# Radial Dependance of the Strong Nuclear Force

## Tidbits

From an online chat:

• From the study of the spectrum of quarkonium (bound system of quark and antiquark) and the comparison with positronium one finds as potential for the strong force,
 $~V(r)$ $~=$ $~ - \frac{4}{3} \cdot \frac{\alpha_s(r) \hbar c}{r} + kr \, ,$

where, the constant $~k$ determines the field energy per unit length and is called string tension. For short distances this resembles the Coulomb law, while for large distances the $~kr$ factor dominates (confinement). It is important to note that the coupling $~\alpha_s$ also depends on the distance between the quarks.

This formula is valid and in agreement with theoretical predictions only for the quarkonium system and its typical energies and distances. For example charmonium: $~r \approx 0.4~\mathrm{fm}$.

• Of course, the "breaking of the flux tube" has no classical or semi-classical analogue, making this formulation better for hand waving than calculation.
• This is fine for the quark-qark interaction, but people reading this answer should be careful not to interpret it as a nucleon-nucleon interaction.
• At the level of quantum hadron dynamics (i.e., the level of nuclear physics, not the level of particle physics where the real strong force lives) one can talk about a Yukawa potential of the form,
 $~V(r)$ $~=$ $~ - \frac{g^2}{4\pi c^2} \cdot \frac{e^{-mr}}{r} \, ,$

where $~m$ is roughly the pion mass and $~g$ is an effective coupling constant. To get the force related to this you would take the derivative in $~r$.

This is a semi-classical approximation, but it is good enough that Walecka used it briefly in his book.

• The nuclear force is now understood as a residual effect of the even more powerful strong force, or strong interaction, which is the attractive force that binds particles called quarks together, to form the nucleons themselves. This more powerful force is mediated by particles called gluons. Gluons hold quarks together with a force like that of electric charge but of far greater power. Marek is talking of the strong force that binds the quarks within the protons and neutrons. There are charges, called colored charges on the quarks, but protons and neutrons are color neutral. Nuclei are bound by the interplay between the residual strong force, the part that is not shielded by the color neutrality of the nucleons, and the electro magnetic force due to the charge of the protons. That also cannot be simply described. Various potentials are used to calculate nuclear interactions.

## Pointers from Richard Imlay circa 1983

When I asked Richard Imlay (high-energy experimentalist at LSU) for a reference to high-energy physics articles in which quark-quark interactions have been expressed in terms of a radially dependent (e.g., logarithmic ) potential, he pointed me to the following:

Figure 1 from Tuts (1983)

References Cited in Figure Caption:

1. [curve 3] E. Eichten, E. et al. (1980), Phys. Rev. D21, pp. 203-233, Charmonium: Comparison with experiment
2. [curve 2] W. Buchmuller, G. Grunberg, & S.-H. H. Tye (1980), PRL, 45, pp. 103-106, Regge slope and the Λ parameter in quantum chromodynamics: An empirical approach via quarkonia;
W. Buchmuller & S.-H. H. Tye (1981), Phys. Rev. D24, pp. 132-156, Quarkonia and quantum chromodynamics
3. [curve 1] A. Martin (1980), Physics Letters B93, pp. 338-342; A fit of upsilon and charmonium spectra
A. Martin (1981) Physics Letters B100, pp. 511-514, A simultaneous fit of $~b\bar{b}$, $~c\bar{c}$, $~s\bar{s}$ (bcs Pairs) and $~c\bar{s}$ spectra
4. [curve 4] G. Bhanot & S. Rudaz (1978), Physics Letters B78, pp. 119-124, A new potential for quarkonium

Additionally, my handwritten notes, circa 1983, point to:

# Cosmologies

## Standard Presentation

• Derivation of the Friedmann Equations in the context of our discussion of Newtonian free-fall collapse.

Newtonian Description of Pressure-Free Collapse

 $~\biggl( \frac{\dot{R}}{R} \biggr)^2$ $~=$ $~\frac{8}{3}\pi G \rho - \frac{k(R_i, v_i)}{R^2} \, ,$ $~\frac{\ddot{R}}{R}$ $~=$ $~- \frac{4}{3}\pi G \rho \, ,$ where,     $~k(R_i,v_i)$ $~=$ $~\frac{8}{3}\pi G \rho_i R_i^2 - v_i^2 \, .$
• Frieman, Turner & Huterer (2008, ARAA, 46, 385 - 432) provide an excellent, very readable review of dark matter and dark energy in the context of various cosmologies; see also, chapter 29 of Carroll & Ostlie (2007, 2nd Edition). Their equations (2) and (3) are written in the following table — with factors of $~c^2$ inserted to explicitly clarify how the dimensional units are the same for every term in each equation.

Friedmann equations:
Field equations of GR applied to the FRW metric

 $~H^2 = \biggl( \frac{\dot{a}}{a} \biggr)^2$ $~=$ $~\frac{8}{3}\pi G \rho - \frac{k}{a^2} + \frac{\Lambda c^2}{3}\, ,$ $~\frac{\ddot{a}}{a}$ $~=$ $~- \frac{4}{3}\pi G \biggl[\rho + \frac{3p}{c^2} \biggr] + \frac{\Lambda c^2}{3} \, .$

## ASTR4422 Class Notes

Homework set #3 that was assigned to my ASTR4422 class in the spring of 2005 explored how solutions to the Newtonian free-fall collapse problem can be mapped directly to cosmological models of the expanding universe. The stated objective was to match the "closed universe," $~\Omega_0 = 2$ model presented in Figure 27.4 (p. 1230) of the 1st edition of Carroll & Ostlie. (In the spring of 2009, this was assignment #5, and the aim was to match Figure 29.5 from the 2nd edition of Carroll & Ostlie.)

In the free-fall model, the collapse starts from rest at initial radius and density, $~r_0$ and $~\rho_0$, respectively, in which case — see, for example, our discussion of the role of the integration constant

 $~k_i$ $~=$ $~\frac{2G}{r_i} \biggl[ \frac{4}{3} \pi \rho_i r_i^3 \biggr] \, .$

Hence, we have,

 $~H^2 = \biggl( \frac{\dot{R}}{R} \biggr)^2$ $~=$ $~\frac{8}{3}\pi G \rho - \frac{2G}{r_i} \biggl[ \frac{4}{3} \pi \rho_i r_i^3 \biggr] \frac{1}{R^2}$ $~=$ $~\frac{8}{3}\pi G \rho_i \biggl[ \frac{\rho}{\rho_i} - \frac{r_i^2}{R^2} \biggr]$ $~=$ $~\frac{8}{3}\pi G \rho_i \biggl[ \biggl(\frac{r_i}{R}\biggr)^3 - \biggl(\frac{r_i}{R}\biggr)^2 \biggr]$ $~=$ $~\frac{8}{3}\pi G \rho_i \biggl[ \sec^6\zeta - \sec^4\zeta \biggr] \, .$

Now, adopting the terminologies, $~\Omega \equiv \rho/\rho_\mathrm{crit}$ and, for any $~H$, $~\rho_\mathrm{crit} \equiv 3H^2/(8\pi G) ~~\Rightarrow ~~ H^2 = 8\pi G \rho/(3\Omega)$, we have,

 $~\frac{8\pi G \rho}{3\Omega}$ $~=$ $~\frac{8}{3}\pi G \rho_i \biggl[ \sec^6\zeta - \sec^4\zeta \biggr]$ $~\Rightarrow ~~~\frac{1}{\Omega}$ $~=$ $~\frac{\rho_i}{\rho} \biggl[ \sec^6\zeta - \sec^4\zeta \biggr] = 1 - \cos^2\zeta \, .$

Hence, if in the present epoch [denoted by subscript 0], $~\Omega = \Omega_0 = 2$ (as in the Carroll & Ostlie figure that we're trying to match), then in our "free-fall" model, the present epoch occurs at the dimensionless time given by,

 $~1 - \cos^2\zeta_0$ $~=$ $~\frac{1}{2}$ $~\Rightarrow ~~~ \cos^2\zeta_0$ $~=$ $~\frac{1}{2}$ $~\Rightarrow ~~~ \zeta_0$ $~=$ $~\frac{\pi}{4} \, .$

This, in turn, implies that,

 $~H_0^2$ $~=$ $~\frac{8}{3}\pi G \rho_i \biggl[ \sec^6\zeta_0 - \sec^4\zeta_0 \biggr]$ $~=$ $~\frac{8}{3}\pi G \rho_i \biggl[ 2^3 - 2^2\biggr] = \frac{32}{3}\pi G \rho_i$ $~=$ $~ \frac{1}{\tau_\mathrm{ff}^2} \biggl[\frac{3\pi}{32G\rho_i}\biggr] \frac{32}{3}\pi G \rho_i = \biggl(\frac{\pi}{\tau_\mathrm{ff}} \biggr)^2 \, .$

As our parametric solution of the Newtonian free-fall problem details, quite generally we can write,

 $~t$ $~=$ $~\frac{2\tau_\mathrm{ff}}{\pi} \biggl[ \zeta + \frac{1}{2}\sin(2\zeta)\biggr]$ $~=$ $~\frac{2}{\pi} \biggl[\frac{3\pi}{32G\rho_i} \biggr]^{1 / 2} \biggl[ \zeta + \frac{1}{2}\sin(2\zeta)\biggr]$

## With Logarithmic Potential Included

Let's return to the Newtonian expression for the acceleration equation and replace the time-dependent density, $~\rho$, with the time-independent mass, that is,

 $~\frac{\ddot{R}}{R}$ $~=$ $~- \frac{4}{3} ~\pi G\rho = - \frac{GM_R}{R^3}$ $~\Rightarrow ~~~ \ddot{R}$ $~=$ $~- \frac{GM_R}{R^2} \, .$

This is the form of the equation that has been integrated analytically in our separate discussion of Newtonian free-fall collapse. Now, in our published speculation about a modified force-law to explain flat rotation curves, we proposed (see that publication's equation 1) a gravitational acceleration of the form,

 $~\ddot{R}$ $~=$ $~- \frac{GM_R}{R^2} \biggl[1 + \frac{R}{a_\mathrm{T}}\biggr] \, .$

This was intended to represent the modified gravitational acceleration felt by a (massless) test particle moving outside of a point-mass, $~M_R$. When considering a position inside of a spherical mass distribution whose radius, $~R_2 > R$, the first term remains the same because material outside of the location, $~R$, does not exert a net gravitational acceleration. But the second term cannot be treated that way. Following our separate discussion of a 1/r force law, we propose the following acceleration due to such an extended mass source:

 $~\ddot{R}$ $~=$ $~ - \frac{G}{R^2} \biggl[\frac{4}{3}\pi \rho R^3\biggr] - \frac{G}{a_T} \biggl[ \frac{4}{3}\pi\rho R_2\biggr] R \biggl\{ 1 - 3 \sum_{n=1}^{\infty} \biggl( \frac{R}{R_2} \biggr)^{2n} \biggl[(2n-1)(2n+1)(2n+3) \biggr]^{-1} \biggr\} \, .$

Furthermore, let's equate $~R_2$ with the "size of the universe," namely, $~ct$; and let's again define the mass inside of the Lagrangian $~R$ as $~M_R$. Then we have,

 $~\ddot{R}$ $~=$ $~ - \frac{GM_R}{R^2} - \frac{GM_R }{R^2} \biggl( \frac{R_2}{a_T}\biggr) \biggl\{ 1 - 3 \sum_{n=1}^{\infty} \biggl( \frac{R}{R_2} \biggr)^{2n} \biggl[(2n-1)(2n+1)(2n+3) \biggr]^{-1} \biggr\}$ $~=$ $~ - \frac{GM_R}{R^2} - \frac{GM_R }{R^2} \biggl( \frac{ct}{a_T}\biggr) \biggl\{ 1 - 3 \sum_{n=1}^{\infty} \biggl( \frac{R}{ct} \biggr)^{2n} \biggl[(2n-1)(2n+1)(2n+3) \biggr]^{-1} \biggr\} \, .$

# Potentially Useful References

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