The main drawback in Rutherford's model of the atom as pointed out by Niels Bohr was that according to Maxwell's equations, a revolving charge experiences a centripetal acceleration hence it must radiate energy continuously as a result of which its kinetic energy is reduced and it eventually should fall into the nucleus. Bohr then proposed his own model in which he explained this through quantum theory stating that the general laws of electrodynamics and classical mechanincs do not apply in the case of electrons. Does this imply that if a body is revolving around another, then it will tend to spiral in towards the centre? If so, then how is the stability of the solar system explained?
$\begingroup$
$\endgroup$
asked Jul 23, 2013 at 2:48
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
11
If so, then how is the stability of the solar system explained?
The solar system bodies are not electrically charged to any significant extent so there is no significant electromagnetic radiation.
However, there is a somewhat analogous mechanism where gravitational radiation can cause orbital decay.
Although this is not significant in our solar system, orbital decay has been detected in relativistic systems as a binary neutron star system.
answered Jul 23, 2013 at 3:00
-
$\begingroup$ my question on stability of the solar system was based on the analogy between gravitational and electric forces. can u explain why the energy radiation occurs? $\endgroup$ Commented Jul 23, 2013 at 3:26
-
$\begingroup$ Be careful about making the analogy between EM and gravitation because there is a fundamental difference. In EM oscillating dipoles radiate EM waves but in GR oscillating dipoles do not radiate gravity waves. The lowest pole that can produce gravity waves is an oscillating quadrupole. $\endgroup$ Commented Jul 23, 2013 at 6:50
-
$\begingroup$ @JohnRennie Great point: the Liénard-Wiechert potentials have almost dipolar shape: deviations from this arise from the small "harmonic distortion" (deviation from the $\sin\theta$ variation) coming from the $\frac{1}{1-\mathbf{n}.\mathbf{\beta}}$ terms (Wikipedia page notation), so the quadrupole term is very small. Naturally there must be GR analogues of the Larmor formula and the Liénard-Wiechert potentials - could you point me there? $\endgroup$ Commented Jul 23, 2013 at 8:16
-
$\begingroup$ As yet, I believe that orbital decay is the only experimental evidence for gravitational waves. Hopefully this will change soon. $\endgroup$ Commented Jul 23, 2013 at 8:17
-
$\begingroup$ @Alfred Centauri: The net charge of planets is almost zero but each individual electron and proton (which are many) must radiate? $\endgroup$– richardCommented Jul 23, 2013 at 8:34
$\begingroup$
$\endgroup$
2
Well the answer to the question asked:
does a charged particle revolving around another charged particle radiate energy?
is simply yes.
Perhaps the "error" in the Rutherford-Bohr-Sommerfeld atoms is the assumption that the electrons are in fact "revolving" about the nucleus.
Remove that revolving, and the need for Hertz-Maxwell radiation, goes away.
Doesn't quantum mechanics remove any revolving; or any other form of charge acceleration in the atom.
-
$\begingroup$ @ user26165...Without "revolving" there will be no Nuclear Magnetic Resonant (NMR) imaging!!?? $\endgroup$– barryCommented Oct 28, 2021 at 16:36
-
$\begingroup$ @ user26165..What would we do without Hertz-Maxwell radiation? Apart from my cell phone and TV, it has kept me in sustenance for 50 years. QM has just given me headaches. $\endgroup$– barryCommented Oct 28, 2021 at 16:43