How do electrons keep out of the nucleus?

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Electrons do not crash into the nucleus due to the balance of potential and kinetic energy, as described by quantum mechanics. The Coulombic interaction creates an equilibrium where electrons in orbitals have lower potential energy but higher kinetic energy, preventing them from falling into the nucleus. The analogy of planets orbiting the sun is misleading, as atomic orbitals do not imply constant velocity or well-defined trajectories. Instead, electrons are better understood as wavefunctions that are spread out around the nucleus, lacking precise positions until measured. The discussion emphasizes the need for a quantum mechanical explanation for electron behavior, moving beyond classical analogies.
  • #31
Thanx Marlow and i will goodle it to find out about how they prove fluctuations must exist and how they actually do exist. But my question is if a field exist in QFT which is defined as being massless, not comprised of particles and just force, then what is formula to justify that? How can QFT say that fields aren't particles if they don't have a formula to justify that force, which is what fields are, doesn't have to be made of particles i.e. have mass.
 
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  • #32
Dragongod said:
Thanx Marlow and i will goodle it to find out about how they prove fluctuations must exist and how they actually do exist. But my question is if a field exist in QFT which is defined as being massless, not comprised of particles and just force, then what is formula to justify that? How can QFT say that fields aren't particles if they don't have a formula to justify that force, which is what fields are, doesn't have to be made of particles i.e. have mass.

Look the problem really is that you are thinking too much in classical terms here. Now, i don't want to explain all this because i want to go watch TV but i really urge you too browse through my journal. You'll find many texts there as to why fields are actually used and how particles arise as fluctuations of these fields. You will also find many links to online QFT-courses that are reliable. Other then repeating myself i urge you to read those before making such claims. I am not saying you are wrong to ask those questions, it's just that you really need to know what you are dealing with, prior to start making those claims

regards
marlon
 
  • #33
I have looked at your journal, maybe i didn't look hard enough, but i didn't really find an answer to my question. I don't care much about the fluctuation thing. My question is if QFT uses the label FIELD and defines a field as having force without mass, then there should be a formula in QFT that represents this relationship. As far as i know, there is no formula in QFT that explicitly describes what FORCE is. If there is please let me know.
 
  • #34
There's no such thing as force:neither in special relativity*,nor in quantum mechanics,ergo not in QFT.


Daniel.

P.S.*U could build in relativistic physics a 4 vector f^{\mu}=mw^{\mu} [/tex],using the acceleration 4 vector,however it's not really used...(okay,2 examples i can think of:Lorentz 4-force and Abraham-Lorentz 4-force,however the name & the notation f^{\mu} doesn't appear).
 
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  • #35
See I knew it. That was my whole point. How can they say a field has force if they don't have a definition of force! Does anyone see a little problem with that?

"In QFT, particles arise as fluctuations of fields but the fields themselves ARE NOT particles. Particles arise (and also forces) as actual vinrations of these fields."
-----how can QFT use the word field without saying what it is and how its possible for it to exist without particles. IT DOESN'T MAKE SENSE
 
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  • #36
A field (classical em.one) has both momentum & energy.It has a 4 tensor:energy-momentum.9 of its 16 components form the famous Maxwell tensor and describe the ability of the classical em field to transmit momentum...

Daniel.
 
  • #37
Dragongod said:
See I knew it. That was my whole point. How can they say a field has force if they don't have a definition of force! Does anyone see a little problem with that?

"In QFT, particles arise as fluctuations of fields but the fields themselves ARE NOT particles. Particles arise (and also forces) as actual vinrations of these fields."
-----how can QFT use the word field without saying what it is and how its possible for it to exist without particles. IT DOESN'T MAKE SENSE

It does.A classical em.field can exist in vacuum.Please study the classical theory of fields.A purely relativistic theory.You'll understand many things.

Daniel.
 
  • #38
I read a few thing on classical theory of fields. It still doesn't make sense to me. It never answers what is a vacuum. It gives a definition but it doesn't ANSWER what it is(how its possible to exist and what its properties are and how that fits in with the rest of the laws of physics). Also it doesn't explain how a Field can have momentum and energy but somehow not have mass. I don't get how they justify that.
 
  • #39
maybe i didn't read well enough but if it does explain how a field has that ability (can exert momentum and have energy without mass) can you explain to me how.
 
  • #40
Einstein's formula can account for massless particles (or fields)

E^{2}=m^{2}c^{4}+\left|\vec{p}\right|^{2}c^{4}

Daniel.

P.S.Mass equal zero doesn't mean energy =0,nor momentum =0.
 
  • #41
E^2 = m^2*c^2 + [p]^2*c^2? does this translate to Energy squared equals mass squared multiplied by the speed of light squared plus (momentum) squared multiplied by the speed of light squared. If it does then all its really saying is that energy equals mass and momentum. Therefore it would be proving that a Field cannot exist without mass i.e. particles.
 
  • #42
Particles can be massless.Period.Photons,gluons,gravitons and their SUSY partners are all massless...

Daniel.
 
  • #43
LOL they are not massless. Photons are said to have relativistic mass for some reason that i don't remember but its mass nonetheless. Glouns do have mass they are the particles that generate the force to keep leptons and quarks "stable." And as for gravitons, as far as i know, nobody really discusses or have proved whether or not they do or don't have mass. As far as i know its still very controversial as to whether they actually exist. According to me, I do believe gravitons exist but as far as science goes they still debate it.
 
  • #44
Nope.Please read.You're definitely out of it completely.

Daniel.
 
  • #45
HAHAH if you say so. but that formula you posted in no way shows that fields can exist without particles.
 
  • #46
Classical fields exist without particles...Period.


Daniel.

P.S.Classical fields are irreductible representations of the \mbox{SO(3,1)}.Typically,functions

\mathcal{Q}^{a}:\mathbb{M}_{4}\rightarrow \mathcal{A}

,where \mathbb{M}_{4} is the flat Minkowski space and \mathcal{A} is an algebra...
 
  • #47
thanx for the formulas but what i am trying to get at is that these formulas are based on assumptions. None of the theories have actually been "observed" such as Minkowski space. Its completely based on space-time, a term Einstein coined but didn't actually prove.
 
  • #48
It never answers what is a vacuum.

Say what? I suppose you need a definition of "is" aswell? It's empty space, devoid of matter, nothingness. What is a force? Something that accelerates a mass.

LOL they are not massless.

He means they have no rest mass. Techniquely they just have momentum because having mass implies that they have infite energy because it moves at the speed of light. Plus the term "mass" is simply a quality of something that resists change in motion. Since light always moves at c it is hard to say it has mass.

If it does then all its really saying is that energy equals mass and momentum. Therefore it would be proving that a Field cannot exist without mass i.e. particles.

No, from your tone I getting the idea that you think mass and momentum are one and the same. Mass is a quality that resists change in motion. While momentum is the product of mass and velocity.
 
  • #49
ZapperZ said:
Well, since we are trying to "be careful" here, let's also make sure we be extra careful in saying that the atomic orbitals somehow implies a "velocity" or speed of anything. It doesn't. By saying such things, we are already implicitly implying a well-defined charged particle moving around. You don't have such things until a position measurement is done. Before then, an electron in an s-orbital, for example, has no well-defined position and identity. Rather, based on the wavefunction alone, it is "spread out" in a uniform sphere around the nucleus. So the electron is everywhere simultaneously (which is connected to the Schrodinger Cat-type puzzlement - another illustration that things in QM are interconnected). This is how we get an angular momentum of zero for the s-orbital - from the geometry of the orbital itself.

This is another illustration where our social language can cause many confusion in trying to describe things that have no linguistic equivalent. As soon as we say "electron moves in an orbit", a whole range of implications kick in. We automatically imply that there is this well-defined object that we can track along the way and moving in a well-defined trajectory. QM implies no such thing, at least as far as atomic orbitals are concerned. We have seen a whole zoo of evidence where an "electron" can simultaneously spread itself into many locations to produce unclassical effects (bonding-antibonding bands, etc.) Zz.

>> You don't have such things until a position measurement is done. Before then, an electron in an s-orbital, for example, has no well-defined position and identity.

Is this the so-called Copenhagen interpretation? Why is the Copenhagen interpretation being taken for granted here?
 
  • #50
feynmann said:
>> You don't have such things until a position measurement is done. Before then, an electron in an s-orbital, for example, has no well-defined position and identity.

Is this the so-called Copenhagen interpretation? Why is the Copenhagen interpretation being taken for granted here?

Don't take it for granted. Why don't you find <x> for the such a situation and draw up your own conclusion?

Zz.
 
  • #51
RE question on how do electrons keep out of the nucleus: Quantum mechanics allows only certain radial and angular distributions of electrons in the central Coulomb field of a nucleus. These are defined by integer quantum numbers n and l (l = little L), with l <= n-1. The probability distribution of the n=1, l=0 state has a very very small (but non zero) probability of being inside the nucleus at any given instant in time.

Some nuclei are radioactive. For nuclei lighter than lead, the radioactvity is mostly beta decay (positron and electron decay, with neutrinos) when the nucleus is either neutron deficient or neutron excessive. Sometimes a proton "wants" to decay to a neutron, but does not have enough free energy to create both a neutron and a positron. In this case it will capture a n=1. l=0 electron (K shell) (see above), emit a neutrino, and change to a neutron. This is called K capture or electron capture. An example is beryllium 7, which decays to lithium 7 with a half life of about 53 days.

So the electron does not always keep out of the nucleus, but it needs to stay away from hungry protons.
 
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  • #52
Bob S said:
The probability distribution of the n=1, l=0 state has a very very small (but non zero) probability of being inside the nucleus at any given instant in time.

Actually the nucleus (r=0) is the most probable location in space of any. But not the most probable radius.
 
  • #53
aren't there energy levels that the electrons can stay in, and it can only go from one energy level to another
 
  • #54
Vals509 said:
aren't there energy levels that the electrons can stay in, and it can only go from one energy level to another

Yes, and the s-type orbitals have a non-zero value at the nucleus, so any ground-state atom will have a non-zero electron density at the nucleus.

There's also a cusp in the density at that point, since the coulomb potential 1/r has a singularity there.
An important result (Kato's theorem) states that, for a closed shell atom, that the spherical average of the density gradient at the nucleus is:

\frac{\partial\rho(r)}{\partial r}|_{r=0}=-\frac{2Z}{a_0}\rho(r=0)

Which is something density-functional theorists spend a great deal of time pondering about, because it's one of very few exactly known properties of the electronic density.
 

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