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 International Skeptics Forum Finite Theory: Historical Milestone in Physics

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 20th December 2019, 10:24 AM #1001 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Lunar Laser Ranging Experiment (Correction of Correction) According to FT, here is the speed of light reversed (or the time it takes for a photon to travel 1 meter) at each position of a photon between the surface of the Earth and the surface of the Moon, which is directly proportional to the ratio of the gravitational potential of observed photon and the gravitational potential of the observer: $y'=\frac{1}{c} \times \left. \frac{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| x-{p_i}\right| }\right.}}{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| {p_i}\right| }\right.}}\right.$ $y'=\frac{\frac{{m_s}}{\left| x-{p_s}\right| }+\frac{{m_m}}{\left| x-{p_m}\right| }+\frac{{m_e}}{\left| x-{p_e}\right| }+h}{c\, \left( \frac{\mathit{ms}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Or (example): The integral of the aforementioned equation will return the time it'll take for a photon to travel a certain distance while considering 3 perfectly aligned masses is: $y=\frac{1}{c} \times \int {\left. \frac{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| x-{p_i}\right| }\right.}}{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| {p_i}\right| }\right.}}dx\right.}$ $y=\frac{{m_s} \log{\left( \left| x-{p_s}\right| \right) }+{m_m} \log{\left( \left| x-{p_m}\right| \right) }+{m_e} \log{\left( \left| x-{p_e}\right| \right) }+h x}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }-\frac{{m_s} \log{\left( \left| {p_s}\right| \right) }+{m_m} \log{\left( \left| {p_m}\right| \right) }+{m_e} \log{\left( \left| {p_e}\right| \right) }}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Or (example): Where m, p & r are the mass, position and radius of the Earth, Sun & Moon respectively: $m_e= 5.9736 \times {{10}^{24}}kg$ $m_s= 1.98892 \times {{10}^{30}}kg$ $m_m= 7.348 \times {{10}^{22}}kg$ $p_e= -6.371 \times {{10}^{6}}m$ $p_s= 1.52 \times {{10}^{11}}m$ $r_m= 1737500m$ Also: $c=299792458 m/s$ $G=6.67408 \times {{10}^{-11}} m^3 kg^{-1} s^{-2}$ $h=\frac{{{c}^{2}}}{2 G}$ Since the position of the Moon is the unknown we would like to solve then the following equation will be the one to be used: $y=\frac{{m_s} \log{\left( \left| x-{r_m}-{p_s}\right| \right) }+{m_e} \log{\left( \left| x-{r_m}-{p_e}\right| \right) }+h\, \left( x-{r_m}\right) +{m_m} \log{\left( \left| {r_m}\right| \right) }}{c\, \left( \frac{{m_m}}{\left| x\right| }+\frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) } + C$ Where: $C$ is neglected. Given the time for the laser to travel between the surface of the Earth and the surface of the Moon is: $y = 1.25 s$ Then FT predicts the position of the Moon to be (approximately): $x = 3.764020321921518 \times {{10}^{8}} m$ While GR predicts the position of the Moon to be: $x = 3.764780725 \times {{10}^{8}} m$ So there is a discrepancy of $76040.3 m$ between the two theories at that scale already.
 20th December 2019, 10:33 AM #1002 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Maxima Code For the record I am using the following code in Maxima to perform the numerical analysis of the Lunar Laser Ranging Experiment: Code: load("newton1"); float(newton((m_s * log(abs(x - r_m - p_s)) + m_e * log(abs(x - r_m - p_e)) + m_m * log(abs(r_m)) + h * abs(x - r_m)) / (c * (m_m / abs(x) + m_s / abs(p_s) + m_e / abs(p_e) + h)) - 1.25, x, 1, 0.0001));
 20th December 2019, 01:11 PM #1003 phunk Illuminator     Join Date: Aug 2007 Posts: 4,107 Originally Posted by philippeb8 Definition 1 A 'kinetic frame of reference' moves coherently with the source of the local gravitational field where the latter is in turn defined to be the strongest gravitational acceleration. For example, if the observer and the observed object are nearby a planet then the kinetic frame of reference is set on the planet's surface, rotating with the same angular speed. Note that this can be a non-inertial frame. Explain how spinning planets and stars are oblate spheroids rather than spheres, if their reference frame is spinning with them so there is nothing like centrifugal force.
 20th December 2019, 01:33 PM #1004 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 This is getting really annoying! Originally Posted by philippeb8 Given the time for the laser to travel between the surface of the Earth and the surface of the Moon is: $y = 1.25 s$ Where is the evidence! Quote: Then FT predicts the position of the Moon to be (approximately): $x = 3.764020321921518 \times {{10}^{8}} m$ The laser beam passes through the Earth's atmosphere, twice. Where is the FT-based estimate of how this affects the time? As I understand it, FT says the speed of light (or the wavelength?) is affected by the rotation (or spin) of both the Earth and Moon (and Sun?). Yet you have not included this effect in your calculations. Why not? Quote: While GR predicts the position of the Moon to be: $x = 3.764780725 \times {{10}^{8}} m$ Please, just stop with this nonsense. Alternative: present a derivation of this result, from first GR principles. And include all relevant effects and factors. Quote: So there is a discrepancy of $76040.3 m$ between the two theories at that scale already. There is only one theory, GR. FT is not a theory, it is a joke.
 20th December 2019, 02:57 PM #1005 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by phunk Explain how spinning planets and stars are oblate spheroids rather than spheres, if their reference frame is spinning with them so there is nothing like centrifugal force. This is exactly the line I was drawing with the thought experiment: the star alone won’t have this effect but the Earth within the context of the solar system will because of the presence of a “parent frame of reference”.
 20th December 2019, 03:19 PM #1006 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate This is getting really annoying! Where is the evidence! The laser beam passes through the Earth's atmosphere, twice. Where is the FT-based estimate of how this affects the time? As I understand it, FT says the speed of light (or the wavelength?) is affected by the rotation (or spin) of both the Earth and Moon (and Sun?). Yet you have not included this effect in your calculations. Why not? Please, just stop with this nonsense. Alternative: present a derivation of this result, from first GR principles. And include all relevant effects and factors. There is only one theory, GR. FT is not a theory, it is a joke. - When you measure the time dilation vertically then you use gravitational time dilation - When you measure the time dilation horizontally then you use kinetic time dilation
 20th December 2019, 03:57 PM #1007 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 Where m, p & r are the mass, position and radius of the Earth, Sun & Moon respectively: $m_e= 5.9736 \times {{10}^{24}}kg$ $m_s= 1.98892 \times {{10}^{30}}kg$ $m_m= 7.348 \times {{10}^{22}}kg$ $p_e= -6.371 \times {{10}^{6}}m$ $p_s= 1.52 \times {{10}^{11}}m$ $r_m= 1737500m$ Also: $c=299792458 m/s$ $G=6.67408 \times {{10}^{-11}} m^3 kg^{-1} s^{-2}$ $h=\frac{{{c}^{2}}}{2 G}$ Since the position of the Moon is the unknown we would like to solve then the following equation will be the one to be used: $y=\frac{{m_s} \log{\left( \left| x-{r_m}-{p_s}\right| \right) }+{m_e} \log{\left( \left| x-{r_m}-{p_e}\right| \right) }+h\, \left( x-{r_m}\right) +{m_m} \log{\left( \left| {r_m}\right| \right) }}{c\, \left( \frac{{m_m}}{\left| x\right| }+\frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) } + C$ Where: $C$ is neglected. Given the time for the laser to travel between the surface of the Earth and the surface of the Moon is: $y = 1.25 s$ Here's a thought experiment: we do this in deep space, using a laser, a reflector, and a detector. All masses are zero. So: $y= \frac{hx}{ch} + C$ or: $x = cy - cC$ From an earlier post (#997): $C = -2.536918459882221 \times 10^{-4} s$ So, assuming my arithmetic is not wrong, the difference between x = cy and FT is -76054 m Now where have I seen that number (or one very close to it) before?
 20th December 2019, 04:01 PM #1008 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Bump ... Originally Posted by JeanTate Quote: In FT's context this means: $h_{solar} = \frac{c^2}{2G} = 6.75\times 10^{26} kg/m$ Because it is impossible to predict the light bending without knowing the perihelion shift in advance (or vice-versa), then we could also redo the light bending experiment with greater precision in order to find a more precise "h" for the solar system. How much of this is FT and how much actual observation? For example, is "perihelion shift" observation? If so, of what? How about "the light bending", is it observation? If so, what, when, etc?
 20th December 2019, 04:05 PM #1009 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Double bump ... Originally Posted by JeanTate Bump. Let me make this explicit, in terms of questions: 1) In the context of FT, what is the definition of “kinetic time dilation”? 2) In the context of FT, what is the definition of “gravitational time dilation”? 3) In the context of FT, what is the definition of “gravitational frame of reference”? 4) In the context of FT, what is the definition of "gravitons"? 5) How, in detail, does "its definition implies the existence of gravitons emitted by the source"?
 20th December 2019, 04:08 PM #1010 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Another double bump ... Originally Posted by JeanTate Bump. More questions: 1) is a "gravity field" a scalar? a vector? something else? 2) how can two "gravity fields" be "added together"? 3) how does the "summed up field" get "the same rotating properties of its constituents"?
 20th December 2019, 04:11 PM #1011 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Another bump .... Originally Posted by JeanTate Two expressions. With ~ten undefined symbols. Otherwise known as "whatever you think that is, it sure ain't a definition!" (my hilite) Dude, where's your glasses? How about Betelgeuse? Are you really so ignorant about extra-galactic astronomy? The Magellanic Clouds are satellite galaxies of our own, the Milky Way. M31 is the most massive member of the Local Group (our own Milky Way is the second most massive). The Local Group is gravitationally bound (kinda like how the Orion Nebula is gravitationally bound). In fact, M31 seems to be "incoming" (i.e. blueshifted) and will likely collide with the Milky Way in a few billion years' time. How about a straight answer to my questions?
 20th December 2019, 04:15 PM #1012 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 - When you measure the time dilation vertically then you use gravitational time dilation - When you measure the time dilation horizontally then you use kinetic time dilation (my hilite) And how are these two things - vertically and horizontally - determined? Take, for example, various positions around 67P/Churyumov–GerasimenkoWP.
 20th December 2019, 04:19 PM #1013 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 This is exactly the line I was drawing with the thought experiment: the star alone won’t have this effect but the Earth within the context of the solar system will because of the presence of a “parent frame of reference”. Indeed. And many, many questions have been asked about this. By several posters here. Including: real stars are made of lots of particles. Why must the "star" be treated as a solitary object and not a composite of a great many separate, possibly interacting, objects?
 20th December 2019, 05:36 PM #1014 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Initial value problem? Just sayin' __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 20th December 2019, 05:48 PM #1015 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by Robin Initial value problem? Just sayin' My maths is rusty but shouldn't you solve it for $x_0=0$ and find the constant, then solve for y=1.25 with the constant in place? __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 20th December 2019, 06:01 PM #1016 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Alternately, do it as a definite integral like you are supposed to. Or integrate the initial expression numerically which, as I pointed out earlier, mysteriously gives almost exactly the same answer as ct. Funny that. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 20th December 2019, 06:14 PM #1017 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 As far as I know, solving for the indefinite integral and ignoring the constant of integration is what is called the wrong way of doing it. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax" Last edited by Robin; 20th December 2019 at 06:35 PM.
 20th December 2019, 06:53 PM #1018 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by Robin As far as I know, solving for the indefinite integral and ignoring the constant of integration is what is called the wrong way of doing it. The constant seems to be important for the speed gun experiment but I neglected it for the lunar laser ranging experiment. I’ll double check again later.
 20th December 2019, 08:24 PM #1019 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by philippeb8 The constant seems to be important for the speed gun experiment but I neglected it for the lunar laser ranging experiment. I’ll double check again later. The constant is always important, you can't just neglect it. You just did the maths wrong. You have to either do it as a definite integral (ie y(x)-y(0)) or else as a boundary value problem where y(0)=0 to find the constant. Or else integrate the original expression numerically. There are three correct ways to do it. You did it the wrong way. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 20th December 2019, 08:46 PM #1020 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by philippeb8 The constant seems to be important for the speed gun experiment but I neglected it for the lunar laser ranging experiment. I’ll double check again later. Oh well, ignore what I said until I have time to say it properly I am in holiday mode and not thinking straight, I will revisit later when I have time. What I said is right, you do have to do it one of those ways and if you do then I suspect you will get rid of that 76,050 m distance Let me set it out properly when I have time. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax" Last edited by Robin; 20th December 2019 at 08:57 PM.
 20th December 2019, 09:02 PM #1021 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by philippeb8 The constant seems to be important for the speed gun experiment but I neglected it for the lunar laser ranging experiment. I’ll double check again later. Yes I have double checked what I said before, solve it with the constant of integration added and see what you get. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 20th December 2019, 09:42 PM #1022 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 As the constant of integration depends on the position of the Moon, the best bet would be to solve the definite integral with $p_m=x+r_m$ substitution. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 20th December 2019, 11:08 PM #1023 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Lunar Laser Ranging Experiment (Correction of Correction of Correction) According to FT, here is the speed of light reversed (or the time it takes for a photon to travel 1 meter) at each position of a photon between the surface of the Earth and the surface of the Moon, which is directly proportional to the ratio of the gravitational potential of observed photon and the gravitational potential of the observer: $y'=\frac{1}{c} \times \left. \frac{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| x-{p_i}\right| }\right.}}{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| {p_i}\right| }\right.}}\right.$ $y'=\frac{\frac{{m_s}}{\left| x-{p_s}\right| }+\frac{{m_m}}{\left| x-{p_m}\right| }+\frac{{m_e}}{\left| x-{p_e}\right| }+h}{c\, \left( \frac{\mathit{ms}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Or (example): The integral of the aforementioned equation will return the time it'll take for a photon to travel a certain distance while considering 3 perfectly aligned masses is: $y=\frac{1}{c} \times \int {\left. \frac{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| x-{p_i}\right| }\right.}}{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| {p_i}\right| }\right.}}dx\right.}$ $y=\frac{{m_s} \log{\left( \left| x-{p_s}\right| \right) }+{m_m} \log{\left( \left| x-{p_m}\right| \right) }+{m_e} \log{\left( \left| x-{p_e}\right| \right) }+h x}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }-\frac{{m_s} \log{\left( \left| {p_s}\right| \right) }+{m_m} \log{\left( \left| {p_m}\right| \right) }+{m_e} \log{\left( \left| {p_e}\right| \right) }}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Or (example): Where m, p & r are the mass, position and radius of the Earth, Sun & Moon respectively: $m_e= 5.9736 \times {{10}^{24}}kg$ $m_s= 1.98892 \times {{10}^{30}}kg$ $m_m= 7.348 \times {{10}^{22}}kg$ $p_e= -6.371 \times {{10}^{6}}m$ $p_s= 1.52 \times {{10}^{11}}m$ $r_m= 1737500m$ Also: $c=299792458 m/s$ $G=6.67408 \times {{10}^{-11}} m^3 kg^{-1} s^{-2}$ $h=\frac{{{c}^{2}}}{2 G}$ Since the position of the Moon is the unknown we would like to solve then the following equation will be the one to be used*: $y=\frac{{m_s} \log{\left( \left| x-{r_m}-{p_s}\right| \right) }+{m_e} \log{\left( \left| x-{r_m}-{p_e}\right| \right) }+h\, \left| x-{r_m}\right| +{m_m} \log{\left( \left| {r_m}\right| \right) }}{c\, \left( \frac{{m_m}}{\left| x\right| }+\frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }-\frac{{m_m} \log{\left( \left| x\right| \right) }+{m_s} \log{\left( \left| {p_s}\right| \right) }+{m_e} \log{\left( \left| {p_e}\right| \right) }}{c\, \left( \frac{{m_m}}{\left| x\right| }+\frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Given the time for the laser to travel between the surface of the Earth and the surface of the Moon is: $y = 1.25 s$ Then FT predicts the position of the Moon to be (approximately): $x = 3.76478087122961 \times {{10}^{8}} m$ While GR predicts the position of the Moon to be: $x = 3.764780725 \times {{10}^{8}} m$ So there is a discrepancy of $14.623 m$ between the two theories at that scale already. Note that FT predicts the Moon to be farther away, which does make sense. *Thanks Robin. Last edited by philippeb8; 20th December 2019 at 11:18 PM.
 21st December 2019, 03:35 AM #1024 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Approximately 15 metres or 0.000006% different to ct is what I get too using my suggested substitution. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 21st December 2019, 04:43 AM #1025 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by Robin Approximately 15 metres or 0.000006% different to ct is what I get too using my suggested substitution. I’m going to have to put your name for sure in the acknowledgment section of the next version of my book.
 21st December 2019, 05:57 AM #1026 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by philippeb8 I’m going to have to put your name for sure in the acknowledgment section of the next version of my book. Please don't. __________________ The non-theoretical character of metaphysics would not be in itself a defect; all arts have this non-theoretical character without thereby losing their high value for personal as well as for social life. The danger lies in the deceptive character of metaphysics; it gives the illusion of knowledge without actually giving any knowledge. This is the reason why we reject it. - Rudolf Carnap "Philosophy and Logical Syntax"
 21st December 2019, 06:50 AM #1027 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by Robin Please don't. As you wish.
 21st December 2019, 08:35 AM #1028 Blue Mountain Resident Skeptical Hobbit     Join Date: Jul 2005 Location: Waging war on woo-woo in Winnipeg Posts: 5,843 phillipeb8, It's sad to see someone wasting their intellectual resources and time pursuing a chimera. I'm a decent programmer and computer system administrator. However, I'm not a computer scientist and haven't studied things like compiler design, tail recursion, code optimization, proper random number generation, IEEE floating point, red/black trees, etc. If one day in a sudden burst of inspiration I designed a really, really neat algorithm to solve a problem, I know I wouldn't say "this algorithm is so cool that that last 60 years of computer science are bunk." Rather, I'd assume the following, in order: My novel algorithm isn't that novel at all, but rather has been studied by people more expert in the subject than I am and have rejected it because it isn't that good. My novel algorithm is actually more or less standard in the industry, and all I did was reinvent the wheel. My algorithm actually is novel, but may suffer from an "edge case" flaw (that is, it will work 99% of the time, but 1% of the time it will fail and lead to incorrect results.) Lastly, that I may have actually discovered something novel that will make a contribution to computer science. For number three, I would do what you did: post the algorithm on a computing site and let people examine it to see if it truly is novel and/or is flawed. And I would actually listen to the criticisms. Your continued insistence on the correctness of FT and assertion that special and general relativity are wrong, despite all the evidence posted here and on CQ that FT is flawed, plus three hundred years of physics by a lot of very, very smart people, and the fact SR and GR have stood up to practically every test we've thrown at it, really should cause you to think that perhaps FT really is flawed. __________________ The social illusion reigns to-day upon all the heaped-up ruins of the past, and to it belongs the future. The masses have never thirsted after truth. They turn aside from evidence that is not to their taste, preferring to deify error, if error seduce them. Gustav Le Bon, The Crowd, 1895 (from the French) Canadian or living in Canada? PM me if you want an entry on the list of Canadians on the forum. Last edited by Blue Mountain; 21st December 2019 at 08:36 AM.
 21st December 2019, 09:08 AM #1029 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by Blue Mountain phillipeb8, It's sad to see someone wasting their intellectual resources and time pursuing a chimera. I'm a decent programmer and computer system administrator. However, I'm not a computer scientist and haven't studied things like compiler design, tail recursion, code optimization, proper random number generation, IEEE floating point, red/black trees, etc. I did + parallel programming + automata + basic AI + image synthesis / analysis + advanced probabilities and statistics + ... Quote: If one day in a sudden burst of inspiration I designed a really, really neat algorithm to solve a problem, I know I wouldn't say "this algorithm is so cool that that last 60 years of computer science are bunk." I did also rewrite the last 60 years of computer science with a deterministic memory manager. Some ambition is actually what is required to accomplish such things. Quote: Rather, I'd assume the following, in order: My novel algorithm isn't that novel at all, but rather has been studied by people more expert in the subject than I am and have rejected it because it isn't that good. My novel algorithm is actually more or less standard in the industry, and all I did was reinvent the wheel. My algorithm actually is novel, but may suffer from an "edge case" flaw (that is, it will work 99% of the time, but 1% of the time it will fail and lead to incorrect results.) Lastly, that I may have actually discovered something novel that will make a contribution to computer science. For number three, I would do what you did: post the algorithm on a computing site and let people examine it to see if it truly is novel and/or is flawed. And I would actually listen to the criticisms. - I am out of the academia; - Theoretical physics are arrogantly ignoring my requests; But: - I do have a contact at the ISS National Lab that could refer me to other points of contact; - I will work for a military-based company in January and I could easily test my theory there. Quote: Your continued insistence on the correctness of FT and assertion that special and general relativity are wrong, despite all the evidence posted here and on CQ that FT is flawed, plus three hundred years of physics by a lot of very, very smart people, and the fact SR and GR have stood up to practically every test we've thrown at it, really should cause you to think that perhaps FT really is flawed. The deadlocked conversation I have with Reality Check is not a threat. Remember FT can solve: - The perihelion shift; - The light bending; - The time dilation cancelation altitude; - The rotation curve without dark matter; - The expansion of the universe without dark energy; - The mass of the invisible universe encompassing the visible one. And the difference between you and I is I don’t get blinded by the hype of certain individuals in mainstream physics.
 21st December 2019, 09:12 AM #1030 The Man Unbanned zombie poster     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 14,473 Originally Posted by philippeb8 Ok I’ll change “gravitational frame of reference” to “kinetic frame of reference”. This is way less confusing! Doesn't help, as simply renaming your "Gravitational frame of reference" can't remove the obvious dependence on your "gravitational acceleration amplitude", your "Gravitational potential" and thus the relation to your "gravitational time dilation". Heck, that's probably why you called them all "Gravitational" in the first place. Deliberate obfuscation by renaming can't help. __________________ BRAINZZZZZZZZ
 21st December 2019, 09:20 AM #1031 The Man Unbanned zombie poster     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 14,473 Originally Posted by philippeb8 Please forget GR’s equation. I should have swapped X and Y. See, there ya go. A simple notation mistake. Don't try to excuse it, just note it and move on. __________________ BRAINZZZZZZZZ
 21st December 2019, 09:36 AM #1032 The Man Unbanned zombie poster     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 14,473 Originally Posted by tusenfem Originally Posted by philippeb8 Just add: “/ 1 m” yes sure! and where is that supposed to come from? (Maybe while you are at it, you might just replace all those pesky variables with the numbers you want to get out, even more easy.) if you do that in the equation for y', then you have problems with your magical h-term. if you do it willy-nilly in the equation for y then you are just trying to make your equations fit the result you want to get. Maybe, think first and then retry integrating y' and see where you have gone wrong? Or maybe even earlier, where you suddenly get your magical h-term out of a summation? What happens when units of distance are furlongs and the units of time are fortnights? __________________ BRAINZZZZZZZZ
 21st December 2019, 09:39 AM #1033 The Man Unbanned zombie poster     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 14,473 Originally Posted by philippeb8 I wrote this equation a long time ago and I applied it for the lunar laser ranging experiment and I get a decent result apparently. I just need to review correctly tusenfem’s claims during the day... What you claimed to get was an approximation. Which you've claimed helped support your notions because it is different than what you say GR gets. Perhaps your FT approximation is just crap and when you get an actual FT result it is the same you claim to get with GR? __________________ BRAINZZZZZZZZ
 21st December 2019, 09:53 AM #1034 The Man Unbanned zombie poster     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 14,473 Originally Posted by philippeb8 You’re doing a negative evaluation of what is being presented. The equations I’m showing are very consistent. The lunar laser ranging experiment is just a recent application of the maths involved. And I’m not a professor so I do not have experience in presenting theories or concepts... If you'd do "a negative evaluation of what is being presented" your presentation might improve. Unfortunately, an improved presentation tends to mean problems with the notions themselves are more readily apparent, even to you. __________________ BRAINZZZZZZZZ
 21st December 2019, 09:54 AM #1035 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 Remember FT can solve: - The perihelion shift; - The light bending; - The time dilation cancelation altitude; - The rotation curve without dark matter; - The expansion of the universe without dark energy; - The mass of the invisible universe encompassing the visible one. That’s it? Dude, I solved all those years ago. Most of them before breakfast. Child’s play. You should devote your time and energy to solving world hunger, curing cancer, and reversing global warming.
 21st December 2019, 09:56 AM #1036 The Man Unbanned zombie poster     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 14,473 Originally Posted by philippeb8 I was ultimately result-oriented and I didn’t pay enough importance on its aesthetic. In mathematics those "aesthetic"s are results themselves. This belies your assertion of being "result-oriented". __________________ BRAINZZZZZZZZ
 21st December 2019, 10:04 AM #1037 The Man Unbanned zombie poster     Join Date: Jun 2007 Location: Poughkeepsie, NY Posts: 14,473 Originally Posted by philippeb8 Like I was saying Einstein was only considering time dilation but did not include the length contraction until the full GR was released in 1915. But the 1911 equation is exactly like FT's with $h = -c^2/G$. As I recall it wasn't length contraction that was lacking in the error but the gravitational potential of the light itself. __________________ BRAINZZZZZZZZ
 21st December 2019, 10:06 AM #1038 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by The Man What you claimed to get was an approximation. Which you've claimed helped support your notions because it is different than what you say GR gets. Perhaps your FT approximation is just crap and when you get an actual FT result it is the same you claim to get with GR? The first approximation was but after that I use standard numerical analysis algorithms, like I was saying previously in the “Maxima” post.
 21st December 2019, 10:09 AM #1039 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by The Man If you'd do "a negative evaluation of what is being presented" your presentation might improve. Unfortunately, an improved presentation tends to mean problems with the notions themselves are more readily apparent, even to you. Now I know why professors need to be perfectionist, because otherwise they would generate confusion.
 21st December 2019, 10:14 AM #1040 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate That’s it? Dude, I solved all those years ago. Most of them before breakfast. Child’s play. You should devote your time and energy to solving world hunger, curing cancer, and reversing global warming. So I do have the following solutions: - Finite Theory; - Deterministic C++ memory manager; - Hydrogen engines; - Real-time concealed weapon detection; - Open source social media alternative. Curing cancer is not on my list yet but I believe we are on the edge of finding a cure. Same with AI.

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