﻿ The Radial Charge Distributions of Nucleons
San José State University

applet-magic.com
Thayer Watkins
Silicon Valley,
& the Gateway
to the Rockies
USA

Distributions of Nucleons

## Background

The quarkic model of nucleon structure has the neutron being made up of two Down quarks and one Up quark. The Down quarks have an electrostatic charge of −1/3 and the Up quarks one of +2/3 so the neutron is electrostatically neutral. In this model a proton is made up of two Up quarks and one Down quark and thus has an electrostatic charge of +1.

The density of the charge distribution of neutrons and protons has been determined experimentally by sending beams of electrons into collections of the nucleons. This is shown below.

The charge distribution of the neutron is consistent with the quarkic model but that of a proton is not. It should have a radial range of negative charge just as the one for the neutron has a range of positive charge. The purpose of this material is to use the radial charge distribution of the neutron to construct the radial charge distribution of the proton.

## The Flawed Conventional Model of Quarks

The conventional model take quarks to be point particles. A charged point particle would have infinite energy. There is not enough energy in the entire finite Universe to create even one. Therefore there does not exist even one charged point particle.

Furthermore there is the Larmor Proposition that accelerated or decelerated charges emit electromagnetic radiation. This proposition does not apply to spatially distributed charges but it does apply to point particles. Thus if quarks were point particles they would be emitting electromagnetic energy due to acceleration and deceleration due to thermal aggitation as well as rotation within nucleonic structureS. This emission would be from an infinite source so effectively it woudl be the creation of new energy in the Universe. The notion of quarks as charged point particles is just shear nonsense.

## An Alternate Model of Quarks

Quarks can be spherical shells. Outside of their spherical shell they would have the same effect as if their charges were concentrated at their centers. Thus they are in the nature of being pseudo-point particles.

Nucleons are structures of three concentric spherical shells, each shell being a quark.

Quarks thus come in large, medium and smalll varieties.

From the magnetic moments of neutrons and protons it can be shown that the scale of an Up quark is three quarters the scale of the correspondimg Down quark. The radius of the small Up quark is approximately 1/4 fermi. The small Down quark is thus spread over the range from 0 to (4/3)(1/4)=(1/3) fermi.

Let ρU(r) and ρD(r) be the radial charge density functions for the Up and Down quarks, respectively. Then the integral conditions are

#### ∫01/4ρU(r)(4πr²)dr = (2/3) ∫01/3ρD(r)(4πr²)dr = −(1/3)

Consider a change in the variable of integration in the first equation from r to z, where z=4r and hence r=z/4. The integral condition becomes

#### ∫01ρU(z/4)(4πz²/64)dz = (2/3) or, equivalently ∫01(3/128)ρU(z/4)(4πz²)dz = 1

Now consider a change in the variable of integration in the second equation from r to z, where z=3r and hence r=z/3. The second integral condition becomes

#### ∫01ρD(z/3)(4πz²/27)dz = −(1/3) or, equivalently ∫01(−1/9)ρD(z/3)(4πz²)dz = 1

The constraints will be simultaneously satisfied if their integrands are identically equal; i.e.,

#### (−1/9)ρD(z/3) = (3/128)ρU(z/4) or, equivalently ρD(z/3) = (−27/128)ρU(z/4)

Let z/3=s. Then z/4=3s/4=(3/4)s. Thus

#### ρD(s) = (−27/128)ρU((3/4)s)

In a neutron the two Down quarks are spread over a range from (1/4) fermi to 1.1133 fermi. The upper limit can be taken to be ∞ with the charge density beyond 1.1133 fermi being zero. The constraint is then

#### ∫1/4∞ρD(r)(4πr²)dr = −(2/3)

A change in the variable of integration from r to z where z=4r and r=z/4 gives

#### ∫1∞ρD(z/4)((4πz²)/64)dz = −(2/3) and hence ∫1∞(−3/128)ρD(z/4)((4πz²)/64)dz = 1

In a proton the two Up quarks are spread over a range from (1/3) fermi to 0.84 fermi. The constraint to be satisfied is thus

#### ∫1/3∞ρU(r)(4πr²)dr = (4/3)

A change in the variable of integration from r to z where z=3r and r=z/3 gives

#### ∫1∞ρU(z/3)((4πz²)/27)dz = (4/3) and hence ∫1∞(1/36)ρU(z/3)(4πz²)dz = 1

The two constraints will be simultaneously satisfied if their integrands are identically equal; i.e.;

#### (1/36)ρU(z/3)(4πz²) = (−3/128)ρD(z/4)(4πz²) or, equivalently ρU(z/3) = (−108/128)ρD(z/4)

Let z/3=s. Then z/4=3s/4=(3/4)s. Thus

#### ρU(s) = (−108/128)ρD((3/4)s)

Here are the results of the above constructions

## Scale of the Medium Size Quarks

The radius of the large Up quark is 0.84 fermi and that of the small Up quark is about 0.25 fermi. The radius of the medium Up quark is likely to be the geometric mean of these two values; i.e., 0.46 fermi. The radius of the large Down quark is 1.11331 fermi and that of the small Down quark is about 1/3 fermi. The geometric mean of these two values is 0.61 fermi. Thus the radius of the medium Down quark is about 0.61 fermi.