atoms also involves accelerating charged particles (in this case, protons within the nucleus).

Also note that it is theoretically possible to accelerate charged particles sufficiently that they would cause EM radiation at gamma-ray frequencies. Many believe this is the mechanism that causes gamma-ray bursts.

However strictly speaking gamma-ray-frequency EM radiation caused by such a mechanism would be more correctly called 'high frequency X-ray radiation', while 'gamma-rays' come from the nucleus of an atom.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
1
$\begingroup$

Technically both emission of photons from an atom and emission of gamma rays from nuclei involve acceleration of charges. In both cases charge distributions are rearranged in the process of EM emission. Hence charges are accelerated during the emission process. For atoms it is the electrons jumping from a higher energy orbital to a lower energy orbital. For nuclei, it is protons.

Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ I'm not sure I'd really call the act of transitioning levels (electronic or nuclear) to get directly tied to accelerating the charge. Compared with, say, moving electrons in an antenna for radio wave generation, it really is quite different. $\endgroup$
    – Jon Custer
    Oct 17, 2016 at 3:19
  • $\begingroup$ @Jon Well, there is a (brief!) time-derivative of the position distribution. It's an appropriate popularization at some level. $\endgroup$
    – dmckee --- ex-moderator kitten
    Oct 17, 2016 at 17:27
  • $\begingroup$ @dmckee - fair enough I suppose, particularly for electron rearrangement in an atom during photon emission. Gamma emission starts being much fuzzier to me though, but I never did that much modern nuclear physics. $\endgroup$
    – Jon Custer
    Oct 17, 2016 at 17:45
  • $\begingroup$ @JonCuster The electron in an antenna is moving in a conduction band of a metallic lattice. It is being driven by an oscillating electric field. The radiation that this electron emits is analogous to the stimulated emission that occurs in a collection of atoms with a population inversion. $\endgroup$
    – Lewis Miller
    Oct 18, 2016 at 0:44
  • $\begingroup$ @LewisMiller - I beg to differ that it is analogous to stimulated emission. I'd be happier with spontaneous emission, but even that isn't a great analogy. The motion of the electrons in the antenna couples directly to propagating EM modes and does not depend on discrete (atomic or band) levels. (And, one can make a 'non-metallic' antenna - a plasma can radiate quite nicely!). $\endgroup$
    – Jon Custer
    Oct 18, 2016 at 13:58
1
$\begingroup$

The emission of photons from atoms and nuclei has much in common. Whilst the classical view of the production of radiation involves the acceleration of charged particles, or more usefully, the second time derivative of a dipole moment, the quantum mechanical picture involves evaluating the expectation of a perturbation operator, due to the electromagnetic field, between the final and initial states of the atom/nucleus. a.k.a. Fermi's Golden Rule

Since the wavelength of the emitted radiation is usually much larger than the emitting object in each case (think photons from atomic processes with wavelengths of $\sim 10^{-7}$ m and nuclear gamma rays with wavelengths of $\sim 10^{-13}$ m, then it turns out that it is the electric dipole operator that is most important (unless quantum mechanical selection rules forbid a transition via the electric dipole operator).

A classical way of thinking about this would be to consider the emission of radiation (from nuclei or atoms) as involving an oscillating (and therefore accelerating) electric dipole moment. The classical analogy does not work in detail though because it does not explain why the ground states of atoms/nuclei are stable or why certain radiative transitions are forbidden, or at least much less likely than others. Ultimately to explain the behaviour of photons and their interactions with atoms, nuclei and other particles, a quantum mechanical approach is necessary.

Improve this answer
$\endgroup$

Your Answer

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

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