Vmedvil 11 Posted November 13, 2017 Share Posted November 13, 2017 (edited) Well, on another forum dubblesoix had posted his Gravity model and I started to use Super-gravity equations on it, what is wrong with this, it turned into something strange which I am starting to call a "Wormhole metric" upon E transformation, this has to be wrong I still think what is your opinion on this, the equations that compose it are not wrong but this just seems odd. Dubblesoix's toy model taken way too far, I think. And Before Hand thanks, I know this is a long post to read, just to tell me how wrong the three of us are I blame myself and Polymath for not just dropping it at dubblesoix's solution. Edited November 13, 2017 by Vmedvil Link to post Share on other sites

Mordred 1372 Posted November 14, 2017 Share Posted November 14, 2017 took me a bit to go through this post but looks to me you arrived at the solution on your last pot on that thread. String theory happens to be one of my weaker topics however. (never really had much interest in it) so how SO(32) is handled to the unitary group reductions I can't be of much help other than stating any orthogonal group regardless of number of dimensions can be expressed with unitary groups. An orthogonal group is a double cover group. However simply do not have familiarity with SO(32) Link to post Share on other sites

Vmedvil 11 Posted November 14, 2017 Author Share Posted November 14, 2017 (edited) Ya, then your really not going to like that one being SO(32)/ SO(32) = 1 , but here is a paper on it. SO(32) Explained for Heterotic String. But it is composed of two E_{8} groups or E_{8} x E_{8}' = SO(2 * n) , n = 8 + 8 = 16 where Dubblesoix's E_{8} * t_{p}^{2} = 1 and my E_{8} * C^{2} = 1 That being my reasoning that it is balanced in that last form. E_{8} x E_{8} also equaling P^{2}(O x O) = P^{2}O^{2} E8 Decomposition Edited November 14, 2017 by Vmedvil Link to post Share on other sites

Mordred 1372 Posted November 14, 2017 Share Posted November 14, 2017 (edited) Yes but Dubblesix isn't working under SU(3) Hermitean he is working under SU(2) hermitean. If I picked up the exceptional E_8 group its irreducible is SU(3) ie the Jordon identities are 3×3 hermitean matrix. Granted that is a quick study of the E_8 group ie it uses the Gell Mann matrixes . The other problem I can surmise on this quick study is E_8 has a triple inner product which certainly isn't what Dubblesix is applying. Edited November 14, 2017 by Mordred Link to post Share on other sites

Vmedvil 11 Posted November 14, 2017 Author Share Posted November 14, 2017 (edited) 1 hour ago, Mordred said: Yes but Dubblesix isn't working under SU(3) Hermitean he is working under SU(2) hermitean. If I picked up the exceptional E_8 group its irreducible is SU(3) ie the Jordon identities are 3×3 hermitean matrix. Granted that is a quick study of the E_8 group ie it uses the Gell Mann matrixes . The other problem I can surmise on this quick study is E_8 has a triple inner product which certainly isn't what Dubblesix is applying. Oh, so I messed up my E_{7} to E_{8} transformation on my equation too, see i didn't realize that, I thought SU(2) and SU(3) were compatible., Yes which would make that 2 x 2 meaning F_{4} which could be used with a E_{8} which that E_{7} is like that too, I dunno how to fix that tho. So, his is a F_{4}, mine is a E_{7} and we are trying to cancel it with E_{8} x E_{8}, how do you fix that? Octonionic vs Quaterooctonic vs octooctonionic symmetry. the octonionic projective plane – FII, dimension 16 = 2 × 8, F_{4} symmetry, Cayley projective plane P^{2}(O), the bioctonionic projective plane – EIII, dimension 32 = 2 × 2 × 8, E_{6} symmetry, complexified Cayley projective plane, P^{2}(C ⊗ O), the "quateroctonionic projective plane"^{[2]} – EVI, dimension 64 = 2 × 4 × 8, E_{7} symmetry, P^{2}(H ⊗ O), the "octooctonionic projective plane"^{[3]} – EVIII, dimension 128 = 2 × 8 × 8, E_{8} symmetry, P^{2}(O ⊗ O). See, now I am confused wouldn't F_{4} x E_{7} = E_{8} Which F_{4} x E_{7} /E_{8} x E_{8} = 1/ E_{8 } Yes? Which I can think of a E_{8} that would work being 1/ħ ħ = Planck action. Which I am going to barrow this, sorry Dubblesiox [ latex ] \frac{ds}{dt} = \frac{1}{\hbar} \frac{c^4}{8 \pi G}\sqrt{<\psi|R_{ij}|\psi>} = \frac{1}{\hbar} \sqrt{<\psi|H|\psi>} [ /latex ] Edited November 14, 2017 by Vmedvil Link to post Share on other sites

Vmedvil 11 Posted November 14, 2017 Author Share Posted November 14, 2017 (edited) I dunno why that latex bugged. [latex] \frac{ds}{dt} = \frac{1}{\hbar} \frac{c^4}{8 \pi G}\sqrt{<\psi|R_{ij}|\psi>} = \frac{1}{\hbar} \sqrt{<\psi|H|\psi>} [/latex] Whatever Here Where his equation says that Which my equation says diag[0,0,0,C] or when ω^{2} = C^{2 }[0,0,0,0] like F_{uv} , but actually it is 7 so, [0,0,0,0,0,0,0] after absorbing T_{uv} x F_{uv } which T_{uv} = G_{uv} ∇'(x',y',z') = ∇(1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2} Then E_{b}(t,ω,R,M,I) = ∇(1/((1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2}))M_{b}C^{2} Which is odd upon ω_{s}^{2} = k^{2} + m^{2} Does go E_{8} so I dunno where a reverse solve goes into Schrodinger EQ ∇ = ∇'(x',y',z')(1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2} Then ∇ = E_{b}(t,ω,R,M,I) (1/((1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2}))M_{b}C^{2} R = ∇ which the kerr-child metric agrees with my ω^{2} Edited November 14, 2017 by Vmedvil Link to post Share on other sites

SuperPolymath 6 Posted November 14, 2017 Share Posted November 14, 2017 (edited) 43 minutes ago, Vmedvil said: [math]\frac{ds}{dt} = \frac{1}{\hbar} \frac{c^4}{8 \pi G}\sqrt{<\psi|R_{ij}|\psi>} = \frac{1}{\hbar} \sqrt{<\psi|H|\psi>}[/math] Was not latex Edited November 14, 2017 by SuperPolymath Link to post Share on other sites

Vmedvil 11 Posted November 14, 2017 Author Share Posted November 14, 2017 (edited) 2 hours ago, Vmedvil said: I dunno why that latex bugged. [latex] \frac{ds}{dt} = \frac{1}{\hbar} \frac{c^4}{8 \pi G}\sqrt{<\psi|R_{ij}|\psi>} = \frac{1}{\hbar} \sqrt{<\psi|H|\psi>} [/latex] Whatever Here Where his equation says that Which my equation says diag[0,0,0,C] or when ω^{2} = C^{2 }[0,0,0,0] like F_{uv} , but actually it is 7 so, [0,0,0,0,0,0,0] after absorbing T_{uv} x F_{uv } which T_{uv} = G_{uv} ∇'(x',y',z') = ∇(1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2} Then E_{b}(t,ω,R,M,I) = ∇(1/((1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2}))M_{b}C^{2} Which is odd upon ω_{s}^{2} = k^{2} + m^{2} Does go E_{8} so I dunno where a reverse solve goes into Schrodinger EQ ∇ = ∇'(x',y',z')(1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2} Then ∇ = E_{b}(t,ω,R,M,I) (1/((1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2}))M_{b}C^{2} R = ∇ which the kerr-child metric agrees with my ω^{2} Where a = fine structure constant "Mass" being e^{2}/ ħC which turns acos(Θ) into mass where r - acos(2Θ) = 0 which means 2M - r = 0 where 2MG - IC^{2 }= 0 So, I don't think the error is with the quaternionic equation or even dubblesiox's equation which it is definately not the D-brane, Kerr-Schwarzchild, or Schrodinger Equation, so I don't know at this point. Edited November 14, 2017 by Vmedvil Link to post Share on other sites

Vmedvil 11 Posted November 14, 2017 Author Share Posted November 14, 2017 (edited) 1 hour ago, Vmedvil said: Where a = fine structure constant "Mass" being e^{2}/ ħC which turns acos(Θ) into mass where r - acos(2Θ) = 0 which means 2M - r = 0 where 2MG - IC^{2 }= 0 So, I don't think the error is with the quaternionic equation or even dubblesiox's equation which it is definately not the D-brane, Kerr-Schwarzchild, or Schrodinger Equation, so I don't know at this point. Hell, No GR and QM are just not compatible I tend to believe now. I dunno, how all those equations can be correct and still get a wrong SO(n) group where you get SU(2) and SU(3) where somehow F_{4} x E_{7} ≠ E_{8 }I think somewhere along the line the mathematics guys/girls screwed up, physics blames math for this. In any case, yes there is a wormhole where you sync ω_{1} - ω_{2} = 0 , which links Hilbert space A and B which is the same as weak Quantum Entanglement, see this is the same problem again, when you link these it just generates anomalies bad even in supposedly anomaly free groups. WTF?, everything else points to this being right no anomalies in the linkage EQ's or any of the defining EQ's but somehow it still doesn't group correctly. After, my rant i realized yes E_{8} can be composed of lesser groups was your error math people, it does go E_{8} which F_{4} x E_{8} = SO(32) which with that 1/ħ = 1 Edited November 14, 2017 by Vmedvil Link to post Share on other sites

Vmedvil 11 Posted November 14, 2017 Author Share Posted November 14, 2017 (edited) 1 hour ago, Vmedvil said: Hell, No GR and QM are just not compatible I tend to believe now. I dunno, how all those equations can be correct and still get a wrong SO(n) group where you get SU(2) and SU(3) where somehow F_{4} x E_{7} ≠ E_{8 }I think somewhere along the line the mathematics guys/girls screwed up, physics blames math for this. In any case, yes there is a wormhole where you sync ω_{1} - ω_{2} = 0 , which links Hilbert space A and B which is the same as weak Quantum Entanglement, see this is the same problem again, when you link these it just generates anomalies bad even in supposedly anomaly free groups. WTF?, everything else points to this being right no anomalies in the linkage EQ's or any of the defining EQ's but somehow it still doesn't group correctly. After, my rant i realized yes E_{8} can be composed of lesser groups was your error math people, it does go E_{8} which F_{4} x E_{8} = SO(32) which with that 1/ħ = 1 I typed that wrong, I meant to say how does F_{4} x E_{7} = SO(28) not SO(32) then realized that math theory was wrong and it does actually go E_{8} which would make that SO(32) Otherwise it doesn't match up to cancel which means that "E_{8} cannot be composed of lesser groups" is WRONG!. Why because if that doesn't go E_{8} it says all of math is wrong past like Calculus 1 or 2 due to those equations being composed of nearly everything after that. Edited November 14, 2017 by Vmedvil Link to post Share on other sites

Mordred 1372 Posted November 14, 2017 Share Posted November 14, 2017 Ok I suggest you look specifically at the generators for the SU(2) then the 3^2=8 generators for SU(3) the previous uses the Pauli matrices ie Dirac. While the latter uses the Gell-Mann matrices. The strong force for example uses the latter via the eightfold Wayen. Link to post Share on other sites

Mordred 1372 Posted November 14, 2017 Share Posted November 14, 2017 (edited) Here this will help with the above Exceptional lie algebra SU(3) and Jordon pairs it details the group reductions https://www.google.ca/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/1403.5120&ved=0ahUKEwiniNOlnL_XAhUSFuwKHURyAckQFgghMAI&usg=AOvVaw3bvor0uH-6fTf3Hp9JxCyR Note it is arxiv on phone atm This will help as it provides the Jordon Schrodinger equation "On a Jordan-algebraic formulation of quantum mechanics: Hilbert space construction" https://arxiv.org/abs/hep-th/9304124 Edited November 14, 2017 by Mordred Link to post Share on other sites

Vmedvil 11 Posted November 15, 2017 Author Share Posted November 15, 2017 (edited) 5 hours ago, Mordred said: Ok I suggest you look specifically at the generators for the SU(2) then the 3^2=8 generators for SU(3) the previous uses the Pauli matrices ie Dirac. While the latter uses the Gell-Mann matrices. The strong force for example uses the latter via the eightfold Wayen. See, I thought that initially that it was Strong Quantum entanglement like space , but then I got a speed of 600 something C for speed instead of around 10,000 C, I will look into otherwise, I am done trying to GR+SR with QFT + QM, it and just drawing the model like this. Physics people go on strike against math until math people can unify their four models without anomalies. You know how math majors make fun of art majors well art can do something math cannot currently produce the UFT in physics with no problems. Edited November 15, 2017 by Vmedvil 1 Link to post Share on other sites

Mordred 1372 Posted November 15, 2017 Share Posted November 15, 2017 (edited) Well a large part of the problem with changing metrics is that you also change the axioms associated with each metric. For example in QM and QFT you have completely different operators. The problem isn't as pronounced under GR to SR primarily in GR all frames are inertial which is a bit different than SR with a rest frame. One must examine the axioms behind each metric before trying to mix them. It would be like trying to mix canonical and conformal treatments, particularly since several of the transposed vectors themselves may be different. In particular when dealing with complex conjugates etc... By the way naturally if you apply the Strong force to photons as per c you will get the wrong answer. I included the above to demonstrate the two primary types of generators used in particle physics ie the standard model itself. Photons don't interact with the strong force so are not part of the SU(3) group. It is described under the SU(2) and but not SU(3). Different fields can readily overlap the same volume without influencing one or the other. Edited November 15, 2017 by Mordred Link to post Share on other sites

Vmedvil 11 Posted November 15, 2017 Author Share Posted November 15, 2017 (edited) 1 hour ago, Vmedvil said: See, I thought that initially that it was Strong Quantum entanglement like space , but then I got a speed of 600 something C for speed instead of around 10,000 C, I will look into otherwise, I am done trying to GR+SR with QFT + QM, it and just drawing the model like this. Physics people go on strike against math until math people can unify their four models without anomalies. You know how math majors make fun of art majors well art can do something math cannot currently produce the UFT in physics with no problems. Lol....... We are gonna do it because of Math Strike. Now, Computers are Striking against math. Edited November 15, 2017 by Vmedvil Link to post Share on other sites

Mordred 1372 Posted November 15, 2017 Share Posted November 15, 2017 (edited) Group theory itself can be tricky two completely different objects that have absolutely nothing to do with one another can be described by the same symmetry group depending on how the vectors, scalars and spinors. That is all a group is "a representation" it is a means to organize all the degrees of freedom upon its vectors being applied. x posted lol nice picture Edited November 15, 2017 by Mordred Link to post Share on other sites

Vmedvil 11 Posted November 15, 2017 Author Share Posted November 15, 2017 (edited) 53 minutes ago, Mordred said: Group theory itself can be tricky two completely different objects that have absolutely nothing to do with one another can be described by the same symmetry group depending on how the vectors, scalars and spinors. That is all a group is "a representation" it is a means to organize all the degrees of freedom upon its vectors being applied. x posted lol nice picture All joking aside, thanks Mordred, I will go through the entire calculation again later, I don't doubt it is a simple math error that needs to be corrected. For right Now, I am on strike, Cuban Cigar time, Fidel Castro's favorite brand Cohiba Edited November 15, 2017 by Vmedvil Link to post Share on other sites

Mordred 1372 Posted November 15, 2017 Share Posted November 15, 2017 (edited) Good luck but what Dubblesix has is missing a key the k coordinate basis for the Levi Cevita. He has the Kronecker affine connections [math]\delta_{ij}[/math] which works under [math]\eta_{\mu\nu}[/math] you need the Levi-Cevita [math]\delta_{ijk}[/math] to include tidal force under GR in order to get the full Schwartzchild Edited November 15, 2017 by Mordred Link to post Share on other sites

Vmedvil 11 Posted November 15, 2017 Author Share Posted November 15, 2017 (edited) 1 hour ago, Mordred said: Group theory itself can be tricky two completely different objects that have absolutely nothing to do with one another can be described by the same symmetry group depending on how the vectors, scalars and spinors. That is all a group is "a representation" it is a means to organize all the degrees of freedom upon its vectors being applied. x posted lol nice picture ∇'(x',y',z') = ∇(1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2} ∇g = 0 isn't that the Levi-Cevita Connection. Edited November 15, 2017 by Vmedvil Link to post Share on other sites

Mordred 1372 Posted November 15, 2017 Share Posted November 15, 2017 (edited) Close but not quite under Banack spaces which if I understand correctly you want. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www2.physics.umanitoba.ca/u/khodr/Publications/inner-product-ANS-2015.pdf&ved=0ahUKEwi3_Zroy7_XAhWC0hoKHRBLBYUQFgggMAA&usg=AOvVaw2uUcHQ_2e_rfYt0L6ZYYbQ start with ijk coordinate basis ie i follows x, to value 1 on x coordinate, j follows the y axis to value 1 while k follows the z coordinate axis l. See last link on Cartesian Calculus 1 chapter uses Kronecker Here see 1.57 https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www2.physics.umanitoba.ca/u/khodr/Publications/inner-product-ANS-2015.pdf&ved=0ahUKEwi3_Zroy7_XAhWC0hoKHRBLBYUQFgggMAA&usg=AOvVaw2uUcHQ_2e_rfYt0L6ZYYbQ Here Vector calculus see section on cross product and its association to Levi-Civita 1.58 Chapter Calculus II applying Levi Civita for spherical and polar coordinates https://www.google.ca/url?sa=t&source=web&rct=j&url=http://physics.csusb.edu/~prenteln/notes/vc_notes.pdf&ved=0ahUKEwiJ1qns1b_XAhVBiRoKHYFfABYQFggwMAU&usg=AOvVaw2yHnhKwx8JCDebWO7Wxz5i This last should give you the details you need to get the triple inner product for your [math]E_8[/math] ie the dyad and quaturnions which should lead to your octonionic projective plane Edited November 15, 2017 by Mordred Link to post Share on other sites

Vmedvil 11 Posted November 15, 2017 Author Share Posted November 15, 2017 (edited) 5 hours ago, Mordred said: Close but not quite under Banack spaces which if I understand correctly you want. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www2.physics.umanitoba.ca/u/khodr/Publications/inner-product-ANS-2015.pdf&ved=0ahUKEwi3_Zroy7_XAhWC0hoKHRBLBYUQFgggMAA&usg=AOvVaw2uUcHQ_2e_rfYt0L6ZYYbQ start with ijk coordinate basis ie i follows x, to value 1 on x coordinate, j follows the y axis to value 1 while k follows the z coordinate axis l. See last link on Cartesian Calculus 1 chapter uses Kronecker Here see 1.57 https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www2.physics.umanitoba.ca/u/khodr/Publications/inner-product-ANS-2015.pdf&ved=0ahUKEwi3_Zroy7_XAhWC0hoKHRBLBYUQFgggMAA&usg=AOvVaw2uUcHQ_2e_rfYt0L6ZYYbQ Here Vector calculus see section on cross product and its association to Levi-Civita 1.58 Chapter Calculus II applying Levi Civita for spherical and polar coordinates https://www.google.ca/url?sa=t&source=web&rct=j&url=http://physics.csusb.edu/~prenteln/notes/vc_notes.pdf&ved=0ahUKEwiJ1qns1b_XAhVBiRoKHYFfABYQFggwMAU&usg=AOvVaw2yHnhKwx8JCDebWO7Wxz5i This last should give you the details you need to get the triple inner product for your E8 ie the dyad and quaturnions which should lead to your octonionic projective plane So, what your saying is I need to cross product instead of dot product, so this has a curl like Maxwell's equations which is funny the original idea behind this model was the form of Maxwell's equations for electromagnetism applied to SR, but instead of electromagnetism, it is Gravity-Spin being the two parts. How do you do that for a set of equations because there is time and space. ∇'(x',y',z') = ∇(1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2} Then ∇E_{b}(t,ω,R,M,I) = ∇(1/((1-(((2M_{b}G / R_{s}) - (I_{s}ω_{s}^{2}/2M_{b}))^{2}/C^{2}))^{1/2}))M_{b}C^{2} and I noticed something odd. ∇^{2 } will cancel with ()^{1/2} Maybe this isn't the finished form like I thought for these when it fit Schrodingers EQ and Matched Kerr-Schwarzchild, I thought it was done. That Second Part will take some time with all the e elements, actually both will. Edited November 15, 2017 by Vmedvil Link to post Share on other sites

Mordred 1372 Posted November 15, 2017 Share Posted November 15, 2017 (edited) Yes very much like curl more accurately rotatons under symmetry where the symmetry applies directly the applicable conservation laws etc. Ie for GR freefall has conservation of energy/momentum under a constant velocity and a change in velocity via acceleration is a rotation (rapidity). For conservation of angular momemtum under the closed angular momemtum system where the conservation law applies torque is zero. When torque is applied you undergo a rotation of the symmetry groups. The generator matrixes are rotations under symmetry. Recall under electromagnetism the magnetic field is 90 degrees out of phase to the electric field. So you first define the symmetry between the two fields then apply the rotation translation. For time you need to apply a time translation which will again be a rotation under symmetry. Hence cross products Edited November 15, 2017 by Mordred Link to post Share on other sites

Mordred 1372 Posted November 15, 2017 Share Posted November 15, 2017 Oh forgot the most important detail. The dot product of two vectors is a scalar. The cross product of two vectors is a vector. Link to post Share on other sites

Vmedvil 11 Posted November 15, 2017 Author Share Posted November 15, 2017 (edited) 6 hours ago, Mordred said: Yes very much like curl more accurately rotatons under symmetry where the symmetry applies directly the applicable conservation laws etc. Ie for GR freefall has conservation of energy/momentum under a constant velocity and a change in velocity via acceleration is a rotation (rapidity). For conservation of angular momemtum under the closed angular momemtum system where the conservation law applies torque is zero. When torque is applied you undergo a rotation of the symmetry groups. The generator matrixes are rotations under symmetry. Recall under electromagnetism the magnetic field is 90 degrees out of phase to the electric field. So you first define the symmetry between the two fields then apply the rotation translation. For time you need to apply a time translation which will again be a rotation under symmetry. Hence cross products Ya, time is 90 degrees out of phase of space in the light cone like magnetism so that makes sense. This makes sense too due to the graviton being its own "Antiparticle" like the photon in special or the Z boson always. You know what I am just going to describe this the exact same way as Maxwell did for light and electromagnetism because this secondly makes sense of gravitational waves are like constructive and deconstructive interference of the double slit experiment, but for two null cones. Edited November 15, 2017 by Vmedvil Link to post Share on other sites

Vmedvil 11 Posted November 15, 2017 Author Share Posted November 15, 2017 (edited) 1 hour ago, Vmedvil said: Ya, time is 90 degrees out of phase of space in the light cone like magnetism so that makes sense. This makes sense too due to the graviton being its own "Antiparticle" like the photon in special or the Z boson always. You know what I am just going to describe this the exact same way as Maxwell did for light and electromagnetism because this secondly makes sense of gravitational waves are like constructive and deconstructive interference of the double slit experiment, but for two null cones. Where is places the graviton as something very odd Time-Space Radiation with a spin of ω_{s}^{2 }= 4π^{2}f_{s}^{2 } where E_{p} = hf_{p} then E_{s} = h(ω_{s}/2π) which does something screwed up change the causality of space where C is negative because it travels sideways in time being space which is gravity if Where Edited November 15, 2017 by Vmedvil Link to post Share on other sites

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