Super Polymath Posted November 11, 2017 Report Share Posted November 11, 2017 (edited) So then it would be a Quantum entanglement like system. What's ironic is that what we designed entanglement to explain (particle pairs) actually wouldn't be the result of quantum entanglement in my theory (it would be the result of gravitational interactions occurring within wave-particles [bodies smaller than a planck length]). But yet, black body radiation is caused by a QE-like system. Edited November 11, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
Vmedvil Posted November 11, 2017 Report Share Posted November 11, 2017 (edited) So then it would be a Quantum entanglement like system. Then all you have to do if that is correct is sync the ω1 = ω2 to generate a Black-hole Entanglement and generate a stable wormhole between two BH which is a D-Brane between them. E7 = ∇Eb(t,ω,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isω12/2Mb)+V)2/C2))1/2))MbC2 E7 = ∇Eb(t,ω,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isω22/2Mb)+V)2/C2))1/2))MbC2 Which solves our L problem. Edited November 11, 2017 by Vmedvil Super Polymath 1 Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 11, 2017 Report Share Posted November 11, 2017 (edited) Then all you have to do if that is correct is sync the ω1 = ω2 to generate a Black-hole Entanglement and generate a stable wormhole between two BH which is a D-Brane between them. E7 = ∇Eb(t,ω,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isω12/2Mb)+V)2/C2))1/2))MbC2 E7 = ∇Eb(t,ω,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isω22/2Mb)+V)2/C2))1/2))MbC2 Which solves our L problem. Let's see what 006 thinks about this modification. Thoughts? Objections? Edited November 11, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
Vmedvil Posted November 11, 2017 Report Share Posted November 11, 2017 (edited) Let's see what 006 thinks about this modification. Thoughts? Objections? Hold on let me fuse everything. [math] Planck-Curvature = \frac{2}{3} <\psi|R_{ij}|\psi> \pi (\sqrt \frac{E_8}{L_p^2 L})^3 [/math] Then [math] \Delta <E> = \frac{C^4}{12 G } \int \Delta <R_{ij}> (\sqrt \frac{E_8}{L_p^2 L})^3 = \frac{C^4}{12 G } \int <\psi|(R_{ij} - <\psi |R_{ij}| \psi>)|\psi> (\sqrt \frac{E_8}{L_p^2 L})^3 [/math] [math] \sqrt \frac{E_8} { t_p^2 C^2 L}= \sqrt \frac{E_8} { L_p^2 L} =\sqrt \frac{E_8}{ E_8 E_8 L} [/math] [math] t_p^2= O^2 [/math][math] c^2 = P^2[/math] Then all you have to do if that is correct is sync the ωA = ωB to generate a Black-hole Entanglement and generate a stable wormhole between two BH which is a D-Brane between them, which E8 is the top E8 in this situation. E8 = ∇Eb(t,k,m,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (IsωA2/2Mb)+V)2/C2))1/2))MbC2 E8 = ∇Eb(t,k,m,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (IsωB2/2Mb)+V)2/C2))1/2))MbC2 Where Which If this holds true. explaining Dq which is basically where the wormhole lands or Strong Quantum Entanglement. This being an Exact solution to Supergravity or Quantum Gravity for Rs in Heterotic String SO(32) as the Schwarzchild metric and Kerr Metric for GR in non Planck level gravity. Edited November 11, 2017 by Vmedvil Super Polymath 1 Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 11, 2017 Report Share Posted November 11, 2017 (edited) This being an Exact solution to Supergravity or Quantum Gravity for Rs in Heterotic String SO(32) as the Schwarzchild metric and Kerr Metric for GR in non Planck level gravity. Do you think you could take a 3 hour road-trip & have the boys at Fermilab Muon Tevatron run a few test to confirm this for you now? Just explain some of the concepts behind this idea of this 5 dimensional hypersphere eternally absorbing itself to them - you know how non-discretely continuous (zeno's paradox) spacetime allows for two infinite quantities of temporally negative parallel universes to be perpendicular to one another at infinite points in space without one ever being able to fully consume the other. That self-absorption is behind expansion & gravity, & literally leads to electromagnetism, the weak nuclear force, & the strong nuclear force - but gravity & expansion are even more fundamental than those, which makes this self-absorption the most fundamental & really the only force behind all the forces of nature. You see with matter & energy existing as just fluctuations within the density medium of reality caused by this self-absorption, without a perpendicular anti de sitter space there would only be one solid clump with no movement or change. Edited November 11, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 12, 2017 Report Share Posted November 12, 2017 (edited) E8 = ∇Eb(t,k,m,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (IsωA2/2Mb)+V)2/C2))1/2))MbC2 E8 = ∇Eb(t,k,m,R,M,I,x,y,z) = ∇(1/((1-(((2MbG / Rs) - (IsωB2/2Mb)+V)2/C2))1/2))MbC2 I'm limited to basic algebra but as far as I can comprehend from what you SAY, or what you understood when you made these, is that there should be for of these, two positive, & two negative (de sitter & anti-de sitter space) written as a relation - & somehow you have to make it so the positive & negative D branes are parallel to themselves & perpendicular to each other. What you have here is just where the parallel positives touch. Also, remember:I explained entanglement in the toy model of gravity thread & it's not the same as the first link you posted. That's the result of particle pairs being causally connected. We're not causally connected to zero point energy arising from parallel white holes the size of a particle, that has nothing to do with the quantum zeno effect. Particle pairs are causally connected beyond the speed of light because of gravity beneath the planck length. Wave particles are made out of a billion billion billion smaller particles, moving a non-wave particle (displacement) via the observer effect will cause superluminal gravitational interactions when that particle breaks up into a wace because the speed of light crosses a distance beneath the planck length at a faster rate than C. This causes a domino effect that moves (polar) displacement particle B the same state (polarization) as particle A. The billions of smaller particles that compose a wave particle were gravitationally linked over all the wave particles between particle A & B. Changing the polarization of one link in the chain changes the polarization of all of the links in the chain (zeno) ftl. Weak qe use this math form Edited November 12, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
Vmedvil Posted November 12, 2017 Report Share Posted November 12, 2017 (edited) I'm limited to basic algebra but as far as I can comprehend from what you SAY, or what you understood when you made these, is that there should be for of these, two positive, & two negative (de sitter & anti-de sitter space) written as a relation - & somehow you have to make it so the positive & negative D branes are parallel to themselves & perpendicular to each other. What you have here is just where the parallel positives touch. Also, remember: use this math form It is in that form just V1 = C and C1-C2 = 0 at Rs , I am taking it that form angular momentum versus gravity, where V2 = Velocity which is the V term you see. This is the Schwarzchild Metric and Kerr Metric as a single equation along with SR. C1and C1 - C2 Though, they are transformed to equal C and not guv Then as V solution. Edited November 13, 2017 by Vmedvil Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 12, 2017 Report Share Posted November 12, 2017 (edited) Idk, you should be able to tell the speed of weak entanglement with these equations & you do didn't seem to know it in the other thread. We know it's over 10,000C Edited November 12, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
Vmedvil Posted November 12, 2017 Report Share Posted November 12, 2017 Idk, you should be able to tell the speed of weak entanglement with these equations & you do didn't seem to know it in the other thread. We know it's over 10,000C It still errors above C, so no actually I cannot. Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 12, 2017 Report Share Posted November 12, 2017 More importantly & the point of all of this, we've defined missing elements like de & dm, we have 5 dimensions & 5 forces of nature, so ultimately we need that exact survival probability. Then we can get stable fusion, we can bypass C in our communications, & most pertinent of all we know where particles are waves & where they aren't, that has massive implications for nano-robotics-replication-assembly. If we need platinum we can make it from anything. That's what you Esub8 should be telling, the speed of weak entanglement & the survival probability. Graphically it should be telling us that. Though, I understand that requires hella calculations. Quote Link to comment Share on other sites More sharing options...
Vmedvil Posted November 12, 2017 Report Share Posted November 12, 2017 (edited) More importantly & the point of all of this, we've defined missing elements like de & dm, we have 5 dimensions & 5 forces of nature, so ultimately we need that exact survival probability. Then we can get stable fusion, we can bypass C in our communications, & most pertinent of all we know where particles are waves & where they aren't, that has massive implications for nano-robotics-replication-assembly. If we need platinum we can make it from anything. That's what you Esub8 should be telling, the speed of weak entanglement & the survival probability. Graphically it should be telling us that. Though, I understand that requires hella calculations. My Computer would stack on that equation even at four cores to 4.0 ghz https://www.newegg.com/Product/Product.aspx?Item=N82E16819115072 Edited November 12, 2017 by Vmedvil Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 12, 2017 Report Share Posted November 12, 2017 (edited) It still errors above C, so no actually I cannot.It still errors above C, so no actually I cannot.It still errors above C, so no actually I cannot.So we still don't have L. Like I said, you should have 4 Esub8s, two for parallel de sitter D branes at the p's & q's where they merge, & two for anti de sitter space, where the parallel -D branes touch. They're asymmetric but still causally connected beyond C. That heterotic string dimensions that is s in both de sitter & anti de sitter space that connects & inhabits the other four dimensions as the 3rd of 5 total dimensions, has infinite relativistic C transformations as de sitter & anti de sitter perpendicular spaces get closer & closer to each other at points nigh-infinitely shorter than the planck length. Edited November 12, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 12, 2017 Report Share Posted November 12, 2017 My Computer would stack on that equation even at four cores to 4.0 ghz https://www.newegg.com/Product/Product.aspx?Item=N82E16819115072This is why you'll only get approval approximates. But how accurate are these approximates. Good enough for stable fusion reactions or wave collapse predictions is good enough. Which is why it needs to go to the test phase now. Quote Link to comment Share on other sites More sharing options...
Vmedvil Posted November 12, 2017 Report Share Posted November 12, 2017 (edited) So we still don't have L. Like I said, you should have 4 Esub8s, two for parallel de sitter D branes at the p's & q's where they merge, & two for anti de sitter space, where the parallel -D branes touch. They're asymmetric but still causally connected beyond C. That heterotic string dimensions that is s in both de sitter & anti de sitter space that connects & inhabits the other four dimensions as the 3rd of 5 total dimensions, has infinite relativistic C transformations as de sitter & anti de sitter perpendicular spaces get closer & closer to each other & points nigh-infinitely shorter than the planck length. Well, that is not how the equation works remember when we defined C as relative the C is getting longer by a degree but I don't know certain variables of the D brane like when C dilates ω = k + m, but what is curvature and permeability there. Edited November 12, 2017 by Vmedvil Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 12, 2017 Report Share Posted November 12, 2017 Well, that is not how the equation works remember when we defined C as relative the C is getting longer by a degree but I don't know certain variables of the D brane.There are infinite variables down to fractional infinity, you never get to the center of the heterotic string. Treat that as the Mandelbrot set. The smallest fluctuations that lead to gravity (dm) & expansion (de) occur @ infinitely small scales. So just get an approximate. Quote Link to comment Share on other sites More sharing options...
Vmedvil Posted November 12, 2017 Report Share Posted November 12, 2017 There are infinite variables down to fractional infinity, you never get to the center of the heterotic string. Treat that as the Mandelbrot set. The smallest fluctuations that lead to gravity (dm) & expansion (de) occur @ infinitely small scales. So just get an approximate. It is being treated as that set but I still don't know k and m for a wormhole. Quote Link to comment Share on other sites More sharing options...
Super Polymath Posted November 12, 2017 Report Share Posted November 12, 2017 (edited) You have k and m where de sitter & anti de sitter spaces touch, not just where the normal d branes touch (p's & q's) Edited November 12, 2017 by Super Polymath Quote Link to comment Share on other sites More sharing options...
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