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  4. Experiment to test W=mg
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Experiment to test W=mg

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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #300 on: 10/01/2018 01:13:28 »
Quote from: Kryptid on 09/01/2018 21:24:46
Where does the "positive repulsive force from above" come from?
In my theory the positive charge of galaxies, stars, planets, moons, rocks, dust and gas create a positive electric field that permeates the entire universe. The strength of the positive force exerted on an atom depends on the strength of the positive charge of its poles.
Quote from: Kryptid on 09/01/2018 21:24:46
Then we know from the kinetic energy equation Ek = (1/2)mv2
W reduction at increasing T in vacuum, if exists, disproves conservation of mass and this equation too.
Quote from: Colin2B on 09/01/2018 23:25:59
In other words the weight doesn’t change.
Results required.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #301 on: 10/01/2018 06:01:33 »
Quote from: Yaniv on 10/01/2018 01:13:28
In my theory the positive charge of galaxies, stars, planets, moons, rocks, dust and gas create a positive electric field that permeates the entire universe. The strength of the positive force exerted on an atom depends on the strength of the positive charge of its poles.

I'll keep that in mind for future reference.

Quote
W reduction at increasing T in vacuum, if exists, disproves conservation of mass and this equation too.

Forgetting about actual equations and numbers for a moment, surely you agree with these common sense notions:

(1) It takes more energy to lift a 50 kilogram weight than it does to lift a 30 kilogram weight.
(2) It takes more energy to throw a baseball at 100 kilometers per hour than it does to throw it at 60 kilometers per hour.
(3) It takes more energy to climb 10 flights of stairs than it does to climb only 5 flights.
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Offline Bored chemist

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Re: Experiment to test W=mg
« Reply #302 on: 10/01/2018 08:43:42 »
Quote from: Yaniv on 10/01/2018 01:13:28
Results required.
No.
It's a deduction from what you are putting forward, if it's wrong that's because your idea is wrong (spoiler alert- your idea is wrong)
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #303 on: 10/01/2018 10:57:19 »
Quote from: Kryptid on 10/01/2018 06:01:33
Forgetting about actual equations and numbers for a moment, surely you agree with these common sense notions:

(1) It takes more energy to lift a 50 kilogram weight than it does to lift a 30 kilogram weight.
(2) It takes more energy to throw a baseball at 100 kilometers per hour than it does to throw it at 60 kilometers per hour.
(3) It takes more energy to climb 10 flights of stairs than it does to climb only 5 flights.
I agree.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #304 on: 10/01/2018 16:31:50 »
Quote from: Yaniv on 10/01/2018 10:57:19
I agree.

These things show that there is a direct relationship between mass and the energy. The more mass, the more resistance to movement and therefore the more energy that is needed to overcome that resistance. It also takes more energy to stop a heavy mass from moving than it does a lighter mass if they are going the same speed. This means heavier objects must have more kinetic energy than lighter ones moving at the same speed. Inversely, lighter masses are easier to move and easier to stop than heavy ones. Go all the way to zero mass, and the resistance to movement also goes to zero and thus no energy is needed to move it or stop it. This also means zero kinetic energy.

Since you agree that moving electrons have kinetic energy, then they must also have mass. If they were massless, then it would take no energy to stop their movement and hence they would not carry any kinetic energy that needed to be overcome or opposed. Given that you say that electrons fall in a gravitational field, that points towards them having mass as well.
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #305 on: 10/01/2018 21:17:39 »
Quote from: Kryptid on 08/01/2018 19:39:25
(2) Does your model predict that positrons and electrons have electric charge that is equal and opposite in magnitude?
The positive charge of a positron should be higher in magnitude than the negative charge of an electron. This gives a neutron a tiny positive charge (small fraction the charge of a positron). In weak gravitational fields a neutron forms a dipole with a stronger and weaker positive poles and falls at the same rate as other positively charged objects.

Quote from: Kryptid on 10/01/2018 16:31:50
It also takes more energy to stop a heavy mass from moving than it does a lighter mass if they are going the same speed.
In my theory also more force is required to stop a heavy object than a lighter object travelling at the same speed because a heavier object is more positively charged and is pushed by a stronger force.
Quote from: Kryptid on 10/01/2018 16:31:50
Inversely, lighter masses are easier to move and easier to stop than heavy ones.
In my theory also lighter objects are easier to move because they are fixed in place by weaker forces.
Quote from: Kryptid on 10/01/2018 16:31:50
Since you agree that moving electrons have kinetic energy, then they must also have mass. If they were massless, then it would take no energy to stop their movement and hence they would not carry any kinetic energy that needed to be overcome or opposed. Given that you say that electrons fall in a gravitational field, that points towards them having mass as well.
W reduction at increasing T in vacuum falsifies everything you write about mass and energy.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #306 on: 10/01/2018 21:40:24 »
Quote from: Yaniv on 10/01/2018 21:17:39
The positive charge of a positron should be higher in magnitude than the negative charge of an electron. This gives a neutron a tiny positive charge (small fraction the charge of a positron). In weak gravitational fields a neutron forms a dipole with a stronger and weaker positive poles and falls at the same rate as other positively charged objects.

That contradicts what you said earlier:

Quote from: Yaniv on 09/01/2018 01:28:02
The natural motion of a neutron (any nuclear constituent consisting of equal number of positrons and electrons) passing through an electric field of equal and opposite charges is straight. A neutron passing through the field forms a dipole with positrons attracted to cathode and electrons attracted to anode at equal and opposite forces. An additional positron is required to change the straight trajectory of a neutron and may appear as resistance to acceleration.

You said that neutrons are not deflected by electric fields and that the forces on their constituent positrons and electrons are equal and opposite. The forces cannot be equal if the positron is more highly charged than the electron.

Quote
W reduction at increasing T in vacuum falsifies everything you write about mass and energy.

So positrons and electrons are massless in your model.
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Offline alancalverd

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Re: Experiment to test W=mg
« Reply #307 on: 10/01/2018 21:41:02 »
Reading this thread, I am beginning to understand whence the Health and Safety Executive recruits its inspectors.

That is not a compliment.
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Offline Bored chemist

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Re: Experiment to test W=mg
« Reply #308 on: 10/01/2018 22:02:33 »
Quote from: alancalverd on 10/01/2018 21:41:02
Reading this thread, I am beginning to understand whence the Health and Safety Executive recruits its inspectors.

That is not a compliment.
It's probably libellous; you might want to accept that it's false.

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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #309 on: 10/01/2018 22:13:54 »
Quote from: Kryptid on 10/01/2018 21:40:24
That contradicts what you said earlier:
Yes. I thought about it more and concluded the positive charge of a positron has to be higher than the negative charge of an electron.
Quote from: Kryptid on 10/01/2018 21:40:24
You said that neutrons are not deflected by electric fields and that the forces on their constituent positrons and electrons are equal and opposite. The forces cannot be equal if the positron is more highly charged than the electron.
What I said earlier about neutrons is wrong and would like to correct myself here. The natural motion of a neutron passing through an electric field is almost straight. The positive charge of a neutron is a tiny fraction that of a positron or an electron and deflection is minimal and appear straight. The forces on a positron should be tiny bit stronger than forces on an electron.
Quote from: Kryptid on 10/01/2018 21:40:24
So positrons and electrons are massless in your model.
Yes.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #310 on: 10/01/2018 23:51:47 »
Okay, all of that is settled. Now, onto another matter. You say that all particles are actually composite particles made up of electrons and positrons. I would like to know if your model predicts that a single composite particle that is not bound inside of an atom (in other words, a free proton or a free neutron) has a temperature.
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #311 on: 11/01/2018 00:08:18 »
Quote from: Kryptid on 10/01/2018 23:51:47
I would like to know if your model predicts that a single composite particle that is not bound inside of an atom (in other words, a free proton or a free neutron) has a temperature.
A free proton and a free neutron don't have temperature.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #312 on: 11/01/2018 05:48:55 »
Quote from: Yaniv on 11/01/2018 00:08:18
A free proton and a free neutron don't have temperature.
Good. Then that means that the mass of free particles is constant. Now I need to know how the mass of composite particles is determined from their structure. I already know from your page that protons are composed of two positrons and an electron (which I will write as P2E) and that a neutron is composed of three positrons and three electrons (which I will write as P3E3). That being said, I would like for you to tell me the composition of the subatomic particles listed below which have missing information. "P" is for positron and "E" is for electron:

Particle..............Structure......Mass

Proton................P2E..............938.27 MeV
Antiproton..........PE2..............938.27 MeV
Neutron.............P3E3............939.57 MeV
Muon.................P?E?............105.66 MeV
Tau....................P?E?............1,776.82 MeV
Omega Minus....P?E?............1,672.45 MeV
Neutral Pion......P?E?.............134.98 MeV
Positive Pion.....P?E?.............139.57 MeV
Z Boson.............P?E?.............91,187.6 MeV

In case you need to know what any of these are: https://en.wikipedia.org/wiki/List_of_particles
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #313 on: 11/01/2018 07:12:12 »
Quote from: Kryptid on 11/01/2018 05:48:55
Then that means that the mass of free particles is constant.
The charge of a free particle is constant.
Quote from: Kryptid on 11/01/2018 05:48:55
Now I need to know how the mass of composite particles is determined from their structure.
The charge and speed of a free particle impacting a detector determine its energy.
Quote from: Kryptid on 11/01/2018 05:48:55
Proton................P2E..............938.27 MeV
Antiproton..........PE2..............938.27 MeV
Neutron.............P3E3............939.57 MeV
Muon.................P?E?............105.66 MeV
Tau....................P?E?............1,776.82 MeV
Omega Minus....P?E?............1,672.45 MeV
Neutral Pion......P?E?.............134.98 MeV
Positive Pion.....P?E?.............139.57 MeV
Z Boson.............P?E?.............91,187.6 MeV
Proton...............P2E
Antiproton.........PE2
Neutron.............PE....and.....P2E2.....and......P3E3......exc
I am not sure about the rest of the particles in the list but can imagine other particles such as P2E3 and P3E2 and others emerging from the debris of nuclear collisions.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #314 on: 11/01/2018 16:39:02 »
Quote from: Yaniv on 11/01/2018 07:12:12
The charge and speed of a free particle impacting a detector determine its energy.

I'm talking about mass, not energy. You've insisted that E=mc2 and even Ek = (1/2)mv2 are incorrect, so surely you're not trying to equate mass with energy?

Quote
Proton...............P2E
Antiproton.........PE2
Neutron.............PE....and.....P2E2.....and......P3E3......exc
I am not sure about the rest of the particles in the list but can imagine other particles such as P2E3 and P3E2 and others emerging from the debris of nuclear collisions.

So your model can't explain the structures and masses of the other particles? That's a big problem.
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #315 on: 11/01/2018 19:49:05 »

Quote from: Kryptid on 11/01/2018 16:39:02
I'm talking about mass, not energy. You've insisted that E=mc2 and even Ek = (1/2)mv2 are incorrect, so surely you're not trying to equate mass with energy?
I think to equate energy to electric current produced by a particle hitting an opaque detector. A fast particle should register a higher current than a slow particle of equal charge. Also a high charge particle should register a higher current than a low charge particle travelling at the same speed.
Quote from: Kryptid on 11/01/2018 16:39:02
So your model can't explain the structures and masses of the other particles? That's a big problem.
The figures provided are incorrect if any of the predictions raised in this thread are confirmed by experiments and should not be used to discard my theory.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #316 on: 11/01/2018 21:49:58 »
Quote from: Yaniv on 11/01/2018 19:49:05
I think to equate energy to electric current produced by a particle hitting an opaque detector. A fast particle should register a higher current than a slow particle of equal charge. Also a high charge particle should register a higher current than a low charge particle travelling at the same speed.

You're talking about energy again. I am not. You don't think mass changes with energy or velocity anyway so it's irrelevant to what I'm asking. So what makes one particle heavier than another? Will a particle with more charge weigh more than one with less charge? Will a particle containing more electrons be heavier than one with less electrons? Or do more positrons make it weigh more instead? Or is it the total of both that matters? Or is it a combination of the total number of electrons and positrons and the net charge on the composite particle that determines the mass?

Quote
The figures provided are incorrect if any of the predictions raised in this thread are confirmed by experiments and should not be used to discard my theory.

You already said that individual particles don't have a temperature and as such weight changing with temperature will do nothing to affect the known rest masses of individual subatomic particles.
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #317 on: 11/01/2018 22:10:10 »
Quote from: Kryptid on 11/01/2018 21:49:58
So what makes one particle heavier than another? Will a particle with more charge weigh more than one with less charge? Will a particle containing more electrons be heavier than one with less electrons? Or do more positrons make it weigh more instead? Or is it the total of both that matters? Or is it a combination of the total number of electrons and positrons and the net charge on the composite particle that determines the mass?
Maybe (total number of positrons and electrons) to (charge) ratio ?
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Offline The Spoon

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Re: Experiment to test W=mg
« Reply #318 on: 11/01/2018 22:21:35 »
Quote from: Yaniv on 11/01/2018 22:10:10
Quote from: Kryptid on 11/01/2018 21:49:58
So what makes one particle heavier than another? Will a particle with more charge weigh more than one with less charge? Will a particle containing more electrons be heavier than one with less electrons? Or do more positrons make it weigh more instead? Or is it the total of both that matters? Or is it a combination of the total number of electrons and positrons and the net charge on the composite particle that determines the mass?
Maybe (total number of positrons and electrons) to (charge) ratio ?
So it is down to guessing now?
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #319 on: 11/01/2018 23:36:46 »
Quote from: Yaniv on 11/01/2018 22:10:10
Maybe (total number of positrons and electrons) to (charge) ratio ?

Okay, let's see what that predicts:

(1) Electron: contains 1 electron + 0 positrons = 1 total. Total charge is -1. So, 1/1 = Mass of 1
(2) Positron: contains 0 electrons + 1 positron = 1 total. Total charge is slightly more than +1. So 1/slightly more than 1 = Mass of slightly less than 1
(3) Neutrino: contains 1 electron + 1 positron = 2 total. Total charge is slightly more than 0. So 2/slightly more than zero = Mass of some very, very big number (exact number depends on how much more charge positron has than electron)
(3) Proton: contains 1 electron + 2 positrons = 3 total. Total charge is slightly more than +1. So 3/slightly more than 1 = Mass of slightly less than 3
(4) Neutron (four particle version): contains 2 electrons + 2 positrons = 4 total. Total charge is slightly more than 0. So 4/slightly more than 0 = Mass of some very, very big number (higher than neutrino mass)
(5) Neutron (six particle version): contains 3 electrons + 3 positrons = 6 total. Total charge is slightly more than 0. So 6/slightly more than 0 = Mass of some very, very big number (higher than four-particle neutron)

So this predicts the following order of masses (from least to greatest): positron, electron, proton, neutrino, neutron with four particles, neutron with six particles. Does that sound good?
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