The Gravitation Constant

[David F.]

Thread Closed

 

[AndersHermansson]

You're making one mistake that i can see. The neutron decays into a proton an electron and a neutrino, but that doesnt mean that the neutron is made out of them, or "is just an electron and a proton which have combined" as you put it.

I'm having trouble following your line of thought. Can you show mathematically what you mean by:

"G has absorbed a 1/4pi factor"

"If we take the factor back out by muliplying by 4pi"

I don't understand how you got to:
G*4pi = m/(4pi*q)

 

[1100f]

Outside a nucleus, the Neutron decays into a Proton and an Electron (half life about 10s). Within the nucleus, there is a process called "Electron Capture" which combines an Electron and a Proton. In fusing 4 H atoms into He, first combine 2 H into Deuterium which is one Neutron, one Proton and one Electron (one H combines into a neutron), then combine two Deuterium into He. This seems to be a very reversable process (there are probably some neutrinos involved).

The basic process in neutron decay us that one of its d quark "radiates" a W^- boson. The boson carrying flavor, the d quark changes also its flavor and is transformed int a u quark. The W^- then decays into an electron and an electron-antineutrino, the same way that a charged particle can radiates a photon and that photon decays into an electron-positron pair. you don't say that the photon contains an electron. The same way, the neutron does not contain any electron.

I don't know how to calculate the mass/charge ratio of a Neutron except to work with its components. But forget all this... the equation is very close without it.

Math:

F=G*M1*M2/r^2
F=G*4pi*M1*M2/(4pi*r^2)
F=G'*M1*M2/(4pi*r^2)

G'=4pi*G=m/(4pi*q)

I didn't derive this last formula, I just noticed that the G' factor was equal to the mass/charge ratio of the particles, error less than 1%. I don't know why this is true but it is so close, I have trouble believing it is coincidence.

G is a universal constant. It does not depend on the mass that generates the gravitational field. The same is for the constant in the electric field.
The constant in Coulombs law does not depend on the electric charge of the particle that generates the electric field.
look for example at 2 point particles: the electron and the muon, they have the same electric charge and different masses, so you cannot say that G is different for each of these particles. They generate the same electric field because they have the same charge, but different gravitationnal field, not because they have different values of G, but because they have different masses.

BTW. In general relativity, instead of G being the gravitational constant, it is divided by 4Π and the result is called k.

 

[Zarkov]

>> Does this indicate some connection between electrostatic charge and gravity?

You are getting the hang of what gravity actually is. Good work !!!!!!!!!!!!!!!!!!!!

>> I don't know why this is true but it is so close, I have trouble believing it is coincidence.


Believe it, gravity is the resulting force produced by non-inertial acceleration in a poynting like field (orthogonally crossed electric-magnetic spherical energy field).
"the charge" in this case is a magnetic electric-charge (proton/neutron) that by superposition manifests the displayed spherical energy field around all material bodies.... however it is the spin of this field that results in 'gravity'.

 

[1100f]

This equation suggests that G might vary for masses with different composition. However, for any mass made up of normal matter (protons, neutrons, electrons...) the varience will certainly be less than 1%. This may not be true of sub-atomic particles, but the difference is still many orders of magnitude less than the electric fields the particles produce. Nothing would change in the electrostatic interactions between subatomic particles since gravity is simply not strong enough to be a consideration.

it was you who suggested that F = GM1M2/r^2 and not G(M1,M2)M1M2/r^2.

 

[1100f]

G = (mp + mn + me)/2 /[(4pi)^2*q]

Which q do you put in this equation?
the electric charge of the neutron? Or maybe the electric charge of the proton? In that case why not the electric charge of the electron?

 

[James R]

David:

Pay no attention to Zarkov. He has his own half-baked "theory" of gravity, which has been proved incorrect and useless.

 

[Zarkov]

>> "theory" of gravity, which has been proved incorrect and useless.

If only James R, there is only positive results, mmmmh, but beyond your understanding, sigh.

Not one error has been highlighted....

 

[James R]

Zarkov, your post is off-topic. Take it somewhere else.

 

[1100f]

There is only one q in the universe. We add a negative to it for electrons, and there are actually two in a Neutron and they cancel each other.

Why only one q? Maybe it has to do with something like the maximum quantum energy density of the aether -- who knows? I'll bet it is something fundemental about space-time.

No, quarks do not have the same electric charge as the electron.
Furthermore, in a neutron, there is no electron, so you do not have two charges that cancel, there are three quarks in a neutron. No electron.
As I allready told you, in the decay of a neutron, what happens is that e u quark radiates a W^- boson which decays into an electron and an electron-antineutrino.

Finally, this is not related to the "maximum quantum energydensity of the aether" since there is no aether.

 

[James R]

David F.,

The Standard Model of particle physics would not work if the quark charges only had values of +1 and -1, as you suggest. Certain reactions would not work out in the way that is actually observed experimentally.

I can't see any particular philosophical reason for objecting to a charge on a particle of, say, (2/3)e, where e is the electron charge. If we can have a charge of e (=1.6 × 10^-19 C), then why not (2/3)e? What's so special about that particular value of e?

 

[1100f]

Yes, I can read Feynman diagrams. This does not mean they are correct, they are a theoretical model -- and a pretty messy one at that.

Feynmann diagrams are not a model, they are a tool to calculate physical quantities within the framework of relativistic quantum field theory.

We don't actually know what the charge on a quark is. It has never (to my knowledge) actually been measured -- it is just a theory.

Look for example at the ratio σ(e⁺e^- --> hadrons)/σ(e⁺e⁺ --> e⁺e^-).
This definitely measure the charges of the quarks.

In many circles, there is a philosophical belief that there can't be such a thing as a +2/3 charge. You can get the same result by assigning a full +1 or -1 to any quark and then adding +1+1-1 to get +1 or -1-1+1 to get -1.

I agree with you, these are philisophical beliefs, not based on experimental results.

This is just a model and has not been proved -- don't talk like it is law... it doesn't warrant that yet.

Earth is not flat, it is a sphere. This is also a model, backed up by scattering experiments where light from the sun is scattered by the earth.

I don't know if there is an aether or not. Maybe the aether is an energy field -- like G fields. I do know light is a wave and a wave has to travel through something. I don't think we know everything quite yet -- and spouting theory as if it is law does not help.

Not all waves need a medium to propagate. What they need is a mechanism to propagate. For EM waves, the mechanism is provided by Maxwell's equations.

 

[mhobbs_bbt]

Hi.

F=GM1M2/r^2 ?

Apologies, fairly new to physics. I've seen lots of this formula for gravitation force. Is there a theory behind where it comes from or is it just based on observations ?

Is G supposed to be exactly 6.67 10^-11 or is this just an approximation ?

 

[mhobbs_bbt]

I have run across a correlation... see what you think.

If we first take Coulomb's law: F=qQ/r^2 or field strength E=q/r^2 and realize that the field dissipation factor (denominator) should be the surface area of a sphere, then we should use E=4pi*q/(4pi*r^2) or the charge should really be 4pi*q.

Next, we take Newtons gravity law F=G*M1*M2/r^2 and realize this also should have a denominator of 4pi*r^2, so we realize G has absorbed a 1/4pi factor. If we take the factor back out by muliplying by 4pi we see that the leftover G is the ratio of the mass to the charge of the subatomic particles (using 4pi*q as the charge).

G*4pi = m/(4pi*q)
or
G = m/[(4pi)^2*q]

If we plug the mass and charge of a Proton into this equation, we get over 99% of G. We can add the Neutron and Electron contributions with:

G = (mp + mn + me)/2 /[(4pi)^2*q]

Since a Neutron is just an Electron and a Proton which have combined (maybe with a neutrino in there to make up the additional mass) the sum mp+mn+me is really just two Protons and two Electrons. I don't know how to add all the other little subatomic particles to this equation but it is very close to the observed value just as it stands.

Does this indicate some connection between electrostatic charge and gravity?

David,

The formula, F=GM1M2/R^2, looks too closely related to the basic formula's for volume / surface area of a sphere to be unrelated. Especially when the constant looks like it should, somehow, be close to 2/3 (albeit at a different scale).

I've managed to tie, rather clumsily, this formula for gravity to those for two sphere's and reached a value that works, albeit with 2/3 rather than .667 ?

This would then tie in quite nicely with your suggestion ? However, before I embarras myself any more, I'd like to know if there's already an explanation of F=GM1M2/R^2 ? Do you know of one ?



Mike.

 

[James R]

The formula, F=GM1M2/R^2, looks too closely related to the basic formula's for volume / surface area of a sphere to be unrelated.

Yes.

In fact, the formula can be derived using something called Gauss's law, which related the surface area of a sphere around a point mass to the gravitational field flux through the surface. That goes part way to explaining why there's that R^2 factor in the denominator. (The surface area of a sphere is 4 pi R^2). Notice that we could easily have used G' in the force equation instead of G, where G' would be defined as:

G' = 4 pi G

 

[mhobbs_bbt]

Yes.

In fact, the formula can be derived using something called Gauss's law, which related the surface area of a sphere around a point mass to the gravitational field flux through the surface. That goes part way to explaining why there's that R^2 factor in the denominator. (The surface area of a sphere is 4 pi R^2). Notice that we could easily have used G' in the force equation instead of G, where G' would be defined as:

G' = 4 pi G

Okay. Please bear in mind, I don't come from a scientific background. Attached is my, probably rather naive, theory.

Is it completely wrong ?? Seems to work on paper & suggests that G isn't a random constant ?

 

[Zarkov]

>> mhobbs_bbt: G isn't a random constant ?


I expect you are correct, surface area to volume ratio 2:3
so if 6.667 or 0.667 only depends upon units.

 

[James R]

mhobbs_bbt:

There is some interesting thinking in your attached document, and I must congratulate you on taking the time to investigate this matter. It shows considerable scientific inventiveness on your part. As somebody untrained in science, you certainly seem to have the potential to make a good scientist.

Having said that, your conclusion that the value for G (6.67 × 10^-11) is significant and related to 2/3 is wrong.

First, I note that the 6.67 is an approximation. A more accurate value is 6.67259. If you divide this by 10, you don't get 2/3.

More importantly, the numerical value of G varies depending on what system of units you use. In our SI system, we use the kilogram as a mass standard, and the metre as a distance standard. However, the definitions of the kilogram and the metre are totally arbitrary. A kilogram is simply a lump of platinum-iridium alloy of a particular size stored in a vault in France, and a metre was originally defined arbitrarily as 1/10000th the distance from the North pole to the equator.

Let's say we took our mass standard to be the mass of an electron instead. Then, in the equation F=GMm/r², we'd need to express all masses in terms of electron masses.

The force between two 1 kg masses separated by distance 1 m is 6.67 × 10^-11 Newton, as you know. In terms of electron masses, 1 kg = 1.099 × 10³⁰ emu. (1 emu = 1 electron mass unit). The force between the same two masses, taking the masses in emu, would be:

F = G' (1.099 × 10³⁰)² / 1²

or

F = G' (1.208 × 10⁶⁰) = 6.67 × 10^-11 Newton.

Therefore, we calculate

G' = 5.52 × 10^-41

Clearly, as long as we write all masses in emu, then if we use the value G' given here, we will calculate the correct forces in Newton between any two objects.

Hopefully you can see that G is totally dependent on our choice of units.

However, as I said in my previous post, you have managed to come up with a particular insight in terms of the relationship between the flux of the gravitational field through a closed surface, which is formalised in a mathematical rule known as Gauss's law. It turns out that if we take the total flux of gravitational field through any closed surface surrounding a mass, then the ratio of that flux to the mass enclosed by the surface is a particular constant, equal to 4 pi G. This suggests that Newton's law could have been written as:

F = GMm/(4 pi r²), and the value of G redefined to be equal to 6.67 × 10^-11 × 4 pi.

In a mathematical sense, this would make certain calculations neater, although G would still depend on the particular choice of unit system we used.

As a final note, if you look at the electric force between two charged particles with charges Q and q, the standard expression for the electrostatic force is usually written as:

F = QQ / e(4 pi r²)

where the constant 1/(4 pi e) plays a similar role to G in the gravitational force.

 

[mhobbs_bbt]

James R :

Thanks. At last a logical, sensible, relatively straight forward (?), answer. Something to think about. I'll take a closer look at the details. Might take a while.

In the meantime :

The more accurate value of 6.67259 is presumably based on first assuming 6.67, then observing the real effects of gravity and adjusting the value so that the formula fits better ? ie. an estimate ? If this is the case, then isn't it just as likely that 2/3 is the correct value and our estimates of mass / distance are a little out ?

I'm assuming it went something like, 1) Observations suggest 6.67, 2) 6.67 used as the basis for estimating the mass / velocity of the planets, etc. 3) measurements show discrepancies when applied to other objects, 3) adjust 6.67 to make the measurements fit a higher % of the observations. - ie. the 6.67259 is founded on values suggested by the original 6.67 rather than derived from an original source ? and probably still doesn't work for everything ?

The 'arbitrary' measurements for mass / length are not totally arbitrary. I agree that you could pick any values to use but they are closely related in that 1kg of water = (10cm)^3 of water. The measurements for weight / volume ( in this very particular case ) do have a basic connection.

I realise that the planets / stars are not made entirely of water, but IF the original observations resulted in the correct 'hidden formula' but representing assumptions about the basic points of measurement, then most of what I've listed should still stand ?

Hidden within the constant would then be a hidden assumption that the mass were concentracted within a sphere of 10M, or whatever measurement. This would affect the 'real' value used for the distance between objects which, although small over the long distances involved, would introduce discrepancies that couldn't easily be explained, resulting in arbitrary adjustments to the constant ????

 

[mhobbs_bbt]

mhobbs_bbt:

There is some interesting thinking in your attached document, and I must congratulate you on taking the time to investigate this matter. It shows considerable scientific inventiveness on your part. As somebody untrained in science, you certainly seem to have the potential to make a good scientist.

Having said that, your conclusion that the value for G (6.67 × 10^-11) is significant and related to 2/3 is wrong.

First, I note that the 6.67 is an approximation. A more accurate value is 6.67259. If you divide this by 10, you don't get 2/3.

More importantly, the numerical value of G varies depending on what system of units you use. In our SI system, we use the kilogram as a mass standard, and the metre as a distance standard. However, the definitions of the kilogram and the metre are totally arbitrary. A kilogram is simply a lump of platinum-iridium alloy of a particular size stored in a vault in France, and a metre was originally defined arbitrarily as 1/10000th the distance from the North pole to the equator.

Let's say we took our mass standard to be the mass of an electron instead. Then, in the equation F=GMm/r², we'd need to express all masses in terms of electron masses.

The force between two 1 kg masses separated by distance 1 m is 6.67 × 10^-11 Newton, as you know. In terms of electron masses, 1 kg = 1.099 × 10³⁰ emu. (1 emu = 1 electron mass unit). The force between the same two masses, taking the masses in emu, would be:

F = G' (1.099 × 10³⁰)² / 1²

or

F = G' (1.208 × 10⁶⁰) = 6.67 × 10^-11 Newton.

Therefore, we calculate

G' = 5.52 × 10^-41

Clearly, as long as we write all masses in emu, then if we use the value G' given here, we will calculate the correct forces in Newton between any two objects.

Hopefully you can see that G is totally dependent on our choice of units.

However, as I said in my previous post, you have managed to come up with a particular insight in terms of the relationship between the flux of the gravitational field through a closed surface, which is formalised in a mathematical rule known as Gauss's law. It turns out that if we take the total flux of gravitational field through any closed surface surrounding a mass, then the ratio of that flux to the mass enclosed by the surface is a particular constant, equal to 4 pi G. This suggests that Newton's law could have been written as:

F = GMm/(4 pi r²), and the value of G redefined to be equal to 6.67 × 10^-11 × 4 pi.

In a mathematical sense, this would make certain calculations neater, although G would still depend on the particular choice of unit system we used.

As a final note, if you look at the electric force between two charged particles with charges Q and q, the standard expression for the electrostatic force is usually written as:

F = QQ / e(4 pi r²)

where the constant 1/(4 pi e) plays a similar role to G in the gravitational force.

Another thought.

If you change the mass to being measured in EMU's, then 1 unit of measurement of volume is 1 EMU. This unit would represent the electron mass unit for a volume based around a fixed radius ( approx. 1.56*10^-10 ) assuming a sphere. One unit of volume would then still represent 4/3 * PI * (1 unit of length)^3.

The rest of the equations would then still cancel out to give :

2/3 M1*M2 / R^2

units would then of course be in EMU^2 / (unit of length) ^ 2, and R would have to be calculated as a number of 'units of length'. The formula would then represent the force at a distance of 1 'unit of length' from the centre of an object as a base point ??

But still the 2/3 ??

If your unit of length is representative of the unit of volume used, you always come back to the 2/3, no matter what units you pick, the constant is then shifted to the units being used, which would make sense as the units are 'arbitrary'.

Is 1kg of mass always 1.099 * 10^30 EMU's, no matter what the material ?