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?
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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.
Alfred Centauri
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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? – Ajaykrishnan Jayagopal Jul 23 '13 at 03:26
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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. – John Rennie Jul 23 '13 at 06:50
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@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? – Selene Routley Jul 23 '13 at 08:16
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As yet, I believe that orbital decay is the only experimental evidence for gravitational waves. Hopefully this will change soon. – Selene Routley Jul 23 '13 at 08:17
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@Alfred Centauri: The net charge of planets is almost zero but each individual electron and proton (which are many) must radiate? – richard Jul 23 '13 at 08:34
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@JohnRennie, that's correct thus the somewhat qualifier. Thanks for making the point explicit. – Alfred Centauri Jul 23 '13 at 11:00
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@Riza One has to understand that when we are talking of electrons and protons we are talking of elementary particles,and when we are talking of elementary particles we are talking of quantum mechanics, the basic underlying framework of nature. Quantum mechanics says the neutrality comes from quantum mechanical bindings of electrons and protons into nuclei and atoms and molecules. Only the few electrons freed naturally from these bounds, for example during lightning will display the classical acceleration profile. – anna v Jul 25 '13 at 07:41
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@annav this is true for relative motion of electron and proton but what if you accelerate the whole atom? (please see also this http://physics.stackexchange.com/questions/72005/electromagnetic-radiation-of-charged-particles/72046#72046) – richard Jul 25 '13 at 11:30
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The whole atom is neutral and to first order it does not radiate. To higher orders there is a mechanism displayed in black body radiation where kinetic energy is transformed to electromagnetic energy in three stages, quantum mechanically, see my answer here:http://physics.stackexchange.com/questions/72174/why-do-moving-particles-emit-thermal-radiation/72225#72225 – anna v Jul 25 '13 at 12:49
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@ anna v…..Electrons orbiting protons radiate in all the excited states but do not radiate in the ground state. Zero point is a sea of energy with a mean energy density of half Planck’s constant per mode. This is similar to a dipole’s inability to radiate into a cavity. The cavity walls act as mirrors that reflect energy back to the dipole, this reduces the real component of the radiation resistance to zero, dipole becomes purely reactive. – barry Oct 28 '21 at 16:29
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. A cavity is equivalent to a three dimensional sea of dipoles all trying to radiate. It is claimed there are approx 10^80 hydrogen atoms in the observed universe, all of them with electrons attempting to approach the proton, These attempts give rise to the zero point energy that we observe and ponder its origin!? – barry Oct 28 '21 at 16:29
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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.
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@ user26165...Without "revolving" there will be no Nuclear Magnetic Resonant (NMR) imaging!!?? – barry Oct 28 '21 at 16:36
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@ 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. – barry Oct 28 '21 at 16:43