[guest39538]

a.jpg (17.83 kB . 740x464 - viewed 3348 times)

[Kryptid]

Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.

[guest39538]

Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.

I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. 


Coulombs constant , might as well be Mandarin , I can't read it.

[Kryptid]

Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.

I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. 


Coulombs constant , might as well be Mandarin , I can't read it.

Then I'll try to do it for you. I'll just need some information first. How far apart are the proton and electron in your model of the hydrogen atom? Is it in agreement with the current model (i.e. approximately the Bohr radius) or is it some different value?

[guest39538]

Added- What I do know is this, your calculation uses M1 and M2


My formula uses     a+b and a+b   . 

I have 4 values to start with where you only have 2 values, my 4 values being your 2 values but 2 values per mass.

Don't worry about the gravitational equation. I only used that to show what the magnitude of attraction should be that you are trying to match by using electrical attraction. You should concentrate on Coulomb's equation instead. Given that this is your model, you should know all of the electric charges and distances involved. Putting them into the equation and calculating the resulting force should be a simple matter.

I wish it was that simple, I wish I had a Maxwell to help me, it is not that I can't do equations, it is a matter of knowing what all the symbols mean and how to present them, additionally what the numbers mean when they have -10 but the - is high or low etc, additionally when the number ends with an E, I have no idea what that suppose to mean. 


Coulombs constant , might as well be Mandarin , I can't read it.

Then I'll try to do it for you. I'll just need some information first. How far apart are the proton and electron in your model of the hydrogen atom? Is it in agreement with the current model (i.e. approximately the Bohr radius) or is it some different value?

r=0

In my model the Proton and Electron are merged and have a void in the center which is a dense field.   Sort of like a maltesers.

[Kryptid]

In my model the Proton and Electron are merged and have a void in the center which is a dense field.   Sort of like a maltesers.

...okay then.

I can already tell you that the attractive force will be exactly zero in that case. If the proton in Atom 1 is the exact same distance from the proton in Atom 2 as it is from the electron in Atom 2, then the attractive and repulsive forces will be exactly equal to each other and cancel out. If you insist on it, I can do the calculation anyway if you don't believe me. It really shouldn't be necessary though...

[guest39538]

In my model the Proton and Electron are merged and have a void in the center which is a dense field.   Sort of like a maltesers.

...okay then.

I can already tell you that the attractive force will be exactly zero in that case. If the proton in Atom 1 is the exact same distance from the proton in Atom 2 as it is from the electron in Atom 2, then the attractive and repulsive forces will be exactly equal to each other and cancel out. If you insist on it, I can do the calculation anyway if you don't believe me. It really shouldn't be necessary though...

No, the attractive force is not 0, the electron and the proton while merged retain their attractive force to retain form, if there was 0 attractive force they would not sustain being merged. 

When r=0 between two atoms , the attractive force remains between electrons and protons of each atom, the fields form bonds.

then the attractive and repulsive forces will be exactly equal to each other and cancel out

They do , an atom cannot pass through an atom because it is equally repulsive as attractive.  They can only be attracted to r=0 .  They may be able to ''squash'' other atoms out of form.

[Kryptid]

No, the attractive force is not 0, the electron and the proton while merged retain their attractive force to retain form, if there was 0 attractive force they would not sustain being merged.

I'm talking about zero attractive force between the two different atoms, not between the proton and the electron in the same atom. I'm also talking about net attraction, not gross attractions. Sure, the proton in Atom 1 will be attracted to the electron in Atom 2, but it is also being repelled by the proton in Atom 2. Since protons have electric charge equally strong to that of electrons, and since you have defined that protons and electrons are at the same place in the same atom, then the attraction and repulsion between the atoms have to be equal to each other. They are pulling on each other just as strongly as they are pushing away from each other. The net force has to be zero.

When r=0 between two atoms , the attractive force remains between electrons and protons of each atom, the fields form bonds.

The repulsive forces remain between the protons in different atoms and between the electrons in different atoms as well.

They do , an atom cannot pass through an atom because it is equally repulsive as attractive.

Then you agree that two stationary hydrogen atoms separated by a distance of 1 nanometer would not move each other. The only way they could move towards each other would be if the attraction between the two atoms is greater than the repulsion between them. That can't happen if the forces of attraction and repulsion are equal.

[guest39538]

I'm talking about zero attractive force between the two different atoms, not between the proton and the electron in the same atom. I'm also talking about net attraction, not gross attractions. Sure, the proton in Atom 1 will be attracted to the electron in Atom 2, but it is also being repelled by the proton in Atom 2. Since protons have electric charge equally strong to that of electrons, and since you have defined that protons and electrons are at the same place in the same atom, then the attraction and repulsion between the atoms have to be equal to each other. They are pulling on each other just as strongly as they are pushing away from each other. The net force has to be zero.

I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging.

[Kryptid]

I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging.

So how do the attractive forces between the two atoms become stronger than the repulsive forces between the same two atoms? The repulsive force between two protons should be exactly equal to the attractive force between a proton and an electron (at the same distance apart).

Let me do the equations. I'll consider a pair of hydrogen atoms one nanometer apart. I'll label the proton in the first atom P1, the proton in the second atom P2, the electron in the first atom as E1 and the electron in the second atom as E2. Between these two atoms, the possible interactions are P1P2 (proton-proton repulsion), P1E2 (proton-electron attraction), P2E1 (proton-electron attraction) and E1E2 (electron-electron repulsion). Take note that P1E1 and P2E2 are not considered because they are interactions within the atoms, not between the atoms. Now let's calculate the force involved in each interaction. Positive numbers indicate repulsion, negative numbers indicate attraction:

F_P1P2 = k((Q₁Q₂)/r²)
F_P1P2 = (8.9875 x 10⁹ N m² C⁻²)((1.602 x 10^-19 C)(1.602 x 10^-19 C)/(10^-9 m)²)
F_P1P2 = (8.9875 x 10⁹)((2.566 x 10^-38)/(10^-9)²)
F_P1P2 = (8.9875 x 10⁹)((2.566 x 10^-38)/10^-18)
F_P1P2 = (8.9875 x 10⁹)(2.566 x 10^-20)
F_P1P2 = 2.306 x 10^-10 newtons

F_P1E2 = k((Q₁Q₂)/r²)
F_P1E2 = (8.9875 x 10⁹ N m² C⁻²)((1.602 x 10^-19 C)(-1.602 x 10^-19 C)/(10^-9 m)²)
F_P1E2 = (8.9875 x 10⁹)((-2.566 x 10^-38)/(10^-9)²)
F_P1E2 = (8.9875 x 10⁹)((-2.566 x 10^-38)/10^-18)
F_P1E2 = (8.9875 x 10⁹)(-2.566 x 10^-20)
F_P1E2 = -2.306 x 10^-10 newtons

F_P2E1 = k((Q₁Q₂)/r²)
F_P2E1 = (8.9875 x 10⁹ N m² C⁻²)((1.602 x 10^-19 C)(-1.602 x 10^-19 C)/(10^-9 m)²)
F_P2E1 = (8.9875 x 10⁹)((-2.566 x 10^-38)/(10^-9)²)
F_P2E1 = (8.9875 x 10⁹)((-2.566 x 10^-38)/10^-18)
F_P2E1 = (8.9875 x 10⁹)(-2.566 x 10^-20)
F_P2E1 = -2.306 x 10^-10 newtons

F_E1E2 = k((Q₁Q₂)/r²)
F_E1E2 = (8.9875 x 10⁹ N m² C⁻²)((-1.602 x 10^-19 C)(-1.602 x 10^-19 C)/(10^-9 m)²)
F_E1E2 = (8.9875 x 10⁹)((2.566 x 10^-38)/(10^-9)²)
F_E1E2 = (8.9875 x 10⁹)((2.566 x 10^-38)/10^-18)
F_E1E2 = (8.9875 x 10⁹)(2.566 x 10^-20)
F_E1E2 = 2.306 x 10^-10 newtons

Now let's add up all of the forces and see what the net attraction or repulsion is:

F_net = F_P1P2 + F_P1E2 + F_P2E1 + F_E1E2
F_net = (2.306 x 10^-10 newtons) + (-2.306 x 10^-10 newtons) + (-2.306 x 10^-10 newtons) + (2.306 x 10^-10 newtons)
F_net = 0 newtons

There's the math for you. There is no net electrostatic force between two hydrogen atoms when they have the structure that you propose that they have.

[guest39538]

I was talking about two different atoms. The force between the two atoms is not 0, both atoms want to merge but can't because the repulsive force stops it merging.

So how do the attractive forces between the two atoms become stronger than the repulsive forces between the same two atoms? The repulsive force between two protons should be exactly equal to the attractive force between a proton and an electron (at the same distance apart).

Let me do the equations. I'll consider a pair of hydrogen atoms one nanometer apart. I'll label the proton in the first atom P1, the proton in the second atom P2, the electron in the first atom as E1 and the electron in the second atom as E2. Between these two atoms, the possible interactions are P1P2 (proton-proton repulsion), P1E2 (proton-electron attraction), P2E1 (proton-electron attraction) and E1E2 (electron-electron repulsion). Take note that P1E1 and P2E2 are not considered because they are interactions within the atoms, not between the atoms. Now let's calculate the force involved in each interaction. Positive numbers indicate repulsion, negative numbers indicate attraction:

F_P1P2 = k((Q₁Q₂)/r²)
F_P1P2 = (8.9875 x 10⁹ N m² C⁻²)((1.602 x 10^-19 C)(1.602 x 10^-19 C)/(10^-9 m)²)
F_P1P2 = (8.9875 x 10⁹)((2.566 x 10^-38)/(10^-9)²)
F_P1P2 = (8.9875 x 10⁹)((2.566 x 10^-38)/10^-18)
F_P1P2 = (8.9875 x 10⁹)(2.566 x 10^-20)
F_P1P2 = 2.306 x 10^-10 newtons

F_P1E2 = k((Q₁Q₂)/r²)
F_P1E2 = (8.9875 x 10⁹ N m² C⁻²)((1.602 x 10^-19 C)(-1.602 x 10^-19 C)/(10^-9 m)²)
F_P1E2 = (8.9875 x 10⁹)((-2.566 x 10^-38)/(10^-9)²)
F_P1E2 = (8.9875 x 10⁹)((-2.566 x 10^-38)/10^-18)
F_P1E2 = (8.9875 x 10⁹)(-2.566 x 10^-20)
F_P1E2 = -2.306 x 10^-10 newtons

F_P2E1 = k((Q₁Q₂)/r²)
F_P2E1 = (8.9875 x 10⁹ N m² C⁻²)((1.602 x 10^-19 C)(-1.602 x 10^-19 C)/(10^-9 m)²)
F_P2E1 = (8.9875 x 10⁹)((-2.566 x 10^-38)/(10^-9)²)
F_P2E1 = (8.9875 x 10⁹)((-2.566 x 10^-38)/10^-18)
F_P2E1 = (8.9875 x 10⁹)(-2.566 x 10^-20)
F_P2E1 = -2.306 x 10^-10 newtons

F_E1E2 = k((Q₁Q₂)/r²)
F_E1E2 = (8.9875 x 10⁹ N m² C⁻²)((-1.602 x 10^-19 C)(-1.602 x 10^-19 C)/(10^-9 m)²)
F_E1E2 = (8.9875 x 10⁹)((2.566 x 10^-38)/(10^-9)²)
F_E1E2 = (8.9875 x 10⁹)((2.566 x 10^-38)/10^-18)
F_E1E2 = (8.9875 x 10⁹)(2.566 x 10^-20)
F_E1E2 = 2.306 x 10^-10 newtons

Now let's add up all of the forces and see what the net attraction or repulsion is:

F_net = F_P1P2 + F_P1E2 + F_P2E1 + F_E1E2
F_net = (2.306 x 10^-10 newtons) + (-2.306 x 10^-10 newtons) + (-2.306 x 10^-10 newtons) + (2.306 x 10^-10 newtons)
F_net = 0 newtons

There's the math for you. There is no net electrostatic force between two hydrogen atoms when they have the structure that you propose that they have.

That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.

Imagine weighing two identical mass on pan scales to measure 0. 


force scales.1.jpg (15.89 kB . 731x461 - viewed 3310 times)


The Universe works on a critical balanced system.

[Kryptid]

That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.

The net force absolutely is zero. If there is more force on the attractive side than the repulsive side, please explain where it comes from. The numbers certainly don't show it.

Imagine weighing two identical mass on pan scales to measure 0.

That's a great analogy, actually. Even though both masses are experiencing a force pulling them down towards the Earth, they are keeping each other from tipping the scales in either direction. Their forces are equal so that the scale doesn't move. Same thing with two hydrogen atoms. The attractions and repulsions are equal, so there is no movement.

[guest39538]

That looks complex and the answer should be 0 , but 0 means a balance of 0 not no force.

The net force absolutely is zero. If there is more force on the attractive side than the repulsive side, please explain where it comes from. The numbers certainly don't show it.

Imagine weighing two identical mass on pan scales to measure 0.

That's a great analogy, actually. Even though both masses are experiencing a force pulling them down towards the Earth, they are keeping each other from tipping the scales in either direction. Their forces are equal so that the scale doesn't move. Same thing with two hydrogen atoms. The attractions and repulsions are equal, so there is no movement.

The attractive force and repulsive force is equal of the atoms, if the field density of the atoms were greater the atom would not displace through the field to adjoin the other atom. The fields repulsion is not as great as the atoms attraction, the atoms repulsion solidity means 0 radius but solidity.

Yes you understood the analogy , hopefully now can see why I think gravity be related to this. 

I think you measure the net charge as zero, but you measure the net force of the charge as gravity.  I did say ages ago I thought science already had the gravity mechanism answer.

Added- Hence N-field particle and n-field.

[guest39538]

A simple experiment to show something

Equipment

Two people

One pen


Method

Person one - On their left hand draw a plus sign

On their right hand draw a negative sign

Lock their thumbs together so hands are joined


Person Two - On their left hand draw a negative sign

On their right hand draw a positive sign

Lock their thumbs together joining hands


Person one and two

Lock your hands with your fingers to the other persons hands,


You both are now observing bonding.


I give permission for schools to use this to show children how bonding works , an easy way to explain molecular construction.


You left hand is attracted to your right hand and vice versus.  Your left hand is attracted to their right hand and your right hand is attracted to their left hand.  So on......

[Kryptid]

The fields repulsion is not as great as the atoms attraction

This is the part that you have not demonstrated. The equations show quite clearly that the attractions and repulsions are equal. Where are you getting this extra attraction from? It certainly isn't from electrostatic attraction.

[guest39538]

The fields repulsion is not as great as the atoms attraction

This is the part that you have not demonstrated. The equations show quite clearly that the attractions and repulsions are equal. Where are you getting this extra attraction from? It certainly isn't from electrostatic attraction.

You are not considering this correctly , consider two rubber balls and squeeze them together, that is the repulsion because of field density N-field particle.   


Move them a length apart , the spacial field between them is not dense enough to stop them being drawn together by force to a 0 radius.

Now consider magnetic suspension, the field is dense enough to stop the object falling.   

added - i.e Quantum field physicality.  ( Q.F.P)

[Kryptid]

You are not considering this correctly

So you're saying that Coulomb's law is wrong? That's quite a bold assertion.

consider two rubber balls and squeeze them together, that is the repulsion because of field density N-field particle.

Atoms aren't rubber balls, but I don't really care about repulsion at this time. I'm trying to understand why you think two forces acting in equal and opposite directions somehow adds up to make net attraction...

Move them a length apart , the spacial field between them is not dense enough to stop them being drawn together by force to a 0 radius.

What force? It isn't electrostatic. I've already shown mathematically via Coulomb's law that it can't be.

[guest39538]

So you're saying that Coulomb's law is wrong? That's quite a bold assertion.

No, you read that wrong

What force? It isn't electrostatic. I've already shown mathematically via Coulomb's law that it can't be.

It is , no net charge does not mean no net force.  You must be reading wrongly.

[guest39538]

22.jpg (26.08 kB . 740x464 - viewed 3215 times)

[guest39538]

0.5² = 0.25 = 1/4
*2+*2 =*4
0.5²/4=0.0681

No idea at this time what this means.