Thornhill's Latest Gravity Presentation

Plasma and electricity in space. Failure of gravity-only cosmology. Exposing the myths of dark matter, dark energy, black holes, neutron stars, and other mathematical constructs. The electric model of stars. Predictions and confirmations of the electric comet.

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willendure
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Re: Thornhill's Latest Gravity Presentation

Bengt Nyman wrote: So you think you understand your computer, because you know how to use it.
In my case, this is completely mis-directed... I graduated in computer science at a top university; I understand how my computer works from the silicon up...

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

willendure wrote: In that case, please enlighten us. How does the geometry work, so that the hydrogen atoms you are modelling attract one another, no matter what their orientation?
Take one hydrogen atom, free from any external influences.
Take a second hydrogen atom and introduce it to the first hydrogen atom.
At first the atoms look nothing like my diagram. They are round, centered and pretty.
However, with the speed of gravity the electron in atom 1 senses the electron in atom 2 and attempts to push it away. But doing so the proton in atom 1 objects, it likes the electron in atom 2.
What's atom 1 to do ?
"My proton wants to be a little closer to your electron, but please keep your proton out of the way," says atom 1.
"Ok!" agrees atom 2.

The rest is history, and gravity.

I realize that you are asking about the bigger picture, with more atoms influencing each other. It becomes very complicated very fast with each atomic relationship creating one set of at least 4 vectors and where all the vectors affecting one atom combines into one resultant dipole gravity vector for that specific atom. See strong force simulations etc.
Now move on to a multi-atom body surrounded by several other bodies, and It very quickly exceeds our visualization capabilities.

I hope Querious is listening as he ponders his stubborn aluminum foil.

kaublezw
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Re: Thornhill's Latest Gravity Presentation

Bengt - Your hypothesis is intriguing, and I enjoyed watching your simuluations. How do you propose testing it experimentally?

willendure
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Re: Thornhill's Latest Gravity Presentation

Bengt Nyman wrote: Take one hydrogen atom, free from any external influences.
Take a second hydrogen atom and introduce it to the first hydrogen atom.
At first the atoms look nothing like my diagram. They are round, centered and pretty.
However, with the speed of gravity the electron in atom 1 senses the electron in atom 2 and attempts to push it away. But doing so the proton in atom 1 objects, it likes the electron in atom 2.
What's atom 1 to do ?
"My proton wants to be a little closer to your electron, but please keep your proton out of the way," says atom 1.
"Ok!" agrees atom 2.
Continuing on this flight of fancy...

Attraction = e^2 (0.81 - 1.21 - 1 - 1) = e^2(0.9*0.9 + 1.1*1.1 - 1*1 - 1*1)

In the case where the distance between the atoms is 1 (unit). Lets suppose the distance is actually r units:

Attraction = e^2 ((r-0.1)*(r-0.1)+(r+0.1)*(r+0.1) - 2(r*r))
= e^2 (r^2 -0.2r + 0.01 + r^2 + 0.2r + 0.01 - 2r^2)
= e^2 (2r^2 - 2r^2 + 0.2r - 0.2r + 0.02)
= e^2 (0.02)

Giving rise to a gravity that does not drop off as 1/r^2, as we observe in reality, and instead giving rise to a constant force, irrespective of distance.

If that were true, Newton would have written f = GmM not f = GmM/r^2.

I do believe this matter is settled.

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

kaublezw wrote:Bengt - Your hypothesis is intriguing, and I enjoyed watching your simulations. How do you propose testing it experimentally?
Perhaps if we could measure the distance between hydrogen protons in a small cloud of hydrogen atoms, distant from the earth and in a vacuum, we could show:
1. They like to stick together.
2. The distance between protons on the surface of the cloud is smaller tangentially than radially and smaller than in the middle of the cloud, illustrating dipole elongation pointing toward the middle of the cloud.

It might also be possible to do it with larger bodies on the surface of the earth. The problem is, how do you prove that it is the cause of gravity and not just the result of gravity and subsequent body deformation.

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

willendure wrote: I do believe this matter is settled.
Are you saying that I made a mistake writing up the formula ?
Or are you objecting to the fact that 1^2 = 1
OOPS, I see it now:
1/0.81 + 1/1.21 - 1/1 - 1/1 = 1.2345679 + 0.8264462 - 1 - 1 = 0.0610141
The formula is right, just as Coulomb's law stipulates it. Look again.
However, sloppy calculus on my part.
The gravitational vector is 0.0610141, not 0.0201.

Your grandiose conclusion is again premature. But please keep trying.
Last edited by Bengt Nyman on Fri Sep 11, 2015 8:46 am, edited 2 times in total.

willendure
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Re: Thornhill's Latest Gravity Presentation

Bengt Nyman wrote:
kaublezw wrote:Bengt - Your hypothesis is intriguing, and I enjoyed watching your simulations. How do you propose testing it experimentally?
Perhaps if we could measure the distance between hydrogen protons in a small cloud of hydrogen atoms, distant from the earth and in a vacuum, we could show:
1. They like to stick together.
2. The distance between protons on the surface of the cloud is smaller tangentially than radially and smaller than in the middle of the cloud, illustrating dipole elongation pointing toward the middle of the cloud.

It might also be possible to do it with larger bodies on the surface of the earth. The problem is, how do you prove that it is the cause of gravity and not just the result of gravity and subsequent body deformation.
Who cares, because as I already pointed out, two metal spheres attract horizontally when positioned next to each other on the earth, thus showing that gravity does not only pull downwards.

Your dipoles can only pull one way, and gravity pulls all ways.

willendure
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Re: Thornhill's Latest Gravity Presentation

Bengt Nyman wrote:
willendure wrote: I do believe this matter is settled.
Are you saying that I made a mistake writing up the formula ?
Or are you objecting to the fact that 1^2 = 1
All I did was to use your formulas to calculate that you are wrong.

Enough of this, I'll let the arguments that I have already made stand by themselves.

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

willendure wrote: All I did was to use your formulas to calculate that you are wrong.
Enough of this, I'll let the arguments that I have already made stand by themselves.
No, the formula is right so you can not have calculated anything or found anything wrong with the theory.
I made a mistake in the calculation. Thank you for pointing that out. But the result comes out qualitatively the same.

Your grandiose conclusion is still wrong, or at least premature. Please try again.
Last edited by Bengt Nyman on Fri Sep 11, 2015 8:55 am, edited 1 time in total.

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

willendure wrote: ... gravity pulls all ways.
I agree.
BUT, you are still confusing my dipole gravity with Wallace. They are not the same. http://www.dipole.se
The error you found is being corrected. Thank You.

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

Dipole attraction and subsequent dipole gravity is always positive though very weak. See arbitrary example below. For a more complete description see http://www.dipole.se

Attraction = e^2/0.9^2 + e^2/1.1^2 - e^2/1^2 - e^2/1^2
= e^2(1/0.81 + 1/1.21 - 1/1 - 1/1
= e^2(1.23456790 + 0.82644628 - 1 - 1)
= e^2(0.06101418)
= 0.061e^2
Attachments

querious
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Re: Thornhill's Latest Gravity Presentation

Bengt Nyman wrote:Dipole attraction and subsequent dipole gravity is always positive though very weak. See arbitrary example below. For a more complete description see http://www.dipole.se

Attraction = e^2/0.9^2 + e^2/1.1^2 - e^2/1^2 - e^2/1^2
= e^2(1/0.81 + 1/1.21 - 1/1 - 1/1
= e^2(1.23456790 + 0.82644628 - 1 - 1)
= e^2(0.06101418)
= 0.061e^2
Ignoring for now the other copious & obvious problems with this theory, just demonstrate, like above, that:

1) A small change in overall separation results in a force change of 1/r2, like gravity, rather than 1/r4, as dipoles are known to do.

2) How adding a 3rd atom results in the same forces between them.

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

"querious"
1. This is a calculation based on coulombs law treating charges separately involving /r^2 as it should.
Calculations of magnetic dipoles present a totally different situation involving /r^4.

2. Adding a third atom does not change the force vectors between A1 and A2, however it adds force vectors between A1 and A3 and between A2 and A3. The result on for example A1 is a compound force vector which totally changes the compound force acting on A1, in direction as well as in amplitude.

Bengt Nyman
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Re: Thornhill's Latest Gravity Presentation

willendure wrote: In my case, this is completely mis-directed... I graduated in computer science at a top university; I understand how my computer works from the silicon up...
Excellent, and good for you.
I choose physics.

querious
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Joined: Mon Jun 23, 2008 8:29 pm

Re: Thornhill's Latest Gravity Presentation

Bengt Nyman wrote:
"querious"
1. This is a calculation based on coulombs law treating charges separately involving /r^2 as it should.
Calculations of magnetic dipoles present a totally different situation involving /r^4.
As the dipoles move further apart, the distortion they cause in each other must ALSO reduce. So you have that field strength reduction ON TOP OF the reduction in force caused by increasing overall separation.

I don't see that taken into account in your simple calculation. The loss would be bigger than 1/r2.

Gravity doesn't do that.

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