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Boundless
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if a hydrogen atom has only 1 proton and 1 electron what is the reason that the electron does not collapse onto the proton if there is a + and - charge?
Boundless said:if a hydrogen atom has only 1 proton and 1 electron what is the reason that the electron does not collapse onto the proton if there is a + and - charge?
I don't think it's taken as a fiat nowadays. If we take for granted that the Schrödinger's equation is valid, then you naturally get a ground state for the energy for the electron.Matterwave said:Essentially there is a hierarchy of states that the charges can occupy, and there are no more states below the ground state, so the particles can't decay and collapse into each other. This is more or less taken as fiat, one of the assumptions of the QM theory.
fluidistic said:I don't think it's taken as a fiat nowadays. If we take for granted that the Schrödinger's equation is valid, then you naturally get a ground state for the energy for the electron.
But I agree that you have to take the Schrödinger's equation validity as a fiat. :)
Boundless said:Thank you DaleSpam
tom.stoer said:I don't agree that non-rel. QM with Schrödinger equation provides a full explanation of the stability of atoms. In principle one would have to prove this using QED, but this is out of reach, unfortunately.
The reason I disagree is the following: in classical electrodynamics the electron would radiate away energy as electromagnetic waves; so we have to explain the mechanism which forbids this radiation. In non-rel. QM the electromagnetic field is not a dynamical, quantum mechanical degree of freedom (like the electron) but a fixed, classical field. So the quantum mechanical setup cannot explain why no photons are emitted b/c it does not contain photons by construction! It's an approximation to the full theory obtained by integrating out the photon d.o.f.
So I would say that non-rel. QM provides a strong hint that atoms are stable b/c they have a stable ground state; but this is not a rigorous proof.
Correct - but this is not the point.vanhees71 said:Here, I disagree. Starting with non-relativistic atomic physics is well justified by the fact that it is much simpler than the fully relativistic picture and also a good approximation as long as you study not too heavy atoms.
Particularly the hydrogen atom is a prime example for the fact that the non-relativistic approximation gives an excellent description, at least on a quantitative level.
No. In this model the atom does not radiate b/c it simply can't radiate. There is no radiation degree of freedom! The fact that there is no radiation is not a result but an input of the model!vanhees71 said:Also qualitatively you get an approximately correct answer: The hydrogen atom doesn't radiate in this model, because you are considering a stationary state, i.e., nothing is moving in the atom.
Correct - but again this is not the point. The Dirac eq. does not contain a dynamical photon field, either.vanhees71 said:The right hint is that you get an even better description of the hydrogen spectrum by simply using the Dirac equation in a semiclassical way, i.e., putting an external static Coulomb field for the nucleus. This gives you the correct fine structure of the spectrum.
This is the correct starting point. The remaining problem is that the perturbation series is not well-defined mathematically :-(vanhees71 said:To get this in a full QED treatment the right ansatz is to work in Coulomb gauge, use the soft-photon resummation to justify the Coulomb field as substitute for the proton and then do systematically perturbation theory around this configuration, including the quantized em. field. This leads to a perfect description of the spectrum, including radiative corrections like the Lamb shift. This program has been driven to at least 4 (or even 5?) loops now, and the agreement between theory and experiment is overwhelming.
tom.stoer said:Correct - but this is not the point.
But I think you understand what I want to say: to prove the absence of photons you must not start with a model w/o photons.
This is what I mean. Results derived from perturbation theory do not have the status of a mathematical theorem.vanhees71 said:I disagree with the statement that the perturbative treatment is mathematically not well defined ... It's to be seen as an asymptotic series (in the sense of small coupling constants or small [itex]\hbar[/itex]).
Boundless said:This is a very interesting article I found :
This is a great question! I will give you a little bit of data, and I think you will have an "aha" moment.
Experiments have shown, to date, that the lifetime of an electron is more than 4.6X10^26 years, or essentially infinite. (http://pdg.lbl.gov/2006/tables/lxxx.pdf) Experiments have also shown that the lifetime of a proton is > 1.9X10^29 years, or also essentially infinite. (http://pdg.lbl.gov/2006/tables/bxxx.pdf)
The lifetime of a free neutron, however, is only 885.7 seconds, or 14 minutes and 52 seconds. (http://pdg.lbl.gov/2006/tables/bxxx.pdf) By free, I mean that the neutron is wandering around by itself and is not part of a nucleus. Have that "aha" moment?
What this means, as I'm sure you figured out, is that the free neutrons decayed away in the very early universe before they could "meet up" with a proton and an electron and make a hydrogen with a neutron (also known as a deuterium). In the early universe, so much hydrogen was made that it began to coalesce into clumps due to gravitational attraction. These clumps grew and grew, and eventually, became stars. Inside stars, hydrogen atoms "fuse" to become helium, and then some of the helium fuses with other hydrogens or other atoms to make some of the heavier atoms, on up to iron. The so-called "binding energy" of each nucleus is less and less as you go up in atomic number, meaning it is more energetically favorable to be in that state, again, until you hit iron. See http://en.wikipedia.org/wiki/Nuclear_fusion for more details on how fusion works.
Could this mean that the proton electron don't really have an affinity for each other until the neutron comes into the picture?
tom.stoer said:Think about a potential which does not provide a well-defined ground state, i.e. V(r) ~ 1/rn. There is no radiation in this model, either!
As far as I remember the Hamiltonian for 1/r^n with n≥2 is not well-defined (or self-adjoint) w/o additional constraints.geoduck said:Why doesn't the uncertainty principle provide a ground state for 1/r^n potential? If the particle falls into origin, then it gets localized so it's momentum should get large and be kicked out of origin.
The attraction between positive and negative charges is a fundamental force of nature known as electric force. This force is responsible for holding atoms and molecules together, as well as for creating electric fields.
The strength of the attraction between positive and negative charges is directly proportional to the amount of charge present. This means that the greater the charge, the stronger the attraction will be.
Yes, the attraction between positive and negative charges can be broken by introducing an external force. This force can be strong enough to overcome the electric force and separate the charges.
When positive and negative charges come into contact, they will neutralize each other and cancel out the electric force between them. This can result in the release of energy in the form of heat or light.
The attraction between positive and negative charges decreases as the distance between them increases. This is known as the inverse square law, which states that the force between two charges is inversely proportional to the square of the distance between them.