Jump to content

Simple Unification Theory


ColinJ

Recommended Posts

Hi,

Completely new to the forum.  Please be kind.  Just looking for some feedback on an idea that a single equation to calculate both the force of molecular bonds and the force of gravity is possible if a sphere is not approximated as a point at the molecular level using conventional Coulombic forces.

 

Anyway, my attempt at explaining it is here https://newtontweaked.com/

 

Thanks in advance,

 

Colin.

Link to comment
Share on other sites

Hi. Welcome to the forums. Please, take a look at Section 2 point 7 of the guidelines.

Quote

Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos. Videos and pictures should be accompanied by enough text to set the tone for the discussion, and should not be posted alone.

https://www.scienceforums.net/guidelines/

A sphere is never "approximated by a point" when calculating gravitational forces outside the sphere. It's just that the field of both objects happens to be the same. That's called Newton's theorem, and in the case of General Relativity, Birkhoff's theorem. A different case would be if the matter distribution were not spherically symmetric.

Molecular bonds are a completely different matter, and you really need some quantum mechanics to tackle them...

Link to comment
Share on other sites

On 10/10/2021 at 11:01 PM, joigus said:

Hi. Welcome to the forums. Please, take a look at Section 2 point 7 of the guidelines.

!

Moderator Note

Thanks joigus! Let's make this official. ColinJ, can you copy/paste from your document and post it here? It's important to make sure each step in an idea is sound before taking another.

 
Link to comment
Share on other sites

Apologies for the delay – was on the road for a few days.

The entire thing is a strange concept.  My argument is that a simple numerical approximation using a spreadsheet disagrees with the shell theorem.  Below is the approach for the numerical approximation.  Also attached is a spreadsheet implementing the diagram.

 

 

CircleToCircle_Nice_20210211_Colombic-10

 

The numerical approximation in the spreadsheet over n rows is shown as a Riemann sum below.

image.png.69a2a1db733fa45b15fb19d7f77f1785.png

Which evaluates to

image.png.721007ab28ae043e224d6d49fed23684.png

If you use that equation, and not the classic q1q2/r^2 equation, for point to sphere interaction a Hydrogen molecule now has four different electrostatic forces at play, well three in the case of Hydrogen gas.

 

four.png.9842b24b55497d006b18e7ffb21202b8.png

And the weird part - if you graph the net force over distance of all forces you get a Lennard Jones curve.

LennardJones-1024x443.thumb.png.55093803a0f4d781e849bfb5f0000929.png

 

And the peak is about where it should be at stable molecular bond distance and the tail trails off infinitely with an ever decreasing attractive force ( a bit like gravity).

Lots of people have looked at this and said it cannot be correct but nobody can tell me what is actually incorrect other than it disagrees with pretty much everything physics is based on.

Cheers,

Colin.

 

Riemann.png

eqn.png

Paper_PtoC_Sample_20210627.xlsx Paper_CtoC_Sample_20210627.xlsx

Link to comment
Share on other sites

It is confusing.  I'm saying that the Coulombic force of q1q2/r^2 does not apply between a point and a hollow sphere.  Although conventionally a force calculation between a point and a hollow sphere is approximated as a point and another point and q1q2/r^2 is used.  But, if you run a very simple numerical approximation between a point and a hollow sphere where the sphere is represented as n points, using the standard Coulombic equation between the n points and sum the forces it does not equal the force of q1q2/r^2 when the sphere is represented as point.

 

Inequality.png.1c0076d24a9e68862e4098556e0a66d4.png

Assuming distances d1 and d2 are identical and the charges on points 1, 2 and the sphere are identical the electrostatic force using Coulomb’s Law between points 1 and 2 does not equal the electrostatic force between point 1 and the charged sphere.

Link to comment
Share on other sites

Is nobody replying because I haven't explained this properly or because this concept means a lot of physics is wrong?  I am reminded of Monty Python - I came here for an argument :-)  But, I could be arguing my spare time.

 

Link to comment
Share on other sites

On 10/17/2021 at 10:38 PM, ColinJ said:

It is confusing.  I'm saying that the Coulombic force of q1q2/r^2 does not apply between a point and a hollow sphere.  Although conventionally a force calculation between a point and a hollow sphere is approximated as a point and another point and q1q2/r^2 is used.  But, if you run a very simple numerical approximation between a point and a hollow sphere where the sphere is represented as n points, using the standard Coulombic equation between the n points and sum the forces it does not equal the force of q1q2/r^2 when the sphere is represented as point.

 

The equivalence between a charged sphere and a point charge (and similarly with mass and a uniform sphere for gravitation) is shown in the shell theorem. Can you point to the error in its derivation?

Link to comment
Share on other sites

A bit busy with work so this might take a while but I started as you suggested.  I only got as far as confirming my approach to calculating the distance in the initial force vector produces the same value as the shell theorem.  Namely that Compare.gif.914be68192e0041698eb466bf0a9d18f.gif with my version on the left and the classic interpretation on the right of that equaity.  On to the next step - the bands.  Although I don't really understand the necessity of a band in the classic version when the intent is an infinitely narrow band which is a slice through the sphere which is a point on a cross section of a sphere which is my approach.  But, after I do my taxes I will resume the process of determining at which point the two approaches differ.  And in the meantime if anyone can tell me what's wrong with my far simpler approach I'm still all ears.  Ideally if anyone knows of a numerical approximation of the classical shell theorem.  Thanks. 

CircleToCircle_Nice_20210211_Colombic.thumb.png.95b05e0c5c0925d1d4e2c17ace46bd5c.png

 

 

Link to comment
Share on other sites

  • 3 months later...

I have not found a breakdown of classical shell theory that makes any sense.  The original article by Newton is now a screen saver.  It refers to corpuscles, circular superficies and sphaerical superficies and a bunch of other stuff that may take a while to figure out in contemporary English. Has anyone got a decent breakdown of classical shell theory that does not involve corpuscles and actually has all the necessary steps?  I've looked at a few presentations and they all seem to make giant leaps in thinking without any real justification.

Link to comment
Share on other sites

On 10/11/2021 at 4:50 AM, ColinJ said:

Hi,

Completely new to the forum.  Please be kind.  Just looking for some feedback on an idea that a single equation to calculate both the force of molecular bonds and the force of gravity is possible if a sphere is not approximated as a point at the molecular level using conventional Coulombic forces.

 

Anyway, my attempt at explaining it is here https://newtontweaked.com/

 

Thanks in advance,

 

Colin.

 

Why would you want to do this ?

Lennard-Jones is one of a large number of empirical energy potential functions, obtained by fitting power law equations to observations, where the independant variable is distance.

But all these have at least two parts a repulsive part and an attractive part.

So the result is an interplay between the repulsive and attractive forcesand their variations ove the distance.

 

On the other hand Gravity has ony an attractive part and additionally operates over a very different distance range.

 

https://en.wikipedia.org/wiki/Interatomic_potential

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.