The Universe and it's Intrinsic Curvature Given Two Ways, One of Them New

  • 8 Replies
  • 3061 Views

0 Members and 1 Guest are viewing this topic.

*

Offline Koki

  • First timers
  • *
  • 3
    • View Profile
First read newbielink:http://www.as.utexas.edu/astronomy/education/spring05/komatsu/lecture15.pdf [nonactive]

The Gaussian curvature for a 3-sphere system with a proper density is actually identified with the square of the reciprocal of the length [tex]\ell[/tex]


[tex]8 \pi \rho_0 (\frac{G}{c^2})[/tex]


A three-dimensional hypersphere is


[tex]\frac{K}{6} = (\frac{\hbar}{mc})^{-2}[/tex]


Where [tex]K[/tex] is the Gaussian curvature [Appendix] which can be re-written as


[tex]K = \rho_0 (\frac{\ell}{M})[/tex]


Our factor of [tex]6[/tex] magically disappears when you plug in classical values. Iíll quickly show this proof:

the density of a three dimensional hypersphere should be


[tex]\rho_0 = \frac{M}{(\frac{4}{3} \pi R^3)}[/tex]


The general relativistic relationship to the principle curvatures  is found as


[tex]8 \pi \rho_0 (\frac{G}{c^2})[/tex]


sidenote *the term [tex]8\pi[/tex] arises often because it simplified several of the field equations describing gravity


The expression above was noted by Motz on his paper of quantization, and by a reference I will have below this post. Let us plug in and solve the equation. Be careful to remember your algebra when solving, when dividing a whole number by a fraction. I calculate this as


[tex]\frac{M}{6 R^3}[/tex]


A three dimensional hypersphere is normally written as  when equated with the Compton wavelength. Equating my expression with this term we have


[tex]\frac{M}{6R^3} (\frac{G}{c^2}) = \frac{K}{6}[/tex]
 

Cancelling out the 6's, we end up with a general relativistic interpretation of how the geometry of a 3-sphere relates to the proper The The proper density and the Gaussian Surface Curvature are related as

[tex]K = \rho_0 (\frac{G}{c^2}) = \rho_0 (\frac{L_p}{M_p})[/tex]


Where the last equality comes about from a knowing that it has dimensions of length over mass. What is the Gaussian curvature of a metric? Refer back now to the cited reference at the beginning:

[tex]ds^2 = c^2dt^2 - [\frac{dR^2}{1 - KR^2} + R^2(d\alpha^2 + sin\ \alpha d\phi][/tex]


Where [tex]K[/tex] is the curvature of the universe.


[tex](\frac{\dot{a}}{a})^2 =  \frac{8 \pi}{3} \rho \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}[/tex]


where


[tex]\sum_{i=(1,2..n)}^{N} m_i[/tex]


is the many body system and now allow [tex]H^2 = (\frac{\dot{a}}{a})^2[/tex] (the Hubble Constant) and multiply through by our Schwarzschild constant


[tex]H^2(\frac{G}{c^2}) =  \frac{8 \pi K}{3}  \frac{Rc^2}{(\sum_{i=(1,2..n)}^{N} m_i)} - \frac{k}{a^2}(\frac{G}{c^2})[/tex]


because [tex]K = k_1k_2 = \rho_0 (\frac{G}{c^2})[/tex] as we proved before. We also showed that this equation was actually a length over the mass, in exchange we may think of this as the radius of the universe and entire ensemble of statistical averages so that the equation doesn't hold two identical features on either side of the equal sign, it's just more aesthetic


[tex]H^2(\frac{G}{c^2}) =  \frac{8 \pi K}{3}  \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}(\frac{R}{(\sum_{i=(1,2..n)}^{N} m_i)})[/tex]


There are of course limits we can look at for variety, the equation above can satisfy a cosmological model. If [tex]R \rightarrow 0[/tex] then the proper density goes to [tex]\rho \rightarrow \infty[/tex] which implies a breakdown in the presence of a singular region within this non-space. In other words, it can be argued for a universe, the disappearance of space itself is unthinkable. Making time disappear is another matter and one completely acceptable. The thing we must keep in mind however is that the equation


[tex](\frac{\dot{a}}{a})^2 =  \frac{8 \pi}{3} \rho \frac{R c^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}[/tex]


satisfies the matter-dominated region of the universe. If we really where to make the radius go to zero, with neglecting the range of density to infinite size, we still have the problem of an environment application, it wasn't until 100,000 give or take (according to Susskind) that our universe first appeared to have matter. In later work when investigating Pensrose, he concluded for me a finding I too independently came to, that if matter fields [tex]\chi[/tex] where to vanish in the universe, time ceases to exist. Because matter is in fact emergent, especially under the Wheeler model of Geometrogenesis, matter isn't and must not be fundamental, meaning the equation I derived


[tex]H^2(\frac{G}{c^2}) =  \frac{8 \pi K}{3}  \frac{R c^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}(\frac{R}{(\sum_{i=(1,2..n)}^{N} m_i)})[/tex]


Satisfies the low energy range and emergent geometry with matter... and most of all, time itself. I can prove in a quantum approach how matter is ''given'' time, by finding a relationship between the zitter motion of particle clocks and with local gravity. To show how time emerged when matter is considered, we begin with an equation which is in the strong gravity range using the gravitational fine structure and a timeless action:



[tex]\alpha_G \cdot \hbar = \int p_{\gamma} \cdot q[/tex]


[tex]\int \hbar\ dr_s = \lambda \int [m_{\gamma} v \cdot q][/tex]


We know this because of Maurapitus' principle


[tex]\int mv\ ds = \int p \cdot q[/tex]


[tex]ds = cdt[/tex]


You can find an energy


[tex]\int E = \int p \cdot \dot{q}[/tex]


Our generalized coordinate [tex]\dot{q}[/tex] has absorbed the time term. You find the clock to matter by distributing a frequency


[tex]\int \hbar\ dr_s = \lambda \int [mv \cdot q][/tex]


The mass term requires a coefficient: [tex](\frac{c^2}{\hbar})[/tex] to become a frequency term itself. This would imply a relativistic change in the wavelength [tex]\lambda_2 - \lambda_1 = \Delta \lambda[/tex] since frequency and wavelength are related


[tex]\nu = \frac{c}{\lambda}[/tex]


[tex]\lambda = \frac{c}{\nu}[/tex]


distributing the coefficient we have


[tex]\int d r_s\hbar\ (\frac{c^2}{\hbar}) = \lambda \int [m(\frac{c^2}{\hbar})v \cdot q][/tex]


This gives time to matter! The equation simplifies to


[tex]\int c^2 \ dr_s = \lambda \int [m(\frac{c^2}{\hbar})v \cdot q][/tex]


[tex]\int c^2 r_s = \mu[/tex]


where [tex]\mu[/tex] is the gravitational parameter [tex]GM[/tex]. To finish off, we return to my equation of cosmology in the low energy range


[tex]H^2(\frac{G}{c^2}) =  \frac{8 \pi K}{3}  \frac{Rc^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}(\frac{R}{(\sum_{i=(1,2..n)}^{N} m_i)})[/tex]


The term of mass over length is in fact a reciprocal of the linear density expression


[tex]\bar{\rho} = \frac{M}{\ell}[/tex]


Before I go, there are some interesting things we can think upon. A dominated matter universe would have a relation allowed


[tex]8 \pi G \rho = 3 H^2[/tex]


Notice we can make


[tex]\frac{8 \pi G \rho}{3} = H^2[/tex]


Is the same LHS expression for the equation


[tex]\frac{8 \pi}{3} \rho [/tex] for equation


[tex]H^2(\frac{G}{c^2}) =  \frac{8 \pi K}{3}  \frac{R c^2}{(\sum_{i=(1,2..N)}^{N} m_i)} - \frac{k}{a^2}(\frac{R}{(\sum_{i=(1,2..n)}^{N} m_i)})[/tex]

except for a factor of [tex]G[/tex].
« Last Edit: 20/01/2015 16:48:52 by Koki »

*

Offline Koki

  • First timers
  • *
  • 3
    • View Profile
The latex is a bit sensitive here, I will give you another link to see the equations a bit more clearly and some of them more correctly

newbielink:http://www.physicsdiscussionforum.org/post6906.html#p6906 [nonactive]

*

Offline PmbPhy

  • Neilep Level Member
  • ******
  • 2804
    • View Profile
Quote from: Kok
First read http://www.as.utexas.edu/astronomy/education/spring05/komatsu/lecture15.pdf

The Gaussian curvature for a 3-sphere system with a proper density is actually identified with the square of the reciprocal of the length [tex]\ell[/tex]


[tex]8 \pi \rho_0 (\frac{G}{c^2})[/tex]
Welcome to the forum! :)

Were did that relationship come from? I can't make heads or tails out of the rest because it makes no sense. I also don't see what Gaussian curvature has to do with Gaussian curvature. It's Riemann curvature that pertains to space and spacetime.

By the way. It's a bad idea to expect members to read a paper that long.

*

Offline Koki

  • First timers
  • *
  • 3
    • View Profile
Hi!


It is derived from a scale length due to a proper density. You obtain it by related a wavelength of reciprocal 2 with the dimensions given.

You said ''It's Riemann curvature that pertains to space and spacetime.''


Yes under the Minkowskian four dimensional symmetry of the theory. But there is now more evidence suggesting that there is no fundamental four symmetry, instead Machian relativity agrees more with General Relativity than Minkowski relativity does with the general. A clear example of this is because General relativity doesn't have a fundamental clock - at least not in any high energy origin range where no matter fields existed. General relativity uses diffeomorphisms to explain the new definition of ''time'' if one can be bold enough to call it that. As Mach himself said, ''Time is an abstraction derived from the changes of things,'' so space-time is likely faulty as unification. Dirac also held this position.

*

Offline JohnDuffield

  • Hero Member
  • *****
  • 507
    • View Profile
Hi Geo. Sorry I've not replied at length. Can I just point out Preliminaries on Baez? See this:

"Similarly, in general relativity gravity is not really a 'force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial".

Also see Einstein and this and this. Your pencil falls down because spacetime is "tilted" in the room you're in. The spacetime isn't curved, and nor is the space. Instead the space is inhomogeneous.

*

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • 12339
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
Relativity has a very simple clock. It's related to you local measurement of a 'light speed'. Split that measurement in even chunks and you come to a clock. Wheter it breaks down at Planck scale I don't know, but it's still there, locally measured.
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."

*

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • 12339
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
And if you think that you can take some 'average' of some 'common universe' defining it , you just lost a constant, called the speed of light in a (perfect) vacuum.
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."

*

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • 12339
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
My point is that Planck scale is a try for unifying. It that is incorrect you have a 'smooth' universe, if it is correct we have a 'pixilated' universe, or consisting of 'quanta' if you like. But that only means as far as we can measure, or assume. It depends on how you think of 'backgrounds' too. Is that needed? Or can 'quanta' create it?
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."

*

Offline yor_on

  • Naked Science Forum GOD!
  • *******
  • 12339
  • (Ah, yes:) *a table is always good to hide under*
    • View Profile
In the end, in what way do you consider it acceptable to ignore time? From personal experience? (I can go with that naively, doubt I'll ever be a grown up :)
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."