EM radiation seems to come from two different sources:
- According to the Maxwell, by acceleration of electrons
- According to the Bohr, by jumping of electrons between energy levels?
Are these two, completely different things?
$\begingroup$
$\endgroup$
2
-
4$\begingroup$ Maxwell probably did not knew anything about electrons -they were discovered after his death in 1897 by J. J. Thomson. Maxwell probably knew of radiation due to oscillating currents in wires, like those in present-day antennae. Radiation due to accelerated electrons is a later idea, I believe also by Thomson, since he studied scattering of radiation and used electrons to explain it. $\endgroup$– Ján LalinskýJun 11, 2014 at 15:55
-
2$\begingroup$ What about photons produced from nuclear reactions? $\endgroup$– thisJun 11, 2014 at 19:18
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
2
These are different causes of radiation, but they both produce the same kind of radiation: electromagnetic radiation.
When an electron is accelerated, it loses energy and emits one or more photons. This is a purely classical view of what happens.
When an electron falls from a higher energy state in an atom to a lower energy state, it emits a photon. This isn't really viewed as an acceleration in quantum mechanics; the "jump" is instantaneous. The speed of the electron before and after the jump are determined entirely by the states the electron is in, rather than by any details of the jump. In addition, there really isn't any force present that would cause the jump; it just happens.
In addition to those two mechanisms for creating EM radiation, there is also matter-antimatter annihilation. If an electron and a positron collided, they would annihilate each other, producing gamma rays. I can't think of any other mechanisms for causing radiation, but there might be others.
answered Jun 11, 2014 at 15:09
-
$\begingroup$ Don’t agree with a promoted “pure f… magic” attitude towards Quantum Mechanics. There are different languages to speak about microscopic objects: corpuscular (i.e. positions/velocities/paths/world_lines) and wave functions. If we speak wave functions, then one can say “the "jump" is instantaneous”, but “really isn't any force present” is a pure nonsense since the word “force” belongs to another vocabulary. One also may not say “the speed of an electron in the quantum state … equals …”. If we speak wave functions, then there are operators, their spectra, and mean values, not equalities. $\endgroup$ Nov 23, 2014 at 8:52
-
$\begingroup$ But if we speak world lines (that is indeed possible in QED), then we ought to admit that the force exerted on an electron behaves like δ function (is infinite at some t ₀ and zero out of it), and likewise behaves its acceleration. $\endgroup$ Nov 23, 2014 at 8:57
$\begingroup$
$\endgroup$
3
According to the Maxwell, by acceleration of electrons
This is the classical electromagnetic theory describing light. Accelerated charges radiate electromagnetic radiation and the description is classical.
When speaking of photons we are talking quantum mechanics, as the photon is the quantum of light, and the classical wave emerges from a zillion photons in coherence/synergy. Both views of light are valid. Macroscopically it is better to use the classical formulation which is compact and elegant and describes light in macroscopic dimensions, and use the quantum mechanics forms when the dimensions are commensurate with h_bar.
If one wants to describe the radiation from a single electron in an electric field one has to use the quantum mechanical forms, and again it is an interacting system that apportions the loss of energy from the electron and emission of a photon, and can be calculated with Feynman diagrams.
According to the Bohr, by jumping of electrons between energy levels?
The Bohr atom, similar to the planetary model for gravity, is an interim proposal to explain the spectra observed from atoms, i.e. in the quantum mechanical framework. Using the classical maxwell equations the electron orbiting the atom would radiate away and fall into the nucleus. For stable atoms quantized orbits were posited and the transitions in energy became quantized, the energy leaving as a photon.
In the correct quantum mechanical theory, the whole system, electron/atom is described by one function, a quantum mechanical state function and the probability of transition expelling a photon can be accurately calculated. That photon, joining its brothers as in lasing, will build up an electromagnetic wave that macroscopically is described by the classical equations.
Are these two, completely different things? Or which one is true?
They are the same, physics has continuity going from the quantum to the classical framework.
-
$\begingroup$ To be sure. So "jumping" and "acceleration" are same thing. Electron is accelerating while jumping maybe? Right? $\endgroup$ Jun 11, 2014 at 18:15
-
1$\begingroup$ It is not a good idea to use classical concepts for quantum mechanical concepts. In principle we do not know what the electron is doing, except seeing the transition by the emission of a photon. You could say "it is accelerating in steps" but it is better to start acquiring a quantum mechanical intuition, where everything works with probability distributions , because trying to fit qm behavior to classical concepts leads to confusions. $\endgroup$– anna vJun 11, 2014 at 18:39
-
$\begingroup$ The only really competent answer for a given question in the whole thread, that deals directly with the problem posed and doesn’t debunk a rubbish with a new, higher-order rubbish. Isn’t it? $\endgroup$ Nov 23, 2014 at 9:07
$\begingroup$
$\endgroup$
Both are true. Both processes involve reduction of energy of an electron the total energy list being converted into the released photon. So when an electron jumps from higher to lower energy level in an atom it looses energy which manifests itself as a photon. Similarly an accelerated electron looses energy which is then in turn manifested as a photon.
$\begingroup$
$\endgroup$
Actually both "types" can be approached in the same classical framework.
The "classical" approach to radiation from an atom has an incoming wave exciting an oscillation. This oscillation can be treated as an accelerated dipole and emits radiation accordingly.
This serves adequately to model Rayleigh scattering and stimulated emission, but has major problems in that it does not explain spontaneous emission or the direction of stimulated emission.
The true commonality of these processes can only be seen with QED.
$\begingroup$
$\endgroup$
3
From the quantum picture, both events are equivalent. If you study the quantum evolution of a decaying state, or what is called a "Dipole Transition" in an atom (an atom going from one state to the other), you'll see that it's proportional to an oscillation of the electron involved with the frequency of the photon being emitted.
Refer to the book: Optically polarized Atoms, Auzinsh: Chapter 7, 7.8: Visualization of atomic transitions.
answered Jun 11, 2014 at 16:08
-
$\begingroup$ “you'll see that it's proportional…” which namely quantity is? And to which quantity? $\endgroup$ Nov 23, 2014 at 9:03
-
$\begingroup$ @IncnisMrsi The wave function. It evolves from one state to another. $\endgroup$ Nov 23, 2014 at 15:42
-
$\begingroup$ Do you mean “probability density oscillates with frequency of the photon being emitted”? Rewrite it in less ambiguous language, please. $\endgroup$ Nov 23, 2014 at 15:49