Everything posted by Mordred
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Who derived the above relation you or AI https://www.google.com/search?q=%5Crho+%3D+%5Cfrac%7BR_s+c%5E2%7D%7B8%5Cpi+G+r%5E3%7D+%3D+%5Cfrac%7B%5Ckappa%5E%7B2%7Dc%5E%7B2%7D%7D%7B8%5Cpi+Gr%5E%7B2%7D%7D&rlz=1C1VDKB_enCA1142CA1142&gs_lcrp=EgZjaHJvbWUyBggAEEUYOdIBCDEzNjlqMGo3qAIIsAIB&sourceid=chrome&ie=UTF-8&udm=50&fbs=ADc_l-aN0CWEZBOHjofHoaMMDiKp9lEhFAN_4ain3HSNQWw-mMGVXS0bCMe2eDZOQ2MOTwnMa06_-qUutYsIv5lB1HqB7Pf6gcWqPaZz5tPxChciWAvSSkt6iwnm8RS5iDIeKXMcHR4MMz-Y7wrd2kVxuyN5L5Fi49WMc2CLg4H9YMVUU0sdAvw8c5baTg2adLWst0WJE16c5OvP6B3zpkfcIBdQb4Nxtg&ved=2ahUKEwit6pODlvuSAxUcMDQIHV7TJmkQ0NsOegQIAxAB&aep=10&ntc=1&mstk=AUtExfDA6k-pGEJLfvT-Whh9mpJ8Rv55s-ylm6OHEw-LdTIhitRcSlJP6q0VkARNCS10P7oiM5UFDIQCgw3pu7W5spbj2k80JtUZWU387TL06i_YVqZZoyHKohJizCDlMyWJW76LDOBR95gnlS3L8VhPEMCAtec-seV7vQVCLTko71y0TPrVW0OU4U7s6vopeyp_1j6LnHTvS9J6OuVZBL1-NkV05P5P0KDg5eAfLSoyY58uwWEAp_KG20Jb5-2YJXbjZc6Y5JTUmN1PQjBqwnrD_WpW66SDm5YK-eFff3v_1WdMsidKY-TWO3qUKfl4RYrtqNvRH6xQDFaofg&csuir=1&mtid=gVeiabTpJuzb0PEP_vGJgA4 \[H_{0}=\sqrt{8\pi G\frac{4\sigma_{SB}T_{CMB}^{4}}{3\alpha^{2}c^{3}}}\] what about this equation which I noticed also in the Allen app. https://www.google.com/search?q=H_%7B0%7D%3D%5Csqrt%7B8%5Cpi+G%5Cfrac%7B4%5Csigma_%7BSB%7DT_%7BCMB%7D%5E%7B4%7D%7D%7B3%5Calpha%5E%7B2%7Dc%5E%7B3%7D%7D%7D&rlz=1C1VDKB_enCA1142CA1142&oq=H_%7B0%7D%3D%5Csqrt%7B8%5Cpi+G%5Cfrac%7B4%5Csigma_%7BSB%7DT_%7BCMB%7D%5E%7B4%7D%7D%7B3%5Calpha%5E%7B2%7Dc%5E%7B3%7D%7D%7D&gs_lcrp=EgZjaHJvbWUyBggAEEUYOdIBCDQ4MTRqMGo5qAIAsAIB&sourceid=chrome&ie=UTF-8
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Insight or just coincidence?
Heisenburg uncertainty principle for starters. The inherent uncertainty that the more you determine the particle position the less certain you are in the momentum and vice versa. Prelude to understand the QM Schrodinger equation.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Its too bad you do not understand linearization. However your right I'm done trying to help you improve the quality of your articles. I can quarantee without linearization applied to specific sets such as S^2 vs S^1 vs SO ( 3.1) aka (spacetime) Lorentz or Poincare group relations none of your articles will ever gain any weight in any peer review publication standard. Llinearizarion also involves specific equations such as your use of the Beta function. However that's obviously too much for you. Good luck Obviously any adherence to math relations beyond simple scalar quantities is too much math for you. For other readers 3P is the spatial sum of Pressures in all three spatial directions. \[P_x, P_y, P_z\]
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
The answer would require use of three fields each describing a specified order of relations Linear acceleration and stress math]f(r) = 4\pi r^2[/math] can be plotted on a 2D graph does not make the expression itself "graphical coordinates Thats a 2 dimensional graph using the double cover of the SO(3) group under SU(2)
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Its too bad you do not understand your mathematics do not match what you verbally describe as your ontology. S^2 You show as a 2 dimensional plane. Which does not produce a 3 dimensional sphere. You lectured me on my proof of critical density because it used spheres yet your very own mathematics also produces spheres. However in case you never noticed not once did that 3 show up in the geometry mapping generated by \[ 4\pi r^2\]. I also asked previous " where is the pressure" Would you like to know why pressure to energy density has a 1 to 3 ratio ? \[ \rho=\frac{3p}{c^2\] Via stress energy momentum tensor \[T_{00}, T_{i,j}\] Pressure has 3 specific mathematical relations to account for. Engineers will love this part being described.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Im applying strictly mathematical rules in my line of questioning that has nothing to do with physics. You do realize the very fact that \[4\pi r^2\] can be graphed it is in effect graphical coordinates?
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
So can't stick to simply the geometry relations. Graph \[4\ pi r^2\] pick any radius. What shape do to you get. Does or does not your S^2 group produce the same graph. Plain and simple
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
I know it's 2 spheres I stated that above. In application to my proof area then compared to area now to determine the Hubble constant. The surface area of both objects are 3 dimensional spheres that is the area being described by whichever radius its at. That is not a circle ( you need 3 dimensions to define area.) Would you like to know the math relations the very term ( orthogonal ) tells me. One projection in a 90 degree relation to the other \[ A \cdot B\] I have no identity of what A or B has. But I can automatically apply the Kronecker Delta to that inner product ( Hilbert Space). Regardless if it's coordinates or not... ( if both A and B are finite and continous) with regards to Hilbert
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Oh my please tell me you know basic geometry relations. One sphere \[A=4\pi r^2\] 2 spheres \[A=8\pi r^2\] One singular geometric object with \{8\pi r^2\] is a cylinder with 4 times the radius in height. Now look at the mathematical proof I posted again. You have the above relations throughout all your documents apparently without realizing those terms automatically describe geometric objects that are not a circle.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Its a simple geometry question nothing more \[ 4\ pi r^2\] Mathematically describes the area of a sphere.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Explain in detail how you get Just and only just the \[4\ pi r^2\] nothing else. Then once you do that maybe just maybe it might help others including myself what your doing to correlate an observable 3d universe without any time coordinate.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
You have switched from 4\pi r^2 to 8 \ pi r^2 in the above one term to the next without explaining why. \[ A= 4 \pi r^2\] is the area of a sphere. The question I asked why are you using that term if your not using spheres.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Follow up question What single geometric object has the relation \{ 8\pi r^2\}.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Without referring to any of your links or articles ( post the math here) Answer the following. Why do you use the \[8\pi r^2\] relation If you are not using the volume comparision between 2 spheres.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Well said, for that matter that's even explained in my 1920 publication textbook on physics rather interesting reading as the atom is only described by protons and electrons and no other particle was known. The textbook also specifically went into classical examples on refrigeration, power generators etc. The amusing part is the term Superposition is included even though it doesn't even refer to QM...not surprised of course but still amusing Just a side note Newtons gravitational law had a different form then. \[F=k\frac{m\acute{m}}{r^2}\] with units given in dynes. K was simply some constant.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Thank you for proving my earlier comment last night. That disregard of the mathematical proof of the critical density formula that you consider invalid despite all its success and clear match to observational evidence simply because of your refusal to apply vectors in understanding that proof is a clear example.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Vectors are not a bookkeeping device beyond mathematic relations directly related to displacement. Your ontology arguments do not matter in the slightest. This is strictly mathematics.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
There is nothing indirect on my last statement and the mathematics I posted has nothing to do with any particular model. Its literally geometry and vectors under a geometry.relating kinetic and potential energy terms without any specified particle contributor. You do know what the dot product or cross product means with regards to vectors do you not ? Those math operations and symbols are included in the above proof for critical density. The mass term has zero model dependency neither does the gravitational terms. It is any arbitrary inward generation of force of attraction to any arbitrary force of expansion in mathematical displacement terms using vectors. I only related those mathematics with regards to the FLRW metric with regards to Hubble constant.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Your wrong look at the inner and cross product vector addition rules. An inner product space is 2 dimensional reread the above relations.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Correct our universe is a 3d sphere for the observable universe. However note that the critical density formula itself only applies 2 dimensions. Or did you miss the R^2 term ? Here is another relation you missed look at dimensionality of the cross product and the dimensionality of the dot product space. ( inner product space) with above \[\mu \cdot \nu= \nu \cdot \mu\] For relevant vector addition rules regarding inner and cross products. The above defines this as an inner product space and the dot product of 2 vectors potential energy (collapse) vs kinetic energy outward ( exoansion). You only require 2 dimensions to describe that relation as shown above. In stress energy tensor terms the critical density enerrgy density and pressure relations are \[\rho =T_{00} ] \[P= T_{i,j}\]
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Ok lets compare by looking at just the kinetic and potential energy terms in just a sphere with gravitational or otherwise pull in terms of the critical density. Its mathematical proof in more detail. kinetic energy \(\frac{1}{2}mv^2=\frac{1}{2}m\dot{R}^2\) potential energy \(U -\frac{GMm}{R}\) mass \(\rho \times volume=\rho \times \frac{4}{3}R^3\) substituting M into U \[U=-\frac{G(\rho \cdot \frac{4}{3}\pi R^3m}{R}=-\frac{4\pi G\rho R^2}{3}\] total energy \[E_t=K+U=\frac{1}{2}m\cdot {R}^2-\frac{4\pi G\rho R^2}{3}\] set curvature to zero energy balance for critically dense universe E=0 the point of when expansion halts or it can be described where the escape velocity is identical to the expansion speed. solve for density \(\rho\rightarrow \rho_c\) \[\frac{1}{2}\cdot{R}^2=\frac{4\pi G\rho R^2}{3}\] incorperate Hubble parameter \[\frac{v}{R}=\frac{ \dot{R}^2}{R^2}\] \[H^2=\frac{\dot{R}^2}{R^2}\] \[\rho_c=\frac{3H^2}{8\pi G}\] that is what you should have
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
No offense buts that's another article your asking to go through this one being 77 pages. Please post how you incorperated pressure into your calculation using the equation above.
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Where is the pressure term
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
Take a closer look at the critical density formula from which all the equations of state for cosmology are related from. Then look at the energy only mathematical proof of the critical density formula. You will find the missing term I mentioned. In so far as the FLRW metric and how the Maxwell Boltzmann equations also ties into the above or for that matter the Raychaudhuri equations under GR. ( note 3 methodologies arriving at the same relations ) I leave it to you to choose to include the missing term ( yes I know precisely which term you dropped ) or not. Thats your choice but without that missing term your equations will not accurately follow the evolutionary history of expansion regardless if your employing Maxwell Boltzmann or any of the above methodologies. ( it would amount to a waste of my time to go through the effort of showing what I described above simply to have it discounted by your ontology reasons ). Hint the energy only mathematical proof uses nothing more than the energy momentum relation you employ. You may wish to keep your ontology constrained but that's not a view I share. You have already mentioned numerous constraints throughout this thread on mathematical operations you cannot deploy ( geometry, mappings etc etc etc.)
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Simplifying SR and GR with Relational Geometry — Algebraic Derivations Without Tensors. Testing and discussion.
From a practical and logical sense changing the definition of work makes sense. By describing the required " ability " to the operation being performed between states/ systems including operators, partition functions ... you can encompass a wider variety of mathematics not restricted to physics. Also by identifying what's changing in the work term also helps identify the mathematical treatment. This is already done. So why would you need to change the definition of energy as well ?