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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 #280 on: 08/01/2018 11:30:27 »
Quote from: Kryptid on 08/01/2018 01:35:48
Surely that isn't the case? If we look at your model of atoms, a diatomic hydrogen molecule has equal numbers of electrons and positrons so it would have to be neutral. As you heat up these hydrogen molecules, they gain more electrons as they get hotter and hotter so they would have to have a net negative charge.
A population of hydrogen molecules could consist of cationic and neutral molecules. As more negative heat particles are added more cationic molecules are converted to neutral molecules and the positive charge of the population and weight decreases. As more negative heat particle are added at some point there will be more negative than positive particles in the volume and the charge of the volume will become negative.
Quote from: Kryptid on 08/01/2018 01:35:48
It sounds like you could make some good mathematical predictions about just how much weight should decrease as a piece of metal gains or loses negative charge, since negative charge comes in discrete, quantifiable amounts.
I don't know how you can make quantitative predictions without results of experiments. If you do the experiment and find say 1 microgram drop in weight per 1 calorie absorbed you can begin to make quantitative predictions. Also interesting is to determine how many negative heat particles are found in 1 calorie.
Quote from: Kryptid on 08/01/2018 01:35:48
I will probably need to know why a proton weighs almost 2,000 times more than an electron if the components of a proton are only two positrons and an electron. What mechanism in your model accounts for all the extra mass?
A proton deflects less than an electron in a mass spectrometer. Neutrons are not deflected. In my theory a proton could deflect less than an electron because some of the force is used to drag the neutral constituent of the proton.

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Offline Bored chemist

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Re: Experiment to test W=mg
« Reply #281 on: 08/01/2018 13:38:21 »
Quote from: Yaniv on 07/01/2018 19:51:13
The article reads "To build its device, the team developed a grid of tiny cavities etched in silicon, forming the photonic crystal. By precisely applying electric current to the grid, the scientists can control, or harmonically tune, the photonic crystal to synthesize magnetism and exert virtual force upon photons". This sounds to me something like an electromagnet.

If it sounds like an electromagnet to you then that just shows that you also don't understand  electromagnets.

Quote from: Yaniv on 07/01/2018 19:51:13
Deflection of light by an MRI magnetic field may be too small to notice. Placing many magnets in sequence such as in particle accelerators could increase the effect.

MRI magnets are some of the biggest strongest magnets we have; if you can't spot the effect with these then you are probably not going to see it elsewhere.


Quote from: Yaniv on 07/01/2018 19:51:13
Results required.
No the results are nor required because they were already posted- initially by you.
What is "required" is that you pay attention to the results.
Quote from: Yaniv on 07/01/2018 22:14:20
In my theory there are no negatively charged materials, only positive.
Well, we know that's wrong by simple experiment.
So what you keep dishonestly referring to as a "theory" is nonsense on yet another basis.

Quote from: Yaniv on 08/01/2018 11:30:27
Also interesting is to determine how many negative heat particles are found in 1 calorie.
Zero, (which is quite an interesting number) seems like a very good assumption.
Quote from: Yaniv on 08/01/2018 11:30:27
A proton deflects less than an electron in a mass spectrometer. Neutrons are not deflected. In my theory a proton could deflect less than an electron because some of the force is used to drag the neutral constituent of the proton.
Deflection of light by an MRI magnetic field may be too small to notice. Placing many magnets in sequence such as in particle accelerators could increase the effect.
I presume you don't  understand antiprotons and positrons.
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Offline Colin2B

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Re: Experiment to test W=mg
« Reply #282 on: 08/01/2018 14:30:17 »
Quote from: Bored chemist on 08/01/2018 13:38:21
.
What is "required" is that you pay attention to the results.

@Bored chemist - this lack of attention to previous evidence seems to be a standard response and replies which merely “keep dishonestly referring to as a "theory" is nonsense on yet another basis”; all makes me think there is no one at the other end listening, which is why i no longer feel it is useful to respond.
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Offline Bored chemist

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Re: Experiment to test W=mg
« Reply #283 on: 08/01/2018 15:06:51 »
Quote from: Colin2B on 08/01/2018 14:30:17
Quote from: Bored chemist on 08/01/2018 13:38:21
.
What is "required" is that you pay attention to the results.

@Bored chemist - this lack of attention to previous evidence seems to be a standard response and replies which merely “keep dishonestly referring to as a "theory" is nonsense on yet another basis”; all makes me think there is no one at the other end listening, which is why i no longer feel it is useful to respond.
In cases like this I respond (some would say "bloody mindedly") in order that the voice of reason has the last word.
That way, any visitors to the page who are less familiar with science won't get misled into thinking that the idea might be right because nobody said it wasn't.
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Offline alancalverd

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Re: Experiment to test W=mg
« Reply #284 on: 08/01/2018 17:09:52 »
Theories are ten a penny. Theories with numbers are very rare because they tend only to live long enough for someone to disprove them. Theories without numbers are worthless.
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Offline Bored chemist

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Re: Experiment to test W=mg
« Reply #285 on: 08/01/2018 19:12:28 »
Quote from: alancalverd on 08/01/2018 17:09:52
Theories are ten a penny. Theories with numbers are very rare because they tend only to live long enough for someone to disprove them. Theories without numbers are worthless.
Scientific theories are as rare as hen's teeth.
Made up drossy hypotheses masquerading as theories are ten-a-penny.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #286 on: 08/01/2018 19:39:25 »
Quote from: Yaniv on 08/01/2018 11:30:27
A population of hydrogen molecules could consist of cationic and neutral molecules. As more negative heat particles are added more cationic molecules are converted to neutral molecules and the positive charge of the population and weight decreases. As more negative heat particle are added at some point there will be more negative than positive particles in the volume and the charge of the volume will become negative.

Alright then.

Quote
I don't know how you can make quantitative predictions without results of experiments. If you do the experiment and find say 1 microgram drop in weight per 1 calorie absorbed you can begin to make quantitative predictions. Also interesting is to determine how many negative heat particles are found in 1 calorie.

I'll have to think about this some, but I'll let you know if I can indeed think of some quantifiable predictions. Knowing the basic properties of some materials in itself might be useful here.

Quote
A proton deflects less than an electron in a mass spectrometer. Neutrons are not deflected. In my theory a proton could deflect less than an electron because some of the force is used to drag the neutral constituent of the proton.

That causes it to look almost 2,000 times heavier than a single electron?

I actually did spend time reading your website and I've come to realize that your model doesn't necessarily predict a violation of conservation of mass. It predicts that weight should change at increasing temperature, but weight and mass are not the same thing. A mass of one kilogram will have a different weight on Mars than it will on Earth, for example. Yes, I know your model doesn't have mass in the conventional sense because it says that gravity is actually an effect of electromagnetism. However, I'm talking about inertial mass, not gravitational mass. Inertial mass is a measure of resistance to acceleration. Even if you hold that inertial mass is an electromagnetic effect, it's still a measurable and important quantity. Adding extra electrons to an object as it becomes hotter might make that object less strongly attracted to the Earth and therefore lighter, but those extra added electrons should still make the inertial mass of the hot object increase because electrons have inertial mass.

So for further clarification for my investigation:

(1) Does your model predict that positrons and electrons have equal inertial mass?
(2) Does your model predict that positrons and electrons have electric charge that is equal and opposite in magnitude?
(3) Does your model have any problems with the accepted inertial mass and charge of the electron (that is, 9.1 x 10-31 kg and -1.6 x 10-19 C respectively)?
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #287 on: 09/01/2018 01:28:02 »
Quote from: Kryptid on 08/01/2018 19:39:25
I actually did spend time reading your website and I've come to realize that your model doesn't necessarily predict a violation of conservation of mass. It predicts that weight should change at increasing temperature, but weight and mass are not the same thing. A mass of one kilogram will have a different weight on Mars than it will on Earth, for example.
Weight and mass are the same thing under the conditions of the proposed experiment carried at a fixed geographical position. W =mg. g is constant. Any change in W is a change in m.
Quote from: Kryptid on 08/01/2018 19:39:25
Inertial mass is a measure of resistance to acceleration. Even if you hold that inertial mass is an electromagnetic effect, it's still a measurable and important quantity.
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.
 
Quote from: Kryptid on 08/01/2018 19:39:25
(1) Does your model predict that positrons and electrons have equal inertial mass?
I think positrons and electrons should react equally fast to changes in electric fields.
 
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?
Yes, but could change if required in future.
Quote from: Kryptid on 08/01/2018 19:39:25
(3) Does your model have any problems with the accepted inertial mass and charge of the electron (that is, 9.1 x 10-31 kg and -1.6 x 10-19 C respectively)?
W reduction at increasing T in vacuum, if exists, implies W is a temperature-dependent variable. How a variable can be used to define constants ?
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #288 on: 09/01/2018 02:27:20 »
Maybe this go more smoothly if I try to understand one issue at a time.

Quote from: Yaniv on 09/01/2018 01:28:02
Weight and mass are the same thing under the conditions of the proposed experiment carried at a fixed geographical position. W =mg. g is constant. Any change in W is a change in m.

So would that mean that a hot object should also be easier to push in a zero-gravity vacuum than a cold object? If that's true, then wouldn't that mean that electrons have to have negative inertial mass? If adding them to an object with positive inertial mass makes that object's inertial mass less positive then their inertial mass has to be negative. However, then we end up with a contradiction because you propose that all objects are made up of electrons and positrons (and you agree that they both have equal inertial mass). This means that all objects should have negative inertial mass too. How does your model account for this seeming contradiction?
« Last Edit: 09/01/2018 03:34:57 by Kryptid »
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #289 on: 09/01/2018 07:36:13 »
Quote from: Kryptid on 09/01/2018 02:27:20
So would that mean that a hot object should also be easier to push in a zero-gravity vacuum than a cold object?
Yes, because a hot object is less positive than a cold object and experiences weaker repulsive forces from all directions. Once the object is in motion changing its temperature should not affect its speed.
Quote from: Kryptid on 09/01/2018 02:27:20
If that's true, then wouldn't that mean that electrons have to have negative inertial mass? If adding them to an object with positive inertial mass makes that object's inertial mass less positive then their inertial mass has to be negative. However, then we end up with a contradiction because you propose that all objects are made up of electrons and positrons (and you agree that they both have equal inertial mass). This means that all objects should have negative inertial mass too. How does your model account for this seeming contradiction?
I don't understand this part of the message.

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Offline Bored chemist

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Re: Experiment to test W=mg
« Reply #290 on: 09/01/2018 11:28:59 »
Quote from: Yaniv on 09/01/2018 07:36:13
I don't understand this part of the message.
That's unfortunate, because it's the bit where Kryptid totally nails your idea as impossible.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #291 on: 09/01/2018 14:48:14 »
Quote from: Yaniv on 09/01/2018 07:36:13
I don't understand this part of the message.

In order to break it down a bit, imagine that you put a single electron on a very, very sensitive scale. Does the scale register a positive weight?
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #292 on: 09/01/2018 15:33:03 »
Quote from: Kryptid on 09/01/2018 14:48:14
In order to break it down a bit, imagine that you put a single electron on a very, very sensitive scale. Does the scale register a positive weight?
I imagine an electron placed on top of a pan should reduce weight of a pan and if placed at bottom of a pan should increase weight of the pan.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #293 on: 09/01/2018 15:43:11 »
Quote from: Yaniv on 09/01/2018 15:33:03
I imagine an electron placed on top of a pan should reduce weight of a pan and if placed at bottom of a pan should increase weight of the pan.

What accounts for this?

As an alternative scenario, imagine that I want to accelerate an electron that is in a zero-gravity vacuum. Do I have to expend energy in order to accelerate it? Will a positive force be necessary?
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #294 on: 09/01/2018 16:02:51 »
Quote from: Kryptid on 09/01/2018 15:43:11
What accounts for this?
In my theory weight of the pan is determined by the difference between opposite positive repulsive forces acting on the pan, stronger repulsive force from above and weaker repulsive force from below. Placing an electron on top of pan will reduce the repulsive force from above and reduce the difference between forces acting on the pan and its weight. Placing an electron at bottom of the pan will reduce repulsive force from below and increase the difference between forces and weight.
Quote from: Kryptid on 09/01/2018 15:43:11
As an alternative scenario, imagine that I want to accelerate an electron that is in a zero-gravity vacuum. Do I have to expend energy in order to accelerate it? Will a positive force be necessary?
The electron can be accelerated by applying a negative charge to push it or a positive charge to pull it.
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #295 on: 09/01/2018 16:09:57 »
Quote from: Yaniv on 09/01/2018 16:02:51
In my theory weight of the pan is determined by the difference between opposite positive repulsive forces acting on the pan, stronger repulsive force from above and weaker repulsive force from below. Placing an electron on top of pan will reduce the repulsive force from above and reduce the difference between forces acting on the pan and its weight. Placing an electron at bottom of the pan will reduce repulsive force from below and increase the difference between forces and weight.

Hmm... so what happens if we add an electron to a single atom? Due to quantum mechanics and the wave nature of particles, the electron should be equally smeared over the surface of the atom and therefore equally above and below the atom at the same time. If the electron is equally spread out over the whole atom, then the weight should not change, right?

Quote from: Kryptid on 09/01/2018 15:43:11
The electron can be accelerated by applying a negative charge to push it or a positive charge to pull it.

And that acceleration will give the electron a positive measure of kinetic energy?
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Offline Yaniv (OP)

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Re: Experiment to test W=mg
« Reply #296 on: 09/01/2018 16:40:40 »
Quote from: Kryptid on 09/01/2018 16:09:57
Hmm... so what happens if we add an electron to a single atom? Due to quantum mechanics and the wave nature of particles, the electron should be equally smeared over the surface of the atom and therefore equally above and below the atom at the same time. If the electron is equally spread out over the whole atom, then the weight should not change, right?
Electrons are equally smeared all around an atom located far away from a gravitational force. Near a positively charged object such as the earth the distribution of electrons shifts towards the positive earth giving the atom polarity with a weak positive pole facing the earth and a strong positive pole facing away from the earth. The weak positive pole decreases the positive repulsive force from the direction of the earth and the strong positive pole increases the positive repulsive force from above pushing the atom towards the earth. The difference between forces gives weight to a stationary atom.
Quote from: Kryptid on 09/01/2018 16:09:57
And that acceleration will give the electron a positive measure of kinetic energy?
Yes.
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Offline Bored chemist

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Re: Experiment to test W=mg
« Reply #297 on: 09/01/2018 16:49:06 »
Quote from: Yaniv on 09/01/2018 16:40:40
Electrons are equally smeared all around an atom located far away from a gravitational force. Near a positively charged object such as the earth the distribution of electrons shifts towards the positive earth
We know from x-ray diffraction experiments that this is not true.

Do you accept that  your idea must be wrong because it relies on things that  are known not to be true?
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Offline Kryptid

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Re: Experiment to test W=mg
« Reply #298 on: 09/01/2018 21:24:46 »
Quote from: Yaniv on 09/01/2018 16:40:40
Electrons are equally smeared all around an atom located far away from a gravitational force. Near a positively charged object such as the earth the distribution of electrons shifts towards the positive earth giving the atom polarity with a weak positive pole facing the earth and a strong positive pole facing away from the earth. The weak positive pole decreases the positive repulsive force from the direction of the earth and the strong positive pole increases the positive repulsive force from above pushing the atom towards the earth. The difference between forces gives weight to a stationary atom.

Where does the "positive repulsive force from above" come from?

Quote
Yes.

Alright. Then we know from the kinetic energy equation Ek = (1/2)mv2 that electrons must have a positive mass. That means when we add electrons to a piece of matter (either by heating it or charging it), we are adding mass to that piece of matter because electrons have a positive mass value. This can only mean one of two things in light of your model:

(1) When objects absorb electrons by heating up, their weight decreases but their mass increases. This means that mass and weight cannot be the same thing in your model.

or

(2) Weight and mass are the same thing, which creates the paradox that adding electrons both increases and decreases the object's weight at the same time.
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Offline Colin2B

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Re: Experiment to test W=mg
« Reply #299 on: 09/01/2018 23:25:59 »
Quote from: Kryptid on 09/01/2018 21:24:46
or

(2) Weight and mass are the same thing, which creates the paradox that adding electrons both increases and decreases the object's weight at the same time.
In other words the weight doesn’t change. Given he has already stated that weight and mass are same phenomenon.

More holes than a leaky sieve.
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