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Natural Science Forum / Physics / Particle Physics / November 2004



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quark rest mass equation

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alistair - 17 Oct 2004 11:05 GMT
for quarks of electric charge -1/3
rest mass  = rest mass of down quark x n^2
where n = 1,3,5,7

for quarks of electric charge +2/3
rest mass = rest mass of down quark x n^2 x 2^N
where N = -1,1,3,5

The values of n^2 represent a circular area of magnetic charges (rest
mass is proportional to number of magnetic charges per quark) on the
circumference of which a single electrical charge orbits an
electrically neutral 3-charge nucleus.At n = 5 the density of magnetic
charges becomes so great that
the Pauli exclusion principle is overcome by gravity and the circular
area can contain ten times more magnetic charges than the equation
says it should at n=5.This is why the top quark and bottom quark have
such large masses.

The values of n^2 in the Bohr equation for the energy levels of atomic
hydrogen are also areas containing magnetic charges.The values of N
probably have their origin in the quantisation of angular momentum of
the orbiting charge or the
quantisation of the magnetic moment.
alistair - 17 Oct 2004 23:26 GMT
If n = radius of orbiting charge and the orbited charges have a net
electric charge and doubling the charge of the orbited particles gives
a doubling of radius then the following force laws arise:

when n=1 force = k1 x 1/r^3
    n=3 force = k2 x 1/r^7
    n=5 force = k3 x 1/r^11
    n=7 force = k4 x 1/r^15

As the force decreases - on doubling the charge and the radius -
the velocity and magnetic moment of the orbiting charge decreases
and the rest mass increases.

This was deduced using rest mass = constant x electric charge/current
x Area
and noting, from experimental data for quark rest masses, that on
doubling the quark charge magnitude:

for n =1 rest mass halves
for n=3  rest mass doubles
for n=5 rest mass increases eight times
for n=7 rest mass increases 32 times*****

*****this is the rest mass increase at n=7 before the Pauli exclusion
principle is overcome by the gravity? of magnetic charges inside the
circle, which give an observed rest mass increase of 10 x 32 = 320
times.
alistair - 18 Oct 2004 19:07 GMT
For the relativistic doppler shift:

change in wavelength = (c - Vs) To / (1 - Vs ^2 /c^2)^1/2

where Vs is emitter velocity, c is speed of light and To is time.

Suppose change in wavelength was equal to just 1 / (1 - Vs ^2 /
c^2)^1/2

then (c - Vs) To = 1
c -Vs = 1 / To

c = Vs + 1 / To
c = Vs + frequency of emitted wave

I now suggest that for M = Mo / (1 - v^2 /c^2) ^ 1/2

that this relation is actually

M = Mo x (c - Vs) To / (1 - Vs ^2 /c^2)^1/2

when (c - Vs) To = 1
and c = Vs + frequency of emitted wave
[SI units are correct because a frequency = a velocity when wavelength
is fixed at 1 metre: v = lambda x f becomes v = f)

In other words a mass emits a wave as it travels  through space at
constant velocity.The slower the mass travels , the greater the
frequency of the
emitted wave.A mass at rest would emit the highest frequency.
Because of this inverse doppler relation for a rest mass,a rest mass
moving through a sea of magnetic charges would keep moving at the same
speed -
Newton's first law!!
alistair - 18 Oct 2004 23:04 GMT
We can form a neutron from each of the quark families:
The quarks in a proton made of up and down quarks have a total rest
mass
(experimental) of 0.02 Gev.
The quarks in a proton made of charm and strange quarks would have a
mass
of about 4 Gev.This means that the quarks of a proton made from charm
and strange quarks are about 200 times (4/0.02) more massive than a
standard proton's quarks.And this is about the  ratio of the mass of a
muon to an electron.
The quarks in a "proton" made of four top quarks and one bottom quark
(top,antitop,top,top,bottom)- a kind of pentaquark - would have a
total rest mass of,say,4x, where x is the mass of one top quark
(assuming bottom quark has a relatively small mass compared to
top).The ratio of a tau's mass to that of an electron is about
3200:1.Comparing the "proton" containing 3 antitop quarks and one top
quark to the standard proton of mass 0.02:
4x/0.02 = 3200
x = 16 Gev

using rest mass of a quark = n^2 x rest mass of down quark x 2^N
and N = 5 and rest mass of down quark = 0.01 Gev :
49 x 0.01 x 2^5 = 16

It seems that the rest mass of the top quark should be at least ten
times smaller than is experimentally observed (174 Gev).

This gives support to the idea that the top quark has a very large
mass because
the Pauli exclusion principle is being overcome.Otherwise its mass
would be 16 Gev as the lepton mass ratios suggest.
A. Larcinese - 20 Oct 2004 16:38 GMT
With regards to your last/second-to-last post, I think your logic is
fine.  I am not learned enough to critique it, but the question comes
to my mind is where does all of this additional mass come from?

You express that theoretically the top quark should be isolated at 16
or 17 GeVs, but you also state that in the labs they have been unable
to do so at any power below 174 GeV.  Could this have anything to do
with a change in energy level while in motion relative to the change
in energy generated by, say, the bottom quark in motion?  I mean,
technically there should be some exponential difference in that
relationship if you are examining a 'pentaquark.'

Maybe I didn't make any sense, but I am interested in this logic that
you have brought to my attention.  What do you think accounts for this
additional experimental mass?
alistair - 21 Oct 2004 16:55 GMT
> With regards to your last/second-to-last post, I think your logic is
> fine.  I am not learned enough to critique it, but the question comes
[quoted text clipped - 11 lines]
> you have brought to my attention.  What do you think accounts for this
> additional experimental mass?

The additional mass comes from the space around the quark - it is in
the form of
mass-causing particles which are real (not virtual vacuum particles).
In the top quark and bottom quark there is just more mass squeezed
into
a given area -raising the density of space.I think this density
increase happens because spin 1/2 fermions team up into spin 1 bosons
which can live in the same space at the same time.Thus the rest mass
around a quark's electric charge could show superfluid behaviour (zero
viscosity) and this would be reflected perhaps in the scattering of
photons electrons etc. off the bottom and top quarks.
alistair - 20 Oct 2004 23:38 GMT
To overcome the Pauli exclusion principle, mass-causing fermionic
particles, inside a circlular area surrounding a nucleus of electric
charges, must become highly ordered and form one big boson like a
superfluid.Because a superfluid is non-viscous we would expect photons
and electrons to scatter off a top quark
and bottom quark differently to how they scatter off the other quarks.
Why does the density of mass-causing fermions increase ten times for
the bottom and top quarks? This must be the limit at which the
superfluidity can survive.We know that the bottom quark has a rest
mass of about 5 Gev and that current particle accelerators could take
the mass up as much as 200 times, to about 1 Tev.
But this doesn't happen, so the density increase of ten times seems to
be the limit.We can speculate that particles in the space around the
top and bottom quarks make superfluidity untenable above a certain
density because the
particles have a small enough wavelength to be absorbed by the
mass-causing fermions, and to split the fermions up.
alistair - 21 Oct 2004 10:42 GMT
The next quark in the series will be at n = 9 and it will have a rest
mass of
2 x 9^2 x 2^7 x 10 x 0.02 Gev = 4150 Gev = 4.1 Tev (if it is part of a
three-quark "proton")or 2.05 Tev (if it is part of a five quark
"proton").
The quarks at n =3 have not been detected.This could be because they
are sometimes mistaken for the strange quark and charm quark but there
is
also the possibility that at n= 3 an electrically neutral "neutron"
forms
but no "proton" (an uncharged "neutron" at n=9 would be undetected in
accelerator experiments).The decay rate of the Zo is not affected by
new quark families (the decay rate depends on the number of different
types of neutrino).
This could be because the non-existence of a "proton" at n=3 or n=9
would suggest neutrinos and antineutrinos do not exist for processes
of the type: "neutron" -> "proton" + "electron" + antineutrino.
alistair - 21 Oct 2004 17:14 GMT
CORRECTION:
At n =11 the quark of charge +2/3 will have a rest mass of 6.2 Tev
(five-quark)
or 4.1 Tev (7-quark) or 12.4 Tev (3-quark).These values will come from
an energy of 24.8 Gev - just out of the range of the next generation
of accelerators!The quark, at n =11, of charge -1/3 would already have
been detected if it could live in isolation from the quark of charge
+2/3 because its rest mass is only about 24.2 Gev.
alistair - 21 Oct 2004 17:52 GMT
Do the leptons overcome the exclusion principle and increase their
rest mass beyond what is expected from the equation?If they do so by a
factor of about ten then we can expect to see them at 2.2 Tev or 4.4
Tev or 13.6 Tev
(using proportionality relationships e.g 2 x rest mass of quark (+2/3)
+ 1 x rest mass of quark(-1/3)from one quark family divided by  2 x
rest mass of quark (+2/3)+ 1 x rest mass of quark(-1/3)from another
quark family =
rest mass of lepton from family 1/rest mass of lepton from family 2.
This might explain why we haven't seen them so far.
alistair - 21 Oct 2004 14:38 GMT
At n =11 we would expect a quark of charge + 2/3 with a rest mass of
12.4 Tev (5-quark) or 24.8 Tev (3 quark).A 7-quark would give a rest
mass of about 37.2 Tev.
alistair - 18 Oct 2004 23:08 GMT
CORRECTION to my previous post:
Comparing the "proton" containing 3 antitop quarks and one top
quark

SHOULD READ:
Comparing the "proton" containing 3 top and one antitop quark..
alistair - 21 Oct 2004 23:30 GMT
A neutrino made from an up quark and an anti-up quark would have a
rest mass
half as great as an antineutrino made from a down quark and an
anti-down quark.
The quarks of opposite electric charge would become so close together
that the
mass-causing particles which surround the charges would be squeezed to
such an extent that many of the mass-causing particles would be
shed.This is probably why the neutrino and antineutrino have such low
masses compared to another
particle that is electrically neutral overall - the neutron.In the
neutron
there are two electric charges of the same sign that repel one another
and so do not get close enough to squeeze their mass-causing
particles.
Muon neutrinos and electron neutrinos and tau neutrinos are each
distinguishable from one another because n will have a different
integer value
for the quark pairs in each neutrino - in other words the neutrinos
all have
different rest masses.Why does an electron have such a small mass
compared to a proton? An electron is made from three quarks -2/3,-2/3
and +1/3 (the electron is an antiproton!!).There are two charges of
-2/3 in the electron and two charges of +2/3 in the proton.The quarks
of charge -2/3 in the electron may be getting pushed together by the
repulsion of negative charges in the vacuum - this would decrease the
colour force between them and the total rest mass of the electron.At
the same time the charges of +2/3 in the proton may be getting pulled
apart by the attraction of negative charges in the vacuum giving the
proton a higher rest mass than it would otherwise have.Negative
charges in the vacuum must be smaller in size than positive charges,
enabling them to get closer to the quarks of the proton and electron.
The decay rate of the Zo boson could be unaffected by new quark
families if the new quarks make new z bosons - in other words an
electrically neutral Zo could be made of three quarks +2/3,-1/3,-1/3.
alistair - 21 Oct 2004 23:48 GMT
Neutrino oscillations could occur when mass-causing fermions
associated with
one electric charge ( a neutrino is made from two charges : +1/3 and
-1/3) move over to the other electric charge and then back again.So
n=5 for one charge can become n=3 while n=3 for the other charge can
become n=5 and vice-versa.
An electron neutrino with n=1 for both electric charges, would not be
able to oscillate,because there is no excited state which can lose
energy.
alistair - 22 Oct 2004 19:35 GMT
The following relates to the magnetic moment associated with the rest
mass
of a quark, and makes the idea seem more plausible because the origin
of relativistic spin can be linked to one charge orbiting another:
If I have two negative electric charges, one of them with a far larger
magnitude than the other, orbiting a positive charge,this system of
charges could represent a rotating electric field vector and a
rotating magnetic field
vector( since magnetism is a form of electric energy).The ratio of the
angular velocity, mvr, for one negative charge compared to the other
could be a unitless quantity of magnitude 1/2 - the same magnitude as
quantum relativistic spin for a fermion.By increasing the radius of
the orbit of one negative charge I could change this ratio to 1/1 = 1
and create a spin
1 particle (a photon perhaps)or I could change the ratio to 2/1 (a
graviton perhaps).
In addition, by changing the magnitude of the charges I could change
the ratio of electric field/magnetic field (which gives a speed).
alistair - 22 Oct 2004 21:43 GMT
The general wavefunction for a free particle is:

Y (x,t) = A cos( kx - wt) + iA sin (kx - wt)

and the complex conjugate is:

Y*(x,t) = A cos( kx - wt) - iA sin (kx - wt)

These are multiplied together to give,along with a proportionality
constant,
the probability of finding a particle at a particular place at a
particular time.It is not known why this should be so.Here is a
suggestion:

If the wavefunction already represents a probability before it is
squared,
then multiplying it by another wavefunction  - the complex conjugate -
suggests that we are dealing with the probability of two events
occurring simultaneously.This could be the probability of one charge
being
in a particular place at a particular time, and  another charge being
at that place at the same time.
alistair - 22 Oct 2004 22:50 GMT
There are six electric charges in a photon in two groups of three -
one group with a charge of -1 and one group with a charge of +1.
When the polarization of a photon is changed,the groups flip over
(like a drumstick with one group on each end) in the direction in
which the photon is travelling.This is an acceleration of charges and
results in the generation of another photon, with a speed greater than
light.Why does the second photon have a speed greater than light?
Because when particles are emitted they travel faster than the
emitter.Charges in a photon moving through a sea of mass-causing
particles have mass by virtue of colliding with the mass-causing
particles (but no permanently attached particles causing rest mass),so
by the law of conservation of momentum,when something with a  big mass
emits something with a smaller mass,the smaller mass moves faster than
the slowly recoiling bigger mass.Accelerated fermions with rest mass
can emit photons travelling at the speed of light and photons
travelling greater than the speed of light.Photons moving faster than
the speed of light are the particles that cause correlations in
Einstein-Podolsky-Rosen paradoxes.
alistair - 01 Nov 2004 01:02 GMT
At n = 5 the density of magnetic
> charges becomes so great that
> the Pauli exclusion principle is overcome by gravity and the circular
> area can contain ten times more magnetic charges than the equation
> says it should at n=5.This is why the top quark and bottom quark have
> such large masses.

Another possibility is this:
The Pauli exclusion principle is not overcome.
The particles that cause mass exist with a wavelength given by the
Planck length.
The Planck length of 1.6 x 10^-35 metres should actually be
about 9 x 10^-35 metres ( using [Gh/c^3]^1/2).
So the speed of a photon with a wavelength close to the Planck length
is 0.3 times smaller than we normally measure it.Therefore E=mc^2
becomes ten times smaller due to the c^2 term and so the mass must
become ten times greater to compensate and keep the energy at the
expected value.
Can a photon have such a small wavelength and therefore such a
colossal
energy which corresponds to 10^114 Joules per cubic metre.Perhaps so,
if the well-known vacuum calculation of quantum mechanics of 10^120
Joules/m^3 is anything to go by.But the force of gravity would also
have to be inhibited or else space-time would be curved to an
unimaginable degree-
these photons could not be allowed to curve space-time and so cannot
emit
or absorb gravitons.
 
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