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 18th December 2019, 09:25 PM #881 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 Please forget GR’s equation. I should have swapped X and Y. Sure. (looks around to see where X and Y are used, comes up empty). Let me guess ... you meant x and y (and maybe y’). Also, if we “forget GR’s equation”, where does the “GR predicts” number/value in both 712 and 718 (“3.74741 x 108m”) come from?
 18th December 2019, 09:27 PM #882 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 I wrote the equation in about half an hour. I will double check it before claiming victory. Dude, you wrote two posts, over an hour apart. The two posts contain the same equation.
 18th December 2019, 09:27 PM #883 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate Sure. (looks around to see where X and Y are used, comes up empty). Let me guess ... you meant x and y (and maybe y’). Also, if we “forget GR’s equation”, where does the “GR predicts” number/value in both 712 and 718 (“3.74741 x 108m”) come from? The results are fine. I just quickly wrote GR’s equation without paying attention to the units.
 18th December 2019, 09:30 PM #884 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate Dude, you wrote two posts, over an hour apart. The two posts contain the same equation. Then I must have spent another half hour arguing with Reality Check...
 18th December 2019, 09:33 PM #885 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 The results are fine. I just quickly wrote GR’s equation without paying attention to the units. Got it! You invented the results (or derived them using Newtonian physics), wrote them down, and claimed - without any supporting evidence, logic, references, etc - that it’s a result derived from using GR. Now that is chutzpah!
 18th December 2019, 09:35 PM #886 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate Got it! You invented the results (or derived them using Newtonian physics), wrote them down, and claimed - without any supporting evidence, logic, references, etc - that it’s a result derived from using GR. Now that is chutzpah! Well GR says the speed of light is always constant, no?
 18th December 2019, 09:38 PM #887 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 Then I must have spent another half hour arguing with Reality Check... Yes of course you did. (checks who posted between 712 and 718, finds three posts by Robin, one each by tusenfem and Belz..., and none/zip/nada by Reality Check). Or maybe not ...
 18th December 2019, 09:42 PM #888 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 Well GR says the speed of light is always constant, no? I can feel a withering post by W.D.Clinger (and likely ones by The Man, RC, Robin, tusenfem, ...) coming on ...
 18th December 2019, 10:20 PM #889 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by philippeb8 The results are fine. I just quickly wrote GR’s equation without paying attention to the units. It isn't the GR equation as far as I know. Let's call it the simple method. Multiply 1.25 by c and add the radius of the moon. Plug that into your final equation and see if what you get. 1.25 seconds, right? __________________ 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"
 18th December 2019, 10:42 PM #890 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Finite Theory: Historical Milestone in Physics Originally Posted by Robin It isn't the GR equation as far as I know. Let's call it the simple method. Multiply 1.25 by c and add the radius of the moon. Plug that into your final equation and see if what you get. 1.25 seconds, right? Oh I forgot to subtract the radius in GR’s equation... thanks Robin! Last edited by philippeb8; 18th December 2019 at 11:00 PM.
 18th December 2019, 11:05 PM #891 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 Oh I forgot to subtract the radius from GR’s result... thanks Robin! Dude, you might want to consider investing some time in learning at least the basics of GR. Some serious time. Otherwise you’ll continue to make real howlers, and further erode any reader here’s confidence that you have something interesting/new to say about physics. (No, whatever FT is, it is not a “Historic Milestone in Physics”)
 18th December 2019, 11:23 PM #892 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate Dude, you might want to consider investing some time in learning at least the basics of GR. Some serious time. Otherwise you’ll continue to make real howlers, and further erode any reader here’s confidence that you have something interesting/new to say about physics. (No, whatever FT is, it is not a “Historic Milestone in Physics”) There’s still a discrepancy and this lunar laser ranging experiment has nothing to do with the particle accelerator blunder and the ISS experiment.
 18th December 2019, 11:36 PM #893 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by philippeb8 There’s still a discrepancy . Actually, by my calculation, using $ct+r_m$ for x gets closer to 1.25 than the value you suggested. __________________ 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; 18th December 2019 at 11:38 PM.
 19th December 2019, 01:07 AM #894 tusenfem Master Poster     Join Date: May 2008 Posts: 2,647 Originally Posted by philippeb8 - I did remove the minus sign in this thread to simplify but it really should be there; Oh yes, that makes complete sense, a fudge factor that suddenly appears in a summation without any reason, and then the minus sign is the complicated part? __________________ 20 minutes into the future This message is bra-bra-brought to you by z-z-z-zik zak And-And-And I'm going to be back with you - on Network 23 after these real-real-real-really exciting messages (Max Headroom) follow me on twitter: @tusenfem, or follow Rosetta Plasma Consortium: @Rosetta_RPC
 19th December 2019, 01:08 AM #895 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by Robin Actually, by my calculation, using $ct+r_m$ for x gets closer to 1.25 than the value you suggested. ... vs FT.
 19th December 2019, 01:35 AM #896 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by philippeb8 ... vs FT. Substitute $ct+r_m$ for x in your final equation and you get closer to 1.25 than the value you suggested. So if you look for a more precise value of x to solve your equation, you will get closer to $ct+r_m$ __________________ 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; 19th December 2019 at 01:39 AM.
 19th December 2019, 01:38 AM #897 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 The main thing is that the substitutions are not valid in the final equation. If $p_m$ is unknown then you only need to make substitutions where $p_m$ occurs, but you are making substitutions where it doesn't occur. So all you need is to substitute $p_m=x+r_m$ in the denominator and $x-p_m=r_m$ once in the numerator and that is all. If you do this then you will get something extremely close to ct for x. You will also find this if you do a numerical integration of the original expression. __________________ 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; 19th December 2019 at 01:39 AM.
 19th December 2019, 01:48 AM #898 tusenfem Master Poster     Join Date: May 2008 Posts: 2,647 Originally Posted by Robin I should point out that your prediction differs from the prediction found by x/c by approximately the radius of the Moon. Well, philippeb8 is taking the logarithm of a dimensional number log(x-p), so the result is nonsense, dimentionally. (but I am sure there is going to be a "but I normalized everything, but for simplicity I decided not to write it down" nonsense. __________________ 20 minutes into the future This message is bra-bra-brought to you by z-z-z-zik zak And-And-And I'm going to be back with you - on Network 23 after these real-real-real-really exciting messages (Max Headroom) follow me on twitter: @tusenfem, or follow Rosetta Plasma Consortium: @Rosetta_RPC
 19th December 2019, 02:01 AM #899 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by tusenfem Well, philippeb8 is taking the logarithm of a dimensional number log(x-p), so the result is nonsense, dimentionally. . He is taking the log of the magnitude of them, so that part is OK. __________________ 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"
 19th December 2019, 02:07 AM #900 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Basically the time found by this method differs from x/c by about $1.2 \times 10^{-9}$ __________________ 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"
 19th December 2019, 02:17 AM #901 tusenfem Master Poster     Join Date: May 2008 Posts: 2,647 Originally Posted by Robin He is taking the log of the magnitude of them, so that part is OK. nope, it is still log(meters) that is coming out of that so "shortened" for dimensional analysis y = (m log(x) + h x) / (c ( m/x + h)) [ .. ] means dimension of .. [y] = [m log(x) +h x] / ( [c] ( [m/x + h]) [y] = (kg log(m) + kg) / ( m/s * kg/m)) The first term on the rhs cannot be dimensionally determined, unless philippeb8 comes up with a solution to make the term inside the log dimensionless. __________________ 20 minutes into the future This message is bra-bra-brought to you by z-z-z-zik zak And-And-And I'm going to be back with you - on Network 23 after these real-real-real-really exciting messages (Max Headroom) follow me on twitter: @tusenfem, or follow Rosetta Plasma Consortium: @Rosetta_RPC
 19th December 2019, 02:56 AM #902 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by tusenfem nope, it is still log(meters) that is coming out of that so "shortened" for dimensional analysis y = (m log(x) + h x) / (c ( m/x + h)) [ .. ] means dimension of .. [y] = [m log(x) +h x] / ( [c] ( [m/x + h]) [y] = (kg log(m) + kg) / ( m/s * kg/m)) The first term on the rhs cannot be dimensionally determined, unless philippeb8 comes up with a solution to make the term inside the log dimensionless. Just add: “/ 1 m”
 19th December 2019, 03:50 AM #903 tusenfem Master Poster     Join Date: May 2008 Posts: 2,647 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? __________________ 20 minutes into the future This message is bra-bra-brought to you by z-z-z-zik zak And-And-And I'm going to be back with you - on Network 23 after these real-real-real-really exciting messages (Max Headroom) follow me on twitter: @tusenfem, or follow Rosetta Plasma Consortium: @Rosetta_RPC Last edited by tusenfem; 19th December 2019 at 03:51 AM.
 19th December 2019, 03:58 AM #904 malbui Beauf     Join Date: Nov 2004 Posts: 2,501 Maybe I’m missing something because I’m not a physicist, but isn’t it good practice to get the mathematical underpinnings of your work - all those troublesome equations - all nicely sorted out before going public with claims of having achieved a major milestone? __________________ "But Master! Does not the fire need water too? Does not the mountain need the storm? Does not your scrotum need kicking?"
 19th December 2019, 04:40 AM #905 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by malbui Maybe I’m missing something because I’m not a physicist, but isn’t it good practice to get the mathematical underpinnings of your work - all those troublesome equations - all nicely sorted out before going public with claims of having achieved a major milestone? 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...
 19th December 2019, 05:26 AM #906 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by Robin The main thing is that the substitutions are not valid in the final equation. If $p_m$ is unknown then you only need to make substitutions where $p_m$ occurs, but you are making substitutions where it doesn't occur. So all you need is to substitute $p_m=x+r_m$ in the denominator and $x-p_m=r_m$ once in the numerator and that is all. If you do this then you will get something extremely close to ct for x. You will also find this if you do a numerical integration of the original expression. Just bumping this. Why are you making substitutions in terms which do not into $p_m$? __________________ 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"
 19th December 2019, 05:43 AM #907 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 3,663 Originally Posted by JeanTate Otherwise you’ll continue to make real howlers, and further erode any reader here’s confidence that you have something interesting/new to say about physics. Is it really possible for confidence to erode below zero? (Checks the US Politics subforum.) Never mind. Originally Posted by tusenfem Well, philippeb8 is taking the logarithm of a dimensional number log(x-p), so the result is nonsense, dimentionally. Suppose y = 1/x, where y is in units of foo and x in units of bar = foo-1. Then the definite integral of y=1/x between a and b islog |b| - log |a|with units foo*bar=foo*foo-1=1 because that definite integral is the area under the curve between x=a and x=b. In other words, the logarithm of a dimensional number, when obtained via integration, can be dimensionless, which is not necessarily nonsense. Originally Posted by tusenfem 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? The magical h-terms that philippeb8 pulled out of thin air do appear to be nonsense. But hey, philippeb8 spent only half an hour coming up with those equations. When you're overturning 300 years of physics, you wouldn't want to waste much time on getting your equations right.
 19th December 2019, 05:47 AM #908 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 This expression $y=\frac{{m_s} \log{\left( \left| x-{p_s}\right| \right) }+{m_m} \log{\left( \left| x-{p_m}\right| \right) }+{m_e} \log{\left( \left| x-{p_e}\right| \right) }+h x}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Just substitute $p_m=x+r_m$ to get $y=\frac{{m_s} \log{\left( \left| x-{p_s}\right| \right) }+{m_m} \log{\left( \left| {r_m}\right| \right) }+{m_e} \log{\left( \left| x-{p_e}\right| \right) }+h x}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {x+r_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Putting $r_m$ in those other terms is invalid. __________________ 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"
 19th December 2019, 05:49 AM #909 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 This expression $y=\frac{{m_s} \log{\left( \left| x-{p_s}\right| \right) }+{m_m} \log{\left( \left| x-{p_m}\right| \right) }+{m_e} \log{\left( \left| x-{p_e}\right| \right) }+h x}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ You can substitute $p_m=x+r_m$ to get $y=\frac{{m_s} \log{\left( \left| x-{p_s}\right| \right) }+{m_m} \log{\left( \left| {r_m}\right| \right) }+{m_e} \log{\left( \left| x-{p_e}\right| \right) }+h x}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {x+r_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ But putting $r_m$ in those other terms is invalid. __________________ 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"
 19th December 2019, 06:05 AM #910 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 3,663 Originally Posted by Robin You can substitute $p_m=x+r_m$ to get As noted earlier, I believe philippeb8 "substituted x-rm for x and then set x equal to pm". At the time, I thought that was poor presentation rather than complete nonsense. As tusenfem noted, however, the introduction of the magical h-terms in the earlier equation appears to be complete nonsense. As you noted, the effect of introducing those terms was to convert the equation into what amounts to multiplying x by a factor that is very close to unity. The deviation of that factor from unity by a function that is slightly related to gravitational potential is analogous to, but far less accurate than, the most significant term of the relativistic correction seen in the linearized gravity approximation.
 19th December 2019, 06:39 AM #911 Robin Penultimate Amazing   Join Date: Apr 2004 Posts: 11,645 Originally Posted by W.D.Clinger As noted earlier, I believe philippeb8 "substituted x-rm for x and then set x equal to pm". At the time, I thought that was poor presentation rather than complete nonsense. Well OK, but he obviously confused even himself with that, because he compared it to a surface to surface measurement and called the difference a "discrepancy". __________________ 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"
 19th December 2019, 06:45 AM #912 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by Robin Well OK, but he obviously confused even himself with that, because he compared it to a surface to surface measurement and called the difference a "discrepancy". ... a discrepancy between GR and FT.
 19th December 2019, 08:26 AM #913 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 I'm going to go through the two main posts by philippeb8 on lunar laser ranging, in a series of posts of my own; they are 712 and 718. I will focus on 718 as it is later in time and seems to be somewhat less ambiguous etc. And s snipped. Most, perhaps even all, of the contents of these posts has already been discussed, to one extent or another, by various posters; however, I will not attempt to provide links to those (maybe an exception or two). Start with this, from 718: Originally Posted by philippeb8 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: $y'=\frac{1}{c} \times \left. \frac{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| x-{p_i}\right| }\right.}}{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| {p_i}\right| }\right.}}\right.$ Where m, p & r are the mass, position and radius of the Earth, Sun & Moon respectively Checking for dimensions, on the RHS (L=length, M=mass, T=time): 1/c: 1/(LT-1), or T/L (consistent with "the speed of light reversed"). next the sum; taking n=1: numerator: m/|x-p|: M/L denominator: m/|p|: M/L which cancel, so we are left with T/L Then we read (in both 712 and 718): Quote: $y'=\frac{\frac{{m_s}}{\left| x-{p_s}\right| }+\frac{{m_m}}{\left| x-{p_m}\right| }+\frac{{m_e}}{\left| x-{p_e}\right| }+h}{c\, \left( \frac{\mathit{ms}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Whoa! The line above has the sum going from n=1 to n=3, yet here we have four terms! Where did h come from? It's not in the intro either. Further down in 718 (but not in 712, there h is undefined): Quote: Also: $c=299792458 m/s$ $G=6.67408 \times {{10}^{-11}} m^3 kg^{-1} s^{-2}$ $h=\frac{{{c}^{2}}}{2 G}$ Dimensionally: (LT-1)2/(L3M-1T-2) or M/L which is consistent with the other three terms. (to be continued)
 19th December 2019, 08:40 AM #914 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Kappa - Excerpt Effect of the time dilation in the gravitational field is a consequence of the difference in gravitational potentials. This effect is described by the relation: $\frac{\Delta \tau}{\Delta t} = \frac{1}{h}(h + \frac{M}{r}) = 1 + \frac{M}{hr}$ where, $M$ is a mass of the gravitating object and $r$ is the distance from its centre. Under $\Delta\tau$ we mean the interval of local time at the point situated at distance $r$ from the centre of the source of gravitation. $\Delta t$ is the interval of time measured by the distant observer, situated at distance $r\rightarrow \infty$. General relativistic time dilation effect is a particular case of if $h=-c^2/G$. Indeed, we know that in the weak field limit of General Relativity, time dilation effect in the gravitational field takes the following form*: $\frac{\Delta \tau}{\Delta t} = 1 - \frac{G M}{c^2 r}$ But due to the hypotheses of the Finite Theory, factor $h$ in is not a universal constant but depend on the superposed gravitational potentials. For example, in solar system experiments, where the gravitational potential of the Sun is the source of the strongest gravitational acceleration, we suppose $h = h_{solar}$. The value of $h_{solar}$ can be determined from the observation of the deflection angle of light in the gravitational field of the Sun, as we will demonstrate in the next subsection. Due to the time dilation effect, we expect to have different speed measurements of the same body by different observers. In particular, the speed of light traveling through the gravitational field will be different from the viewpoint of a local observer and from the viewpoint of a distant watcher. According to, a distant observer notes that the light beam has a velocity, which depends on the position in the gravitational field: $v = \frac{dr}{d t} = \frac{dr}{d\tau} \left( 1 + \frac{M_{sun}}{h_{solar} r} \right) = c\, \left( 1 + \frac{M_{sun}}{h_{solar} r} \right)$ In this relation, the local speed $v_{local} = dr/d\tau = c = 2.998\times 10^{8} m/s$ is constant due to our hypothesis. Also, we neglect the effect of length contraction in the gravitational field, which results in the equal values of length interval $dr$ for both local and distant observers. [...] *Cheng, Relativity, Gravitation and Cosmology. A Basic Introduction, Oxford University Press, 2005 Last edited by philippeb8; 19th December 2019 at 09:13 AM.
 19th December 2019, 08:44 AM #915 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 (continued) Again, looking at 718: Quote: 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: $y'=\frac{1}{c} \times \left. \frac{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| x-{p_i}\right| }\right.}}{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| {p_i}\right| }\right.}}\right.$ $y'=\frac{\frac{{m_s}}{\left| x-{p_s}\right| }+\frac{{m_m}}{\left| x-{p_m}\right| }+\frac{{m_e}}{\left| x-{p_e}\right| }+h}{c\, \left( \frac{\mathit{ms}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Where m, p & r are the mass, position and radius of the Earth, Sun & Moon respectively: $p_e= -6.371 \times {{10}^{6}}m$ $p_s= 1.52 \times {{10}^{11}}m$ $r_m= 1737500m$ (my hilite) Let's look at just two parts of the latter: ${\left| x-{p_m}\right| }$ ${\left| x-{p_e}\right| }$ As x approaches pm, |x-pm| approaches zero, right? Ditto for pe. So what happens in each case, in the second equation, to the numerator (the denominator remains sensible, at least with respect to ps and pe, it could even be a constant)? The intro says "between", but that could be arbitrarily close to zero on each end, right? Blows up, right? Maybe it all gets sorted out in the next step(s), "The integral of the aforementioned equation" ... In any case, we're left with the mystery of what pm is. (to be continued) Last edited by JeanTate; 19th December 2019 at 08:51 AM. Reason: added (continued)
 19th December 2019, 09:22 AM #916 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 (continued) 718 again: Quote: 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: $y=\frac{1}{c} \times \int {\left. \frac{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| x-{p_i}\right| }\right.}}{\sum_{i=1}^{n}{\left. \frac{{m_i}}{\left| {p_i}\right| }\right.}}dx\right.}$ $y=\frac{{m_s} \log{\left( \left| x-{p_s}\right| \right) }+{m_m} \log{\left( \left| x-{p_m}\right| \right) }+{m_e} \log{\left( \left| x-{p_e}\right| \right) }+h x}{c\, \left( \frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_m}}{\left| {p_m}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ 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: $c=299792458 m/s$ $G=6.67408 \times {{10}^{-11}} m^3 kg^{-1} s^{-2}$ $h=\frac{{{c}^{2}}}{2 G}$ "log" (base 10)? or "ln" (base e)? Does it matter which? In post #832, philippeb8 said the perfect alignment is "Earth-Moon-Sun". So x=0 is ~6,300 km from the Earth, in the direction of the Sun. And the Sun is some 152 million km from x=0. So what is "the time it'll take for a photon to travel a certain distance" where that distance is from position pe to position -pe? If I plug this in to the above, I get, for just one term: $log{\left( \left| x-{p_e}\right| \right)$ Which is log(0), right? Does this mean that a photon leaving the surface of the Earth gets trapped ~12,600 km away (in the direction of the Sun, or Moon)? I'm confused. (to be continued)
 19th December 2019, 09:30 AM #917 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate (continued) 718 again: "log" (base 10)? or "ln" (base e)? Does it matter which? In post #832, philippeb8 said the perfect alignment is "Earth-Moon-Sun". So x=0 is ~6,300 km from the Earth, in the direction of the Sun. And the Sun is some 152 million km from x=0. So what is "the time it'll take for a photon to travel a certain distance" where that distance is from position pe to position -pe? If I plug this in to the above, I get, for just one term: $log{\left( \left| x-{p_e}\right| \right)$ Which is log(0), right? Does this mean that a photon leaving the surface of the Earth gets trapped ~12,600 km away (in the direction of the Sun, or Moon)? I'm confused. (to be continued) So the position of the observer is: 0 m; We are looking for the time it takes for a photon to reach position: x - r_m; And "x" is the position of the Moon (p_m) we would like to solve.
 19th December 2019, 09:51 AM #918 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 Originally Posted by philippeb8 So the position of the observer is: 0 m; We are looking for the time it takes for a photon to reach position: x - r_m; And "x" is the position of the Moon (p_m) we would like to solve. Thanks for the clarification. So the p refer to the position of the centre of the Earth, Sun, and Moon? If I try to measure the distance from the Earth to the Moon, using a laser set up at ~-6,300 km from "the position of the observer" - i.e. at the centre of the Earth - I will not get an answer, using your equation? Of course, we need to assume a really narrow tube right through to the centre of the Earth, or a transparent Earth ... Last edited by JeanTate; 19th December 2019 at 09:54 AM. Reason: not opposite side, centre of the Earth; fixed some typos
 19th December 2019, 09:58 AM #919 philippeb8 Muse   Join Date: Sep 2018 Posts: 661 Originally Posted by JeanTate Thanks for the clarification. So the p refer to the position of the centre of the Earth, Sun, and Moon? If I try to measure the distance from the Earth to the Moon, using a laser set up at ~-6,300 km from "the position of the observer" - i.e. at the centre of the Earth from the observer - I will not get an answer, using your equation? If course, we need to assume a really narrow tube right through to the centre of the Earth, or a transparent Earth ... - The position refers to the center of the mass; - If you want to flip the position of the observer on the Earth, you'll have to change the sign of p_e and shift the Moon farther away.
 19th December 2019, 10:13 AM #920 JeanTate Illuminator   Join Date: Nov 2014 Posts: 3,452 (continued) Last parts of 718: Quote: 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: $y=\frac{{m_s} \log{\left( \left| x-{r_m}-{p_s}\right| \right) }+{m_e} \log{\left( \left| x-{r_m}-{p_e}\right| \right) }+h\, \left( x-{r_m}\right) +{m_m} \log{\left( \left| {r_m}\right| \right) }}{c\, \left( \frac{{m_m}}{\left| x\right| }+\frac{{m_s}}{\left| {p_s}\right| }+\frac{{m_e}}{\left| {p_e}\right| }+h\right) }$ Other than to check the equation for y, nothing to add here, that hasn't already been covered. For example: philippeb8 has not yet given unambiguous sources for any of the values. This last part has been fairly well covered in many posts; "cherry picking" is the mildest thing one can say, but bovine excrement of varying strengths is more accurate I feel: Quote: 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): $x = 3.76776 \times {{10}^{8}} m$ While GR predicts the position of the Moon to be: $x = 3.74741 \times {{10}^{8}} m$ So there is a discrepancy between the two theories at that scale already. The physics books of the last 300 years need to be rewritten because the notion of light years is another important blunder. (concluded, maybe)

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