[Kryptid]

Anything that is neutral could be plus and neg

I'm afraid not. Objects which contain both positive and negative charges have magnetic moments and are capable of participating in electromagnetic interactions. The neutron, for example, is neutral but contains internal electric charges which give it a magnetic moment. Such is not true for z-bosons, neutrinos or photons. All of those particles are either without any magnetic moment at all or have one that is too small to measure. Neutral black holes should not have magnetic moments either, as they are not composed of anything more fundamental than themselves. Neutrinos are also known not to interact with the electromagnetic force at all, further showing that they do not contain any electric fields.

[guest39538]

Anything that is neutral could be plus and neg

I'm afraid not. Objects which contain both positive and negative charges have magnetic moments and are capable of participating in electromagnetic interactions. The neutron, for example, is neutral but contains internal electric charges which give it a magnetic moment. Such is not true for z-bosons, neutrinos or photons. All of those particles are either without any magnetic moment at all or have one that is too small to measure. Neutral black holes should not have magnetic moments either, as they are not composed of anything more fundamental than themselves. Neutrinos are also known not to interact with the electromagnetic force at all, further showing that they do not contain any electric fields.

What if our equipment is just simply not advanced enough to measure the electrical properties of any sub atomic particle?

The affects so minute . they are almost negligible.

[Kryptid]

What if our equipment is just simply not advanced enough to measure the electrical properties of any sub atomic particle?

The affects so minute . they are almost negligible.

That has its own consequences. If you propose that gravity is actually the result of electromagnetic attraction, then more total electric charges result in stronger gravitational fields. This, in turn, means that objects that contain more total charge should be more massive than those with less total charge (since mass is seen to correlate with gravity). If this was true, then particles that are more massive than the proton must necessarily contain more electric charges than it does and in turn must have a higher magnetic moment than the proton (more electric charge results in stronger magnetic fields).

The z-boson violates this as it is almost 100 times heavier than the proton. If its very large mass was caused by a large amount of internal charge (as your model posits), then it must also have a magnetic moment much higher than that of the proton. Yet the opposite is true. The proton has a measurable magnetic moment while the z-boson does not.

[guest39538]

What if our equipment is just simply not advanced enough to measure the electrical properties of any sub atomic particle?

The affects so minute . they are almost negligible.

That has its own consequences. If you propose that gravity is actually the result of electromagnetic attraction, then more total electric charges result in stronger gravitational fields. This, in turn, means that objects that contain more total charge should be more massive than those with less total charge (since mass is seen to correlate with gravity). If this was true, then particles that are more massive than the proton must necessarily contain more electric charges than it does and in turn must have a higher magnetic moment than the proton (more electric charge results in stronger magnetic fields).

The z-boson violates this as it is almost 100 times heavier than the proton. If its very large mass was caused by a large amount of internal charge (as your model posits), then it must also have a magnetic moment much higher than that of the proton. Yet the opposite is true. The proton has a measurable magnetic moment while the z-boson does not.

The Z bosons are made up of W bosons?

F²

[Kryptid]

The Z bosons are made up of W bosons?

F²

That doesn't avoid the problem, since W bosons have electric charge equal in magnitude to that of the proton. Therefore, a pair of W bosons should still have a high magnetic moment. Secondly, a pair of W bosons would weigh more than a Z boson (W bosons weigh 80.4 GeV while the Z boson weighs 91.2 GeV).

EDIT: I've actually made a mistake. After doing some further research, it turns out that particles with internal electric charges can actually have a magnetic moment of zero. Alpha particles are an example of such a particle.

As I've said before, I urge you to calculate how strong the attractive force between two objects would be using your proposed model of gravity and see if it matches the actual force of gravity. You can start with something very simple like a pair of hydrogen atoms that are separated by some given distance (say, one nanometer). Use Coulomb's law to calculate the force acting between each electric charge and see what it all adds up to. Here is a link that can help you with your calculations: https://socratic.org/questions/what-is-the-coulomb-force-between-two-electrons-that-are-on-opposite-sides-of-th

[guest39538]

The Z bosons are made up of W bosons?

F²

That doesn't avoid the problem, since W bosons have electric charge equal in magnitude to that of the proton. Therefore, a pair of W bosons should still have a high magnetic moment. Secondly, a pair of W bosons would weigh more than a Z boson (W bosons weigh 80.4 GeV while the Z boson weighs 91.2 GeV).

EDIT: I've actually made a mistake. After doing some further research, it turns out that particles with internal electric charges can actually have a magnetic moment of zero. Alpha particles are an example of such a particle.

As I've said before, I urge you to calculate how strong the attractive force between two objects would be using your proposed model of gravity and see if it matches the actual force of gravity. You can start with something very simple like a pair of hydrogen atoms that are separated by some given distance (say, one nanometer). Use Coulomb's law to calculate the force acting between each electric charge and see what it all adds up to. Here is a link that can help you with your calculations: https://socratic.org/questions/what-is-the-coulomb-force-between-two-electrons-that-are-on-opposite-sides-of-th

I will try to work that out, in the meantime


q1 + q2 = m

How do we measure the force of gravity again?

[Kryptid]

q1 + q2 = m

How do we measure the force of gravity again?

F = G((M₁M₂)/r²)

Where

F is the force of gravitational attraction (in newtons)
G is the gravitational constant (6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²)
M₁ is the mass of the first object (in kilograms)
M₂ is the mass of the second object (in kilograms)
r is the distance between them (in meters)

[guest39538]

G is the gravitational constant (6.674×10−11 m3⋅kg−1⋅s−2)

What does a gravitational constant mean? 

I don't understand  this part.

F=(6.674×10−11 m3⋅kg−1⋅s−2)(m1m2)/r²


Am I reading that correctly?

Or are you saying G = m₁m₂/r²

[guest39538]

My brain is overloading this is not computing,


kilogram * kilogram divided by radius squared is just not computing in my brain as being associative to gravity.  Kilogram is the consequence of G, it sounds messy to me. Scratches head.

[guest39538]

I get


G_F = (F₁ + F₂)  + (F₁ + F₂)


((F₁ + F₂)  + (F₁ + F₂)) > (F₁ + F₂)

My calculation is calculating from a point perspective, your calculation is the force over a radius?

[Kryptid]

What does a gravitational constant mean?

It's a measure of how much gravity is produced by a given amount of mass.

I don't understand  this part.

F=(6.674×10−11 m3⋅kg−1⋅s−2)(m1m2)/r²

Am I reading that correctly?

You don't need to completely understand the units being used here just so long as you can do the calculation. The result will be in newtons anyway, which is easy to understand.

Or are you saying G = m₁m₂/r²

Nope, the first version is the correct one. G is a constant.

My brain is overloading this is not computing,

kilogram * kilogram divided by radius squared is just not computing in my brain as being associative to gravity.  Kilogram is the consequence of G, it sounds messy to me. Scratches head.

The strength of gravity felt at a given location is determined by both mass and distance, hence the kilogram and meters.

At any rate, I already did the calculation for the amount of gravitational force between two hydrogen atoms separated by 1 nanometer. The result was 1.869 x 10^-46 newtons.

I get


G_F = (F₁ + F₂)  + (F₁ + F₂)


((F₁ + F₂)  + (F₁ + F₂)) > (F₁ + F₂)

My calculation is calculating from a point perspective, your calculation is the force over a radius?

Yes, the force over a given distance is what I'm looking for. How much attractive force does your N-field produce between two hydrogen atoms separated by 1 nanometer? That's the calculation you need to do.

[guest39538]

What does a gravitational constant mean?

It's a measure of how much gravity is produced by a given amount of mass.

I don't understand  this part.

F=(6.674×10−11 m3⋅kg−1⋅s−2)(m1m2)/r²

Am I reading that correctly?

You don't need to completely understand the units being used here just so long as you can do the calculation. The result will be in newtons anyway, which is easy to understand.

Or are you saying G = m₁m₂/r²

Nope, the first version is the correct one. G is a constant.

My brain is overloading this is not computing,

kilogram * kilogram divided by radius squared is just not computing in my brain as being associative to gravity.  Kilogram is the consequence of G, it sounds messy to me. Scratches head.

The strength of gravity felt at a given location is determined by both mass and distance, hence the kilogram and meters.

At any rate, I already did the calculation for the amount of gravitational force between two hydrogen atoms separated by 1 nanometer. The result was 1.869 x 10^-46 newtons.

I get


G_F = (F₁ + F₂)  + (F₁ + F₂)


((F₁ + F₂)  + (F₁ + F₂)) > (F₁ + F₂)

My calculation is calculating from a point perspective, your calculation is the force over a radius?

Yes, the force over a given distance is what I'm looking for. How much attractive force does your N-field produce between two hydrogen atoms separated by 1 nanometer? That's the calculation you need to do.

Shrugs shoulders and looks lost

(8.9875×109 N m² C−2)^2 + (8.9875×109 N m² C−2)^2

I have no idea what I am doing with this math , but the above is what I need i think, cant make the  -  ,  go high either or the 9

[Kryptid]

Shrugs shoulders and looks lost

(8.9875×109 N m² C−2)^2 + (8.9875×109 N m² C−2)^2

I have no idea what I am doing with this math , but the above is what I need i think, cant make the  -  ,  go high either or the 9

You do superscripts with [ sup][ /sup] and subscripts with [ sub][ /sub]. Just don't put the spaces in there like I did in this particular post.

When I get back, I'll post a step-by-step of how I would do the calculations.

[guest39538]

Shrugs shoulders and looks lost

(8.9875×109 N m² C−2)^2 + (8.9875×109 N m² C−2)^2

I have no idea what I am doing with this math , but the above is what I need i think, cant make the  -  ,  go high either or the 9

You do superscripts with [ sup][ /sup] and subscripts with [ sub][ /sub]. Just don't put the spaces in there like I did in this particular post.

When I get back, I'll post a step-by-step of how I would do the calculations.

Ok thank you and here is the way I do step by step calculations.


m1.jpg (19.27 kB . 740x464 - viewed 3588 times)

I am not sure what this means yet, but I get 0.25 as a result for m1 by doing 0.5²

1/4 = g ? 

[Kryptid]

Ok thank you and here is the way I do step by step calculations.


m1.jpg (19.27 kB . 740x464 - viewed 3588 times)

I am not sure what this means yet, but I get 0.25 as a result for m1 by doing 0.5²

1/4 = g ? 

So what does that translate to in actual units?

Here are a couple of calculations. The first one is for the force of gravitational attraction between two hydrogen atoms 1 nanometer apart:

F = G((M₁M₂)/r²)
F = (6.674 x 10⁻¹¹ m³⋅kg⁻¹⋅s⁻²)((1.674 x 10^-27 kg)(1.674 x 10^-27 kg)/(10^-9)²)
F = (6.674 x 10⁻¹¹)((2.802 x 10^-54)/(10^-9)²)
F = (6.674 x 10⁻¹¹)((2.802 x 10^-54)/(10^-18))
F = (6.674 x 10^-11)(2.802 x 10^-36)
F = 1.87 x 10^-46 newtons

This second one is the force of electrical attraction between a proton and an electron separated by a distance of 1 Bohr radius (approximately the radius that an electron is most likely to be found at in a hydrogen atom):

F = k((Q₁Q₂)/r²)
F = (8.9875 x 10⁹ N m² C⁻²)((1.602 x 10^-19 C)(1.602 x 10^-19 C)/(5.291 x 10^-11 m)²)
F = (8.9875 x 10⁹)((2.566 x 10^-38)/(5.291 x 10^-11)²)
F = (8.9875 x 10⁹)((2.566 x 10^-38)/(2.799 x 10^-21))
F = (8.9875 x 10⁹)(9.1676 x 10^-18)
F = 8.239 x 10^-8 newtons

[guest39538]

So what does that translate to in actual units?

Honestly I have no idea at this time, I think I need to work out what ''your'' numbers mean firstly.   Before I have took Pi , recalculated Pi to use / work out m/s  . I can't remember now I did that but it is on the internet somewhere in a thread.

I am not sure , perhaps I will have to invent my own values.   

Sorry I could  not be more helpful, my formulas work in my mind but I do struggle to give the formulas an end value.


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.


Added- Eureka,

G= (F=ma)

An object at relative rest is still accelerating ''downwards'', 


G = (q1+q2)a

That might be squared or times 2.

G= (q1+q2)²a

or

G= (q1+q2)a²


r= 0 so I need no radius in my calculation.

added-

F=g(ma²)


added- M1 is only half the mass of the measure.

1kg = 0.5kg


added- invention

Double sided/faced set of scales .

I claim intellectual rights and copy rights to my double sided/faced scales.  Under common law I said it first, so anyone nicking my idea will be sued under this formal disclaimer and oral patent.
My intellectual rights are protected by this thread and forum, anyone using my idea will be using my intellectual rights so therefore cannot copy my idea.

S.P.Leese

23/03/2018





 

[guest39538]

d scales.jpg (43.62 kB . 740x464 - viewed 3402 times)

[guest39538]

Typo error , re-done


d scales.jpg (45.02 kB . 740x464 - viewed 3394 times)

[guest39538]

myth.jpg (27.04 kB . 740x464 - viewed 3396 times)

[guest39538]

A 1 kg object in free fall


F=1/2(m)a2

So at relative rest a 1 kg object is applying 4.905N

From a 1 m fall , F = 9.81N