Is there a 'classical' equation for the strong force?

  • #1

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Sorry if this is a bit stupid, but I have absolutely no grasp on chromodynamics or really any nucleic forces...

Is there a 'classical' way of expressing the strong force in an equation? By classical, I mean extremely simple, approximate way of describing it for two point particles. Like a classical approximation for General Relativity is Newtonian Gravity, F1=F2=Gm1m2/d2, and a classical approximation for electrodynamics is F1=F2=Gq1q2/d2. Is there any sort of way of describing the strong force like that or is it simply too complex to even begin? This doesn't have to be accurate or anything thing, it's just for my understanding, so feel free to cut corners and glide over details.
 

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  • #2
Bill_K
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Ok, crudely and inaccurately, the color force between two quarks has been compared to a harmonic oscillator potential, so that the attraction between them increases as the distance between them increases and they can never escape to infinity.
 
  • #3
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I have such an equation in my book, I will post it in the moment. It was a sum of 2 simple terms. I haven't ever seen a derivation of this equation, though and this is indeed an enigmatic topic. Perhaps the strong force theory is too hard to make such computations.
 
  • #4
Sorry if this is a bit stupid, but I have absolutely no grasp on chromodynamics or really any nucleic forces...

Is there a 'classical' way of expressing the strong force in an equation? By classical, I mean extremely simple, approximate way of describing it for two point particles. Like a classical approximation for General Relativity is Newtonian Gravity, F1=F2=Gm1m2/d2, and a classical approximation for electrodynamics is F1=F2=Gq1q2/d2. Is there any sort of way of describing the strong force like that or is it simply too complex to even begin? This doesn't have to be accurate or anything thing, it's just for my understanding, so feel free to cut corners and glide over details.
Coulomb's law is an approximation? I'm pretty sure it's been proven accurate up to ridiculously small deviations in the power of 2.
 
  • #5
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OK, so the promised equation.

The potential of quark-antiquark interaction is:

[tex]V = - \frac{4}{3} \frac{\alpha}{r} + k r[/tex]

Where [tex]\alpha[/tex] is a coupling constant. As the book reads: the first term comes from single gluon exchange and is similar to Coulomb potential. The second term comes from gluon-gluon interaction, the field force lines attract each other and form a "string".

Donald H. Perkins, "Introduction to high energy physics", chapter 2.7.
 
  • #6
Thanks. To everyone. Those were all a lot better answers than I've received before.
 
  • #7
tom.stoer
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A short remark: the Coulomb potential formula (plus tiny corrections) can be derived strictly in QED: 1/r is nothing else but the Fourier transform of 1/k² which is the propagator of massless particles w/o self interactions. The problem with QCD is that the full propagator is not known analytically due to the self interactions of the gluons; therefore there is are rather complicated object which eventually leads to something like a "potential" with one linear term ~r, but afaik there is no strict derivation of this ~r behavior; instead one has to use e.g. lattice gauge simulations.
 
  • #8
kith
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What about the weak force? Is there a similar approximate law for the interaction of two neutrinos?
 
  • #9
tom.stoer
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Not in the sense of a potential but in the sense of a four-fermion interaction; see "Fermi interaction"
 
  • #10
Ben Niehoff
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One can certainly define classical non-abelian gauge theories. In order to compute anything, you'll have to use finite-difference, discrete time-domain analysis. That is, at every time step, you'll need to keep track of the values of the electric and magnetic fields at every point in space. Note also that classical non-abelian gauge theory is a terrible approximation to the quantum theory...in the low-energy regime, quantum effects totally dominate. However, you can probably still make pretty pictures with it.

The classical QCD equations are as follows. Note that everything here is matrix-valued. For example, the electric field [itex]\vec E[/itex] takes values in the SU(3) Lie algebra, a basis of which is given by the Gell-Mann matrices. So at every point in space, [itex]\vec E[/itex] has three vector components, each of which is a 3x3 complex matrix. Similarly for every other symbol in these equations. A decent-sized simulation will have to store a ton of data and do a lot of matrix multiplication. But anyway:

First the definitions of the Yang-Mills fields:

[tex]\begin{align*}\vec E &= -\nabla \phi + \frac{\partial \vec A}{\partial t} + \phi \vec A - \vec A \phi \\ \vec B &= \nabla \times \vec A + \vec A \times \vec A\end{align*}[/tex]

Next the Yang-Mills equations:

[tex]\begin{align*} \nabla \cdot \vec E + \vec A \cdot \vec E - \vec E \cdot \vec A &= \rho \\ \nabla \times \vec E + \vec A \times \vec E + \vec E \times \vec A &= -\frac{\partial \vec B}{\partial t} - \phi \vec B + \vec B \phi \\ \nabla \cdot \vec B + \vec A \cdot \vec B - \vec B \cdot \vec A &= 0 \\ \nabla \times \vec B + \vec A \times \vec B + \vec B \times \vec A &= \vec J + \frac{\partial \vec E}{\partial t} + \phi \vec E - \vec E \phi \end{align*}[/tex]

And finally the continuity equation,

[tex]\nabla \cdot \vec J + \vec A \cdot \vec J - \vec J \cdot \vec A + \frac{\partial \rho}{\partial t} + \phi \rho - \rho \phi = 0[/tex]

Note that ALL of these fields are in the adjoint of SU(3); this model does not contain quarks (the charge density [itex]\rho[/itex] and current [itex]\vec J[/itex] are objects that would be constructed out of quark bilinears in the quantum theory).

Also, note that [itex]\rho = \text{const}[/itex] is NOT generically a solution of the continuity equation, so you cannot assign a gauge-invariant notion of "charge" in all cases...the charge of a given object will generally change with respect to time, depending on the background field the object is placed in (this is because of the self-interaction of gluons). Charge is gauge-covariant, but not gauge-invariant.
 
  • #11
tom.stoer
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Also, note that [itex]\rho = \text{const}[/itex] is NOT generically a solution of the continuity equation, so you cannot assign a gauge-invariant notion of "charge" in all cases...the charge of a given object will generally change with respect to time, depending on the background field the object is placed in (this is because of the self-interaction of gluons). Charge is gauge-covariant, but not gauge-invariant.
One can show strictly based on the equations of QCD that physical states have to be color singulet states, i.e. Qa|phys> = 0.
 
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  • #12
Ben Niehoff
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One can show strictly based on the equations of QCD that physical states have to be color singulet states, i.e. Qa|phys> = 0.
I agree, but we are talking about classical gauge theory. In the classical theory, charged objects can exist, but the charge is not constant in time. It changes in a complicated way (due to the fact that the gauge fields are also charged).

Furthermore, even in the quantum theory, the fact that a given quantum state is a color singlet does not mean that the state is color-neutral everywhere. It just means that the total color charge, integrated over all space, is zero.

By the way, I am talking about the free charge [itex]\rho[/itex]. The gauge fields are also charged, so even if there are charged objects, it may still be that the total charge in all space is zero.
 
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  • #13
tom.stoer
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I agree, but we are talking about classical gauge theory. In the classical theory, charged objects can exist, but the charge is not constant in time. It changes in a complicated way (due to the fact that the gauge fields are also charged).
You can derive a classical equation which reads dQa/dt = 0 simply by integrating the continuity equation of classical QCD.

Furthermore, even in the quantum theory, the fact that a given quantum state is a color singlet does not mean that the state is color-neutral everywhere.
For a physical state there is no meaning of "everywhere"; but I think I know what you mean; we have Qa|phys> = 0 but not ja0(x)|phys> = 0

It just means that the total color charge, integrated over all space, is zero.
Yes, that's the meaning of charge both in classical and in quantum field theory.

By the way, I am talking about the free charge [itex]\rho[/itex].
I am always talking about total color current density and total color charge which consists of color carried by quarks and gluons. It does not make much sense to seperate these two contributions.
 

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