Updates On A Toy Model Of Gravity

[Dubbelosix]

Some things I found while investigating a toy theory of gravity. The geometry is related to the Hamitonian and I investigated it in a number of different ways. Some things I found along the way:

 

I propose the spacetime non-commutivity is (with eigenvalues),

 

[math]R_{\mu \nu} = [\nabla_x\nabla_0 - \nabla_0 \nabla_x] \geq \frac{(\ell^{2})^{-1}}{\sum_i \sqrt{n_i(n_i + 1)}}[/math]

 

This form above is simply calculated in the normal way from the Christoffel symbols which form two connections of the gravitational field, still following of course, the commutation laws,

 

[math][\nabla_i, \nabla_j] = (\partial_i + \Gamma_i)(\partial_j + \Gamma_j) - (\partial_j + \Gamma_j)(\partial_i + \Gamma_i)[/math]

 

[math]= (\partial_i \partial_j + \Gamma_i \partial_j + \partial_i \Gamma_j + \Gamma_i\Gamma_j) - (\partial_j \partial_i + \partial_j\Gamma_i + \Gamma_j \partial_i + \Gamma_j \Gamma_i)[/math]

 

[math]= -[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j][/math]

 

I adopt a geometric interpretation of the uncertainty principle from the Cauchy Schwarz [math]L^2[/math] space in which the expectation of the uncertainty is found as the mean deviation of the curvature of the system:

 

[math]\sqrt{|<\nabla^2_i> <\nabla^2_j>|} \geq \frac{1}{2}i(<\psi|\nabla_i \nabla_j|\psi> + <\psi|\nabla_j,\nabla_i|\psi>) = \frac{1}{2}<\psi|\nabla_i, \nabla_j|\psi> = \frac{1}{2} <\psi|R_{ij}|\psi>[/math]

 

The previous equation could be rearranged to show how it could satisfy an inequality bound

 

[math]<\psi|\nabla_i, \nabla_j|\psi>\ =\ <\psi|R_{ij}|\psi>\ \leq\ 2\sqrt{|<\nabla^2_i> <\nabla^2_j>|}[/math]

 

I came to realize the following representation was a ''survival probability'' of a system. A survival probability is often associated with the zeno effect. 

 

[math]\Delta \mathbf{H} = \sqrt{|R_{ij}|^2 - <\psi|R_{ij}|\psi>^2}[/math]

 

I came to also realize it was not dissimilar to the difference of energies taken to denote the superposition in Anandan's theory and Penrose. The binding energy could be directly related to the geometry

 

[math]\Delta <E> = \frac{c^4}{8 \pi G} \int \Delta <R_{ij}>\ dV =\frac{c^4}{8 \pi G} \int <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi>\ dV[/math]

 

The difference of geometries, could be understood as in terms of probability. In a paper I. Białynicki-Birula, J. Mycielski paper on the ''Uncertainty relations for information entropy in wave mechanics,''  the relative entropy and the probability distributions between them was provided as:

 

[math]D(p|q) = \sum_lp_l \log_2(\frac{p_l}{q_i}) = \sum_l p_l(\log_2 p_l - \log_2 q_l)[/math]

 

The way to view this equation, is as an information gain between two distributions (under observation), so is an expectation difference of two states, which is exactly what the difference in geometry equation is all about. So it is possible to construct a probabilistic model.

 

The Mandelstam-Tamm inequality can be written in the following way, with [math]c = 8 \pi G = 1[/math] (as usual), we can construct the relationship: (changing notation only slightly)

 

[math]|<\psi(0)|\psi(t)>|^2 \geq \cos^2(\frac{[<R_{ij}> - <\psi|R_{ij}|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{[<H> - <\psi|H|\psi> ]\Delta t}{\hbar})[/math]

 

A curve in a Hilbert space may take on the following appearance, with an understanding of geometry - the following equation also makes use of the Wigner function which allowed me to write it as an inequality

 

[math]\frac{ds}{dt} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>} = \int \int\ |W(q,p)^2|\ \sqrt{<\psi|R_{ij}^2|\psi>}\ dqdp\ \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math]

 

 

 

 

ref

 

http://www.physicsgre.com/viewtopic.php?f=10&t=127412&p=198856#p198856

Edited November 8, 2017 by Dubbelosix

[Vmedvil]

So, when you say,

 

"I adopt a geometric interpretation of the uncertainty principle from the Cauchy Schwarz [math]L^2[/math] space in which the expectation of the uncertainty is found as the mean deviation of the curvature of the system:

 

[math]\sqrt{|<\nabla^2_i> <\nabla^2_j>|} \geq \frac{1}{2}i(<\psi|\nabla_i \nabla_j|\psi> + <\psi|\nabla_j,\nabla_i|\psi>) = \frac{1}{2}<\psi|\nabla_i, \nabla_j|\psi> = \frac{1}{2} <\psi|R_{ij}|\psi>[/math] "

 

 

That would show the Curvature caused by something like this. 

 

Planck Sized Wave-Particle Volume = 4/3 π(1/ Lp^2)^3 

 

 

simply by  [math] Planck-Curvature = \frac{2}{3} <\psi|R_{ij}|\psi> \pi (\frac{1}{L_p^2})^3 [/math] 

 

The field being uncertain in position?

 

Then 

 

 [math] \Delta <E> = \frac{C^4}{12 G }  \int \Delta <R_{ij}>  (\frac{1}{L_p^2})^3 = \frac{C^4}{12 G } \int  <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\frac{1}{L_p^2})^3 [/math]  

 

Which [math] L_p/t_p = C [/math] , So [math] L_p^2 =  t_p^2 C^2 [/math]

 

This would be the change in energy for every cycle of C in a meter of space down to the Planck length per Planck time, In any case, that is the null spherical space. 

Edited November 8, 2017 by Vmedvil

[Vmedvil]

Wheeler was certainly under the impression that gravity became significant at the Planck scale while we certainly adopt a classical view of gravity in the phase space 

 

Well, that equation certainty saves me a bunch of calculations as a summation if you can find the survival of the World-line. I had estimated that would be in the order of 4.1 * 10 ^ 210 calculations for a chaotic unknown system of Particle States per Meter^3 Second^3.

[Vmedvil]

 

 

 [math] \Delta <E> = \frac{C^4}{12 G }  \int \Delta <R_{ij}>  (\frac{1}{L_p^2})^3 = \frac{C^4}{12 G } \int  <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\frac{1}{L_p^2})^3 [/math]  

 

 

 

Ok, Vmedvil. I am trying to follow this, What is [math](\frac{1}{L_P^2})^3[/math], if its a volume term, it doesn't work like that. This is an inverse length squared you have cubed. 

 

 

It puts it into meters from Planck length, this is for a meter sized Null cone [math] \frac{1}{L_p} [/math] then the second term is for a meter of moving particles that are a Planck length each in radius [math] \frac{1}{L_p} [/math] then [math]V = \frac{4}{3} \pi r^3 [/math] which I took times your number in front of the integral then the [math] \frac{\pi}{\pi} [/math]  canceled. is that volume not in meters? I guess I should have put it in this form so people would understand what I was doing  [math] r = \frac{1}{L_p t_p C} [/math]

Edited November 9, 2017 by Vmedvil

[Super Polymath]

Some things I found while investigating a toy theory of gravity. The geometry is related to the Hamitonian and I investigated it in a number of different ways. Some things I found along the way:

 

I propose the spacetime non-commutivity is (with eigenvalues),

 

[math]R_{\mu \nu} = [\nabla_x\nabla_0 - \nabla_0 \nabla_x] \geq \frac{(\ell^{2})^{-1}}{\sum_i \sqrt{n_i(n_i + 1)}}[/math]

 

This form above is simply calculated in the normal way from the Christoffel symbols which form two connections of the gravitational field, still following of course, the commutation laws,

 

[math][\nabla_i, \nabla_j] = (\partial_i + \Gamma_i)(\partial_j + \Gamma_j) - (\partial_j + \Gamma_j)(\partial_i + \Gamma_i)[/math]

 

[math]= (\partial_i \partial_j + \Gamma_i \partial_j + \partial_i \Gamma_j + \Gamma_i\Gamma_j) - (\partial_j \partial_i + \partial_j\Gamma_i + \Gamma_j \partial_i + \Gamma_j \Gamma_i)[/math]

 

[math]= -[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j][/math]

 

I adopt a geometric interpretation of the uncertainty principle from the Cauchy Schwarz [math]L^2[/math] space in which the expectation of the uncertainty is found as the mean deviation of the curvature of the system:

 

[math]\sqrt{|<\nabla^2_i> <\nabla^2_j>|} \geq \frac{1}{2}i(<\psi|\nabla_i \nabla_j|\psi> + <\psi|\nabla_j,\nabla_i|\psi>) = \frac{1}{2}<\psi|\nabla_i, \nabla_j|\psi> = \frac{1}{2} <\psi|R_{ij}|\psi>[/math]

 

The previous equation could be rearranged to show how it could satisfy an inequality bound

 

[math]<\psi|\nabla_i, \nabla_j|\psi>\ =\ <\psi|R_{ij}|\psi>\ \leq\ 2\sqrt{|<\nabla^2_i> <\nabla^2_j>|}[/math]

 

I came to realize the following representation was a ''survival probability'' of a system. A survival probability is often associated with the zeno effect.

 

[math]\Delta \mathbf{H} = \sqrt{|R_{ij}|^2 - <\psi|R_{ij}|\psi>^2}[/math]

 

I came to also realize it was not dissimilar to the difference of energies taken to denote the superposition in Anandan's theory and Penrose. The binding energy could be directly related to the geometry

 

[math]\Delta <E> = \frac{c^4}{8 \pi G} \int \Delta <R_{ij}>\ dV =\frac{c^4}{8 \pi G} \int <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi>\ dV[/math]

 

The difference of geometries, could be understood as in terms of probability. In a paper I. Białynicki-Birula, J. Mycielski paper on the ''Uncertainty relations for information entropy in wave mechanics,'' the relative entropy and the probability distributions between them was provided as:

 

[math]D(p|q) = \sum_lp_l \log_2(\frac{p_l}{q_i}) = \sum_l p_l(\log_2 p_l - \log_2 q_l)[/math]

 

The way to view this equation, is as an information gain between two distributions (under observation), so is an expectation difference of two states, which is exactly what the difference in geometry equation is all about. So it is possible to construct a probabilistic model.

 

The Mandelstam-Tamm inequality can be written in the following way, with [math]c = 8 \pi G = 1[/math] (as usual), we can construct the relationship: (changing notation only slightly)

 

[math]|<\psi(0)|\psi(t)>|^2 \geq \cos^2(\frac{[<R_{ij}> - <\psi|R_{ij}|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{[<H> - <\psi|H|\psi> ]\Delta t}{\hbar})[/math]

 

A curve in a Hilbert space may take on the following appearance, with an understanding of geometry - the following equation also makes use of the Wigner function which allowed me to write it as an inequality

 

[math]\frac{ds}{dt} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>} = \int \int\ |W(q,p)^2|\ \sqrt{<\psi|R_{ij}^2|\psi>}\ dqdp\ \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math]

 

 

 

 

ref

 

http://www.physicsgre.com/viewtopic.php?f=10&t=127412&p=198856#p198856

I feel as though my theory of gravity (causing the other three fundamental interactions and caused by a temporally lateral fractal dimension as the bi-linear third dimension in a 5 dimensional hypersphere with two opposite temporal linearities, the spacetime is transferred back & forth which is gravity in our phase space & dark energy at large) would work for such a classical gravity in the phase space, & would work perfectly. Edited November 9, 2017 by Super Polymath

[Super Polymath]

Well, that equation certainty saves me a bunch of calculations as a summation if you can find the survival of the World-line. I had estimated that would be in the order of 4.1 * 10 ^ 210 calculations for a chaotic unknown system of Particle States per Meter^3 Second^3.

Well of course a Hamiltonian based on gravity would reduce the variables of what quantum mechanical probabilitie will manifest, you're giving a gravitational cause for these wave collapses, as oppossed to no cause, which is all the QM interpretation really offers. Edited November 9, 2017 by Super Polymath

[Vmedvil]

Well of course a Hamiltonian based on gravity would reduce the variables of what quantum mechanical probabilitie will manifest, you're giving a gravitational cause for these wave collapses, as oppossed to no cause, which is all the QM interpretation really offers.

 

Yes, your right that r is supposed to be the Wave-Particle operator for a volume of space filled with particles which is in Speed of light over units of time-space being SO(n₁) over a Null cone as it increases with size it becomes more stable, n₁ = 3 in a volume.

 

 

 

If dubbleosix shows what point in each SO Layer or (n₂C) of the null cone that will be taken that greatly reduces the calculations for this system being a Temporal Field or what you call probability. In Dimensional Harmonic fields or my UFT, this greatly simplifies that calculations of the Time-space causality Harmonic which is like saying the state of the string in Super String Theory.

 

In string theory, a heterotic string is a closed string (or loop) which is a hybrid ('heterotic') of a superstring and a bosonic string. There are two kinds of heterotic string, the heterotic SO(32) and the heterotic E₈ × E_8.

 

 

In any case, you have told me how the hybrid Heterotic String moves.

 

 

• the octonionic projective plane – FII, dimension 16 = 2 × 8, F₄ symmetry, Cayley projective plane P²(O),
• the bioctonionic projective plane – EIII, dimension 32 = 2 × 2 × 8, E₆ symmetry, complexified Cayley projective plane, P²(C ⊗ O),
• the "quateroctonionic projective plane"^[2] – EVI, dimension 64 = 2 × 4 × 8, E₇ symmetry, P²(H ⊗ O),
• the "octooctonionic projective plane"^[3] – EVIII, dimension 128 = 2 × 8 × 8, E₈ symmetry, P²(O ⊗ O).

Which [math] r = \frac{1}{t_p^2 C^2}= \frac{1}{L_p^2} =\frac{1}{E_8 E_8} [/math] 

 

Which that Hybrid Heterotic String  is the Timelike worldline in this model. 

Dubblesoix are your Laplace operators not squared as well that is why.

 

[math]\Delta \mathbf{H} = \sqrt{|R_{ij}|^2 - <\psi|R_{ij}|\psi>^2} [/math]

 

Though, that may not be perfectly correct as I may need to root my operators as well.

 

[math] r = \sqrt \frac{1}  { t_p^2 C^2}= \sqrt \frac{1} { L_p^2} =\sqrt \frac{1}{ E_8 E_8} [/math] 

 

Now, I think that string is in the form you want as [math] Volume = meter^3 [/math] instead of my weird notation for that as [math] Meter^3 Seconds^3  [/math] as a dimensional harmonic.

 

The E₈ root system is a rank 8 root system containing 240 root vectors spanning R⁸. It is irreducible in the sense that it cannot be built from root systems of smaller rank. All the root vectors in E₈ have the same length. It is convenient for a number of purposes to normalize them to have length √2. These 240 vectors are the vertices of a semi-regular polytope discovered by Thorold Gosset in 1900, sometimes known as the 4₂₁ polytope.

 

The E₈ Lie group has applications in theoretical physics and especially in string theory and supergravity. E₈×E₈ is the gauge group of one of the two types of heterotic string and is one of two anomaly-free gauge groups that can be coupled to the N = 1 supergravity in ten dimensions. E₈ is the U-duality group of supergravity on an eight-torus (in its split form).

One way to incorporate the standard model of particle physics into heterotic string theory is the symmetry breaking of E₈ to its maximal subalgebra SU(3)×E₆.

In 1982, Michael Freedman used the E₈ lattice to construct an example of a topological 4-manifold, the E₈ manifold, which has no smooth structure.

Antony Garrett Lisi's incomplete "An Exceptionally Simple Theory of Everything" attempts to describe all known fundamental interactions in physics as part of the E₈ Lie algebra.^[6][7]

R. Coldea, D. A. Tennant, and E. M. Wheeler et al. (2010) reported an experiment where the electron spins of a cobalt-niobium crystal exhibited, under certain conditions, two of the eight peaks related to E₈ that were predicted by Zamolodchikov (1989).^[8][9]

 

 

Let me fix my original post......

 

 

  [math] Planck-Curvature = \frac{2}{3} <\psi|R_{ij}|\psi> \pi (\sqrt \frac{1}{L_p^2})^3 [/math] 

 

 

Then 

 

 [math] \Delta <E> = \frac{C^4}{12 G }  \int \Delta <R_{ij}>  (\sqrt \frac{1}{L_p^2})^3 = \frac{C^4}{12 G } \int  <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\sqrt \frac{1}{L_p^2})^3 [/math]  

 

which 

 

[math] \sqrt \frac{1}  { t_p^2 C^2}= \sqrt \frac{1} { L_p^2} =\sqrt \frac{1}{ E_8 E_8} [/math] 

 

 

I don't think this is a toy model any longer, I think that model you created is correct.

 

 

The parameter r is a geometrical invariant of the hyperbolic space, and the sectional curvature is K = −1/r². The scalar curvature is thus S = −n(n − 1)/r²

 

 the principal radii of the surface. For example, the scalar curvature of the 2-sphere of radius r is equal to 2/r².

 

 

that 1 maybe 2 above the radius but this is for a section so i think that is right since we are going for Quantum Gravity, wait ya N = 1 is supergravity which satisfies dV. 

 

[math] t_p^2= O^2 [/math]

[math] c^2 = P^2[/math]

Edited November 10, 2017 by Vmedvil

[Vmedvil]

Ok, I'll explain what is wrong here. This is based only on dimensional issues. The equation in issue here is without subscripts

 

[math]E =\ \frac{c^4}{G}<\psi|R|\psi>\ dV[/math]

 

 

The object R is an inverse length sqyuared and the object [math]<\psi|\psi>[/math] normally contains the probability density [math]\frac{P}{V}[/math]. The factor

 

[math]\frac{c^4}{G}[/math]

 

is the upper limit of the gravitational force, and has force dimensions so what we really have in an object that looks like - 

 

[math]E = mc^2 = \ \frac{F}{\ell^2}<\psi||\psi>\ dV[/math]

 

Force divided by length squared is equal to the energy of the system. Dividing the volume gives an energy density. So what is being said here, is that it is illegal dimensionally to square the inverse length component. as a volume term as that gives an inverse length to the power of six [math]\frac{1}{V^6}[/math], plus it is inverse when [math]dV[/math] isn't. 

 

The post above your post, this must not have been complete when you posted that, read it again, I fixed it by square rooting it, i noticed the same problem. it took me 6 hours to complete.

 

 

[math] \frac{1}{L_p} =  meters [/math]

 

The Final version after I noticed my stupid mistake is.

 

  [math] Planck-Curvature = \frac{2}{3} <\psi|R_{ij}|\psi> \pi (\sqrt \frac{1}{L_p^2})^3 [/math] 

 

 

Then 

 

 [math] \Delta <E> = \frac{C^4}{12 G }  \int \Delta <R_{ij}>  (\sqrt \frac{1}{L_p^2})^3 = \frac{C^4}{12 G } \int  <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\sqrt \frac{1}{L_p^2})^3 [/math]  

 

[math] \sqrt \frac{1}  { t_p^2 C^2}= \sqrt \frac{1} { L_p^2} =\sqrt \frac{1}{ E_8 E_8} [/math] 

 

[math] t_p^2= O^2 [/math]

[math] c^2 = P^2[/math]

 

Edit: Complete

Edited November 10, 2017 by Vmedvil

[Vmedvil]

Yes you can square root it. That will be ok. 

 

Even I get lost in complex math sometimes of 240 dimensions.

Edited November 10, 2017 by Vmedvil

[Vmedvil]

Look

 

 

[math]E = R\ dV[/math] is still different to

 

[math]E = R\ \sqrt{\frac{1}{V^2}} = R\ \frac{1}{\ell^3}[/math]

 

What root fixes that I don't even know lol, i will have to go through the entire 240 dimensional calculation again.

Edited November 10, 2017 by Vmedvil

[Super Polymath]

The post above your post, this must not have been complete when you posted that, read it again, I fixed it by square rooting it, i noticed the same problem. it took me 6 hours to complete.

 

 

[math] \frac{1}{L_p} = meters [/math]

 

The Final version after I noticed my stupid mistake is.

 

[math] Planck-Curvature = \frac{2}{3} <\psi|R_{ij}|\psi> \pi (\sqrt \frac{1}{L_p^2})^3 [/math]

 

 

Then

 

[math] \Delta <E> = \frac{C^4}{12 G } \int \Delta <R_{ij}> (\sqrt \frac{1}{L_p^2})^3 = \frac{C^4}{12 G } \int <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\sqrt \frac{1}{L_p^2})^3 [/math]

 

 

[math] \sqrt \frac{1} { t_p^2 C^2}= \sqrt \frac{1} { L_p^2} =\sqrt \frac{1}{ E_8 E_8} [/math]

 

[math] t_p^2= O^2 [/math]

[math] c^2 = P^2[/math]

 

Edit: Complete

 

The quantum foam is not smooth like three complete dimensions. It's not discrete, it's continuous, it's 2.nine something something off into infinity, in that dimensionality there's always a shorter length between two points. The rest of the fractional to complete the whole exists in anti de sitter space, which does pass through because it's perpendicular at points, creating the forces of mass, dark matter & dark energy, which causes fluctuations in the density medium of reality (matter & energy) expressed via the fundamental interactions & the cosmological constant. Special relativity can exist beyond the speed of light in such a fractal dimension.

 

You only need five dimensions, not 240, & microverses as opposed to strings.

Edited November 10, 2017 by Super Polymath

[Vmedvil]

The problem is simple, you used an inverse definition of the volume, when it wouldn't fix with the dimensions. 

 

Man, that has to be right otherwise it doesn't sync with the group correctly. It is something very simple, I am missing.

Edited November 10, 2017 by Vmedvil

[Super Polymath]

To understand what I'm talking about you'll have to read the link I gave on the previous page. It's anti gravity (dark energy) aka expansion (the cosmological constant) that causes a particle to break up into a wave, & the microgravity of multiple wave-particles passing through one another is what causes them to collapse into one non-wave particle. It's literally a miniature rendition of what's happening with our universe & the parts of it that exist beyond our particle event horizon (because the CMB outshines merging galactic superclusters).

Edited November 10, 2017 by Super Polymath

[Vmedvil]

What you can do is

 

[math]\frac{E}{\ell^3} = E(\frac{1}{\ell^3}) =\ <\psi|R|\psi>\ =\ <\psi|\mathbf{H}|\psi>[/math]

 

which produces the Hamiltonian density. 

 

You know what, whatever, I will take it. 

 

 

  [math] Planck-Curvature = \frac{2}{3} <\psi|R_{ij}|\psi> \pi (\sqrt \frac{E}{L_p^2})^3 [/math] 

 

 

Then 

 

 [math] \Delta <E> = \frac{C^4}{12 G }  \int \Delta <R_{ij}>  (\sqrt \frac{E}{L_p^2})^3 = \frac{C^4}{12 G } \int  <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\sqrt \frac{E}{L_p^2})^3 [/math]  

 

[math] \sqrt \frac{E}  { t_p^2 C^2}= \sqrt \frac{E} { L_p^2} =\sqrt \frac{E}{ E_8 E_8} [/math] 

 

[math] t_p^2= O^2 [/math]

[math] c^2 = P^2[/math]

 

Does anyone know what that E is defined as?

Edited November 10, 2017 by Vmedvil

[Vmedvil]

What is being said is you can attach an inverse volume to the energy and it becomes a Hamiltonian. That was not intended to mean you had the dimensions right, it was to show an example of where it could be right.

 

Oh, okay then lets revert back to.

 

 

  [math] Planck-Curvature = \frac{2}{3} <\psi|R_{ij}|\psi> \pi (\sqrt \frac{1}{L_p^2})^3 [/math] 

 

 

Then 

 

 [math] \Delta <E> = \frac{C^4}{12 G }  \int \Delta <R_{ij}>  (\sqrt \frac{1}{L_p^2})^3 = \frac{C^4}{12 G } \int  <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\sqrt \frac{1}{L_p^2})^3 [/math]  

 

[math] \sqrt \frac{1}  { t_p^2 C^2}= \sqrt \frac{1} { L_p^2} =\sqrt \frac{1}{ E_8 E_8} [/math] 

 

[math] t_p^2= O^2 [/math]

[math] c^2 = P^2[/math]

Edited November 10, 2017 by Vmedvil

[Super Polymath]

Are you referring to the fractal dimension that exists in both de sitter & anti de sitter space as the heteronic string? The third dimension that partially inhabits & connects two positive dimensions & two negative dimensions? Are you using 240 dimensions that are purely mathematical constructs to understand these 5 physical dimensions? I don't even know.

Edited November 10, 2017 by Super Polymath

[Vmedvil]

Are you referring to the fractal dimension that exists in both de sitter & anti de sitter space as the heteronic string? The third dimension that partially connects two positive dimensions to two negative dimensions? Are you using 240 dimensions that are purely mathematical constructs to understand these 5 physical dimensions? I don't even know.

 

Exactly, Super polymath, but that is a correct form too if someone defines E there which is the Gravity type term....... Other types of gravity..... WHAT?!?!

Edited November 10, 2017 by Vmedvil