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Kindergarden question about the Strong Force: its origin

  1. Aug 26, 2008 #1
    (If you dont feel like reading all three paragraphs just skip to the sentence writen in bold.)

    So I am autodidacting myself through the world of (quantum) physics, just the concepts of course, no math and I finally got to a stage where I can use my present knowledge to (partially) deduce some things from the world of physics. At least I hope I did. My question / request for confirmation is this:

    I read about quarks and the different colours of quarks and gluons associated with them. I got that you cannot seperate the warks due to ever increasing strength of the gluon field as you increase the distance between separate quarks and if you did get the quark very far from the others you could end up creating additional quarks "releasing" the pressure in the gluon rubber band. So I made my hypothesis about the origin of strong force in everyday atoms:
    Say you have two protons. You get them close enough for the colour force from one protons quarks to glue together with the quarks of second proton and thus overcoming the electrical repulsion and creating helium isotope. You could knock them appart which would result in creating additional particles (most likely leptons) and causing radiation like in a nuclear reactor (only with much heavier atoms). I thought I had it all figured out but then I started to see many flaws in my idea about Strong Force. The gluon pull would probably be too strong to ever alow this kind of separation in normal natural conditions without giant atom smashers. Also the separation would most likely create leptons and I am not sure if leptons are often found in radiation decay... in short - Id love if someone could explain to me how does strong force between protons and neutrons etc. work and where does it come from.

    Thanks in advance.
  2. jcsd
  3. Aug 26, 2008 #2


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    You are asking a question which has bothered theoretical physicsists for decades :-)

    You can calculate it approximately by using 'lattice gauge theory'. And you can use effective field theory approach aswell.

    Also, He2 is not bound, so when fusing two protons togheter, you'll get deuterium.

    Here is some starting reading, I can give you more advanced stuff if you want.


    And understanding physics without math is not a hit :-(
  4. Aug 26, 2008 #3


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    Most probably, there are even people claiming/asking to understand triangles without math. Not that math is the only needed tool (if you see math as a language, then you can bet you need linguistics somewhere in the foundational game) but for sure it is a prerequisite.
    But again, some people means "algebra" or "arithmetic" when they say math. In such case it could be attempted some geometric/topological way to part of the physics stuff. Not the hardest, as group theory or, er, algebraic geometry can not be done without algebra. But any student going so long will surely be ready to embrace the rest of math corpus. Also, some combinatorics can be presented in a pre-algebraic way.

    (For instance, "equal areas in equal times" does not sound very algebraic. Same with "straight lines at constant velocity if undisturbed". And even same with "ellipses, one of the focus being the source of force field". It is more difficult to grasp the "squares of periods and cubes of what?". )
    Last edited: Aug 26, 2008
  5. Aug 26, 2008 #4
    Problem : the nucleon (proton-neutron) interaction is a residual force from QCD (analogy with van der Walls force between molecule from QED). Whereas QCD is asymptotically free (no force at small distances) the residual nucleon interaction is repulsive at short distances.

    A "baby step" is to say that nucleons exchange mesons (pions, rhos...) equivalent to a Yukawa potential (massive Coulombian). The pion long range (low mass) explains the attraction (pseudo-scalars) whereas the rho short range (higher mass) explains the repulsion (vectors).
  6. Oct 26, 2008 #5


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    Depends in what sense we say something to be "strong" or "not strong"..

    The nuclear force is said to be strong since it has greater attraction then what electromagnetic are replusive (on short distances of course).

    There is also interesting that you are using energy when talking of forces here, the RANGE of the force is crucial here: EM is long ranged, Strong Nuclear force is short ranged, that is why the EM force plays role in nuclear structure physics.

    If you can, look on the effects of EM force in isobaric nuclei C14, N14 and O14, look on lowest lying energy levels.

    So what you have demonstrated bjschaeffer is that the EM force matters since it is long ranged.. weak.. but long ranged.
  7. Oct 26, 2008 #6
    The problem is not to have a long or short range force, the problem is to have enough force to attract the nucleons together or more precisely to have an adequate binding energy comparable to the measurements. My crude calculation gives a plausible value.
  8. Oct 26, 2008 #7


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    No, the strength of nuclear force is several hundereds of time stronger than the eletctromagnetic. The "problem" for the nucleus is that the nuclear force is short ranged. The nuclear force is said to saturate.

    This is covered in probably every Nuclear physics textbook, I suggest you pick one up.

    Using phrases as "electromagnetic nature of the nuclear force" makes me wonder if you've ever studied nuclear physics. The nuclear force is the nuclear force, it has no electromagnetic component. Protons however, interact both through the nuclear force and the Electromagnetic force, neutrons only nuclear force (well, besides this, both n and p interacts with gravity and weak force (protons only under certain circumstances although))

    Another problem, there are no electromagnetic contribution to the deutron bidning energy since there are no electromagnetic force between neutrons and protons.. so the figure 0.68MeV is? = the electrostatic energy between two unit charges separated 1fm? Why is that relevant? it is not relevant at all.

    I have done graduate courses in Nuclear physics, but will cite a good ol wiki page for you:
    "At short distances, the nuclear force is stronger than the Coulomb force; it can overcome the Coulomb repulsion of protons inside the nucleus. However, the Coulomb force between protons has a much larger range and becomes the only significant force between protons when their separation exceeds about 2.5 fm."


    Quantify: "enough force"
  9. Oct 26, 2008 #8
    There is no calculation in Wiki

    The force between nucleons is not measurable, only the binding energy is measurable.
    At a separation of 1 fm, the electrostatic energy is of the same order of magnitude as the binding energy as I have shown earlier.
  10. Oct 26, 2008 #9


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    and as I asked, who cares what the electrostatic separation energy for two point particles separated 1fm is? Iron have binding energy almost 9MeV/nucleon..

    It is very well known that you can't put too many protons in a given A-nucleon configuration. But that just reflects the fact that the nuclear force is short ranged, but stronger than the EM force at those small distances. That is WHY nucleus are bound, if the nuclear force would be weaker at the EM force at all distances, then we wouldnt exist and have this discussion.

    Okay, so what to you measure in scattering experiments? Well, scattering lenghts, which is related to the strength of the potential - hence the force... How do you think one gets the conclusion that the nuclear force is strong but short ranged in the first place? You must test your theories against empiricial data.

    the Nuclear physics book by krane is quite a good starter.. get a copy. It even covers quantum mechanical scattering.
  11. Oct 26, 2008 #10

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    I understand what you are trying to say, and largely agree with you, but the above isn't quite true. There is no electric force between a proton and neutron, but there is a magnetic force. Even though a neutron is neutral, it has a magnetic moment.
  12. Oct 26, 2008 #11


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    Yes, that is absolute true!
  13. Oct 26, 2008 #12
    Semi-empirical binding energy formula...

    I believe what your are philosophically searching for is the nuclear 'liquid drop model'. More specifically, the semi-empirical binding energy formula, which corresponds to the 'residual' strong nuclear force.

    Semi-empirical binding energy formula:
    [tex]E_{B} = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z-1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^{2}}{A} + \delta(A,Z)[/tex]

    Coulomb energy: The electric repulsion between each pair of protons in a nucleus contributes toward decreasing its binding energy.

    The Coulomb energy component of the semi-empirical binding energy formula:
    [tex]E = a_{C} \frac{Z(Z-1)}{A^{1/3}}[/tex]

    Note that with least-squares fit, [tex]a_{C} = 0.714 \; \text{Mev}[/tex], which is in close agreement with bjschaeffer's value of [tex]a_{C} = 0.68 \; \text{Mev}[/tex], with a peer measured range of [tex]a_{C} \propto \int_{0.595}^{0.711} \; \text{Mev}[/tex].

    The basis for this term is the electrostatic repulsion between protons. To a very rough approximation, the nucleus can be considered a sphere of uniform charge density. The potential energy of such a charge distribution can be shown to be:
    [tex]E = \frac{3}{5} \left( \frac{1}{4 \pi \epsilon_{0}} \right) \frac{Q^{2}}{R}[/tex]

    Where Q is the total charge and R is the radius of the sphere. Identifying Q with Ze, and noting as above that the radius is proportional to A^(1 / 3), we get close to the form of the Coulomb term. However, because electrostatic repulsion will only exist for more than one proton, Z^2 becomes Z(Z − 1).

    Empirical nuclear radius:
    [tex]R = r_0 A^{\frac{1}{3}}[/tex]

    [tex]Q = Ze[/tex]
    [tex]Z^2 = Z(Z - 1)[/tex]

    Potential energy of charge distribution:
    [tex]E = \frac{3}{5} \left( \frac{1}{4 \pi \epsilon_{0}} \right) \frac{Q^{2}}{R} = \frac{3}{5} \left( \frac{1}{4 \pi \epsilon_{0}} \right) \frac{(Ze)^{2}}{(r_0 A^{\frac{1}{3}})} = \frac{3 e^2 Z^2}{20 \pi \epsilon_{0} r_0 A^{\frac{1}{3}}} = \frac{3 e^2 Z(Z - 1)}{20 \pi \epsilon_{0} r_0 A^{\frac{1}{3}}} = a_{C} \frac{Z(Z-1)}{A^{1/3}}[/tex]

    Potential energy of charge distribution:
    [tex]\boxed{E = \frac{3 e^2 Z(Z - 1)}{20 \pi \epsilon_{0} r_0 A^{\frac{1}{3}}}}[/tex]

    [tex]\boxed{a_{C} = \frac{3 e^2}{20 \pi \epsilon_{0} r_0}}[/tex]

    [tex]\boxed{a_{C} = 0.691 \; \text{Mev}}[/tex]

    Semi-empirical_mass_formula - Wikipedia
    Nuclear size - Wikipedia
    Last edited: Oct 26, 2008
  14. Oct 27, 2008 #13
    The magnetic force may be attractive, isn't it?
  15. Oct 27, 2008 #14


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    Depends on spin configuration, you can compare situation to first order approximation with the splitting of energy levels in your hydrogen atom which you like to do comparisons with.

    Here is some valuable information about the nucleon nucleon force:
  16. Oct 27, 2008 #15

    Vanadium 50

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    It can be. In the deuteron, it is. But this is not the strong nuclear force (which also has a spin-dependent piece). It's ordinary electromagnetism.
  17. Oct 28, 2008 #16
    This complements the above posts...
    Where does the strong force come from:
    It's believed the strong,weak and electromagnetic force come from a common high energy (unified) force early in the universe when temperatures and energies were really high. The gravitational force was already separated at this point.

    This separation of forces theoretically results from a spontaneous symmetry breaking when the universe underwent a phase transition from a high energy unstable state to the current lower energy but far more stable configuration. I believe the unified force had infinite reach like the electromagnetic force still does. The Higgs boson (particle) which theoretically produces mass is also a consequence of spontaneous symmetry breaking. To read more, try guage theory, Higgs phenomena or spontaneous symmetry breaking or Weinberg Salaam electroweak force.
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