Tom O'Neil Posted October 20, 2016 Author Share Posted October 20, 2016 There is several proofs showing group and phase velocity not transferring information at greater than c. You really need to understand the math properly to get a proper grasp on it. These superluminal group/phase velocities have been detected and around for quite some time. Plenty of professional scientists have seen these tests/papers. Yet they agree no superluminal information exchange is violated and that there is proof that SR isn't violated in regards to group velocity/phase velocity. However like I stated the detail is not easily explained outside the math. Light is not the last conduit of information exchange and I have theorized on neutrino cannons in the past which far exceed the speed of light. This may sound sci-fi ish but a neutrino collector can be built then so can a cannon and then pulsed for information exchange. Link to comment Share on other sites More sharing options...
Mordred Posted October 20, 2016 Share Posted October 20, 2016 (edited) Then show the math. Without that you have no theory. Only a postulation. Not even a model. present your mathematical proof Don't resort to the pop media literature, I've seen most of them. You have no idea how often I've seen these come up on similar arguments lol. I also suggest you study what is meant by information exchange in physics. While your at it study the difference between the folliwing terms under SR. peculiar velocity apparent velocity proper velocity Edited October 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
Strange Posted October 20, 2016 Share Posted October 20, 2016 Light is not the last conduit of information exchange and I have theorized on neutrino cannons in the past which far exceed the speed of light. This may sound sci-fi ish but a neutrino collector can be built then so can a cannon and then pulsed for information exchange. There is no evidence to think that neutrinos can travel faster than light. Their speed has been measured by creating a source of neutrinos (your "neutrino cannon") and then measuring how long they take to get to the destination. Link to comment Share on other sites More sharing options...
Mordred Posted October 20, 2016 Share Posted October 20, 2016 We know neutrinos don't exceed c, though CERN through their calibration error confused the population on that subject. Link to comment Share on other sites More sharing options...
swansont Posted October 20, 2016 Share Posted October 20, 2016 [/size] I believe Evil Liar this may state otherwise for the half point. If the half point moves relative to the front photon of the pulse, this would be expected. As I've already explained. Link to comment Share on other sites More sharing options...
Mordred Posted October 20, 2016 Share Posted October 20, 2016 (edited) 1st assertion is that on a Grand and Quantum scale Gt varies as mc2/e. This to say the greater the energy < time and low gravity threshold. Likewise with less energy >Gt! 2) 2nd assertion defies common logic regarding four known laws, but reduced them to 3 known laws(i.e. electromagnetism, strong nuclear and weak nuclear)!1 Graviton ∝ 10 femto Joules An assumption of mine is that light, energy and gravity share properties unknown yet to man on a quantum and grand scale. This is nothing, especially point 3. We don't physics on personal assumptions without showing the math. Ok well lets pick this apart. Spin statistics. 1) All three forces have spin 1 statistics, gravity matches spin 2. Explain how we get spin 2 from 3 or seperately from one spin 1 fields. 2) calculate the coupling constant of all 4. 3) use the three forces you mentioned and explain the coupling constant for gravity. Edited October 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
Tom O'Neil Posted October 20, 2016 Author Share Posted October 20, 2016 I feel if I go off topic this post may get trashed if its OK with the mods I will continue with both the skeleton assertions and neutrino gun but that is what I should stick to for now. Link to comment Share on other sites More sharing options...
Mordred Posted October 20, 2016 Share Posted October 20, 2016 (edited) Well start with the math, properly define your model. Words are meaningless in physics. Your OP post is near incomprehensible in terms of a model structure. Not trying to be rude, however the formula you presented is meaningless unless you can define it properly. You need far greater detail to convince the scientific community they are wrong. For example you posted the formula then show a spacetime 2d manifold image without any correlation math to show how your formula works under freefall If you seriously want your formula to gain weight, you will need to show your geodesic equations for the following. -spacetime geodesic -null geodesic also how to determine worldlines under your formula Edited October 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
Strange Posted October 20, 2016 Share Posted October 20, 2016 I feel if I go off topic this post may get trashed if its OK with the mods I will continue with both the skeleton assertions and neutrino gun but that is what I should stick to for now. You could start by explaining the connection between your initial assertions and neutrinos. You could then show the mathematics that supports your assertions. And then show how the idea would be tested: what quantitative (i.e. mathematical) predictions does your model make that allow it to be tested and distinguished from existing theory. Link to comment Share on other sites More sharing options...
Mordred Posted October 20, 2016 Share Posted October 20, 2016 (edited) You could start by explaining the connection between your initial assertions and neutrinos. You could then show the mathematics that supports your assertions. And then show how the idea would be tested: what quantitative (i.e. mathematical) predictions does your model make that allow it to be tested and distinguished from existing theory. I would add how your theory can lead up to the well tested predictions of the standard models ie explains those tests better. What are the comparisions from your model to standard Edited October 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
Tom O'Neil Posted October 21, 2016 Author Share Posted October 21, 2016 GUT Axiom G t ∝ mc²/e >Mordred “This is nothing, especially point 3. We don't physics on personal assumptions without showing the math. Ok well lets pick this apart. Spin statistics. 1) All three forces have spin 1 statistics, gravity matches spin 2. Explain how we get spin 2 from 3 or separately from one spin 1 fields. 2) calculate the coupling constant of all 4. 3) use the three forces you mentioned and explain the coupling constant for gravity.” >O’Neil 1) 1st assertion is that on a Grand and Quantum scale Gt varies as mc2/e. This to say the greater the energy < time and low gravity threshold. Likewise with less energy >Gt! 2) 2nd assertion defies common logic regarding four known laws, but reduced them to 3 known laws(i.e. electromagnetism, strong nuclear and weak nuclear)! 1 Graviton ∝ 10 femto Joules 3) An assumption of mine is that light, energy and gravity share properties unknown yet to man on a quantum and grand scale. G t ∝ mc²/e Mordred, I'm not a string theory fan! What I see in space time is an unfolding of matter which exploded long ago. Everything was derived from pure energy with anti-matter hydrogen colliding with its opposite hydrogen element in the expanse of ether. So in the case of high energy you would have low gravity and a time frame which operates upon matter differently than now. The geodesic three dimensional universe was created as a tetragrammaton and instantaneously exploded to form what we see today. Forces which work upon us are not separate but interlock and work in cohesion. The ongoing cosmic spin, I call it with uniformity and order as is a geometric wonder. From vibrations and spin of a neutrinos to the spin of galaxies and all matter interwoven with fractals we see an equation that seeks balance. You can't label gravity as a 2 for spin for a theory built on its statistics; as much as I cannot label, (G t ∝ mc²/e) as the Axiom of the Universe! In fact, what I have derived does look a bit more eloquent, but without the math to support it with its basis being the center indicator of an equation as the key to the universes behavior how can it is trusted? 1) Gravity is not constant from the neutrino to the pulsar, but customary to our time frame! 2) Time is a reference point! 3) Matter is the clay 4) Light is the wonder 5) And everything is energy (that being said gravity is a derivative of energy) _________________________________________________ This dimension has shared and continues to share its properties! 3600 seconds*G ∝ 5.972 × 10^24 kg*299792^2/5.4×10^41 Joules https://en.wikipedia.org/wiki/Orders_of_magnitude_(energy) a) Standard Constant for (G) 6.674 08(31) x 10-11 m3 kg-1 s-2 Page 21 b) With these numbers you should be able to derive a closer constant to the respective gravity for earth in 1 hours’ time using my Grand Unification Theorem 3600 seconds*G ∝ 5.972 × 10^24 kg*299792^2/5.4×10^41 Joules = See attachment for outputs which are nearly equal, but I believe the Grand Unification Formula I use more accurately predicts gravity for earth in 1 hours' time span. GUT Axiom.pdf Link to comment Share on other sites More sharing options...
Ophiolite Posted October 21, 2016 Share Posted October 21, 2016 I guess you see little value here but I will attempt to correlate this with my assertions with greater logic Frankly, it would be a great improvement if you were to use proper English. You know, the kind that involves more than juxtaposing a range of words that relate to important concepts in such a way that, to the inexperienced eye, they seem to convey something profound. You may be a knock down, totally decent guy who is kind to small children and animals, and loves his grandma, but boy you don't half post a bunch of meaningless gibberish. I repeat my advice - get a proper education. Link to comment Share on other sites More sharing options...
Tom O'Neil Posted October 21, 2016 Author Share Posted October 21, 2016 Frankly, it would be a great improvement if you were to use proper English. You know, the kind that involves more than juxtaposing a range of words that relate to important concepts in such a way that, to the inexperienced eye, they seem to convey something profound. You may be a knock down, totally decent guy who is kind to small children and animals, and loves his grandma, but boy you don't half post a bunch of meaningless gibberish. I repeat my advice - get a proper education. Why so academia can close my mind to the possibilities and thankyou I am kind, I was a lifeguard for 7 years and I truly love to help people. I agree its funny my English does need help, but I do convey meaning and you cannot deny my work here unless academia has really got to you Ophiolite! Anyway next time Link to comment Share on other sites More sharing options...
Mordred Posted October 21, 2016 Share Posted October 21, 2016 (edited) well I can see this thread isn't going far... by the way nothing I mentioned has anything to do with string theory. Secondly I can accurately state gravity has spin 2 statistics. Gravity wave detection confirmation via the quadrapole characteristics confirm spin 2. Thirdly I agree with Ophiolite, take some basic physics. Your posts show a very poor understanding. This is what I asked you to do with your equation. Derive the geodesic equation from your equation. Here is how it is done (standard gravitational force) In the presence of matter or when matter is not too distant physical distances between two points change. For example an approximately static distribution of matter in region D. Can be replaced by tve equivalent mass [latex]M=\int_Dd^3x\rho(\overrightarrow{x})[/latex] concentrated at a point [latex]\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})[/latex] Which we can choose to be at the origin [latex]\overrightarrow{x}=\overrightarrow{0}[/latex] Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex] [latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex] Where [latex] G_n=6.673*10^{-11}m^3/KG s^2[/latex] and [latex]r\equiv||\overrightarrow{x}||[/latex] According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element. [latex]ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2[/latex] Where [latex]d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2[/latex] denotes the volume element of a 2d sphere [latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere. The general relativistic form is. [latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex] By comparing the last two equations we can find the static mass distribution in spherical coordinates. [latex](r,\theta\varphi)[/latex] [latex]G_{\mu\nu}=\begin{pmatrix}1+2\phi_N/c^2&0&0&0\\0&-(1+2\phi_N/c^2)^{-1}&0&0\\0&0&-r^2&0\\0&0&0&-r^2sin^2(\theta)\end{pmatrix}[/latex] Now that we have defined our static multi particle field. Our next step is to define the geodesic to include the principle of equivalence. Followed by General Covariance. Ok so now the Principle of Equivalence. You can google that term for more detail but in the same format as above [latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex] [latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex] Denotes the gravitational field above. Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism) [latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex] Provided ds^2 is invariant [latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation [latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex] With the line element invariance [latex]\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x[/latex] The inverse of the metric tensor transforms as [latex]\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x[/latex] In GR one introduces the notion of covariant vectors [latex]A_\mu[/latex] and contravariant [latex]A^\mu[/latex] which is related as [latex]A_\mu=G_{\mu\nu} A^\nu[/latex] conversely the inverse is [latex]A^\mu=G^{\mu\nu} A_\nu[/latex] the metric tensor can be defined as [latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=diag(1,1,1,1) which denotes the Kronecker delta. Finally we can start to look at geodesics. Let us consider a free falling observer. O who erects a special coordinate system such that particles move along trajectories [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex] Specified by a non accelerated motion. Described as [latex]\frac{d^2\xi^\mu}{ds^2}[/latex] Where the line element ds=cdt such that [latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex] Now assunme that the motion of O changes in such a way that it can be described by a coordinate transformation. [latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex] This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation. [latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex] Where the new line element is given by [latex]ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu[/latex] and [latex] g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}[/latex] and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex] Denote the metric tensor and the affine Levi-Civita connection respectively. I want you to take your equation and get this geodesic equation. [latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex] The above shows how to get the geodesic equation from [latex]f=G\frac {Mm}{r^2}[/latex] though it isn't obvious from the way its represented above. See Newton potential. It might even help if you understood how the formula [latex]e=mc^2[/latex] is derived. This site has not too bad an example. http://www.askamathematician.com/2011/03/q-why-does-emc2/ This is what is required. YOU MUST SHOW HOW you derived [latex]Gt\propto \frac{mc^2}{e}[/latex] this is like saying Gt=1 which is garbage as Strange already pointed out. Edited October 21, 2016 by Mordred 2 Link to comment Share on other sites More sharing options...
Sensei Posted October 21, 2016 Share Posted October 21, 2016 (edited) If we are not sure of where the electron is in time on an xyz access to its precise location, and we are positive that light bends around matter; then the use of this detector may or may not validate the standard model of physics by witnessing the first experiment which would or would not weigh the neutrino. (...) Neutrinos don't have fixed rest-mass, nor fixed relativistic-mass/energy. You should learn how to create neutrinos in the first place. The most common way is fusion of two protons: [math]p^+ + p^+ \rightarrow D^+ + e^+ + v_e + 0.42 MeV[/math] Neutrino made in this reaction can be any energy between 0 to 420 keV. If electron has more kinetic energy, neutrino has less, and vice versa. Old method of detection of neutrinos is reaction with Chlorine-37: [math]^{37}Cl + v_e + 0.814 MeV \rightarrow ^{37}Ar + e^-[/math] As we can see neutrino made in previous reaction cannot trigger this reaction, as they have up to 420 keV, while Chlorine-37 must have neutrinos with energy above 814 keV! Electron capture is example reaction which creates neutrinos always with fixed relativistic-mass/energy. f.e. [math]^{37}Ar + e^- \rightarrow ^{37}Cl + v_e + 0.814 MeV[/math] (but it must be exclusive decay mode of isotope) Edited October 21, 2016 by Sensei Link to comment Share on other sites More sharing options...
Strange Posted October 21, 2016 Share Posted October 21, 2016 but I do convey meaning Barely. I had assumed English is not your native language. and you cannot deny my work here unless academia has really got to you Academia has never got to me, I'm afraid. But I can confidently state that your posts on this subject are meaningless nonsense. Link to comment Share on other sites More sharing options...
imatfaal Posted October 21, 2016 Share Posted October 21, 2016 Good grief. In your "paper" you provide a calculation include Newtons Gravitational Constant which is known to about five parts in a hundred thousand and give the answer with multiple pages of decimal places. All but the first bit of the first line are complete and utter garbage - even if the equation is correct! Get this thread straight and start answering questions with meaningful answers or I will report and request closure. Link to comment Share on other sites More sharing options...
Strange Posted October 21, 2016 Share Posted October 21, 2016 Standard Constant for (G) 6.674 08(31) x 10-11 m3 kg-1 s-2 Page 21 b) With these numbers you should be able to derive a closer constant to the respective gravity for earth in 1 hours’ time using my Grand Unification Theorem 3600 seconds*G ∝ 5.972 × 10^24 kg*299792^2/5.4×10^41 Joules We seem to be back to your inability to explain your ideas with any sort of detail or clarity. 1. What is "5.972 × 10^24 kg" ? (I think I know, but you should be explaining what you are doing at each step) 2. What is "299792^2" ? (Again, I think I know, but you should be explaining what you are doing at each step) 3. If the answer to (2) is c then why have you not shown the units? 4. What is "5.4×10^41 Joules" ? Where did this number come from? 5. What is the result of this calculation (0.00000099) supposed to represent? 6. What does "3600 seconds*G" mean? (This comes to 2.4 × 10-7 m3 kg-1 s-1, which doesn't seem to relate to anything else in your "equation" or have much physical meaning.) 7. Why do you have a proportionality symbol (∝) there? Or is that supposed to be the fine structure constant? My scientific prediction based on evidence so far is that you will either not answer these questions at all or you will provide random non-sequiturs as a response. Prove me wrong! (as the crackpots like to say) Link to comment Share on other sites More sharing options...
Ophiolite Posted October 21, 2016 Share Posted October 21, 2016 I should have tracked the number of times on many forums I have felt constrained to say that I know almost no physics, but I damn well no bullshit when I see it. Tom, you have a bunch of people giving you excellent advice, showing you your errors and trying to guide you. However, in your arrogance, or stupidity (you decide which) you ignore it all and sail on with even more nonsense. Please stop it. Just because you were also a lifeguard is no reason for me to give you an easy time. Your posting content does not merit that. Link to comment Share on other sites More sharing options...
Tom O'Neil Posted October 27, 2016 Author Share Posted October 27, 2016 (edited) well I can see this thread isn't going far... by the way nothing I mentioned has anything to do with string theory. Secondly I can accurately state gravity has spin 2 statistics. Gravity wave detection confirmation via the quadrapole characteristics confirm spin 2. Thirdly I agree with Ophiolite, take some basic physics. Your posts show a very poor understanding. This is what I asked you to do with your equation. Derive the geodesic equation from your equation. Here is how it is done (standard gravitational force) In the presence of matter or when matter is not too distant physical distances between two points change. For example an approximately static distribution of matter in region D. Can be replaced by tve equivalent mass [latex]M=\int_Dd^3x\rho(\overrightarrow{x})[/latex] concentrated at a point [latex]\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})[/latex] Which we can choose to be at the origin [latex]\overrightarrow{x}=\overrightarrow{0}[/latex] Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex] [latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex] Where [latex] G_n=6.673*10^{-11}m^3/KG s^2[/latex] and [latex]r\equiv||\overrightarrow{x}||[/latex] According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element. [latex]ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2[/latex] Where [latex]d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2[/latex] denotes the volume element of a 2d sphere [latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere. The general relativistic form is. [latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex] By comparing the last two equations we can find the static mass distribution in spherical coordinates. [latex](r,\theta\varphi)[/latex] [latex]G_{\mu\nu}=\begin{pmatrix}1+2\phi_N/c^2&0&0&0\\0&-(1+2\phi_N/c^2)^{-1}&0&0\\0&0&-r^2&0\\0&0&0&-r^2sin^2(\theta)\end{pmatrix}[/latex] Now that we have defined our static multi particle field. Our next step is to define the geodesic to include the principle of equivalence. Followed by General Covariance. Ok so now the Principle of Equivalence. You can google that term for more detail but in the same format as above [latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex] [latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex] Denotes the gravitational field above. Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism) [latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex] Provided ds^2 is invariant [latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation [latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex] With the line element invariance [latex]\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x[/latex] The inverse of the metric tensor transforms as [latex]\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x[/latex] In GR one introduces the notion of covariant vectors [latex]A_\mu[/latex] and contravariant [latex]A^\mu[/latex] which is related as [latex]A_\mu=G_{\mu\nu} A^\nu[/latex] conversely the inverse is [latex]A^\mu=G^{\mu\nu} A_\nu[/latex] the metric tensor can be defined as [latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=diag(1,1,1,1) which denotes the Kronecker delta. Finally we can start to look at geodesics. Let us consider a free falling observer. O who erects a special coordinate system such that particles move along trajectories [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex] Specified by a non accelerated motion. Described as [latex]\frac{d^2\xi^\mu}{ds^2}[/latex] Where the line element ds=cdt such that [latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex] Now assunme that the motion of O changes in such a way that it can be described by a coordinate transformation. [latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex] This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation. [latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex] Where the new line element is given by [latex]ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu[/latex] and [latex] g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}[/latex] and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex] Denote the metric tensor and the affine Levi-Civita connection respectively. I want you to take your equation and get this geodesic equation. [latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex] The above shows how to get the geodesic equation from [latex]f=G\frac {Mm}{r^2}[/latex] though it isn't obvious from the way its represented above. See Newton potential. It might even help if you understood how the formula [latex]e=mc^2[/latex] is derived. This site has not too bad an example. http://www.askamathematician.com/2011/03/q-why-does-emc2/ This is what is required. YOU MUST SHOW HOW you derived [latex]Gt\propto \frac{mc^2}{e}[/latex] this is like saying Gt=1 which is garbage as Strange already pointed out. I can use wiki too. 1 is proportional to 1! Remember I'm using proportionality in my equation. G t ∝ mc²/e In mathematics, two variables are proportional if a change in one is always accompanied by a change in the other, and if the changes are always related by use of a constant multiplier. The constant is called the coefficient of proportionality or proportionality constant. If one variable is always the product of the other and a constant, the two are said to be directly proportional. x and y are directly proportional if the ratio y/x is constant. If the product of the two variables is always a constant, the two are said to be inversely proportional. x and y are inversely proportional if the product xy is constant. https://en.wikipedia.org/wiki/Proportionality_(mathematics) Edited October 27, 2016 by Tom O'Neil Link to comment Share on other sites More sharing options...
Mordred Posted October 27, 2016 Share Posted October 27, 2016 (edited) has nothing to do with using wiki lol. Your equation is garbage. For the reasons already mentioned. Try actually studying a subject before trying to reinvent it. Or better yet post precisely how you derived your equation. As previously requested. For example your not even using the full e=mc^2 formula as you have no momentum term. You don't just randomly add in terms to existing formulas without going through the correct derivitave's that are used to generate that formula. If you can't show your work in how you derived that equation then its a useless formula. Edited October 27, 2016 by Mordred Link to comment Share on other sites More sharing options...
Tom O'Neil Posted October 27, 2016 Author Share Posted October 27, 2016 Prove to me that the equation is not proportional. You can't, because its beyond your comprehension. Also there is no known Grand Unification Equation as simple and straight forward as this. Link to comment Share on other sites More sharing options...
Mordred Posted October 27, 2016 Share Posted October 27, 2016 (edited) roflmao show your work. Try looking at the remaining units on the rhs and lhs of the equation bud. Do you even know how the formula [latex] e=mc^2[/latex] was derived? You might find the gravitational constant is already incorperated Edited October 27, 2016 by Mordred Link to comment Share on other sites More sharing options...
Tom O'Neil Posted October 27, 2016 Author Share Posted October 27, 2016 can you show me a link where the gravitational constant is used in e=mc^2? Link to comment Share on other sites More sharing options...
Mordred Posted October 27, 2016 Share Posted October 27, 2016 (edited) Prove to me that the equation is not proportional. You can't, because its beyond your comprehension. Also there is no known Grand Unification Equation as simple and straight forward as this. Your absolutely right, but then you have no idea what a unification formula is... Where is your strong force, weak force, electromagnetism? How do you account for 196 degrees of freedom from the standard model particles? Where is your symmetry group tensors? How do you define SO(10)×SO(5)×SO(3)×SO (2)×U (1) groups? where is your 4 momentum 3 velocity correlations? can you show me a link where the gravitational constant is used in e=mc^2? derive it from f=ma to e=mc^2 Or simply look at the post you quoted under Newton potential.. What did you think that term means? Edited October 27, 2016 by Mordred Link to comment Share on other sites More sharing options...
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