Perhaps there is a case for NO Quarks?
The notion of quarks with partial charges constituting the proton is
easily replaced by a simple particle concept that creates the effect
of partial charges, but makes them unable to be removed from a
particle.  The concept also shows both the electron and proton are
comprised of components of the same energy, and a simple equation is
given for the proton-electron mass ratio.

Consider the proton as a self trapped EM-like wave that is propagating
in a helical-toroidal path.  The wave is of spin-1/2, i.e. E = hbar.c/
2r, and propagates twice around the torus axis of symmetry to form a
quantum state and thus has 4pi symmetry.  The rotating E vector of the
wave projects everywhere radial from the 'boundary' of the torus and
the particle thus appears as a point charge to a distant observer.
The E field is also rotating around the torus equatorially (i.e. about
the axis of symmetry), and produces an orthogonal magnetic field,
creating the effect of a magnetic dipole.
Due to the waves double pass the wave energy in the rotating frame is
equivalent to energy hbar.c/r, but due to its being wound around the
torus the particle 'boundary' curvature is 137 times greater than the
energy wavelength would imply, and the particle volume is lower by
alpha. (= 1/137) the fine structure constant.  The lower particle
curvature effectively reduces the localized potential to alpha.hbar.c,
creating the effect of unit charge.

What does this have to do with quarks?
The boundary of a torus has a curvature about its z axes that differs
from those about the x and y axes.  For this particle the curvature
about the z axis is decreased by alpha^1/3 as the radius is increased
by the same factor due to the long path length around the helix, and
is negative, i.e. concave.  The convex curvature about the other two
axes is increased as the radius is decreased by alpha^2/3 for each.
The total particle volume is therefore decreased by
(alpha^2/3.alpha^2/3.alpha^1/3) = alpha.
With the same electric flux density emanating from every where on the
torus 'boundary', in the near field the charge of the particle will
appear different depending on the proximate curvature.  About the x
and y axes the particle will appear to have a charge intensity of
alpha^2/3.hbar.c.  About the z axis it will appear to have a charge of
alpha^-1/3. hbar.c.  Hence the particle overall will appear to have
unit charge to a distant observer, but appears as partial charges of
2/3, 2/3, -1/3 in the near field.
        Obviously this concept shows the quark partial charge effect without
there being any individual components or partial charges, which means
there is no possibility of separating a 'quark' from the particle as
quarks do not exist except as a convenient mathematical concept.

What does this all mean?
Consider a quantum fluctuation in 3-space that produces three spin-1/2
quantum loops, one about each of three orthogonal axes, and each of
about 6.8MeV.  The three loops separate into two particles, a pair of
loops and a single loop.  The single loop falls into a relativistic
state where its volume is decreased by alpha and the curvature of the
'boundary' is effectively decreased to the same degree, i.e. the
apparent mass relates to the curvature so the evident loop energy
increases by 137.  For a single the single loop to localize in 3-space
(i.e. become three dimensional) it must rotate out of plane to a
minimal degree, requiring another 6.8MeV. So to an observer the loop
energy appears to increase by 137+1 =138, and the particle will appear
as though of mass energy 6.8 x 138MeV, about 938MeV.  The proton mass
energy.
        The other two loops form a particle wherein they are mutually
orthogonal, permitting 3-space localization of the particle whose
volume effectively increases by 1/alpha.  But each loop can only
occupy two dimensions and increase by alpha^2/3 as the third dimension
is occupied by the other loop.  The two loop particle therefore
appears as of mass energy 2 x 6.8 x alpha^2/3.MeV.  Now alpha ^2/3 > 1/26.58, so the two loop particle is 2 x 6.8 x 1/26.58 = 0.511MeV, the
same as the electron.

This suggest three space quantum fluctuations can create protons and
electrons in equal numbers, one of each in close proximity to the
other, i.e. forming hydrogen throughout the Universe.

Note the 6.8MeV component is common to both particles so the
gravitational constant should be nominally the same for each if it
acts via the components.

This also allows calculation of the proton-electron mass ratio. The
mass electron is 2mq x alpha^2/3 ke, where mq is about 6.8MeV, and ke
is the established QED factor of 1.001159652 for the electron and
relates to the anomalous magnetic moment.  The proton mass is mq x 1/
alpha x (1+alpha) kp, where kp is a similar factor for the proton.

The proton -electron mass ratio is thus (1/2)(alpha^-5/3)(1+alpha) ke/
kp.   Using the known values of alpha and ke this ratio is 1836.63/
kp.  The empirical mass ratio is 1836.153, so the predicted value of
kp is 1.0002606.

A question: Is there any way via QED of calculating this value?

This concept of an EM torus is consistent with all known aspects of
the quark notion, but produces particles of the correct mass, charge,
and size for both proton and electron, and explains the opposite
charges of the two particles.  How does that compare to the Standard
Model which provides no equation for the mass ratio?

If this interests you, please visit my web page at RethinkingPhysics-
V3.net

Perhaps there is a case for NO Quarks?
The notion of quarks with partial charges constituting the proton is
easily replaced by a simple particle concept that creates the effect
of partial charges, but makes them unable to be removed from a
particle. The concept also shows both the electron and proton are
comprised of components of the same energy, and a simple equation is
given for the proton-electron mass ratio.

Consider the proton as a self trapped EM-like wave that is propagating
in a helical-toroidal path. The wave is of spin-1/2, i.e. E = hbar.c/
2r, and propagates twice around the torus axis of symmetry to form a
quantum state and thus has 4pi symmetry. The rotating E vector of the
wave projects everywhere radial from the 'boundary' of the torus and
the particle thus appears as a point charge to a distant observer.
The E field is also rotating around the torus equatorially (i.e. about
the axis of symmetry), and produces an orthogonal magnetic field,
creating the effect of a magnetic dipole.
Due to the waves double pass the wave energy in the rotating frame is
equivalent to energy hbar.c/r, but due to its being wound around the
torus the particle 'boundary' curvature is 137 times greater than the
energy wavelength would imply, and the particle volume is lower by
alpha. (= 1/137) the fine structure constant. The lower particle
curvature effectively reduces the localized potential to alpha.hbar.c,
creating the effect of unit charge.

What does this have to do with quarks?
The boundary of a torus has a curvature about its z axes that differs
from those about the x and y axes. For this particle the curvature
about the z axis is decreased by alpha^1/3 as the radius is increased
by the same factor due to the long path length around the helix, and
is negative, i.e. concave. The convex curvature about the other two
axes is increased as the radius is decreased by alpha^2/3 for each.
The total particle volume is therefore decreased by
(alpha^2/3.alpha^2/3.alpha^1/3) = alpha.
With the same electric flux density emanating from every where on the
torus 'boundary', in the near field the charge of the particle will
appear different depending on the proximate curvature. About the x
and y axes the particle will appear to have a charge intensity of
alpha^2/3.hbar.c. About the z axis it will appear to have a charge of
alpha^-1/3. hbar.c. Hence the particle overall will appear to have
unit charge to a distant observer, but appears as partial charges of
2/3, 2/3, -1/3 in the near field.
Obviously this concept shows the quark partial charge effect without
there being any individual components or partial charges, which means
there is no possibility of separating a 'quark' from the particle as
quarks do not exist except as a convenient mathematical concept.

What does this all mean?
Consider a quantum fluctuation in 3-space that produces three spin-1/2
quantum loops, one about each of three orthogonal axes, and each of
about 6.8MeV. The three loops separate into two particles, a pair of
loops and a single loop. The single loop falls into a relativistic
state where its volume is decreased by alpha and the curvature of the
'boundary' is effectively decreased to the same degree, i.e. the
apparent mass relates to the curvature so the evident loop energy
increases by 137. For a single the single loop to localize in 3-space
(i.e. become three dimensional) it must rotate out of plane to a
minimal degree, requiring another 6.8MeV. So to an observer the loop
energy appears to increase by 137+1 =138, and the particle will appear
as though of mass energy 6.8 x 138MeV, about 938MeV. The proton mass
energy.
The other two loops form a particle wherein they are mutually
orthogonal, permitting 3-space localization of the particle whose
volume effectively increases by 1/alpha. But each loop can only
occupy two dimensions and increase by alpha^2/3 as the third dimension
is occupied by the other loop. The two loop particle therefore
appears as of mass energy 2 x 6.8 x alpha^2/3.MeV. Now alpha ^2/3 > 1/26.58, so the two loop particle is 2 x 6.8 x 1/26.58 = 0.511MeV, the
same as the electron.

This suggest three space quantum fluctuations can create protons and
electrons in equal numbers, one of each in close proximity to the
other, i.e. forming hydrogen throughout the Universe.

Note the 6.8MeV component is common to both particles so the
gravitational constant should be nominally the same for each if it
acts via the components.

This also allows calculation of the proton-electron mass ratio. The
mass electron is 2mq x alpha^2/3 ke, where mq is about 6.8MeV, and ke
is the established QED factor of 1.001159652 for the electron and
relates to the anomalous magnetic moment. The proton mass is mq x 1/
alpha x (1+alpha) kp, where kp is a similar factor for the proton.

The proton -electron mass ratio is thus (1/2)(alpha^-5/3)(1+alpha) ke/
kp. Using the known values of alpha and ke this ratio is 1836.63/
kp. The empirical mass ratio is 1836.153, so the predicted value of
kp is 1.0002606.

A question: Is there any way via QED of calculating this value?

This concept of an EM torus is consistent with all known aspects of
the quark notion, but produces particles of the correct mass, charge,
and size for both proton and electron, and explains the opposite
charges of the two particles. How does that compare to the Standard
Model which provides no equation for the mass ratio?

If this interests you, please visit my web page at RethinkingPhysics-
V3.net