# TornadoCreator's theory: graviational effect of quarks and strong force

1. Dec 16, 2003

Wouldn't it be true to say that as E=mc(sqared) that a high engergy partical also has high mass simply because if we revert to the therory of imcertainty (i think its called that) in which the errors to which we record everything reach such a large value that they are infact exceptionally larger that the value recorded this tends to only affect things on a quantum level.

This would mean that a high enegy partical could change energy from energy to mass and back to energy over a minute period of time making seem as though both are present simultaneously.
This would cause the wave properties of the fundamental particles (ie electron, positron, tau, quark etc.) would be constently changing wavelenth and would caonstitute the colour change property of many particles.

I do beleive however that a quark in not fundamental. I plan to prove my theories in the future. I hope to (but secretly know i won't) be bigger that Einstien.

2. Dec 25, 2003

### mormonator_rm

High-energy means high-mass. Remember that in Special Relativity, the total energy of a particle is, in essence, its apparent mass:

$$E = (\gamma-1)mc^2 + mc^2$$

with the first term being due to momentum energy, and the second term due to the rest mass,

$$E = \gamma mc^2$$

where $\gamma$ is the scalar time-dilation factor, always greater than one. Hence, the apparent mass is $\gamma m$. The rest-mass of the particle always remains the same, but the apparent mass is what increases with total energy. Note that we have a tendency to express mass in terms of energy; this is because they can be treated as one and the same, especially where we normalize $c\rightarrow 1$. The Heisenburg Uncertainty Principle has nothing to do with this effect.

The Uncertainty Principle only affects our ability to measure the mass of particles. If you look in the Physical Review, you will find particles listed with both masses and "widths". The width $\Gamma$ is related to the mean life $\tau$ of the particle by the Heisenburg Uncertainty Principle, such that;

$$\Gamma \cdot \tau \geq \hbar$$

or in other words the width $\Gamma$ is the uncertainty in the measurement of the mass due to the limited time that the particle exists.

First of all, total energy includes both kinetic energy and rest-mass energy. You can turn momentum into mass, and mass into momentum, but you cannot change the total energy in the process. The momentum of the particle will be related to the wavelength of the particle thanks to the DeBroglie Principle. I believe the effect you are trying to refer to here is the wave-function $\psi$ of a particle.

However, the momentum of a particle has nothing to do with the colour charge of any particle. Only quarks and gluons have colour charge, not leptons like neutrinos, electron, muons, etc.

You may be interested in looking into the theory on Rishons. These are claimed to be particles that make up quarks. I personally do not agree with the theory, but you may find some interesting details in it that will inspire some more thought. As for the last comment, I wish you good luck...

3. Dec 26, 2003

### Mk

Wow, long post

4. Jan 7, 2004

quarks .... split

I have noticed that during radioactivity in which an electron is emmited from a radioactive nucleus a neutron becomes a proton and an electron

This means that at a quark level
Two down quarks and one up quark --> One down quark, Two up quarks and one electron.

(half equation .... bit like the ones in chemistry.)
One down quark --> One up quark + One electron

This means the down quark is made of one up quark and one electron and therefor is not fundamental.

shock horror shock horror.

5. Jan 12, 2004

### mormonator_rm

correct weak interactions

The weak interaction involved in neutron decay to proton and electron involves an additional decay product, the neutrino. Also, this reaction is mediated by one of the weak massive vector bosons as;

$$n \rightarrow p + W^- \rightarrow p + e^- + \nu_{e}$$

Hence, the down quark does not consist of an up quark and an electron, but rather the exchange group $d u\bar$ results in the formation of $W^-$, which then decays further into $(e^- + \nu_e)$. At best, the only equivalence one may find from this is that an electron plus its anti-neutrino is equivalent to a down quark and an anti-up quark.

Last edited: Jan 12, 2004