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There is this famous example about the order difference between gravitational force and EM force. All the gravitational force of Earth is just countered by the electromagnetic force between the electrons on the surface of my feet and the ground.

But also wave-function of electrons(fermions) do not overlap each other due to Pauli exclusion principle.

So which one is the true reason of me not flopping inside the earth forever? Is it only one of them(my guess is Pauli exclusion) or is it both?

Qmechanic
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OK, I'll make a complete revision of my original reply since it was quite sloppy.

First, I originally confused two issues that are nevertheless related, I confused stability of matter and the impenetrability of matter.

But, it must be clear that the two questions are related. If I have two chunks of matter of the same type on top of one another, one can't imagine that the explanation for the fact that these chunks don't "fall through" each other would be unrelated to the explanation of why we do not fall through the ground. So, in ultimate analysis, the question is tied to the question of stability of matter.

Now, there are several steps in the problem. To explain the stability of matter, one has to explain why atoms are stable (and before that why nuclei are stable), then one has to explain why aggregates of atoms like solids or liquids can be stable, i.e. why bulk matter is stable. The stability of bulk matter will then serve as the basis to explain why "we can stand on the ground".

Starting with the last step and assuming we already know about the stability of bulk matter, we can imagine that when we exert a pressure on stable bulk matter, we can expect by what it means to be in a stable equilibrium, that the piece of matter would exert an opposing pressure, trying to restore itself to its most stable configuration, provided stresses are not too great. So solving the problem of stability of bulk matter will help us understand what the nature of the restoring force will be.

Now, as is well known, electromagnetic forces can not be the sole explanation. There exist no stable equilibria when there are only electric charges interacting electromagnetically. I won't go through the proof here, but it is accessible to undergraduates, it can be found in the Feynman Lectures, book 2, chapter 5. It's an application of Gauss' law in the static case. The dynamic case only complicates matters in the wrong direction. As we know, accelerated charges radiate energy away, thus an electron orbiting a nucleus would soon fall inward if nothing prevented it, to take the classic example.

Enters Elliott Lieb and his paper 'The stability of matter' that can be easily found online. So, I'll quote a lot from there. It reviews a lot of the results in the field of mathematical physics of the problem of stability of matter.

So what does Lieb essentially say about stability of atoms: that it is a consequence of a principle introduced by Sobolev. Sobolev's inequality states in a mathematically precise form that if one tries to compress the wavefunction anywhere, the kinetic energy will increase. It's a kind of stronger version of the HUP. (Note that at this point, Lieb does not use Pauli's exclusion principle. This is to be expected, take a hydrogen atom, it is stable, since there's only one electron, Pauli's exclusion principle can't be invoked here to explain its stability.)

Then, Lieb goes on to explain the stability of bulk matter by using Sobolev's inequality again. But this time around, he extends the inequality, and takes into account the fact that matter is made up of fermions. So, the Pauli exclusion principle is indeed used. So, again a lower bound for the kinetic energy is found, the interesting thing is that this lower bound is proportional to $N^{5/3}$ where $N$ is the number of fermions. If the particles were not fermions, the proportionality would have been $N$, which we can see by using the previous bound for 1 atom and multiplying by the number of atoms. So it is really the Pauli exclusion principle that contributes the factor $N^{2/3}$.

Lieb goes then to show that this factor is crucial. He uses Thomas-Fermi theory as a relevant approximation of the behaviour of bulk matter to demonstrate this. This is were the analysis becomes very intricate. I have no time to summarize it in more detail. So I'll just say that some theorems about the nature of TF theory are derived, these are then combined in the end to show that the minimum energy or ground state energy of the system is bounded from below. A numerical value for this bound is derived which is −23 Ry/particle, (1 Ry ≈ 13.6 eV).

The important take-away message is though that Fermi statistics or the Pauli exclusion principle is indeed essential to explain the stability of bulk matter.

In Lieb's paper, there is an extra chapter which tackles the question of why matter doesn't explode, instead of implode. The interesting thing is that pure EM is sufficient to answer this question.

woky
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Raskolnikov
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  • I don't understand your first sentence. Electrons are fermions, so they wouldn't "collapse like bosons" ordinarily (ignoring superconductivity for the moment). And degenerate electron gases are an excellent approximation to the behavior of electrons in metals at ordinary temperatures. I certainly think of that as "ordinary matter". – j.c. Nov 18 '10 at 18:46
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    OK, what I wanted to say is that if electrons were bosons instead of fermions, then atoms would still have a minimal radius, courtesy of the Heisenberg uncertainty principle. Your remark about degenerate electron gases in metals is well-received though. I'll adjust my comment accordingly. – Raskolnikov Nov 18 '10 at 18:48
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    I think this answer is quite good but needs to be reformulated a little. After that I am willing to give my vote :-) – Marek Nov 18 '10 at 19:02
  • @Raskolnikov: Could you expand on your statement that the HUP provides a lower bound on atomic size? I don't think I follow the reasoning behind that. – David Z Nov 18 '10 at 19:12
  • @David: virial theorem gives you $T = - V/2$. Writing $V = -{\gamma \over r}$ then means that ${p^2 \over m} = {\gamma \over r}$. Now from HUP we get crude estimate $p \dot r \approx \hbar$ so that smallest $r$ that doesn't violate HUP is about $r \approx {\hbar^2 \over m \gamma} = {\hbar \over m \alpha c}$. This agrees to an order of magnitude with Bohr's radius. – Marek Nov 18 '10 at 20:41
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    Perhaps I'm not understanding the argument. It seems you're saying atoms have a finite size. But the reason I don't fall through the floor isn't because atoms have size, it's because my atoms can't pass in between the floor's atoms. There's something pushing back, whether it's Coulomb repulsion or degeneracy pressure. – Tim Goodman Nov 18 '10 at 21:10
  • Unless you're saying that my atoms have a finite size which is greater than the inter-atomic spacing between floor atoms... and even then, if not for exclusion and Coulomb repulsion, what's to stop two atoms from overlapping? – Tim Goodman Nov 18 '10 at 21:11
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    Well, it's not enough to look at atoms really, so in that sense, I agree my answer is incomplete yet. There's more going on in materials than the atoms, there's binding between atoms, in molecules or in crystals, etc... I think I should expand a bit on my post. The most important work concerning stability of matter has been done by Eliott Lieb and coworkers. I'll see if I can find some comprehensive summary. – Raskolnikov Nov 18 '10 at 22:14
  • @Raskolnikov: so are you going to update your answer or not? My vote awaits :-) – Marek Dec 03 '10 at 14:32
  • Woops, sorry, forgot about it. – Raskolnikov Dec 03 '10 at 14:42
  • Very nice! I wish I could up-vote again :-) And I am glad that the answer agrees with my gut feeling that exclusion principle's contribution is essential. Though as you say, it's probably hard to explain this in simple terms; one has to actually compute the result. Going to look at the paper. – Marek Dec 04 '10 at 14:34
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    Yes, but it's still only the answer to the question of stability of matter, which is only a subproblem within the larger question that the OP asked. I think sigoldberg1 made a good job of analysing the question at a higher level, but it is clear that explaining each step is a huge task in itself. – Raskolnikov Dec 04 '10 at 14:37
  • http://128.112.100.2/~kirkmcd/examples/QM/lieb_rmp_48_553_76.pdf requests a username and password, and the IP doesn't provide any context for what might work, so the link on its own isn't terribly helpful. – Walter Mundt Jun 17 '12 at 17:07
  • I posted that link almost two years ago, I don't remember needing a password back then. I can try to update the link though. – Raskolnikov Jun 17 '12 at 17:27
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    It is NOT essentially Pauli exclusion... you need much higher pressures to get to close-range for Pauli to be dominant. It is electrostatic / van der Waals, and my fellow condensed matter theorists agree. – Chris Gerig Jul 15 '12 at 21:53
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    @ChrisGerig Why are you so sure? Philip Hoffmann, for example, disagrees with you: "A repulsive force, on the other hand, is required in order to keep the atoms from getting too close to each other. [...] The reason for the strong repulsion at short distances is the Pauli exclusion principle.". – mmc Jul 16 '12 at 02:18
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    at SHORT distances, that's the key part.... we never really get to such distances, especially for just standing on the ground or having two objects collide non-relativistically. Pauli exclusion is mainly or stability of matter, i.e. so that a neutron star does not collapse on itself. – Chris Gerig Jul 16 '12 at 02:35
  • @ChrisGerig Please read the whole two first pages of that chapter. It's talking about ordinary equlibrium interatomic distances. – mmc Jul 16 '12 at 02:41
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    I don't get what you don't understand: It's talking about bonding at very small distances. I read it and it's 1) irrelevant to this discussion, 2) consistent with what I have been stating (which is fact). – Chris Gerig Jul 16 '12 at 03:12
  • @ChrisGerig It's talking about ordinary atomic bonding at ordinary distances. But maybe we are misunderstanding each other... Do you agree that interatomic repulsion, the one that is modeled with the $r^{-12}$ term in a typical Lennard Jones potential, is due to "Pauli repulsion"? – mmc Jul 16 '12 at 03:51
  • @ChrisGerig: I'm willing to concede your point. I'm no expert on the matter. But I can understand a reasoned argument, so please, instead of posting comments on a message that's nearly one year old, why don't you try to formulate your own reply with the necessary references? You're free to downvote my answer if it displeases you. – Raskolnikov Jul 16 '12 at 06:46
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    @ChrisGerig: Both here and at http://mathoverflow.net/questions/119495/mathematical-proof-of-the-stability-of-atoms/119528 , you've made arguments in comments that I find hard to puzzle out because they're not spelled out into a systematic answer. I agree with Raskolnikov that it would be very helpful if you could formulate your ideas as an answer. –  May 26 '13 at 19:40
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    @BenCrowell: actually, over time I have come to understand it more and talk with more people (such as Witten himself). The distances we are concerned with are small enough, and electrostatic-repulsion isn't as dominant as I previously understood. It still plays a big part, but so does Pauli. For instance, as long as their is nucleus-repulsion, we can try to make sense of matter without electron-electron repulsion but not without Pauli exclusion. – Chris Gerig May 28 '13 at 14:20
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Holy moly, there are a lot of confused parts of answers out there. Here is a way to begin to sort out the different related principles, physically.

  1. The question currently asks "EM or/and Pauli?". Short answer: Neither, although it is true that electromagnetism is the force involved (rather than the strong or weak nuclear forces, the only other candidates), and it is true that the size of atoms is determined by the uncertainty and exclusion principles, as it were.

  2. Neglect or remove gravity from the situation. Then the question really becomes, "Why are solids, at the earth's crust say (but it doesn't matter), solid, i.e. why do they resist strain, to the quantitative degree that they do?"

  3. We can idealize the problem. Why do single crystals resist strain? We replace rocks and dirt by interlaced microcrystals, or mixtures bound by friction. In other words, why are crystals rigid? Why can't we walk on water until it freezes?

  4. What are the energy scales of the problem? Well the typical force to be considered will be the force required to disrupt the crystal lattice. This is about the thermal energy at melting, = k*T(melting). So for water (since I like to think in electron volts) about 0.025 ev/molecule, for rocks about 10 times as much. This is because the water molecules in an ice crystal are bound by hydrogen bonds with strength about 0.1 e.v. and atoms in quartz are bound by covalent bonds of about 1 ev. So yes, electromagnetism is the involved force. To be specific, the force on the bottom of a shoe, normally about (1kg*9.8 m/sec^2)/cm^2 = 2 x 10 -15 kg m s-2/(the area corresponding to a molecule of water on the surface of the crystal) times the distance such a force must move the molecule (maybe 10-10 m?) = 2 x 10^-25 J or (using 6.25 x 10^18 ev/joule) 1.2 x 10-6 ev gets compared to kT as above. So the force of the shoe walking on ice supplies only about 4.8 x 10-5 of the required energy to melt or deform a crystal of ice. So we don't sink through solid ice.

  5. Here we learn an important lesson. In physics, to talk about "a cause" is to secretly talk about the about the calculations and equations underlying computing the magnitude of the phenomenon. So, Weisskopf, (see below) points out that the same equations with different numbers, this time using the pressure developed at the bottom of a mountain actually does reach the energy scale required to deform or melt crystals of rock (say quartz), and therefore computes the maximal height of mountains on earth (or any planet), using only a few fundamental constants. This then carries over. To compute the maximal height of mountains on white dwarf stars (fanciful, I think), or importantly on the surface of neutron stars (real) use the same principles, but now invoke the exclusion principal, or other forces, since these are now relevant to the energy scales involved. On a more mundane scale and as a check, if we concentrate one's weight on a much smaller area like the blade of an ice skate, to yield a higher force, we indeed can deform ice crystals, which is said to be the reason for the low friction we experience when ice skating.

  6. Back to earth, shoes. At these very small energies, compared to any atomic energy scale, to say nothing of the scales involved in the weak or strong interactions, we can should just consider atoms as single units, not as separate electrons or nuclei. Quantum mechanics is required to allow us to neglect all the higher energy degrees of freedom which have been "frozen out" at the low (300K) average energy of the molecules involved. Weisskopf has a good exposition of this in his popular book "Knowledge and Wonder", see esp chap. 7 on "The quantum ladder"

  7. We are left with a more well defined problem: why do liquids freeze and become solid, rigid crystalline shapes, and why are crystals rigid in the first place? Rephrased: why does lowering the temperature a little bit, lead to "freezing out" of the translational degrees of freedom of individual molecules or atoms, with the simultaneous aquisition of a single rigid global translational phase (lattice) for the positions of all the molecules/atoms in a crystal. In other words, why, after the phase transition, this change in the microscopic symmetry of the material, is there rigidity?

  8. This also suggests the answer to the original question. In order for strain to become large, molecules need to break from the global phase and leave their position in the crystal lattice, and the lattice resists this change cooperatively, i.e. many molecules must get disrupted, so the energy required is large, considerably larger than the energy available from the tiny pressures involved from the pressure from your weight on the sole of your shoe. There is more, but I'll stop here for now.

P.S. All of the Weisskopf popular physics books are terrific

sigoldberg1
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    +1: My line of thinking exactly. But one has to notice that you only provided a strategy to attack the problem rather than a complete answer. I think the best that can be done now is to ask some of these questions separately so that one can focus directly on some well-defined part of the problem. – Marek Dec 04 '10 at 10:44
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    Agreed, love the clarity, but the answer is missing. I will love to +1 this if the answer is completed. – Sklivvz Dec 04 '10 at 13:55
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    This focuses on why solids are nondeformable. I don't think that's the crucial question here. The crucial question is why bulk atomic matter (whether solid or liquid) is stable against compression. –  May 26 '13 at 18:42
  • This detailed analysis of the different aspects inherent in the question is probably crucial for understanding. – Stilez Jun 16 '18 at 20:03
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Ok, I'll bite the bullet and get the downvotes, but my answer is EM.

Why? Well, you can't stand on water and you can't stand on air. The Pauli principle applies to those cases but doesn't make matter solid. It's the (fundamentally) cristalline structure that makes an object solid enought to stand on. This is yes, related to QM (what is not) but certainly EM in nature.


Edit: Marek asked me to specify a bit more and add some meat to my answer. Fair enough.

The question itself is not a well-posed one. Electro-magnetism is quantum mechanical in nature, and it's basically impossible to speak about atoms without speaking about Pauli's principle and EM. Actually one could say that Pauli's principle is at the basis of chemistry, together with EM. So in this sense, what is left to answer? There can't be an atom without either of the two. In this sense the answer is BOTH.

I choose therefore to interpret the question differently: is the exclusion principle acting between the soles of my shoes and the ground, that which is holding me up?

Now, the reason why we can stand on ground is undoubtedly chemical in nature - there are chemical bonds between atoms and molecules in the ground that make it solid (as opposed to, say, liquid). Frozen water is governed by EM, by the fact that the molecule is electrically charged, and not by chemical covalent bonds.

In this sense, for me, EM is probably more important than Pauli's into making something solid (and not liquid or gaseous).

Sklivvz
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    I won't down-vote you but this answer is definitely too short and incomplete. It's not quite obvious how important Pauli's principle is to make solid matter solid. For that you'd have to consider something like bosonic electrons (but just for purposes of inter-molecular interactions; they would still occupy the same orbitals as ordinary electrons so that we could talk about atoms in the first place) and calculate the stability of the lattice in this setting. I have no idea what result would you obtain. But I have a feeling that you would need Pauli to obtain results that agree with nature. – Marek Dec 03 '10 at 16:15
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    I won't down-vote you either, but let it be known that the reason you can't stand on water is because it's a fluid, i.e. the inter-molecular binding is a lot weaker than in solids! – Noldorin Dec 03 '10 at 22:18
  • @Noldorin, you are absolutely correct. – Sklivvz Dec 04 '10 at 09:30
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    I have some trouble with how you mention 3 physical concepts, Pauli's principle, EM, and QM, and then discuss 2. Maybe people actually in physics feel like QM is ubiquitous or that the universe couldn't do without it, but it seems important to me as a major part of the reason that we don't fall through the ground. I don't understand how it lets moving charge (electron) stay there without EM radiation, but I know it does and it wouldn't be possible to live without that. I also have a strange desire to say that QM creates a outward pressure between atoms, although I know that's wrong. – Alan Rominger Nov 05 '11 at 16:28
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    I have some trouble with "you can't stand on water and you can't stand on air". Wouldn't hydraulics & pneumatics be counter-examples? Drop a steel cylindar half-way down an open ended tube that is the same diameter as the inner-diameter of the tube. Fill one side with water then stop that side up. Stand the tube on end, water on bottom. Now you have a steel cylindar sitting on top of water. To me, this is the same (in terms of opposing forces) as standing on solid ground. Or the more general liquid-equivalent to "standing on ground" is simply hydrostatic pressure. No? – CoolHandLouis Aug 15 '16 at 12:35
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I would just like to point out that even if the earth's crust were a liquid, although we couldn't walk on it's surface, the liquid would still be subject to pressure. The only reason one can "fall through" a liquid is because the molecules are moving out of the way; thus, one is not penetrating/passing through the matter, one is simply pushing the matter aside, and it is he fluid nature of the liquid medium that allows this to occur. When you think about the crust in terms of a solid, for in reality it IS a solid, you must hear in mind that atoms are mostly empty space. So, if there is so much empty space, why can't the molecules of your body and the molecules of the earth's surface just sort of slip past one another, allowing you to effectively pass through the matter? Or better still, why can't the Particles of your body actually and physically move through other matter/particles? This is where the PEP comes into play. The electrons as well as protons and neutrons that make up ordinary matter cannot get close enough together to allow this to happen, especially under ordinary circumstances. So yes, the PEP is at work here. Without it, two bits of matter could conceivably occupy the exact same space simultaneously!

As for the "states of matter portion" that is attributed to EM, because of the electromagnetic bonds involved in the chemical makeup of ordinary matter.

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You do not fall through the floor because of both Coulomb repulsion and the Pauli exclusion principle.

Electrons are attracted to atomic nuclei and repelled from each other. Pauli is what requires that the electrons sit in different energy states, and not all in the low-lying, centre-biased s shells overlapping the nucleus (see picture). It is largely these higher states, and the delocalised states involved in compounds, which are responsible for the various types of bonding which hold matter together. The mutual repulsion of electrons between atoms holds them apart, and prevents one object passing through another.

Quantum Electro-Dynamics (containing both Pauli and Coulomb effects) is a sufficient theory to explain the interactions underlying all of Chemistry, Materials Science, and Why Things Don't Fall Through Each Other Studies. That includes all crystal behaviour (metals, semiconductors, etc), all organic behaviour (hydrocarbons, polymers), liquids, gasses, plasma.

Pauli doesn't prevent overlapping

Pauli is responsible for a very interesting effect, called energy level splitting. As two hydrogen atoms are brought together, the electron each atom has becomes 'aware' of the other, and the previously equal energy levels split into pairs; now the electrons are either in a symmetric or an anti-symmetric configuration, and these have slightly different energies. This splitting happens all over the place, and is responsible for the band gap that makes semiconductors possible, spin polarisation, etc. It is this symmetry/antisymmetry that Pauli is all about.

Pauli does not preclude two electron's wavefunctions from overlapping; far from it. Electrons overlap all the time. It only precludes two electrons from being in the same state. So if I have a hydrogen atom, I can put two electrons into the s orbital, so their wavefunctions overlap completely, but only if they are in different spin states.

Matter is not mostly empty

Questions about things falling through things are a natural response to silly offhand remarks about matter being largely empty, with comparisons of an atom to a pea at the length of a football field etc. In reality, it is more reasonable to consider the electrons to each be in one of a number of cloud shapes centred on the nucleus; the lowest energy states actually involve the electron having some probability of being inside the nucleus, and are more likely to be close to the nucleus than further away. So actually the nucleus is surrounded by a dense cloud of electrons (probability-wise), and as two atoms are pushed closer together, these clouds repel each other.

Atomic orbitals

Phil H
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I've even heard different explanations from my physics profs with Phd-s on this question. My EM teacher said it's EM and my statistical mechanics prof said it was PEP. I think myself it's PEP and EM together but the reason behind EM must be clarifed. What keeps an object solid is molecular and crystal bonds, which involve electrons being stuck in a potential well. In order to break the structure we'd have to break these strong bonds which are oriented because of the shape of the orbitals involved. So this accounts for solidity of normal matter like crystals. That being said it's not what keeps us afloat however, it's is an essential requirement of floating nonetheless. Electrons respond to EM fields, so if materials have EM forces on them there must be an EM field present. But materials are generally electrically neutral and so they neither generate nor respond to EM fields. Some of you may say but when you get down to a very small scale there maybe nonzero fields between electrons. Yes but those fields are compensated by the positive charges, if this wasn't so charges would move, especially in conductors to compensate any field. So what's left really is the PEP to account for us not falling through the floor I think.

xalin
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  • Your E&M teacher is definitely wrong. It has been proved that an electromagnetically interacting system of bosons is unstable. This is discussed in Lieb, Rev Mod Phys 48 (1976) 553, p. 563, available at http://www.pas.rochester.edu/~rajeev/phy246/lieb.pdf . Lieb attributes the original proof to Dyson and Lenard, "Ground-state energy of a finite system of charged particles," J Math Phys 8 (1967) 1538. –  May 26 '13 at 18:47
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I wonder if we are not all short-changing the Heisenberg Uncertainty Principle in this discussion. It is certainly the HUP, not electrostatics or Pauli, that stops the hydrogen atom from collapsing. And it is obviously false to say that in the absence of Pauli, all the electrons in bigger atoms would crowd into exactly the same wave Hydrogen ground state wave function...because the Helium atom certainly doesn't do that, and Pauli places no restrictions on the wave function there. In the absence of Pauli, multiple electron atoms would be spherically symmetric but the electron distribution within would still show some structure, as does Helium. (Although it is arguable that the structure is simply an artifact of the electron-basis-state representation, and that in the density matrix representation the charge distribution is spherically symmetric. But that's another story.)

What prevents two hydrogen atoms from passing right through each other? They could if the electron from A could neutralize the proton from B. And it is not Pauli that stops them from doing so, nor electrostatics which positively wants to neutralize the charges, but simply Heisenberg that stops them from doing so.

Marty Green
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  • The Helium atom does exactly that. Its ground state is $1s^2$. – firtree Aug 29 '14 at 05:04
  • Yes, you're right about that. I think what I meant was that it's not exactly the same s-orbital as the hydrogen atom. It's a different shape, because the electrons push each other apart. But regardless, I'm still don't know why I'm the only guy who thinks Heisenberg is more important than Pauli when it comes to stopping us from falling through the floor. – Marty Green Aug 29 '14 at 12:10
  • Heisenberg and Pauli work together on this, don't make them quarrel :-) For example, Heisenberg determines the cell size in the momentum space, and Pauli makes electrons fill the cells two by two. // For the shape of the Helium ground state orbital: it becomes exactly the same if you disregard the antisymmetry of the wave function, which is the same as the absence of Pauli. – firtree Aug 29 '14 at 18:42
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The interaction between the nucleus (the core of the atoms) and the electrons is electromagnetic, but the Pauli exclusion principle keeps them from falling into the core. This is why matter has volume, and why different objects cannot occupy the same point in space.

You can find more details about the Pauli exclusion principle here.

Then there is the interaction between atoms that make up the different structures in nature, the rocks and the whole Earth (solids) beneath you, for instance. In these structures the atoms are bound together, mostly by electromagnetic interactions, and to pass some other atoms between them you have to invest some energy against that bound. For certain objects this energy is a lot.

Think of crystals. Our puny human fingers are way too weak to "pass between" the crystal atoms. If we build some strong machines, those can pass between the atoms in a crystal - that's called cutting. Like when the silicon crystal is sliced in a computer chip factory. Needs a lot of energy...

Then there other objects that are much less bound - you can e.g. pass between the molecules of water, it's called swimming (could be swimming in other liquids too). Those molecules are not interacting with each other much. You still cannot make your atoms occupy the same space as the H or O atoms in the water, and that's because of Pauli. And of course, you are walking around on the surface of the Earth - passing between the atoms of the air, because atoms and molecules in gasses are not bound to each other at all, you just have to move them out of your way (Pauli) but no bounding (EM) to work agains.

So in this case you mentioned, the bounding force between the atoms that make up the material beneath you is stronger than the gravitational force pushing you and the ground together, thus not enough to break those bounds and let you "slide down".

Gergely
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  • Greg, Kivanc is quite obviously extremely familiar with these subjects, you should not explain in such an elementary matter. And I think you miss the actual point. What keeps him from breaking those bonds, i.e. which force/principle keeps those bonds together? – Cem Nov 24 '10 at 05:27
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    Pauli exclusion has nothing to do with why electrons do not reside in the nucleous, it only explain why no more than two of them occupy the 1s orbital. It is the conjugate relationship between position and momentum (i.e. the Heisenberg Uncertainty Principle) that makes the electron orbitals so much more extensive than the nucleus. – dmckee --- ex-moderator kitten Sep 14 '11 at 14:41
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    It's not true in general that atoms are stable because "the Pauli exclusion principle keeps them from falling into the core." If so, then hydrogen would collapse. However, it is true that Fermi statistics are necessary in order to explain the stability of bulk matter. See Lieb, Rev Mod Phys 48 (1976) 553, p. 563, available at pas.rochester.edu/~rajeev/phy246/lieb.pdf . –  May 26 '13 at 18:50
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Coulomb repulsion including the inverse square dependence has been experimentally verified down to nuclear lengths (by Rutherford's gold foil experiment) and even the fm (femtometer) scale (Breton V et al 1991 Phys. Rev. Lett. 66 572–5).

To get 1000 N of force from a mol of electrons (seems like a reasonable order of magnitude for a contact area) we need to get them $ r = \sqrt{10^{23}k e^2 / 10^3} \approx 10^{-4} $m apart. This shows that the Coulomb repulsion is more than sufficient to hold us up before we get even close to the quantum mechanical regime.

More importantly, the Pauli Exclusion Principle is not a force. It simply says two fermions, in this case electrons, cannot occupy the same quantum state. It doesn't say that fermions necessarily push apart if you were to get them very close to the same quantum state (whatever that means), nor does it say that fermionic wavefunctions cannot occupy the same space (as in, x,y,z). I think if you look at the charge distribution in orbitals of any molecule or atom, you will find plenty of non-negligible spatial overlap.

Pete
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    I don't see how you arrived at expression $10^{23}$ in that expression. You know, macroscopic objects are not just a huge charge sitting at a point. Rather, they form lattice that is macroscopically neutral. To calculate a total force between two such lattices is quite non-trivial and your high school calculation doesn't even get close ;-) – Marek Dec 07 '10 at 09:24
  • The point, which you've missed, is the model will be Maxwell not Schrodinger. – Pete Dec 07 '10 at 14:59
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    I am talking just about electrostatics and nothing more. To calculate correct force between to 2D lattices composed of equal amount of $+$ and $-$ charges is certainly something completely different than your plain Coulomb force between two $10^23 e$ charges. – Marek Dec 07 '10 at 17:50
  • You should check out the Journal of Chemical Physics. No one models these kinds of interface interactions using the Pauli Exclusion Principle. Or, as a mol of monopoles for that matter. You might refine the back-of-the-envelope monopole number by modeling the interaction as a mol of dipoles or quadrupoles. And that is what OP is asking. What model can show that I cannot just pass through a wall? My answer is Maxwell because it is clear that there are seriously large, non-negligible E&M forces at distances many orders of magnitude greater than where PEP could be relevant. – Pete Dec 07 '10 at 18:46
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    @Pete: and why do you think the force should be repulsive anyway? Consider two cubic lattices (e.g. $Na^+$ $Cl^-$) and adjoin them in such a way that sodium from one lattice meets chloride from the other lattice. It's pretty clear that there will be a net attractive force. In general, when adjoning two (probably distinct) lattice, even the overall sign is not clear to me. If it is clear to you, please do explain. – Marek Dec 07 '10 at 20:32
  • @Pete: so what makes you stand on the ground in my opinion is not this net force (which can probably be close to zero) but the fact that when you put lattices really close together, they'll start to see each others electron shells and then both EM and PEP will be relevant. – Marek Dec 07 '10 at 20:34
  • http://physics.stackexchange.com/questions/1743/what-is-the-pressure-between-two-electric-dipole-sheets-of-infinite-extent – Pete Dec 08 '10 at 03:05
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    Purely according to electrostatics, all of the charges in a solid want to crash together into a point. Every + wants to annihilate every -. The fact that some object is spread into the volume that it occupies means that charges are separated: a position of high energy. Without something else to prevent all those particles from collapsing, they would merge into a single massive prticle that carries no net charge. – Kaz Jun 27 '13 at 21:39
  • @Kaz: is Heisenberg’s uncertainty principle still a mystery to you? ☺ – Incnis Mrsi Oct 25 '14 at 18:15
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My first answer contained a very sloppy statement that EM interaction is irrelevant because of neutrality of matter which had to be corrected. I consider interaction of two neutral atoms and leave out the question of why some matter is solid and other is not.

  1. Higher-order EM interaction does exist between atoms. What is maybe counterintuitive, it is an attractive interaction called van der Waals force.
  2. Electrons repel each other but manage to stay together even if there are 50 of them in an atom. We could imagine moving two hydrogen atoms together until they (almost) form a helium atom. Electric repulsion is not going to become infinite, the total energy would be in fact lower. Electric repulsion does not prevent one atom going through another.
  3. However, two electrons in the same state just can not exist close to each other (Pauli exclusion) and the atomic orbitals would have to distort to avoid overlap. There is no special force that does that so eventually it comes back to an electromagnetic interaction but the fact is that without Pauli exclusion this interaction would not happen.

I should also mention the name of Lennard-Jones whose potential describes an interaction between two atoms. Note the $r^{-12}$ term for the short-range Pauli repulsion.

gigacyan
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  • Proton-electron attraction is mainly relevant for keeping the atoms stable. But interaction between atoms comes mainly from their electron shells. So it is E-M interaction that keeps you falling through ground. But of course it is implied that atoms are stable and condensed in a solid lattice of your shoes. – Marek Dec 03 '10 at 14:31
  • I meant that since both bodies are electro-neutral, the total coulombic interaction between them, which is (e-e)&(p-p) repulsion and (e-p) attraction, is very small but agree that it exists. However, I will quote Kittel's "Introduction to solid state physics" where he says that "At sufficiently close separations the overlap energy is repulsive, in large part because of the Pauli exclusion principle." – gigacyan Dec 03 '10 at 14:58
  • Well, okay but you are talking about the lattice ions (e.g. $Na^+$ $Cl^-$ in salt). This has nothing to do with protons (atoms themselves are already neutral), it is just about electron shells. I agree that Pauli exclusion (of overlapping electron orbitals) does play role but E-M is still essential for stability. But to say more a lot deeper answer is needed than your three lines, I am afraid :-) – Marek Dec 03 '10 at 15:54
  • This line of thinking ignores many of physical phenomena like stability of solids. EM force is dominant at the scale of atom-atom or molecule-molecule interactions. Although I don't think that my question doesn't boil down to simply stability of solids problem, I also don't think that EM force is simply ignorable just because of net electrical charge is zero. – Kivanc Uyanik Dec 04 '10 at 11:38
  • @Kivanc: EM force is certainly not ignorable just because net charge is zero. There are higher moment (dipole and quadrupole) interactions that give you e.g. chemical bonds and this is of course very relevant. I think the best (short) answer is that you need both. – Marek Dec 04 '10 at 14:33
  • @Marek: Two atoms will attract and not repel each other because the system tries to minimize its energy and the +-+- system is more stable than +--+, this is where van der Waals force comes from. – gigacyan Dec 09 '10 at 10:41
  • @gigacyan: totally untrue. Have you ever heard of double bond? That's precisely what you have written as the second option ;-) Which of the two options is more stable depends on many many things, including (but not exhausting) precise electron configuration of the atom, whether the electrons have the same spin and the temperature of the system (there are phase transitions between single and double bond ordering in some materials). Also, it depends on the surrounding atoms. So your statement is a huge oversimplification. – Marek Dec 09 '10 at 11:14
  • @Marek: covalent bonding is something different, lets stay with physics. I see that people try to invent some induced dipole moments that would make two solid bodies repel from each other. What I try to explain is that two bodies brought close enough will actually stick to each other. – gigacyan Dec 09 '10 at 12:28
  • @gigacyan: doesn't matter whether it's different. What matters is that you've completely ignored it. Along with lots of other effects. And if bonding isn't physics (on elementary level it's just EM + quantum theory) then I don't know what is ;-) By the way, please do elaborate on your last sentence; it's completely contrary to what I observe. E.g. my body doesn't stick to anything else ;-) – Marek Dec 09 '10 at 12:31
  • @Marek: these are $r~{-6}$ interactions and the bodies have to be very close, it works on two polished glass plates. And yes, of course everything is EM and there is no special Pauli force (I corrected my answer) but bringing two atoms together is not just like bringing two electrons together where you can explain just by Coulomb's law why they would never collide. – gigacyan Dec 09 '10 at 13:14
  • @gigacyan: sure, but you have to have two exactly same materials. How is that relevant to anything we discussed before? As for the last sentence: I completely agree and this is what I've been repeating all over under these "answers". – Marek Dec 09 '10 at 15:25