Jump to content
Science Forums

Recommended Posts

Posted (edited)

Why is gravity so weak ?

 

And how do you define gravity ?

At the energy density of a black hole within the particle it splits into positive and negative space as evident as one of the solutions of Einsteins equations for gravitational curvatures:

 

+y * -y = 1 - (2GM)/(Rc^2)

 

Where y is the length delta or r/r0...

 

When you consider the radius of my strong force as both positive and negative, positively you get the strong force density of a quantum black hole, but negatively you get the gravitation of the particle.

 

The gravitation calculations I covered in post #6 if you're interested. But basically the same value as positive or negative at the event horizon energy density yields both gravitation and the strong force, so it splits into gravitation and the strong force, or did I get something wrong?

Edited by devin553344
Link to post
Share on other sites
  • Replies 92
  • Created
  • Last Reply

Top Posters In This Topic

Top Posters In This Topic

Popular Posts

Particles and waves are part of the wave-particle duality, all electrons and protons and all mass, even planets have a DeBroglie wavelength (https://en.wikipedia.org/wiki/Matter_wave#de_Broglie_hypoth

Posted (edited)

Define Energy Density , devin .

I did already in post #6, its Joules  per meter for gravitation. Gravitation is slightly different than standard pressure and energy density, which is Joules per meter cubed. This since gravitation, in the Einstein physical way, relates velocity squared, ie 2GM/r = v^2.

Edited by devin553344
Link to post
Share on other sites

I did already in post #6, its Joules  per meter for gravitation. Gravitation is slightly different than standard pressure and energy density, which is Joules per meter cubed. This since gravitation, in the Einstein physical way, relates velocity squared, ie 2GM/r = v^2.

But basically the energy density of a black hole can be described as the Einstein constant or c^4/G.

Link to post
Share on other sites

I did already in post #6, its Joules  per meter for gravitation. Gravitation is slightly different than standard pressure and energy density, which is Joules per meter cubed. This since gravitation, in the Einstein physical way, relates velocity squared, ie 2GM/r = v^2.

What is the source of these " Joules/per/meter " ?

Link to post
Share on other sites
Posted (edited)

I defined that in post #5 and #6, its the charge energy mixed with the matter energy, you would really have to look at the definition of my fine structure constant strain definition to understand that I think. But basically the charge and matter do something like:

 

Energy per meter = Ke^2/r^2 * (1/g)^(rC/r)

 

Where K is the electric constant, e is the elementary charge, g is the electromagnetic coupling constant, rC is the charge radius of the particle, r is the radius position inward of the particle's field.

 

Edit: I've changed the PDF  file in the OP to the current solution, which I do periodically, but so far it appears valid.

Edited by devin553344
Link to post
Share on other sites

I defined that in post #5 and #6, its the charge energy mixed with the matter energy, you would really have to look at the definition of my fine structure constant strain definition to understand that I think. But basically the charge and matter do something like:

 

Energy per meter = Ke^2/r^2 * (1/g)^(rC/r)

 

Where K is the electric constant, e is the elementary charge, g is the electromagnetic coupling constant, rC is the charge radius of the particle, r is the radius position inward of the particle's field.

 

Edit: I've changed the PDF file in the OP to the current solution, which I do periodically, but so far it appears valid.

What, charge energy ? What is this " charge energy " , what energy is charged ? ( Energy is a Physical thing , not a mathematical concept ).

Edited by current
Link to post
Share on other sites

What, charge energy ? What is this " charge energy " , what energy is charged ?

It's a curvature force of the self energy of the charge of the proton. Sort of a matter energy. The strong force reduces to standard electric force at a distance with other charges. But within the charge field it uses a curvature, and perhaps charge is a curvature. This is strongly evident from electron ionization energy (https://en.wikipedia.org/wiki/Ionization_energy) which reduces the matter field of the hydrogen atom itself. The mass is reduced by the amount of the electron orbital binding energy. And the only way it could do that is if that electric energy was a curvature like matter itself. Which is Einsteins idea of E=mc^2

 

My strong force matches empirical data exactly.

Link to post
Share on other sites

current, on 30 Aug 2020 - 2:48 PM, said:

 

What, charge energy ? What is this " charge energy " , what energy is charged ?

It's a curvature force of the self energy of the charge of the proton.

What is curving the self energy charge of the proton ?

Edited by current
Link to post
Share on other sites
Posted (edited)

current, on 30 Aug 2020 - 2:48 PM, said:

 

 

 

What is curving the self energy charge of the proton ?

The radius strain deformation, basically r/r0 is a strain or curvature, and I've spelled out one for composite particles and one for point particles, the point particle is:

 

hc/(120 * (rc+re)^4) = Ke^2/re^4

 

Which is pressure or Joules per meter cubed. The strain is in the point nature of Riemann zeta -3 and the radius difference which is r/r0, remember:

 

r^2/r0^2 = 1 - 2GM/Rc^2

 

Basically I've spelled out the curvature of the particle as it relates charge to matter, or the curvature differential?

Edited by devin553344
Link to post
Share on other sites

The radius strain deformation, basically r/r0 is a strain or curvature, and I've spelled out one for composite particles and one for point particles, the point particle is:

 

hc/(120 * (rc+re)^4) = Ke^2/re^4

 

Which is pressure or Joules per meter cubed. The strain is in the point nature of Riemann zeta -3 and the radius difference which is r/r0, remember:

 

r^2/r0^2 = 1 - 2GM/Rc^2

 

Basically I've spelled out the curvature of the particle as it relates charge to matter, or the curvature differential?

What composite particles? Point particles are meaningless . They have no physical reality . Point particles are one , maybe two dimensional . Neither of which , even in combination , exist in reality .

Link to post
Share on other sites

What composite particles? Point particles are meaningless . They have no physical reality . Point particles are one , maybe two dimensional . Neither of which , even in combination , exist in reality .

I have an idea that is your hypothesis about point particles, although I've been able to mathematically establish an argument that they exist three dimensional.

 

The composite particle I've already posted in post #12 like the point particle:

 

hc/(2π^2 * (rc+rp)^4) = Ke^2/rc^4

 

Keeping in mind that these are measured values for the charge radius of the proton(rc) and Deuteron(rc+rp) extend of the strong force field.

Link to post
Share on other sites

I have an idea that is your hypothesis about point particles, although I've been able to mathematically establish an argument that they exist three dimensional.

Now physically explain how point particles exist three dimensionally .

Edited by current
Link to post
Share on other sites

Now physically explain how point particles exist three dimensionally .

Yes of course, which I did in post #12...

 

The energy of a point particle can be described as 1/2 Kx^2 or Hooke's Law (https://en.wikipedia.org/wiki/Hooke%27s_law) for springs, which also relates to Schrodinger equation (http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html):

 

mc^2 = mf^2(r-0)^2

 

Where m is the mass, c is the speed of light, f is the frequency of emitted light during annihilation of the particle, which is one over period, r is the wavelength which is subtracted from zero.

 

And that mass relates to curvature and electric forces as a deformation and energy density that is infinite using Riemann zeta -3. Which it is strained from the wavelength down to zero which represents an infinite strain, and therefore requires that Riemann zeta -3:

 

hc/(120 * (rc+re)^4) = Ke^2/re^4

 

Riemann zeta -3 = 1/120. And if the strain is strained to zero then it is a point particle mathematically and has energy per meter cubed.

Link to post
Share on other sites

Yes of course, which I did in post #12...

 

The energy of a point particle can be described as 1/2 Kx^2 or Hooke's Law (https://en.wikipedia.org/wiki/Hooke's_law) for springs, which also relates to Schrodinger equation (http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html):

 

mc^2 = mf^2(r-0)^2

 

Where m is the mass, c is the speed of light, f is the frequency of emitted light during annihilation of the particle, which is one over period, r is the wavelength which is subtracted from zero.

 

And that mass relates to curvature and electric forces as a deformation and energy density that is infinite using Riemann zeta -3. Which it is strained from the wavelength down to zero which represents an infinite strain, and therefore requires that Riemann zeta -3:

 

hc/(120 * (rc+re)^4) = Ke^2/re^4

 

Riemann zeta -3 = 1/120. And if the strain is strained to zero then it is a point particle mathematically and has energy per meter cubed.

What mass ?

Link to post
Share on other sites
Posted (edited)

What mass ?

Yes, I haven't actually got that far yet, I'm still working on what determines an electrons specific mass. What I've done is to describe a wave-particle duality for composite and point particles. But as far as why the electrons mass for the stable point particle, I'm still working on that.

 

But I have spelled out the mass for the proton and neutron using 3 point particles.

Edited by devin553344
Link to post
Share on other sites
Posted (edited)

What mass ?

What you're probably asking about is the electromagnetic radiation momentum, E=hv=mc^2, but the mass would be in the form of momentum until the strain from wavelength down to zero distance is creating a curvature for the mass of the momentum.

 

So its p/c

Edited by devin553344
Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...

×
×
  • Create New...