Four fundamental forces. Which one is this?

Antiphon
Messages
1,685
Reaction score
4
If you slowly drop an electron onto a proton, you will form a hydrogen atom. In the lowest energy state, it's stable.

Why isn't the lowest energy state an electrostatic-driven collapse of proton and electron?

More specifically, which of the four fundamental forces pushes back on the electron when forced near the proton?

It's not the strong, electric or gravitational forces That only leaves the weak force. But the interaction range for the weak force is much shorter than that.

I assume it is a force that can store and release potential energy or it couldn't oppose the electrostatic attraction.
 
Physics news on Phys.org
Avodyne said:
It's the uncertainty principle that prevents the collapse of the electron onto the proton.
That's one way of thinking about it, but not the only one. In the Bohmian interpretation of QM, this effect is due to the quantum potential, i.e. a quantum force that has no a classical analogue. This force depends not only on the positions of the particles, but also on the wave function.
 
Not enough energy. The proton's mass is 938.2 MeV, the electron's is 0.5 MeV. Together that's 938.7. But the neutron's mass is 939.5. Sure, if you slam 'em together with > 0.8 MeV, part of the time they'll form a neutron and emit a neutrino, but it can't happen at rest, solely on energetic grounds.
 
I like Serena said:
I believe the Pauli exclusion principle prevents collapse.

Actually I started a similar thread a couple of months ago.
The best answer I got was a post with a list of similar threads right here on PF:
https://www.physicsforums.com/showpost.php?p=3232991&postcount=19

Thanks. From the last link on the page I Qote Borek:
So if Pauli exclusion a force? No, it's just a constraint on allowed QM states, not a result of a potential energy term. Does it look, feel, and smell like a force? Yes, because it causes a violation of Newton I, and we are trained by experience to call every violation of Newton I a force.

I see that; for example throwing a ball on a ferris wheel. You see fictitious forces. The difference here is that I can actually put a lot if energy into this fictitious force. In that case, what field is stressed with the energy?

I can for example lift a mass against gravity and store energy in the gravitational field.

By squeezing two charged spheres together you can introduce energy into the electric field, stored mostly in the space between the charges.

When forcing an electron to be closer to a proton than it would seek to be, you are putting energy into a conservative field. You could actually use this in principle to make a spring. When you wind that spring and run a clock with it, where is the energy stored?

I deliberately chose proton and electron to avoid the discussion about two electrons sharing quantum numbers.

Edit: based on the threads in the linked post I have to conclude it's the electromagnetic field operating but in some other quantum context such as the way the electrons of an atom don't radiate except in radiative transition.
 
Last edited:
Antiphon said:
I deliberately chose proton and electron to avoid the discussion about two electrons sharing quantum numbers.

Good point!
So it wouldn't be the Pauli exclusion principle here.
Antiphon said:
Edit: based on the threads in the linked post I have to conclude it's the electromagnetic field operating but in some other quantum context such as the way the electrons of an atom don't radiate except in radiative transition.

I was just reading up on beta-decay, positron-emission, and electron-capture (see http://en.wikipedia.org/wiki/Beta_decay" ).
In this wiki article it states that it is the weak interaction that causes a neutron to decay into a proton and an electron.
However, it does not become clear, why a neutron would be a "higher" energy state.

Edit: actually, it says that a down quark decays into an up quark, which has a lower energy-mass.
 
Last edited by a moderator:
How is this view on it?

I would say that there isn't a "force" causing the hydrogen atom to be stable. I see it as a consequence of the fact that a Neutron is less stable than a Proton. Put simply, a particle is likely to decay into a more stable particle. Until it is more favorable to form a neutron instead of a proton, the proton will not combine with an electron.

For example, let's look at a neutron star. The immense force of gravity makes it more favorable to combine the electrons and protons together into Neutrons rather than to resist gravity using electron degeneracy.
 
Drakkith said:
How is this view on it?

I would say that there isn't a "force" causing the hydrogen atom to be stable. I see it as a consequence of the fact that a Neutron is less stable than a Proton. Put simply, a particle is likely to decay into a more stable particle. Until it is more favorable to form a neutron instead of a proton, the proton will not combine with an electron.

For example, let's look at a neutron star. The immense force of gravity makes it more favorable to combine the electrons and protons together into Neutrons rather than to resist gravity using electron degeneracy.

Interesting, but I don't think that's it at all. The H-atom is stable because it represents the quantum mechanical ground state of the coulomb potential between a proton and electron. It cannot relax because there is nowhere for it to go. As far as I can tell there is no need to involve the weak force or strong force to explain this, but perhaps that is only because I am thinking of the non-relativistic version of the problem.

Now, that doesn't mean your analysis above is wrong .. it certainly isn't. I just don't think it answers the question in the original post.
 
  • #10
Yeah, I hear you Spectra. I'm not sure what can actually answer the question.
 
  • #11
I'll support the idea that its the EM field with a historical argument. The weak and strong forces weren't known in the early days of QM. But pretty good solutions to the hydrogen atom can be had using only the coulomb potentials and Schroedingers equation. From these solutions you already have the ground state of the H atom popping out. If another field were acting here besides the EM field, it wouldn't be in this model and we'd expect a poor representation of the H atom to result, no?

In other words I'm agreeing with Spectracat now.
 
Back
Top