# Difference between revisions of "Talk:Counterexamples to Relativity"

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:Frank, your intuition ("seems to me") is wrong here, and the entry explains it clearly. GPS is a work of engineering and any timing discrepancies between the satellite and ground are obviously better handled directly by synchronization rather than asking a physicist what he thinks of relativity. Engineers don't even bother taking general relativity courses, let alone try to build a satellite system using them.--[[User:Aschlafly|Andy Schlafly]] 14:44, 6 January 2010 (EST) | :Frank, your intuition ("seems to me") is wrong here, and the entry explains it clearly. GPS is a work of engineering and any timing discrepancies between the satellite and ground are obviously better handled directly by synchronization rather than asking a physicist what he thinks of relativity. Engineers don't even bother taking general relativity courses, let alone try to build a satellite system using them.--[[User:Aschlafly|Andy Schlafly]] 14:44, 6 January 2010 (EST) | ||

+ | |||

+ | ::The Time Service Department – a department of the U. S. Navy - states: “The Operational Control System (OCS) of the Global Positioning System (GPS) does not include the rigorous transformations between coordinate systems that Einstein’s general theory of relativity would seem to require – transformations to and from the individual space vehicles (SVs), the Monitor Stations (MSs), and the users on the surface of the rotating earth, and the geocentric Earth Centered Inertial System (ECI) in which the SV orbits are calculated. There is a very good reason for the omission: the effects of relativity, where they are different from the effects predicted by classical mechanics and electromagnetic theory, are too small to matter – less than one centimeter, for users on or near the earth.” | ||

+ | |||

+ | ::Sorry, Frank. {{unsigned|PhyllisS}} | ||

+ | |||

+ | :::As far as I can see there is no reason to feel sorry for FrankC: Your article only covers the idea of using the [[Lorentz transformation instead]] of the [[Galileo transformation]] when calculating the position of an object: one could say that it is about the relativistic effects caused by the movement of the GPS receiver, not of the GPS satellites. That's why it's talking about ''fast moving air-planes and satellites''. | ||

+ | :::FrankC (and others) have shown that there are relativistic effects on the satellites which are taken account of: | ||

+ | ::::[http://www.navcen.uscg.gov/pubs/gps/sigspec/gpssps1.pdf Global Positioning System Standard Positioning Service Signal Specification], 2nd edition, June 1995: | ||

+ | |||

+ | ::::p. 13: ''To compensate for relativistic effects, the output frequency of the satellite's frequency standard -- as it would appear to an observer located at the satellite -- is 10.23 MHz offset by a Df/f = -4.4647 x 10-18 or a Df = -4.567 x 10-3 Hz.'' | ||

+ | |||

+ | ::::p. 39: ''The coefficients transmitted in subframe 1 describe the offset apparent to the control segment two-frequency receivers for the interval of time in which the parameters are transmitted. This estimated correction accounts for the deterministic satellite clock error characteristics of bias, drift and aging, as well as for the satellite implementation characteristics of group delay bias and mean differential group delay. Since these coefficients do not include corrections for relativistic effects, the user's equipment must determine the requisite relativistic correction. Accordingly, the offset given below includes a term to perform this function.'' | ||

+ | ::: (From [[Talk:Global Positioning System]]) | ||

+ | ::: [[User:RonLar|RonLar]] 12:44, 3 August 2010 (EDT) | ||

== Several Clarification/Corrections == | == Several Clarification/Corrections == | ||

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::::: The word "relativity" dates from the early 1800s. That's not what is being discussed here. If preceded with "theory of" then there is no need to capitalize; if stand-alone, however, it does add clarification to capitalize as is done for other specific concepts that differ from the generic names.--[[User:Aschlafly|Andy Schlafly]] 23:52, 29 July 2010 (EDT) | ::::: The word "relativity" dates from the early 1800s. That's not what is being discussed here. If preceded with "theory of" then there is no need to capitalize; if stand-alone, however, it does add clarification to capitalize as is done for other specific concepts that differ from the generic names.--[[User:Aschlafly|Andy Schlafly]] 23:52, 29 July 2010 (EDT) | ||

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+ | == Curl of the gravitational field == | ||

+ | |||

+ | Sorry to get over-technical, but the fundamental law of "fictitious forces" (including gravity) is that the force field (divided by the mass of the test object) is | ||

+ | |||

+ | <math>G^i = - \Gamma^i_{00}</math> | ||

+ | |||

+ | Its curl is | ||

+ | |||

+ | <math>(\nabla \times G)^i = \mathcal{E}^{ijk} g_{km} G^m_{;j}</math> | ||

+ | where the semicolon indicates the covariant gradient. | ||

+ | |||

+ | When you work this out, it involves derivatives of the <math>\Gamma\,</math> quantities. In general relativity, the results are zero by symmetries of Riemann's tensor. | ||

+ | |||

+ | [[User:Simeon|Simeon]] 21:33, 30 July 2010 (EDT) | ||

+ | |||

+ | : Perhaps so, but the "twin paradox" in Relativity states that the age of each twin is dependent on his path of travel. For a conservative field, all physical parameters are path independent.--[[User:Aschlafly|Andy Schlafly]] 23:07, 30 July 2010 (EDT) | ||

+ | |||

+ | :: Simeon, your mathematical work is rigorous and correct. However, the twin paradox example is interesting to study here. I am aware that the twin paradox is solved by the non-inertial turn-around of the ship when it is going back home. However, in this solution, it is still noted that there is an age difference between the twins. [http://mentock.home.mindspring.com/twins.htm Wikipedia affirms this] and so do [http://mentock.home.mindspring.com/twins.htm other sites]. Such an age difference in twins shows that there is some sort of path dependence. I understand that traveling at near-c speeds in space is not the same thing as moving from point A to B in a gravitational field, but the concept does seem to be a bit similar. Could you maybe explain this for us a bit? Thanks. [[User:PhyllisS|PhyllisS]] 00:52, 31 July 2010 (EDT) | ||

+ | |||

+ | OK, I think I understand. I assumed that the "conservative field" / "curl is zero" stuff referred to the gravitational force field. If it refers to the passage of time, that's different. It isn't true that "all physical parameters are path independent". An extremely important one that isn't path independent is the arc length of the path or arc. You can draw a short straight line from A to B, or a long loopy line that starts at A, wanders around, and eventually gets to B. Why is this relevant to the twin paradox? Because, in relativity, an observer's own elapsed time ("local time") is really just the arc length of his "world line" in Minkowski space. Minkowski was an extremely smart guy, by the way. The twin that stays home takes a direct route from point A (their birth) to point B (the moment they compare ages and see that one has gray hair and wrinkled skin.) The other twin takes a very roundabout route, getting in a rocket and going to Alpha Centauri and back. Their path lengths are their local times, which are different. (Why is the length of the roundabout path actually shorter, so that that twin ages less? Because, in Minkowski space, using the "timelike convention" that all the best people use :-), motion in space subtracts from the elapsed time. That's just the way it works.) | ||

+ | |||

+ | Now I assume that there is no dispute about the facts of relativistic time dilation. In addition to being predicted exactly by the Lorentz transform, it has been observed in practice in cosmic ray muon decays, as well as countless observations in particle accelerators. The "twin paradox" is just an extreme consequence of this. It has of course never been observed in that form, just as we don't know whether Schrodinger's cat is alive. | ||

+ | |||

+ | The "twin paradox" is a consequence of special relativity, not general, and hence does not relate to gravity. I hate to be the umpteenth person to tell you that general relativity is too hard to explain, but it's kind of true. I barely understand the most rudimentary basics. (When Eddington made his comment about only 3 people in the world who understand gen. rel., I wasn't the third! :-) But I can say that you don't need to worry about general relativity to understand the "twin paradox". You can finesse the Minkowski-space curvature of the path during the turnaround at Alpha Centauri, and just say that the twin went there and came back. So was something physically different, that the twins could observe? You bet. The "younger" twin will remember having experienced 6 months of horrendous acceleration in the ionic-drive rocket, followed by a year of horrendous turnaround, and another 6 months of horrendous deceleration at the end. She will have soft, smooth skin, but at a great cost. :-) | ||

+ | |||

+ | Sorry to be so long-winded. In quick summary, the thing that's different about the paths is their length, and that is exactly the local elapsed time. [[User:Simeon|Simeon]] 00:07, 1 August 2010 (EDT) | ||

+ | |||

+ | : Simeon, time dilation occurs under the Theory of General Relativity also, so your analysis above is not persuasive in resolving this example of a non-conservative effect. Moreover, your repeated claims about how supposedly only geniuses can understand this are getting tiresome. That approach is a recipe for mistaken reliance on unjustified authority. | ||

+ | |||

+ | : If you don't feel this is understandable, then simply say so and stop there; please do not imply that people should just accept what someone of undisclosed political views claims.--[[User:Aschlafly|Andy Schlafly]] 10:58, 1 August 2010 (EDT) | ||

+ | |||

+ | I give up. | ||

+ | *The only scientists I mentioned were Minkowski and Eddington, and the latter just as a joke. I never said anything about their, or anyone else's, politics. | ||

+ | *Time dilation does indeed occur under both general and special relativity. The point I was trying to make is that general relativity is simply not needed to understand the twin paradox. It only takes special relativity, which is much better understood. I'm sorry to hear that, by not analyzing the twin paradox in terms of general relativity, my persuasiveness suffered. | ||

+ | *I apologize if I "talked down" to you and Phyllis with my comments about GR being too complicated. I assume that both of you have heard, many times, that GR is exceedingly complicated. I was simply trying to soften the blow by pointing out that you ''don't need'' GR. And cracking that joke about how Eddington could not have been referring to me. | ||

+ | *In fact, I know a fair amount about GR. I ''could'' analyze the twin paradox in terms of the gravitation of Earth and Alpha Centauri. But there is simply no need to. | ||

+ | *This "non-conservative effect" business simply makes no sense. If the integration of a vector field along different paths gets different final results, then that field is non-conservative. You seem to be saying that the ''passage of time'' is some kind of vector field, and that the final results of "integrations" (the two different values of local time at the end of the experiment) are supposed to be the same, and that the difference shows that this "vector field" is not conservative, and that that is a counterexample to relativity. The passage of time is not a vector field. The different values of time, as seen by different observers, is not a ''counterexample'' to relativity, it is ''one of the principal effects'' of relativity. It's really what the word "relativity" means when discussing the scientific Theory of Relativity. | ||

+ | *If you really think that the non-globality and non-absoluteness of time is a counterexample to relativity, then so be it. | ||

+ | |||

+ | [[User:Simeon|Simeon]] 23:13, 1 August 2010 (EDT) | ||

+ | |||

+ | : Simeon, if you "give up," then that is your own choice. You have not disproved the counterexample. Instead, you first described the twin paradox as being only about special relativity, and when I pointed out that it exists under general relativity too, you then agree yet do not fully address the substantive issue presented by the paradox. For example, the amount of acceleration undertaken by the twin in his journey will affect his age independent of his time spent away. His subsequent age is ''not'' path independent even in time-space coordinates. | ||

+ | |||

+ | : It's easy to search for "general relativity" and "conservative field" on the internet and see how little has been written about this. That is telling in itself. I'm happy to continue to discuss this here with you or anyone else.--[[User:Aschlafly|Andy Schlafly]] 23:56, 1 August 2010 (EDT) | ||

+ | |||

+ | :: Could you clarify what the ages (and path dependence thereof) in the twin paradox have to do with conservative fields? --[[User:KyleT|KyleT]] 14:04, 2 August 2010 (EDT) | ||

+ | |||

+ | ::: Age is scalar physical attribute. It should not be path dependent in a [[conservative field]].--[[User:Aschlafly|Andy Schlafly]] 14:31, 2 August 2010 (EDT) | ||

+ | |||

+ | :::: Yes, but which [[conservative field]] in particular are you talking about here (that implies age is not path dependent)? --[[User:KyleT|KyleT]] 14:37, 2 August 2010 (EDT) | ||

+ | |||

+ | ::::: Gravity.--[[User:Aschlafly|Andy Schlafly]] 14:53, 2 August 2010 (EDT) | ||

+ | |||

+ | :::::: Well, in Newtonian mechanics, the gravitational field is indeed conservative -- it's the negative gradient of the gravitational potential! But what this means is that gravitational potential energy is path-independent: it doesn't say anything about path-independence any other quantities, and in particular it's not the reason for the path-independence of age. --[[User:KyleT|KyleT]] 15:00, 2 August 2010 (EDT) | ||

+ | |||

+ | ::::::: You take a narrow view of the significance of a "conservative field." Independent physical attributes should remain path-independent as well for the field to be conservative. In Newtonian mechanics and most other physical force fields, they do.--[[User:Aschlafly|Andy Schlafly]] 15:41, 2 August 2010 (EDT) | ||

+ | |||

+ | :::::::: By a conservative field, I mean a vector field on space for which there exists a scalar function V with the gradient of V given by that vector field. This doesn't imply the path-independence of any physical quantities other than V itself. If you this view as too narrow, can you tell me what you take to be the definition of a conservative field? --[[User:KyleT|KyleT]] 15:57, 2 August 2010 (EDT) | ||

+ | |||

+ | ::::::::: Your definition is too narrow when discussing the theory of relativity, which describes the framework in which the force operates. To be meaningful, the definition must be broader. It must ensure the path independence of the scalar, as well as other scalars independent of that scalar.--[[User:Aschlafly|Andy Schlafly]] 18:12, 2 August 2010 (EDT) | ||

+ | |||

+ | :::::::::: Can you tell me what the correct definition is, then? I have pretty good background in this stuff, no need to dumb it down, just be precise. Certainly no field at all is going to conserve every scalar function, so I'd like to know which ones you want. --[[User:KyleT|KyleT]] 18:20, 2 August 2010 (EDT) | ||

+ | |||

+ | ::::::::::: Kyle, I have an [[Essay:Quantifying Openmindedness|open mind]] about this, and don't see a precise definition anywhere that would be meaningful with respect to the theory of relativity. It's striking how relativists avoid this issue, and even stop discussing it when it is brought up. | ||

+ | |||

+ | ::::::::::: I can propose a definition that you may be able to improve. How about: a conservative theory of motion is one whereby scalar values of a particle are independent of its path of motion.--[[User:Aschlafly|Andy Schlafly]] 18:36, 2 August 2010 (EDT) | ||

+ | |||

+ | That's an interesting proposal, and I too have an open mind about this. Can you give an example of such a ''conservative theory of motion''? One such would greatly help in devising the correct definition. --[[User:KyleT|KyleT]] 19:29, 2 August 2010 (EDT) | ||

+ | |||

+ | :Newtonian mechanics would be an obvious example. By the way, how do you explain the general lack of discussion and papers about whether the theory of relativity is conservative, including the abrupt departure of User:Simeon from this discussion?--[[User:Aschlafly|Andy Schlafly]] 21:58, 2 August 2010 (EDT) | ||

+ | |||

+ | :: Some scalar values in Newtonian mechanics are conserved because there exist associated conservative fields (or more generally [[Noether's Theorem|symmetries of the Lagrangian]]). What is an example of a scalar value in the Newtonian mechanics that is not of this type, which makes this a conservative theory of motion while relativity is not? | ||

+ | ::I don't know why relativity's defenders won't confront this. Maybe that could be the topic of the debate page -- I'm interested using this discussion to sharpen counterexample 21. --[[User:KyleT|KyleT]] 23:55, 2 August 2010 (EDT) | ||

+ | |||

+ | ::: This is a really interesting discussion. I think I made a gross mistake in my first post. The theory of relativity urges us to think of the three space coordinates (x, y, and z) and the time coordinate (t) as four coordinates of space-time - that is, that space and time are pretty much the same. I extrapolated from this that since there can be a (conservative) gravitational field in space coordinates, there can also be some sort of conservative field depending on the time coordinate. I then extrapolated this notion to special relativity, and the twin paradox; I postulated that maybe time dilation effects were the work of a non-conservative field that was dependent on the t-coordinate. Now I see that this was all somewhat foolish. However, I wanted to ask you all: can you have a conservative or non-conservative field with respect to time? If not, I think time should '''not''' be considered as almost the same thing as x, y, z space. I feel that the ability for a dimension to have a field (conservative or not) is integral to its being considered a space-like dimension. | ||

+ | |||

+ | ::: Aschlafly, the fact that the twin paradox exists in general relativity is '''irrelevant'''. Yes, sure, the twin paradox occurs within space where general relativity is working, but there are no effects acting on the twins that influences the twin paradox in any way. Likewise, User:Simeon 's departure is also '''irrelevant'''. | ||

+ | |||

+ | ::: What about black holes, though? Surely their gravitational fields aren't conservative, since once an object passes the event horizon, you can't retrieve it. [[User:PhyllisS|PhyllisS]] 01:24, 3 August 2010 (EDT) | ||

+ | |||

+ | Phyllis: | ||

+ | |||

+ | You seem to be very curious about this topic. I'm going to try to give an intuitive, but nevertheless scientifically correct, explanation of what is going on with relativity, the "twin paradox", and vector fields, potential functions, and path integrals. This explanation will probably seem long and tedious, for which I apologize in advance. I also apologize if it seems that I am being too "folksy", or talking down to you. Please bear with me, and please pay close attention. | ||

+ | |||

+ | We have a parking lot, and two twins, who are fitness enthusiasts and always wear pedometers wherever they go. There are two spots, "X" and "Y", painted on the parking lot. Both people stand on spot "X", set their pedometers to zero, and start walking. Twin A simply walks directly to spot Y. Twin B, being more into fitness, walks all over the place, eventually arriving at B. | ||

+ | |||

+ | Now there are quite a number of things we can say. First, the temperatures vary all over the place. They are a ''scalar field''. That means that they are associated with ''location on the parking lot'', not with any particular observer. They are objective measurements that everyone agrees on, because they are aspects of space itself. Our fitness enthusiasts are also amateur meteorologists, and carry thermometers around with them. | ||

+ | |||

+ | ::Twin A: "When I was at the green Toyota, I noticed that the temperature was 67 degrees Fahrenheit." | ||

+ | ::Twin B: "By coincidence, I also wandered past the green Toyota, and got the same reading." | ||

+ | |||

+ | By the way, since temperature is a scalar field, it has a gradient, which is a vector field. That field is conservative, according to the theorem of mathematical physics that says that curl grad Φ = 0 always. This gradient is a ''vector field''. Like the scalar of temperature, it is a property of the ''space (parking lot) itself''. If the twins had been measuring this gradient (perhaps they carry around fancy "differential thermometers"), they would have gotten the same vector at the green Toyota. | ||

+ | |||

+ | There is also a theorem of mathematical physics, sort of the opposite of the theorem above, that says that, if a vector field V has a curl of zero: | ||

+ | *You can make a scalar field <math>\Phi\,</math> (a property of the space itself, not tied to any particular observer) that it is the gradient of. That scalar field is called the "potential" for the (conservative) vector field. (By the way, this is very closely related to "exact differential equations" that you wrote about! Do you see the connection?) | ||

+ | *If you integrate that vector field along any path between two points A and B (that is, you calculate | ||

+ | :::<math>\int_A^B \vec{V} \cdot dl</math> | ||

+ | for that path, where "dl" is the "line element" along the path), you will get <math>\Phi(B)-\Phi(A)\,</math>. | ||

+ | *Since <math>\Phi(B)-\Phi(A)\,</math> is a property of the scalar field itself (and the points A and B), it follows that that path integral is the same for all paths. And if the path ends on the same point it started on, the integral is zero. | ||

+ | |||

+ | |||

+ | ::Twin A: "I was measuring the gradient of the temperature as I walked, and calculating its path integral as I went. I got an answer of 4 degrees." | ||

+ | ::Twin B: "I was doing the same. My integral was much harder to calculate, because I was going all over the place. But I also got 4 degrees. Hey, wait a minute! The temperature at the start point was 68 degrees, and at the end point it was 72 degrees. That explains it." | ||

+ | |||

+ | Now someone at the edge of the parking lot was running a Van deGraff generator, so there were electric fields all over the place. The twins are also physics students, and carry electroscopes wherever they go. They measured the electric field, and calculated its path integrals. The electric field is conservative (in the absence of varying magnetic fields), so they got the same integral. That integral was 600 volts (it's only static electricity, so it isn't dangerous). Since the electric field is conservative, there is, by the previous theorem, a potential function. That function was 600 volts higher at point B than at point A. | ||

+ | |||

+ | Now here's the kicker: | ||

+ | |||

+ | ::Twin A: "I walked directly from A to B. My pedometer says 150 feet." | ||

+ | ::Twin B: "I took a long route all over the place. I walked half a mile." | ||

+ | |||

+ | The pedometer readings ''are not a scalar field''. They are not a property of the space itself. They are properties of the observers. Even though they, in some sense, measure an aspect of the parking lot (how many molecules of asphalt one passes), they are artifacts of the twins' actions. | ||

+ | |||

+ | The twins could have been integrating their motion vectors; that's sort of what pedometers do. But those vectors are not a vector field on the space itself. It makes no sense to ask whether that "vector field" is conservative, because it isn't a vector field. A vector (or scalar, or tensor) field has to be a property ''of the space itself''. These "pedometer vectors" are just things that the twins make up as they walk. | ||

+ | |||

+ | |||

+ | Now for the "twin paradox". The parking lot is replaced by "Minkowski space", also called "4-dimensional space-time". "Points" in this space are now "events", complete with a time. Events A and B are now the act of the twins saying goodbye as one of them got into the rocket, and the act of them re-uniting after B returns. Twin A took a direct route (called her "world line") from A to B. She used a coordinate system in which the spatial coordinates of A and B were the same (Cape Canaveral, latitude yada yada, etc.) and the time coordinate differed by 30 years (2010 to 2040.) B went to Alpha Centauri and back. When she returned, they were both using the same coordinate system (location is Cape Canaveral, latitude yada yada, time is 2040.) But she looks at her watch, and only 5 years have elapsed! What the watch shows is ''not a scalar field on spacetime''. It was ''not the path integral of a vector field on spacetime''. What she integrated was the ticking of her watch, nothing more. | ||

+ | |||

+ | The path that A took is called a geodesic. It is the Minkowski-space equivalent of a "straight line". But, because of the peculiarities of relativity, it shows the ''longest'' elapsed time (30 years) of all paths, rather than the shortest. By going to Alpha Centauri, twin B took a shorter path, in terms of the way path length is measured in Minkowski space. | ||

+ | |||

+ | Very interesting fact: The path length in Minkowski space, that is, the sum of the tiny distances as measured by the Lorentz/Minkowski metric, ''is the same as the local time''. That is (assuming you are using the "spacelike convention"), everyone's wristwatch measures path length along their own world line. The twins simply followed paths of different lengths. That's all there is to the "twin paradox". (That is, that's all there is to it, if you analyze it correctly, as I have described above. Most introductory treatments of relativity don't do it this way. They just throw the Lorentz transform at you.) | ||

+ | |||

+ | |||

+ | Now there are a few points about the "twin paradox" that people find confusing. | ||

+ | |||

+ | First, aren't the laws of physics supposed to be the same for everyone? What made twin B's watch run slower? Well, she ''knew'' she was traveling at high speed. She brought an accelerometer with her in the rocket. Just as twin B in the parking lot knew she was walking all over the place, turning around and such, twin B in space knew that her world-line was turning, and therefore wasn't a straight line (geodesic). How did she know? It takes force to make you deviate from a geodesic (this is really pretty much the same as Newton's laws of motion), and she felt the force. | ||

+ | |||

+ | Second, how can we analyze the curvature of B's world line? Here's where general relativity has to come in. As soon as world lines start to curve, you have to measure their curvature, that is "geodesic curvature". You get into complicated issues of curved coordinate systems (you're in one now; it's what you perceive as "gravity"!), and curved spacetime, and so on. And you get into the <math>\Gamma\,</math> symbols, which measure the deviation from a geodesic, and hence the "fictitious forces" that you feel. This is why general relativity is related to the "twin paradox", in that the space ship followed a curved trajectory and experienced acceleration. But, to analyze the plain facts of the "twin paradox", all you really need to know is that twin B followed a crooked line. Place your ruler on a diagonal on the graph of Minkowski space, draw the line out to Alpha Centauri. Turn the ruler, draw the returning line. Ignore the impossibly sharp corners. Use special relativity to analyze the Lorentz transform for each section of B's world line. | ||

+ | |||

+ | Oh, and to try to answer some of your specific questions, the gravitational field, under either Newtonian or relativistic mechanics, is a conservative field. Its curl is zero. If it weren't, conservation of energy would be violated, and we could make a perpetual motion machine by having a planet run around in circles picking up energy. The "curl=0" aspect of gravity under general relativity is more complicated, because true vector fields have to be on Minkowski space, but it still conserves energy. When Mercury orbits the Sun, its perihelion precesses because of relativistic effects, but its energy is conserved. | ||

+ | |||

+ | I'll try to think some more about your black hole question and get back to you. | ||

+ | |||

+ | [[User:Simeon|Simeon]] 00:00, 4 August 2010 (EDT) | ||

+ | |||

+ | :Simeon, thanks so much for your long explanation! All of that made sense to me, and cleared up the issue of field as property of space vs. path specific to person vs. curved spacetime. Also, yes, I noticed the similarity between exact differential equations and deriving potential functions from vector fields - the former I studied in Differential Equations and the latter I studied in Multivariable Calculus. | ||

+ | |||

+ | :So, if I understand you properly, gravity both (a) curves spacetime and (b) creates a conservative field. (I derived (a) from your point: "You get into complicated issues of curved coordinate systems (you're in one now; it's what you perceive as "gravity"!)", and (b) from your point: "the gravitational field, under either Newtonian or relativistic mechanics, is a conservative field. Its curl is zero.") However, you're saying that if you are in an accelerating reference frame (such as a quickly-spinning merry-go-round) only (a) occurs; there is no field. Is this correct? If so, why is there this discrepancy between the two? [[User:PhyllisS|PhyllisS]] 22:54, 4 August 2010 (EDT) | ||

+ | |||

+ | Yes, gravity curves spacetime. It looks as though you're ready to go to the next level. It has to do with curved ''coordinate systems'' vs. curved ''spacetime''. This is why GR is so complicated. But here goes. | ||

+ | |||

+ | First, we have to recognize that, at some level, we could say that energy is *not* conserved. Spacecraft use "slingshot maneuvers" around one planet to gain extra energy on their way to another planet. I assume you've heard of this. Cassini used three such maneuvers, twice around Venus and once around Earth. So what was going on? If you look at a coordinate system centered on Venus, you would see Cassini come in and go out again, with complete conservation of energy. But, in a coordinate system fixed around the solar system, Venus was moving, so Cassini came in at low speed and went out at high speed. We "stole" some energy out of Venus's orbit. So, to be ''really'' correct, we have to say that gravity is a conservative field in the absence of '''moving''' gravitating bodies. | ||

+ | |||

+ | But, in the larger sense, energy is conserved. Always. Newtonian or relativistic. (But in relativity, the mass figures into the equation. Let's not worry about that just now.) | ||

+ | |||

+ | Now we get to the really cool stuff. Your comment above suggests that you are ready for it. | ||

+ | |||

+ | You will feel a "fictitious force" whenever you are in a "curved spacetime coordinate system". A curved coordinate system would include things like polar coordinates on the plane, or spherical coordinates in 3 dimensions. But this is 4-dimensional spacetime. So I'd like you to take my word for this. On a rotating merry-go-round, your 4-dimensional spacetime coordinate system is curved. (This doesn't require relativity, special or general, to formulate this.) The 3-dimensional slice of it is flat, but, when you bring in time, and the spatial coordinates are accelerating, the overall coordinate system is curved. This creates fictitious forces. In the case of the merry-go-round, the forces are the centrifugal force and the Coriolis force. There are also the fictitious "acceleration G forces" in a rocket. But '''the spacetime itself isn't curved'''. It's like polar coordinates on the plane. Yes, the coordinate system is curved, but the plane isn't. There are Cartesian coordinates on the plane also. Similarly, an observer on the ground is in a flat coordinate system, and doesn't see any fictitious forces. He just sees the mechanism of the ride pushing you inward as you go around. Your recoil against that acceleration is what you perceive as the centrifugal force. | ||

+ | |||

+ | So the moral of the story would seem to be: you can choose a flat coordinate system that exposes the fictitious force for what it is. It's a perception from the curved spacetime coordinate system that the observer is operating in. | ||

+ | |||

+ | But how about gravity? In the case of gravity, '''spacetime itself is curved'''. If spacetime (properly called a "manifold") is curved, every coordinate system is curved, and the fictitious force which is gravity is inescapable. | ||

+ | |||

+ | So your statement "gravity curves spacetime", is exactly correct, but you have to distinguish "curving spacetime" and "curving some particular coordinate system". The actual curvature involves things called "Riemann's tensor" and "Ricci's tensor". | ||

+ | |||

+ | The number of people that understand this is way more than the 3 that Arthur Eddington claimed (it was more than 3 but probably less than 10 at the time), but it's still a pretty complicated subject. | ||

+ | |||

+ | [[User:Simeon|Simeon]] 00:03, 5 August 2010 (EDT) | ||

+ | |||

+ | :Simeon: Ah, that does make a lot of sense now. Thanks so much for explaining this to me. Hmm, that really is an elegant concept -- wow. Cool! [[User:PhyllisS|PhyllisS]] 13:50, 5 August 2010 (EDT) |

## Revision as of 13:50, 5 August 2010

**Attention: Please review previous points on the discussion page before adding your own commentary. Many topics have been discussed many, many, times. If you have something new to add, feel free, but it is not necessary or helpful to read the same arguments over and over and over.**

**Raising arguments which have been discussed before wastes the time of valuable editors and repeatedly doing so violates 90/10.**

Andy, can you clarify #4 for me? I'm not sure I understand it. JacobB 21:50, 28 November 2009 (EST)

- Sure, I welcome discussion of these important points. As I've said, I have an open mind about this and if something is true, then I accept it. But if something is false, I'll criticize it.

- The theory of relativity has taught for decades that as the velocity of a mass increases, then its (scalar) relativistic mass increases per the Lorentzian transformation. Now apply a force ORTHOGONAL to the velocity. Does that force encounter the increased mass, as relativity says, or encounter the rest mass, as logic would dictate?--Andy Schlafly 22:02, 28 November 2009 (EST)

- Ah, I see what you mean. May I suggest a re-wording? "The logical problem of a force which is applied at a right angle to the velocity of a relativistic mass." I think that might be a little clearer than it is currently stated. Your thoughts? JacobB 22:06, 28 November 2009 (EST)

- Please do. Your edits are always welcome, and you've suggested an improvement here. Thank you for making this change.--Andy Schlafly 22:20, 28 November 2009 (EST)

- Why would logic dictate that? Mass is a scalar, and a force from any direction should encounter the same increased mass, not different masses from different directions.

- I suppose that under Newtonian mechanics, a moving object has a velocity of 0 within the plane perpendicular to its line of motion, and any forces operating in that plane will act on the object as if it is at rest. But that's not what
*logic*dictates, that's what the*previous theory*dictates.

- I suppose that under Newtonian mechanics, a moving object has a velocity of 0 within the plane perpendicular to its line of motion, and any forces operating in that plane will act on the object as if it is at rest. But that's not what

- Essentially your counterexample to relativity is that it makes a prediction that contradicts Newton's laws. This is neithe r a contradiction nor a logical problem, and it is should be edited out.NgSmith

- No, it's a logical problem. If you're suggesting that one force can affect the inertial in an entirely independent, orthogonal direction, that's illogical. One thing cannot affect something else that is entirely independent.--Andy Schlafly 15:40, 12 December 2009 (EST)

- Why is that illogical? What logical principle does it violate?

- See, in relativity, orthogonal doesn't
*mean*independent. In relativity, velocity vectors*do not add.*In relativity, the effect of a new force is not independent of the object's existing momentum. And there is nothing illogical about that; it's just a new theory that contradicts the intuition from the previous theory.--NgSmith

- See, in relativity, orthogonal doesn't

- Ng, something cannot be independent (orthogonal) and yet dependent at the same time. Unfortunately, you're arguing with your own theory at this point. Even most relativity promoters have abandoned the position you take here.--Andy Schlafly 21:37, 12 December 2009 (EST)

- It seems that his point is that something can be orthogonal and dependent. I agree: The cross-product of two vectors is orthogonal to both and yet obviously dependent on both. --EvanW 21:41, 12 December 2009 (EST)

- OK, good point, an orthogonal vector can be a function of other orthogonal vectors. But that's a bit different from what we're discussing. Here it's an orthogonal force that is not dependent on anything else, and yet Ng says it encounters relativistic mass due to a different orthogonal force.

- I think relativists have abandoned Ng's position, so he's really arguing with his own side at this point. As a result, I urge him to reconsider his views with an open mind once he confirms that.--Andy Schlafly 21:59, 12 December 2009 (EST)

- First of all, relativity has not "abandoned" the prediction we're talking about. The velocity addition formulas for both parallel and perpendicular velocities have not changed, and they still predict that an orthogonal force will have a harder time accelerating a fast-moving object. Physicists may have changed their informal interpretation of this formula, but not the formula itself, nor its predictions.

- Note also that relativity's prediction can't be all that illogical, because this is what we
*actually observe happening to particles at high speeds.*If you think that fast-moving particles commit some terrible offense against basic logic, take it up with God.

- Note also that relativity's prediction can't be all that illogical, because this is what we

- There is a very simple way to settle this matter: write an encyclopedia article where the material is properly sourced. If this is indeed some counterexample or logical flaw in relativity, then one can easily find a book or paper exposing that flaw, and cite it.--NgSmith Sun Dec 13 17:55:04 EST 2009

- OK, I think I see part of the problem you people are having. The word "independent" has two different meanings. Being
*linearly*independent is a concept from pure mathematics. Being*causally*independent is an unrelated metaphysical concept. Whether a force pushing on something causes it to move, and by how much, is completely, umm, independent of whether the vectors involved are linearly independent (orthogonal). Please try to be very careful about the meanings of the terms. SaraT 17:00, 13 December 2009 (EST)

- OK, I think I see part of the problem you people are having. The word "independent" has two different meanings. Being

- I don't think that's the source of our confusion. I think the main problem is that, according to Newtonian mechanics and thus according to our mechanical intuition, orthogonal things tend to operate independently. Not only that, but a force exerted on an object is usually independent of the object's momentum.

- In relativity, none of these things are true, due to the fact that velocities no longer add like vectors (and thus acceleration no longer incurs a cumulative change in velocity in the usual way.) This is seen as some sort of logical flaw or paradox simply because it contradicts the deeply ingrained intuition that came from the previous theory.--NgSmith Sun Dec 13 18:10:46 EST 2009

- Theories that don't produce anything useful are often a waste of time, or simply false. I realize that liberals tend to downplay accountability -- a conservative insight, but theories should be accountable by what value they yield, particularly when taxpayer dollars are spent (wasted) on the theory.--Andy Schlafly 16:55, 7 January 2010 (EST)

- I call gps a pretty darned useful invention but it doesn't work if you don't take into account relativistic effects. I think that not knowing where relativity is used speaks volumes as to how close minded those trying to disprove relativity, which is different from relativism. (a point completely overlooked by the page) Gaurdro 12:31, 24 May 2010 (EDT)

## Contents

## Counterexample 4 (limiting behavior)

For the fourth "counterexample," the author points out that the momentum <math>p=mv\gamma</math> does not approach the momentum of light as <math>m\rightarrow 0</math> and <math>v\rightarrow c</math>

Aside from the mathematical sloppiness of taking two independent variables to a limit at the same time, at unspecified rates, these sorts of "discontinuities" can be found in just about any scientific theory. In Newtonian mechanics, for example, take the orbit of a planet as the planet's mass goes to 0. For any nonzero mass the orbit is an ellipse; at m=0 it is suddenly a straight line. Is this a "counterexample" to Newton's laws?

Or in electronics, I=V/R. The limiting case is no voltage, no resistance, no current; but if someone foolishly took V/R as both V and R go to zero, he would get a nonsensical answer. Let them both go at the same rate and you get I=1. Is this a "counterexample" to basic electronics?

Or more to the point, momentum in Newtonian mechanics is <math>p=mv</math>, and this also fails to give the momentum of a photon at m=0, v=c. Again, is that a "counterexample" to <math>p=mv</math>? Will we see this entry in a corresponding page of "Counterexamples to Newton's laws?"

But none of these are counterexamples or "discontinuities": they are just a misinterpretation of the formulas. You don't get the momentum of a photon by taking the momentum formula for a mass and setting m=0 and v=c. That's just not what the formula means, or what they are for. This item should also be removed.--NgSmith Tue Dec 15 10:16:21 EST 2009

## Counterexample 9 (Jesus action-at-a-distance)

The quoted verse doesn't strongly suggest "action-at-a-distance" in the relativistic sense. Light could travel the distances mentioned in the passage in a fraction of a second, which is well within the precision given in the verse (an hour). The verse and relativity are not in contradiction here. This should be removed.

- I have an open mind about it. In the the healing of the centurion's servant, if the Greek is translated as same "moment" then relativity is impossible, but if translated as the same "hour" then there is no conflict with relativity.

- But the healing of the centurion's servant is probably not the only place where there is action at a distance in the Bible.--Andy Schlafly 14:52, 5 January 2010 (EST)

- Any distance on the earth is less than 20,000km. A force acting with the speed of light takes less than 1/15,000 ≈ 0.0000667 seconds for this distance.

- I don't think how eyewitnesses could spot such a short time...

- So, there may probably be no other places where action at a distance is described in the Bible.

- FrankC aka ComedyFan 16:17, 5 January 2010 (EST)

- You make an interesting point, Frank. But according to this site, it takes 1/7.4 seconds for light to circle the globe, which is much longer than your figure.[1] More generally and more importantly, there is the issue of how this action in the Bible
*isn't*light.--Andy Schlafly 19:30, 5 January 2010 (EST)

- You make an interesting point, Frank. But according to this site, it takes 1/7.4 seconds for light to circle the globe, which is much longer than your figure.[1] More generally and more importantly, there is the issue of how this action in the Bible

- Indeed, an error in my calculation: 20,000,000m / 300,000,000 m/sec = 1/15 seconds.
- Fast enough, still.
- Whether the action in the Bible
*isn't*light doesn't matter: it is indistinguishable from an action happening at the speed of light for the witnesses of the time, so it doesn't say anything about the validity of the theory of relativity... - FrankC aka ComedyFan 19:46, 5 January 2010 (EST)

- Frank you make an interesting point, and I have an open mind about it. But I'm not entirely convinced. When the woman cured herself of bleeding and Jesus felt power leaving him, that sounds more like heat than light. And for heat to travel virtually instantaneously (or at the speed of light) WOULD violate the theory of relativity.--Andy Schlafly 20:48, 5 January 2010 (EST)

- Yes, it would. And it would also violate classical physics, the laws of thermodynamics etc.

- But of course a miracle is going to violate the laws of physics. I don't see how this can be cited to discredit one physical theory over another.--NgSmith

I have to respectfully disagree with you on that point, Andy - I'm not sure this action could comment on relativity any more than the sun stopping for Joshua could comment on the Copernican model of the solar system. If God wanted heat/light to travel at some finite speed except in certain instances, how is that different from the sun and moon moving in the sky, except in certain instances? JacobB 21:32, 5 January 2010 (EST)

- I have an open mind about this. You make good points, Jacob. But your analogy is not perfect because:

- the Joshua account might be understood as the
*perception*of the army that they sun did not set until they completed their job, but the healing in the New Testament cannot be explained as mere perception - if the Joshua account is taken absolutely literally, Newtonian mechanics does not say it is impossible, while relativity does say action-at-a-distance is impossible

I look forward to our translation work on the Joshua passage (and New Testament passages) to see if that brings forth insights.--Andy Schlafly 22:30, 5 January 2010 (EST)

- Your second point is a good one, and I suppose my example wasn't very good. But on a different note, what makes you say that the Joshua account might be understood as only a perception of the army? I think I'm going to go translate that chapeter, I'll be interested to see what Hebrew words are used for that bit. JacobB 22:49, 5 January 2010 (EST)

- Shall we look at it next? Joshua 10:11-14, I think.--Andy Schlafly 23:18, 5 January 2010 (EST)

IMO, the discussion is a little bit bizarre: Following David Hume's definition of a miracle as a "a violation of the laws of nature", for evaluating the *laws of natures*, miracles can't be taken into account.

As I said earlier: we shouldn't try to restrict God with the laws of our logic - or even physics.

FrankC aka ComedyFan 07:27, 6 January 2010 (EST)

- Frank, perhaps what you mean is that you don't want the logic of the Bible to be used to evaluate claims by scientists. If so, I completely disagree. And so would Isaac Newton and most great scientists.

- As our Conservative Bible Translation project is revealing, Jesus said his works were not miracles, but signs. So any definition of miracle by Hume (who, by the way, leaned toward atheistic rather than Christianity) is not terribly helpful.--Andy Schlafly 08:00, 6 January 2010 (EST)

- So, what's the definition of a
*sign*, then? FrankC aka ComedyFan 08:06, 6 January 2010 (EST)

- So, what's the definition of a

- The same as its name suggests: a disclosure of reality, rather than a violation of it.--Andy Schlafly 08:35, 6 January 2010 (EST)

- I took Hume's definition as I found it on conservapedia's page on miracles.
- The page on signs doesn't describe Jesu works - perhaps you can fix this
- If you don't like Hume, what's about Thomas Aquinas:

*Now, there are various degrees and orders of these miracles. Indeed, the highest rank among miracles is held by those events in which something is done by God which nature never could do. For example, that two bodies should be coincident; that the sun reverse its course, or stand still; that the sea open up and offer a way through which people may pass. And even among these an order may be observed. For the greater the things that God does are, and the more they are removed from the capacity of nature, the greater the miracle is. Thus, it is more miraculous for the sun to reverse its course than for the sea to be divided.*

*Then, the second degree among miracles is held by those events in which God does something which nature can do, but not in this order. It is a work of nature for an animal to live, to see, and to walk; but for it to live after death, to see after becoming blind, to walk after paralysis of the limbs, this nature cannot do—but God at times does such works miraculously. Even among this degree of miracles a gradation is evident, according as what is done is more removed from the capacity of nature.*

*Now, the third degree of miracles occurs when God does what is usually done by the working of nature, but without the operation of the principles of nature. For example, a person may be cured by divine power from a fever which could be cured naturally, and it may rain independently of the working of the principles of nature.*

- Acts 2:43
*Everyone was filled with awe, and many wonders and miraculous signs were done by the apostles*(KJB) So, we have*miraculous signs*and*wonders*

- Acts 2:43

- John 2:11
*This was the first of the miracles Jesus did in Cana of Galilee, and by doing showed his glory, and so his disciples believed in him.*(CBP)*Changing water into wine*is something nature never could do: it's an outright miracle, miraculous sign, whatever...

- John 2:11

- FrankC aka ComedyFan 09:00, 6 January 2010 (EST)

- That's great recitation, Frank, but how about simply applying logic yourself? You're a bright guy, why simply hunt and repeat quotes from others? On this site we encourage
*thinking*in a logical way.--Andy Schlafly 09:21, 6 January 2010 (EST)

- That's great recitation, Frank, but how about simply applying logic yourself? You're a bright guy, why simply hunt and repeat quotes from others? On this site we encourage

- I'm trying to use the fact that I'm standing on the shoulder of giants... FrankC aka ComedyFan 09:23, 6 January 2010 (EST)

- How about using "the fact" of simple logic and the power of your
*own*mind?--Andy Schlafly 09:27, 6 January 2010 (EST)

- How about using "the fact" of simple logic and the power of your

- To make it as clear as possible in my own words:
- I won't restrict God by laws which men made or observed. Can I understand God's ways? Can I expect God to act the way I think to be logical? That would be hubris.
- Testing scientific hypotheses using God's miracles or signs seems to be odd!

- But which part of Thomas Aquinas's definition of miraculous events didn't you like? Granted, he had a slightly other view of the
*capacity of nature*than we have today, but his line of reasoning was as valid in the 15th century as it is today! I hoped that his definition would be more*helpful*than that of David Hume.

- To make it as clear as possible in my own words:

- FrankC aka ComedyFan 10:41, 6 January 2010 (EST)

A miraculous healing seems to violate the Second Law of Thermodynamics - whether it happens on a distance or not. Does this mean that John 4:46-54 is a counterexample to the laws of thermodynamics, too? PhilG 09:58, 2 February 2010 (EST)

- How so? Do you think eating an apple to feel better, or taking an aspirin to alleviate a headache, also violates the Second Law of Thermodynamics?--Andy Schlafly 11:02, 2 February 2010 (EST)

## lack of a single useful device

At conservapedia's article on the Global Positioning System, one can read:

*These receivers rely on precisely timing signals sent from GPS satellites, with corrections for atmospheric attenuation and relativistic effects.*

GPS seems to be a useful device!

FrankC aka ComedyFan 10:53, 6 January 2010 (EST)

- Great catch of a misleading statement, Frank! I've corrected it.

- Our theory of relativity entry explains how it did not aid the development of GPS. The repeated attempt by relativists to falsely claim credit for GPS
*reinforces*the lack of any legitimate contributions.--Andy Schlafly 11:29, 6 January 2010 (EST)

- Well, you are consistent! Just another question: What's about particle accelerators? Generally, the theory of relativity is used to explain why it takes more energy to accelerate an electron from 200,000,000 m/sec to 200,002,000 m/sec than from 2,000 m/sec to 4,000 m/sec.

- Have you thought about an explanation for this phenomenon?

- Accelerators have applications beyond basic research!

- FrankC aka ComedyFan 12:02, 6 January 2010 (EST)

- Frank, I have an open mind about this, but I'm not aware of a single benefit from what you describe, nor do you identify one. Do you have an open mind about this?--Andy Schlafly 14:44, 6 January 2010 (EST)

- Synchrotron radiation is used in medicine
- So, may I ask again: what your explanation for the phenomenon? I suppose you are aware of the phenomenon I described above?

- FrankC aka ComedyFan 15:47, 6 January 2010 (EST)

- Frank, inventors and doctors and engineers don't typically even bother learning relativity. Should I repeat that? Complain to engineering departments and medical schools if you think that should change. Nothing useful has even been designed or built using relativity. If you want to look and look and look for a counterexample then you'll be wasting your time. I'm not going to waste mine. This is my final reply on this topic for now. Do something logical, such as editing the Bible, and after benefiting from that experience we can revisit this issue in a month or so.--Andy Schlafly 15:52, 6 January 2010 (EST)

- Why does it matter whether the users of the invention learn relativity? Most users of microwaves never learn Maxwell's equations either. That doesn't mean that the laws are irrelevant to the gadget's operation.

- Likewise, the engineers who correct the clocks of GPS satellites may not know or care that relativistic effects are behind the clock skew. But that dodges the point that relativistic effects are real, observable, and must be corrected for in several useful inventions.--NgSmith

- Here's a good source: | US Navy. As for engineers not bothering to learn relativity, I think that's a mite off the mark. I'm an engineer and I had to take a class dealing with the basics of SR, and I'm just an electrical engineer. Aerospace engineers certainly deal with relativity a great deal, as do nuclear engineers. DanieleGiusto 00:26, 7 April 2010 (EDT)

## GPS revisited

The same Tom von Flandern who is quoted in the article on the theory of relativity saying that the GPS programmers "have basically blown off Einstein", wrote in an article in 1998:

*So we can state that the clock rate effect predicted by GR is confirmed to within no worse than ±200 / 45,900 or about 0.7%, and that predicted by SR is confirmed to within ±200 / 7,200 or about 3%. This is a very conservative estimate. In an actual study, most of that maximum 200 ns/day variance would almost certainly be accounted for by differences between planned and achieved orbits, and the predictions of relativity would be confirmed with much better precision.*

As for how the satellites take into account the relativistic effects, here is his explanation of the so-called *factory offset* of the atomic clocks for the satellites:

*GPS atomic clocks in orbit would run at rates quite different from ground clocks if allowed to do so, and this would complicate usage of the system. So the counter of hyperfine cesium transitions (or the corresponding phenomenon in the case of rubidium atomic clocks) is reset on the ground before launch so that, once in orbit, the clocks will tick off whole seconds at the same average rate as ground clocks. GPS clocks are therefore seen to run slow compared to ground clocks before launch, but run at the same rate as ground clocks after launch when at the correct orbital altitude.*

Seems to me that relativistic effects have to be taken into account.

FrankC aka ComedyFan 13:13, 6 January 2010 (EST)

- Frank, your intuition ("seems to me") is wrong here, and the entry explains it clearly. GPS is a work of engineering and any timing discrepancies between the satellite and ground are obviously better handled directly by synchronization rather than asking a physicist what he thinks of relativity. Engineers don't even bother taking general relativity courses, let alone try to build a satellite system using them.--Andy Schlafly 14:44, 6 January 2010 (EST)

- The Time Service Department – a department of the U. S. Navy - states: “The Operational Control System (OCS) of the Global Positioning System (GPS) does not include the rigorous transformations between coordinate systems that Einstein’s general theory of relativity would seem to require – transformations to and from the individual space vehicles (SVs), the Monitor Stations (MSs), and the users on the surface of the rotating earth, and the geocentric Earth Centered Inertial System (ECI) in which the SV orbits are calculated. There is a very good reason for the omission: the effects of relativity, where they are different from the effects predicted by classical mechanics and electromagnetic theory, are too small to matter – less than one centimeter, for users on or near the earth.”

- As far as I can see there is no reason to feel sorry for FrankC: Your article only covers the idea of using the Lorentz transformation instead of the Galileo transformation when calculating the position of an object: one could say that it is about the relativistic effects caused by the movement of the GPS receiver, not of the GPS satellites. That's why it's talking about
*fast moving air-planes and satellites*. - FrankC (and others) have shown that there are relativistic effects on the satellites which are taken account of:
- Global Positioning System Standard Positioning Service Signal Specification, 2nd edition, June 1995:

- As far as I can see there is no reason to feel sorry for FrankC: Your article only covers the idea of using the Lorentz transformation instead of the Galileo transformation when calculating the position of an object: one could say that it is about the relativistic effects caused by the movement of the GPS receiver, not of the GPS satellites. That's why it's talking about

- p. 13:
*To compensate for relativistic effects, the output frequency of the satellite's frequency standard -- as it would appear to an observer located at the satellite -- is 10.23 MHz offset by a Df/f = -4.4647 x 10-18 or a Df = -4.567 x 10-3 Hz.*

- p. 13:

- p. 39:
*The coefficients transmitted in subframe 1 describe the offset apparent to the control segment two-frequency receivers for the interval of time in which the parameters are transmitted. This estimated correction accounts for the deterministic satellite clock error characteristics of bias, drift and aging, as well as for the satellite implementation characteristics of group delay bias and mean differential group delay. Since these coefficients do not include corrections for relativistic effects, the user's equipment must determine the requisite relativistic correction. Accordingly, the offset given below includes a term to perform this function.*

- p. 39:
- (From Talk:Global Positioning System)
- RonLar 12:44, 3 August 2010 (EDT)

## Several Clarification/Corrections

I am new to Conservapedia, so I don't fully understand exactly how this site is structured; in particular who has the ability to edit protected pages. This page is apparently protected, but in need of dire work even on the formatting/punctuation/style side of things. I hope someone with the required access to protected pages can incorporate some of these changes. In any event, here are some things that need to be clarified or corrected:

1. It is unclear what #6 is referring to by "overall space". Is this a reference to the flatness problem in cosmology? If so, that should be made explicit, and the most commonly accepted solution (inflation) should be mentioned for completeness. If not, the curvature of space locally is clearly demonstrated by the observed phenomenon of gravitational lensing.

2. #7 is just a feature of the incompatibility between general relativity and quantum field theory. In this sense, general relativity is of course wrong (just as Newtonian mechanics is, for example), because it doesn't apply accurately below distance scales where quantum effects emerge.

3. #8 is not a problem because quantum nonlocality doesn't allow for information transfer. Hence special relativity is not violated.

4. #10 is not a counterexample because gravitons are not predicted by general relativity. They are expected to exist and be predicted by a successful *quantum* theory of gravity, but general relativity is not such a theory.

5. #11 is not a counterexample to the theory at all. It may be an argument for why the theory should not be studied, but that doesn't mean it is *false*, and thus is not a counterexample.

6. #13 is presumably a reference to the horizon problem of cosmology. This should be stated, and, as for the flatness problem, the theory of cosmological inflation should be mentioned. (I realize inflation has not been empirically verified, but since the majority of cosmologists believe it is the correct explanation, it deserves a mention in an encyclopedia article.)

7. #14 is again the problem of the incompatibility of general relativity and quantum field theory (namely that QFT is not background-invariant). This is not a problem with general relativity, other than in the sense that it is only an approximation (like, say, Maxwellian electrodynamics are just an approximation to quantum electrodynamics).

8. #15, aside from the obvious grammatical error (*violated* instead of the correct *violate*), is again not a counterexample to general relativity. General relativity predicts wormholes *only* on the assumption that so-called "exotic matter" exists. This is matter that has net negative mass/energy, and so is predicted not to exist for precisely the reasons listed here (time travel and the like). But this is not a counterexample to general relativity itself, merely the observation that a mathematically possible solution does not have a physical manifestation.

9. #16 is again a quantum gravity issue. It is wrong to call black holes "highly ordered (and thus low entropy)", though. The fact is that science does not yet know how to count black hole microstates, so we don't know whether they are highly ordered or extremely disordered. But the best explanation seems to be that general relativity and the Second Law together suggest that black holes should have extremely *high* entropy, not low entropy. But again, this is not a counterexample to general relativity per se, since it makes no predictions about what black hole entropy should be.

10. #18 appears to be a restatement of #11, and is thus both redundant, and not a counterexample for the reasons listed discussion #11.

I apologize for the length of this list of edits, but something really must be done to improve the quality of this article. I hope that someone with the appropriate access sees fit to make the necessary changes soon. Yill 17:12, 30 March 2010 (EDT)

- REPLY BELOW:

- 1. It is unclear what #6 is referring to by "overall space". Is this a reference to the flatness problem in cosmology? If so, that should be made explicit, and the most commonly accepted solution (inflation) should be mentioned for completeness. If not, the curvature of space locally is clearly demonstrated by the observed phenomenon of gravitational lensing.

- I'll clarify the obvious. It's still a counterexample. Science is not done by consensus, and inflation does not explain the overall flatness of space if relativity were true.

- You needn't be so condescending. I wasn't saying that it isn't a problem with general relativity, I was just saying that since this is an encyclopedia, relevant information should be included. Since a proposed solution exists, it should be mentioned, and perhaps debunked if it is flawed. So you could mention inflation, and then say why it fails to solve the flatness problem. Yill 21:19, 30 March 2010 (EDT)

- The theory of inflation does nothing "to solve the flatness problem" with respect its role as a counterexample to relativity.--Andy Schlafly 03:02, 31 March 2010 (EDT)

- Could you clarify this point? Perhaps you could state exactly what you believe the flatness problem is and how it is a counterexample to GR, just to be sure we aren't talking past each other, as I fear we may have been so far. Yill 16:55, 4 April 2010 (EDT)

- 2. 7: is just a feature of the incompatibility between general relativity and quantum field theory. In this sense, general relativity is of course wrong (just as Newtonian mechanics is, for example), because it doesn't apply accurately below distance scales where quantum effects emerge.

- So at what distances do you declare general relativity to be false? Is there a discontinuity at that distance? Such an approach is absurd.

- I mean, technically it is false at
*all*length scales, just like any classical (non-quantum) theory (Newtonian mechanics, Maxwellian electromagnetism, classical statistical mechanics, etc.). But there exists a range of length scales at which it is extremely accurate, and those are the only ones to which it makes claims having any epistemological value. There is no discontinuity, it just gets progressively worse as quantum effects become more and more apparent, which occurs at smaller and smaller length scales. Quantum effects definitely need to be taken into account around the level of a nanometer or so in most systems of interest, so I would say this is about the regime where GR needs to stop being used. But of course, it depends on the system in question. Yill 21:19, 30 March 2010 (EDT)

- I mean, technically it is false at

- Not "technically it is false," but "it is false." So teach it that way.--Andy Schlafly 03:02, 31 March 2010 (EDT)

- See KrisJ's discussion below. Yill 16:49, 4 April 2010 (EDT)

- 3. 8 is not a problem because quantum nonlocality doesn't allow for information transfer. Hence special relativity is not violated.

- Your statement is a non sequitur, and may not be true. Special relativity does deny non-locality.

- It's not a non sequitur; the problem as I thought it was stated on the page is that special relativity does not allow information transfer faster than the speed of light. Since quantum entanglement cannot actually transfer information, this does not violate that provision of special relativity. Yill 21:19, 30 March 2010 (EDT)

- Special relativity does not define "information" nor was it developed in that context.--Andy Schlafly 03:02, 31 March 2010 (EDT)

- It is true that SR does not define information, but it does define causality (only events within each other's lightcones can be causally connected). Physical transfer of information (as defined by Shannon, and encoded in physical systems in Minkowski spacetime) between points in spacetime can only occur if those points are causally connected. (This SR fact is what the horizon problem, which is cited as another GR counterexample, relies on.) Yill 16:49, 4 April 2010 (EDT)

- Will respond to your other points later.--Andy Schlafly 18:11, 30 March 2010 (EDT)

- I appreciate your attention to my concerns, and I hope I have adequately outlined them. Also, I hope I would not be asking too much to request formatting consistency (like adding periods at the ends of nos. 7, 8, and 9). It would make it look more professional, like other articles I've seen on Conservapedia. Yill 21:19, 30 March 2010 (EDT)

- Yill, your grand total of contributions to this site has been 3 edits to this page, all easily refutable. Frankly, I don't think greater efforts at "formatting consistency" are justified.--Andy Schlafly 03:01, 31 March 2010 (EDT)

- Your not going to be able to attract many users if you disparage newcomers with respect to how few edits they've made. I would like to be a positive contributor to this site, but I have to start somewhere. I would appreciate encouragement and constructive criticism, not condescension and personal attacks. Yill 16:49, 4 April 2010 (EDT)

- Yill, good grammar requires "you're", not "your", in your statement above. All your edits have been 100% talk, in violation of our 90/10 rule, and honestly I see no insights in your talk. I suggest you try contributing substantively to Epistle to the Hebrews (Translated); it is on a much higher educational level and you'll benefit enormously from it.--Andy Schlafly 00:15, 5 April 2010 (EDT)

- You're right, I had a typo there; I apologize for the error. And I am well aware of the 90/10 rule, but seeing as the page I'm working on is protected, I'm not actually able to make any edits. If it were unblocked or I were given the ability to edit it, I would be more than happy to stop posting on this talk page and instead edit the article itself. And frankly I don't particularly see how it's relevant whether you personally happen to see any insights in my talk; my understanding is that Conservapedia is shaped and edited by its users, with appropriate oversight from administrators to ensure accuracy and prevent the chaos of Wikipedia. If need be, I'll appeal to those administrators to get the article fixed, since none seem to have come forward to help. I would love it if you would be willing to work with me to improve this article, but as it stands you seem to have little interest in doing so, having made no further contributions to the substance of the discussion. If you change your mind, I would be happy to work with you on this endeavor.

- As for your suggested article for me to work on, I don't really understand what you mean by it being on a "much higher educational level." However, as I have no expertise in Biblical Greek, I don't think I'd be able to make any meaningful contributions to the translation. I'll let the experts in that subject deal with that article. Yill 16:37, 5 April 2010 (EDT)

- Yill, I recommended the Bible because, as Isaac Newton pointed out, working on translating the Bible increases the quality of one's work in other areas, including science. Sure, I could drop everything else I'm doing and spend all day correcting you about this entry, but if you just picked up a Bible and improved your own work, then I could learn from you instead. I'll correct your misunderstandings below but doubt I will spend much more time responding to you if you're not willing to put in open-minded effort on your own.--Andy Schlafly 23:58, 6 April 2010 (EDT)

Yill, you raise excellent points, most of which have not been raised before. We should sharpen those points, here on this page, and then address them on the actual article page. This will take a fair amount of discussion. I could start by bringing up the discussion of point 7, inaccuracy of relativity at the quantum mechanical scale. One question that was raised was "Is there a discontinuity at that [microscopic boundary] distance? Such an approach is absurd.". No. The way quantum mechanics and classical theories interact at the (microscopic) scales where this happens is well known. It is, of course, generally known as the Bohr correspondence principle, described in any textbook on quantum mechanics, and known in more detail as Ehrenfest's theorem, described in more advanced textbooks. (Very briefly, the quantum mechanical realm eases into the classical realm according to the Ehrenfest theorem.) We should make some citations to those, and put in a careful explanation that, under QM, **all** classical theories are incorrect, and QM is the correct theory for everything, from atoms to planets. Classical theories are just extremely good approximations outside of the quantum-mechanical realm. And, of course, we do not know how that quantum-mechanical realm operated immediately after the big bang (that's what inflation theory is about), but that doesn't affect what we *do* know about general relativity in the macroscopic realm.

The item about point 10 is excellent. Gravitons arose *after GR*, from attempts to unify the theories. They have nothing to do with the macroscopic aspects of GR, which is what GR is actually all about.

KrisJ 10:04, 31 March 2010 (EDT)

- Teach that relativity is incorrect, if you concede the point. There are relativists who claim their theory is the most precisely verified theory of all.

- Those relativists claim that with respect to the macroscopic realm, as KrisJ referred to above. We are discussing how it breaks down at the microscopic level, when QM starts to play a role. Yill 16:49, 4 April 2010 (EDT)

- Gravitons are based on GR, and they are non-existent. Enough said.--Andy Schlafly 11:37, 31 March 2010 (EDT)

- No, for gravitons to be a counterexample to GR, they must be predicted by it. But they are not, just as photons are not predicted by Maxwellian electrodynamics. They are the "quantum" of the gravitational field, as photons are for the electromagnetic field, and are quantum
*by definition*. GR is*not*a quantum theory; it manifestly does not predict them. Yill 16:49, 4 April 2010 (EDT)

- No, for gravitons to be a counterexample to GR, they must be predicted by it. But they are not, just as photons are not predicted by Maxwellian electrodynamics. They are the "quantum" of the gravitational field, as photons are for the electromagnetic field, and are quantum

- KrisJ, I appreciate your assistance with this project. I absolutely agree with your suggestions about 7 and 10, and hopefully we can find an editor with the ability to edit protected pages to help us implement them. If you know of any that could help us, you should ask if they would be willing. Yill 16:49, 4 April 2010 (EDT)

I guess I was wrong about not being able to edit this article. I'm going to delete #10, as per above, and make some formatting changes. I may also make some other clarifying edits. Yill 17:45, 5 April 2010 (EDT)

I also deleted the references to relativity being useful, since those have nothing to do with its epistemological validity. Yill 17:52, 5 April 2010 (EDT)

## Curvature of Space

Re this edit: I don't disagree, but the example is a bad one. Based on local observations, one would assume that the Earth itself is flat, but it clearly isn't. My own point of view is that since the Universe can never be proved to be one thing or another, it is part of God's own ineffable being - it is almost folly to inquire further. RobertE 18:24, 30 March 2010 (EDT)

- No, one would not assume the Earth is flat based on local observations, as a ship can be observed to "rise" over the horizon. I don't agree with the "nature is God" view either.--Andy Schlafly 11:34, 31 March 2010 (EDT)
- Funny coincidence(?) that a defender of relativity invokes pantheism, since it was Einstein's (and Spinoza's) "god." DouglasA 13:50, 31 March 2010 (EDT)

- I actually think the edit has merit, as long as the word "initial" is inserted before curvature, since the problem is that any initial curvature should be vastly amplified over time as the universe undergoes its usual expansion. And it is in fact the global curvature that is the issue here;
*any*manifold we use to model the universe is by definition locally flat (since this is a fundamental property of manifolds). The ship and horizon observation is not a local observation, since it is fundamentally predicated on the global curvature of the Earth. "Local" means that it can be done at arbitrarily small distance scales, which that observation cannot. Yill 17:06, 4 April 2010 (EDT)

## Reversion explained

Reversion was necessary for two reasons: first, to restore material that was improperly censored, and second, to revert an imprecise label put on one of the counterexamples.--Andy Schlafly 17:53, 5 April 2010 (EDT)

- I don't want to get into an edit war here, so I won't undo your reversion for now. But I fail to understand your reasoning, so perhaps you could clarify a bit instead of making the one sentence assertions that have made up your discourse so far. There is no censorship here, merely deletion of objectively incorrect statements. Perhaps you could actually bother to respond to my points above, rather than just reverting my edits without justification. In the meantime, I will replace the periods I added at the end of several of the counterexamples for formatting consistency; hopefully you don't consider
*that*to be "censorship" as well. Yill 20:48, 5 April 2010 (EDT)

- You deleted valid information. Gravitons are predicted by an attempt to reconcile GR with quantum mechanics. Without GR gravitons would not be expected; with GR people do expect to find them. The wholesale deletion of reference to this is unwarranted, and simply conceals a real flaw in GR.

- First of all, I want to thank you for actually explaining your claims. Now we can actually have the real discussion KrisJ suggested above. You are perfectly correct in stating the gravitons are predicted by an attempt to reconcile GR with QM; that is precisely the point I was trying to make. But by your logic we could rightly conclude that the flaw is with QM rather than GR--without QM gravitons would not be expected either. On what basis do you claim that the non-observance of gravitons is a counterexample to GR rather than a counterexample to QM? (Also, I should note that just because gravitons have not yet been observed, that doesn't mean they won't be. For example, the non-observation of the Z boson did not constitute a counterexample to the electroweak theory between 1979 and 1983.) Yill 22:55, 5 April 2010 (EDT)

- Gravitons were historically proposed in trying to reconcile GR with QM. Other theories of gravity may not require gravitons at all. Does string theory? Gravitons are thereby attributable to GR, not to the more developed and better verified QM.
*Simply look at the name "gravitons" itself*.--Andy Schlafly 00:23, 6 April 2010 (EDT)

- Gravitons were historically proposed in trying to reconcile GR with QM. Other theories of gravity may not require gravitons at all. Does string theory? Gravitons are thereby attributable to GR, not to the more developed and better verified QM.

- Actually, any quantum theory of gravity, whether it reduces to GR at large scales or not, requires gravitons
*by definition*. Do you even understand what a graviton*is*?*The quantum of a gravitational field.*Just as any quantum theory of electromagnetism*must*include the photon in its particle spectrum, any quantum theory of gravity*must*include the graviton in its particle spectrum. And yes, string theory requires them; the entire reason string theory started being developed as a theory of everything is that gravitons (i.e. massless spin-2 bosons) naturally appear as part of its particle spectrum! Yill 14:57, 6 April 2010 (EDT)

- Actually, any quantum theory of gravity, whether it reduces to GR at large scales or not, requires gravitons

- Yill, do you know what action-at-a-distance is? It doesn't require the fictional gravitons. Newtonian mechanics doesn't require such imaginary particles.--Andy Schlafly 00:00, 7 April 2010 (EDT)

- Do you know what
*quantum*means? Please acknowledge that you do, and that you know Newtonian mechanics is not a quantum theory, and therefore that*your response does not address my concern.*Yill 00:19, 9 April 2010 (EDT)

- Do you know what

- The "flatness problem" refers primarily to curvature expected from inflation, not GR itself. It is misleading to call the counterexample the "flatness problem," and then pretend it has a solution. The counterexample described is not resolved.--Andy Schlafly 21:12, 5 April 2010 (EDT)

- The flatness problem refers to the fact that, in an
*inflation-free*universe, the FRW metric with matter and radiation equation-of-state parameters predicts that any initial nonzero curvature will increase vastly in magnitude, leaving a highly curved universe at present. Inflation is proposed as a*solution*to the flatness problem; it is not the cause of it. The process of inflation drastically flattens any initial curvature in the universe so dramatically that even after the curvature increase undergone under normal evolution, the universe still appears nearly perfectly flat. Yill 22:55, 5 April 2010 (EDT) - Wait, I just realized that I think we may be talking past one another here. I interpreted the counterexample listed on the page to be the flatness problem, but based on your response I guess that it is not. (Obviously the flatness problem is not a counterexample to GR itself, just to the use of the FRW metric for modeling the universe.) This counterexample seems to be more fundamental, namely the claim that space is nowhere curved, as GR says it must be by matter and energy. Is that correct? Yill 23:25, 5 April 2010 (EDT)

- The flatness problem refers to the fact that, in an

- A
*type*of inflation is proposed to try to explain the unexpected flatness. But there's no way around the basic problem: GR says that space is curved by matter, and an overall flatness is impossible under such a model. Yet an overall flatness is what is observed.--Andy Schlafly 00:23, 6 April 2010 (EDT)

- A

- I still don't understand what you're saying. The
*overall*visible universe*is*flat, at scales large enough that it can accurately be modeled as homogeneous and isotropic. (These scales are beyond the sizes of galactic clusters.) But on much smaller scales, where these assumptions obviously break down, matter does indeed curve spacetime; the phenomenon of gravitational lensing is precisely such an example. If you are at all confused by these different notions, I would recommend taking a look at a modern textbook on the subject; Barbara Ryden's*Introduction to Cosmology*is a good place to start. Yill 14:57, 6 April 2010 (EDT)

- I still don't understand what you're saying. The

- Dark matter supposedly permeates the universe, and there's no way it would be flat if GR were true.--Andy Schlafly 00:48, 7 April 2010 (EDT)

- Okay, now
*that*is a total non sequitur. Again, instead of making blanket assertions, perhaps you should learn why, given that they believe dark matter permeates the universe*and*that it is flat on large scales, cosmologists still think GR works. Let me enlighten you. If the universe were evenly filled with a uniformly dense substance, the curvature would be flat. Yet there were would be matter in it! And that's it. On large enough scales, that's how the universe appears. Hence there is no contradiction. Yill 00:19, 9 April 2010 (EDT)

- Okay, now

## Proposed page move

Can someone rename the article so the R is lowercase in the title? Thanks, GregoryZ 22:21, 29 July 2010 (EDT)

- Why? The term refers to a specific theory, and the many counterexamples to it.--Andy Schlafly 22:31, 29 July 2010 (EDT)

- My simple rationale is "relativity" is not a proper noun. Wikipedia uses the lowecase and so does Wester's, so why not here? --GregoryZ 22:36, 29 July 2010 (EDT)

- It's not a traditional proper noun, you're right, but it does satisfy all the conditions underlying why proper nouns are capitalized. It is a unique term-of-art, having a specific meaning other than the general meaning of the word. As used in physics, "Relativity" is different from the generic "relativity".--Andy Schlafly 23:01, 29 July 2010 (EDT)

- However, it is my belief, "relativity" in this case should not be treated differently. Look at the Wikipedia article, it uses "relativity" in that sense. Also, the CP article on the subject uses the lowercase as well, so I still see no point in capitalizing it here. --GregoryZ 23:07, 29 July 2010 (EDT)

- The word "relativity" dates from the early 1800s. That's not what is being discussed here. If preceded with "theory of" then there is no need to capitalize; if stand-alone, however, it does add clarification to capitalize as is done for other specific concepts that differ from the generic names.--Andy Schlafly 23:52, 29 July 2010 (EDT)

## Curl of the gravitational field

Sorry to get over-technical, but the fundamental law of "fictitious forces" (including gravity) is that the force field (divided by the mass of the test object) is

<math>G^i = - \Gamma^i_{00}</math>

Its curl is

<math>(\nabla \times G)^i = \mathcal{E}^{ijk} g_{km} G^m_{;j}</math> where the semicolon indicates the covariant gradient.

When you work this out, it involves derivatives of the <math>\Gamma\,</math> quantities. In general relativity, the results are zero by symmetries of Riemann's tensor.

Simeon 21:33, 30 July 2010 (EDT)

- Perhaps so, but the "twin paradox" in Relativity states that the age of each twin is dependent on his path of travel. For a conservative field, all physical parameters are path independent.--Andy Schlafly 23:07, 30 July 2010 (EDT)

- Simeon, your mathematical work is rigorous and correct. However, the twin paradox example is interesting to study here. I am aware that the twin paradox is solved by the non-inertial turn-around of the ship when it is going back home. However, in this solution, it is still noted that there is an age difference between the twins. Wikipedia affirms this and so do other sites. Such an age difference in twins shows that there is some sort of path dependence. I understand that traveling at near-c speeds in space is not the same thing as moving from point A to B in a gravitational field, but the concept does seem to be a bit similar. Could you maybe explain this for us a bit? Thanks. PhyllisS 00:52, 31 July 2010 (EDT)

OK, I think I understand. I assumed that the "conservative field" / "curl is zero" stuff referred to the gravitational force field. If it refers to the passage of time, that's different. It isn't true that "all physical parameters are path independent". An extremely important one that isn't path independent is the arc length of the path or arc. You can draw a short straight line from A to B, or a long loopy line that starts at A, wanders around, and eventually gets to B. Why is this relevant to the twin paradox? Because, in relativity, an observer's own elapsed time ("local time") is really just the arc length of his "world line" in Minkowski space. Minkowski was an extremely smart guy, by the way. The twin that stays home takes a direct route from point A (their birth) to point B (the moment they compare ages and see that one has gray hair and wrinkled skin.) The other twin takes a very roundabout route, getting in a rocket and going to Alpha Centauri and back. Their path lengths are their local times, which are different. (Why is the length of the roundabout path actually shorter, so that that twin ages less? Because, in Minkowski space, using the "timelike convention" that all the best people use :-), motion in space subtracts from the elapsed time. That's just the way it works.)

Now I assume that there is no dispute about the facts of relativistic time dilation. In addition to being predicted exactly by the Lorentz transform, it has been observed in practice in cosmic ray muon decays, as well as countless observations in particle accelerators. The "twin paradox" is just an extreme consequence of this. It has of course never been observed in that form, just as we don't know whether Schrodinger's cat is alive.

The "twin paradox" is a consequence of special relativity, not general, and hence does not relate to gravity. I hate to be the umpteenth person to tell you that general relativity is too hard to explain, but it's kind of true. I barely understand the most rudimentary basics. (When Eddington made his comment about only 3 people in the world who understand gen. rel., I wasn't the third! :-) But I can say that you don't need to worry about general relativity to understand the "twin paradox". You can finesse the Minkowski-space curvature of the path during the turnaround at Alpha Centauri, and just say that the twin went there and came back. So was something physically different, that the twins could observe? You bet. The "younger" twin will remember having experienced 6 months of horrendous acceleration in the ionic-drive rocket, followed by a year of horrendous turnaround, and another 6 months of horrendous deceleration at the end. She will have soft, smooth skin, but at a great cost. :-)

Sorry to be so long-winded. In quick summary, the thing that's different about the paths is their length, and that is exactly the local elapsed time. Simeon 00:07, 1 August 2010 (EDT)

- Simeon, time dilation occurs under the Theory of General Relativity also, so your analysis above is not persuasive in resolving this example of a non-conservative effect. Moreover, your repeated claims about how supposedly only geniuses can understand this are getting tiresome. That approach is a recipe for mistaken reliance on unjustified authority.

- If you don't feel this is understandable, then simply say so and stop there; please do not imply that people should just accept what someone of undisclosed political views claims.--Andy Schlafly 10:58, 1 August 2010 (EDT)

I give up.

- The only scientists I mentioned were Minkowski and Eddington, and the latter just as a joke. I never said anything about their, or anyone else's, politics.
- Time dilation does indeed occur under both general and special relativity. The point I was trying to make is that general relativity is simply not needed to understand the twin paradox. It only takes special relativity, which is much better understood. I'm sorry to hear that, by not analyzing the twin paradox in terms of general relativity, my persuasiveness suffered.
- I apologize if I "talked down" to you and Phyllis with my comments about GR being too complicated. I assume that both of you have heard, many times, that GR is exceedingly complicated. I was simply trying to soften the blow by pointing out that you
*don't need*GR. And cracking that joke about how Eddington could not have been referring to me. - In fact, I know a fair amount about GR. I
*could*analyze the twin paradox in terms of the gravitation of Earth and Alpha Centauri. But there is simply no need to. - This "non-conservative effect" business simply makes no sense. If the integration of a vector field along different paths gets different final results, then that field is non-conservative. You seem to be saying that the
*passage of time*is some kind of vector field, and that the final results of "integrations" (the two different values of local time at the end of the experiment) are supposed to be the same, and that the difference shows that this "vector field" is not conservative, and that that is a counterexample to relativity. The passage of time is not a vector field. The different values of time, as seen by different observers, is not a*counterexample*to relativity, it is*one of the principal effects*of relativity. It's really what the word "relativity" means when discussing the scientific Theory of Relativity. - If you really think that the non-globality and non-absoluteness of time is a counterexample to relativity, then so be it.

Simeon 23:13, 1 August 2010 (EDT)

- Simeon, if you "give up," then that is your own choice. You have not disproved the counterexample. Instead, you first described the twin paradox as being only about special relativity, and when I pointed out that it exists under general relativity too, you then agree yet do not fully address the substantive issue presented by the paradox. For example, the amount of acceleration undertaken by the twin in his journey will affect his age independent of his time spent away. His subsequent age is
*not*path independent even in time-space coordinates.

- It's easy to search for "general relativity" and "conservative field" on the internet and see how little has been written about this. That is telling in itself. I'm happy to continue to discuss this here with you or anyone else.--Andy Schlafly 23:56, 1 August 2010 (EDT)

- Could you clarify what the ages (and path dependence thereof) in the twin paradox have to do with conservative fields? --KyleT 14:04, 2 August 2010 (EDT)

- Age is scalar physical attribute. It should not be path dependent in a conservative field.--Andy Schlafly 14:31, 2 August 2010 (EDT)

- Yes, but which conservative field in particular are you talking about here (that implies age is not path dependent)? --KyleT 14:37, 2 August 2010 (EDT)

- Gravity.--Andy Schlafly 14:53, 2 August 2010 (EDT)

- Well, in Newtonian mechanics, the gravitational field is indeed conservative -- it's the negative gradient of the gravitational potential! But what this means is that gravitational potential energy is path-independent: it doesn't say anything about path-independence any other quantities, and in particular it's not the reason for the path-independence of age. --KyleT 15:00, 2 August 2010 (EDT)

- You take a narrow view of the significance of a "conservative field." Independent physical attributes should remain path-independent as well for the field to be conservative. In Newtonian mechanics and most other physical force fields, they do.--Andy Schlafly 15:41, 2 August 2010 (EDT)

- By a conservative field, I mean a vector field on space for which there exists a scalar function V with the gradient of V given by that vector field. This doesn't imply the path-independence of any physical quantities other than V itself. If you this view as too narrow, can you tell me what you take to be the definition of a conservative field? --KyleT 15:57, 2 August 2010 (EDT)

- Your definition is too narrow when discussing the theory of relativity, which describes the framework in which the force operates. To be meaningful, the definition must be broader. It must ensure the path independence of the scalar, as well as other scalars independent of that scalar.--Andy Schlafly 18:12, 2 August 2010 (EDT)

- Can you tell me what the correct definition is, then? I have pretty good background in this stuff, no need to dumb it down, just be precise. Certainly no field at all is going to conserve every scalar function, so I'd like to know which ones you want. --KyleT 18:20, 2 August 2010 (EDT)

- Kyle, I have an open mind about this, and don't see a precise definition anywhere that would be meaningful with respect to the theory of relativity. It's striking how relativists avoid this issue, and even stop discussing it when it is brought up.

- I can propose a definition that you may be able to improve. How about: a conservative theory of motion is one whereby scalar values of a particle are independent of its path of motion.--Andy Schlafly 18:36, 2 August 2010 (EDT)

That's an interesting proposal, and I too have an open mind about this. Can you give an example of such a *conservative theory of motion*? One such would greatly help in devising the correct definition. --KyleT 19:29, 2 August 2010 (EDT)

- Newtonian mechanics would be an obvious example. By the way, how do you explain the general lack of discussion and papers about whether the theory of relativity is conservative, including the abrupt departure of User:Simeon from this discussion?--Andy Schlafly 21:58, 2 August 2010 (EDT)

- Some scalar values in Newtonian mechanics are conserved because there exist associated conservative fields (or more generally symmetries of the Lagrangian). What is an example of a scalar value in the Newtonian mechanics that is not of this type, which makes this a conservative theory of motion while relativity is not?
- I don't know why relativity's defenders won't confront this. Maybe that could be the topic of the debate page -- I'm interested using this discussion to sharpen counterexample 21. --KyleT 23:55, 2 August 2010 (EDT)

- This is a really interesting discussion. I think I made a gross mistake in my first post. The theory of relativity urges us to think of the three space coordinates (x, y, and z) and the time coordinate (t) as four coordinates of space-time - that is, that space and time are pretty much the same. I extrapolated from this that since there can be a (conservative) gravitational field in space coordinates, there can also be some sort of conservative field depending on the time coordinate. I then extrapolated this notion to special relativity, and the twin paradox; I postulated that maybe time dilation effects were the work of a non-conservative field that was dependent on the t-coordinate. Now I see that this was all somewhat foolish. However, I wanted to ask you all: can you have a conservative or non-conservative field with respect to time? If not, I think time should
**not**be considered as almost the same thing as x, y, z space. I feel that the ability for a dimension to have a field (conservative or not) is integral to its being considered a space-like dimension.

- This is a really interesting discussion. I think I made a gross mistake in my first post. The theory of relativity urges us to think of the three space coordinates (x, y, and z) and the time coordinate (t) as four coordinates of space-time - that is, that space and time are pretty much the same. I extrapolated from this that since there can be a (conservative) gravitational field in space coordinates, there can also be some sort of conservative field depending on the time coordinate. I then extrapolated this notion to special relativity, and the twin paradox; I postulated that maybe time dilation effects were the work of a non-conservative field that was dependent on the t-coordinate. Now I see that this was all somewhat foolish. However, I wanted to ask you all: can you have a conservative or non-conservative field with respect to time? If not, I think time should

- Aschlafly, the fact that the twin paradox exists in general relativity is
**irrelevant**. Yes, sure, the twin paradox occurs within space where general relativity is working, but there are no effects acting on the twins that influences the twin paradox in any way. Likewise, User:Simeon 's departure is also**irrelevant**.

- Aschlafly, the fact that the twin paradox exists in general relativity is

- What about black holes, though? Surely their gravitational fields aren't conservative, since once an object passes the event horizon, you can't retrieve it. PhyllisS 01:24, 3 August 2010 (EDT)

Phyllis:

You seem to be very curious about this topic. I'm going to try to give an intuitive, but nevertheless scientifically correct, explanation of what is going on with relativity, the "twin paradox", and vector fields, potential functions, and path integrals. This explanation will probably seem long and tedious, for which I apologize in advance. I also apologize if it seems that I am being too "folksy", or talking down to you. Please bear with me, and please pay close attention.

We have a parking lot, and two twins, who are fitness enthusiasts and always wear pedometers wherever they go. There are two spots, "X" and "Y", painted on the parking lot. Both people stand on spot "X", set their pedometers to zero, and start walking. Twin A simply walks directly to spot Y. Twin B, being more into fitness, walks all over the place, eventually arriving at B.

Now there are quite a number of things we can say. First, the temperatures vary all over the place. They are a *scalar field*. That means that they are associated with *location on the parking lot*, not with any particular observer. They are objective measurements that everyone agrees on, because they are aspects of space itself. Our fitness enthusiasts are also amateur meteorologists, and carry thermometers around with them.

- Twin A: "When I was at the green Toyota, I noticed that the temperature was 67 degrees Fahrenheit."
- Twin B: "By coincidence, I also wandered past the green Toyota, and got the same reading."

By the way, since temperature is a scalar field, it has a gradient, which is a vector field. That field is conservative, according to the theorem of mathematical physics that says that curl grad Φ = 0 always. This gradient is a *vector field*. Like the scalar of temperature, it is a property of the *space (parking lot) itself*. If the twins had been measuring this gradient (perhaps they carry around fancy "differential thermometers"), they would have gotten the same vector at the green Toyota.

There is also a theorem of mathematical physics, sort of the opposite of the theorem above, that says that, if a vector field V has a curl of zero:

- You can make a scalar field <math>\Phi\,</math> (a property of the space itself, not tied to any particular observer) that it is the gradient of. That scalar field is called the "potential" for the (conservative) vector field. (By the way, this is very closely related to "exact differential equations" that you wrote about! Do you see the connection?)
- If you integrate that vector field along any path between two points A and B (that is, you calculate

- <math>\int_A^B \vec{V} \cdot dl</math>

for that path, where "dl" is the "line element" along the path), you will get <math>\Phi(B)-\Phi(A)\,</math>.

- Since <math>\Phi(B)-\Phi(A)\,</math> is a property of the scalar field itself (and the points A and B), it follows that that path integral is the same for all paths. And if the path ends on the same point it started on, the integral is zero.

- Twin A: "I was measuring the gradient of the temperature as I walked, and calculating its path integral as I went. I got an answer of 4 degrees."
- Twin B: "I was doing the same. My integral was much harder to calculate, because I was going all over the place. But I also got 4 degrees. Hey, wait a minute! The temperature at the start point was 68 degrees, and at the end point it was 72 degrees. That explains it."

Now someone at the edge of the parking lot was running a Van deGraff generator, so there were electric fields all over the place. The twins are also physics students, and carry electroscopes wherever they go. They measured the electric field, and calculated its path integrals. The electric field is conservative (in the absence of varying magnetic fields), so they got the same integral. That integral was 600 volts (it's only static electricity, so it isn't dangerous). Since the electric field is conservative, there is, by the previous theorem, a potential function. That function was 600 volts higher at point B than at point A.

Now here's the kicker:

- Twin A: "I walked directly from A to B. My pedometer says 150 feet."
- Twin B: "I took a long route all over the place. I walked half a mile."

The pedometer readings *are not a scalar field*. They are not a property of the space itself. They are properties of the observers. Even though they, in some sense, measure an aspect of the parking lot (how many molecules of asphalt one passes), they are artifacts of the twins' actions.

The twins could have been integrating their motion vectors; that's sort of what pedometers do. But those vectors are not a vector field on the space itself. It makes no sense to ask whether that "vector field" is conservative, because it isn't a vector field. A vector (or scalar, or tensor) field has to be a property *of the space itself*. These "pedometer vectors" are just things that the twins make up as they walk.

Now for the "twin paradox". The parking lot is replaced by "Minkowski space", also called "4-dimensional space-time". "Points" in this space are now "events", complete with a time. Events A and B are now the act of the twins saying goodbye as one of them got into the rocket, and the act of them re-uniting after B returns. Twin A took a direct route (called her "world line") from A to B. She used a coordinate system in which the spatial coordinates of A and B were the same (Cape Canaveral, latitude yada yada, etc.) and the time coordinate differed by 30 years (2010 to 2040.) B went to Alpha Centauri and back. When she returned, they were both using the same coordinate system (location is Cape Canaveral, latitude yada yada, time is 2040.) But she looks at her watch, and only 5 years have elapsed! What the watch shows is *not a scalar field on spacetime*. It was *not the path integral of a vector field on spacetime*. What she integrated was the ticking of her watch, nothing more.

The path that A took is called a geodesic. It is the Minkowski-space equivalent of a "straight line". But, because of the peculiarities of relativity, it shows the *longest* elapsed time (30 years) of all paths, rather than the shortest. By going to Alpha Centauri, twin B took a shorter path, in terms of the way path length is measured in Minkowski space.

Very interesting fact: The path length in Minkowski space, that is, the sum of the tiny distances as measured by the Lorentz/Minkowski metric, *is the same as the local time*. That is (assuming you are using the "spacelike convention"), everyone's wristwatch measures path length along their own world line. The twins simply followed paths of different lengths. That's all there is to the "twin paradox". (That is, that's all there is to it, if you analyze it correctly, as I have described above. Most introductory treatments of relativity don't do it this way. They just throw the Lorentz transform at you.)

Now there are a few points about the "twin paradox" that people find confusing.

First, aren't the laws of physics supposed to be the same for everyone? What made twin B's watch run slower? Well, she *knew* she was traveling at high speed. She brought an accelerometer with her in the rocket. Just as twin B in the parking lot knew she was walking all over the place, turning around and such, twin B in space knew that her world-line was turning, and therefore wasn't a straight line (geodesic). How did she know? It takes force to make you deviate from a geodesic (this is really pretty much the same as Newton's laws of motion), and she felt the force.

Second, how can we analyze the curvature of B's world line? Here's where general relativity has to come in. As soon as world lines start to curve, you have to measure their curvature, that is "geodesic curvature". You get into complicated issues of curved coordinate systems (you're in one now; it's what you perceive as "gravity"!), and curved spacetime, and so on. And you get into the <math>\Gamma\,</math> symbols, which measure the deviation from a geodesic, and hence the "fictitious forces" that you feel. This is why general relativity is related to the "twin paradox", in that the space ship followed a curved trajectory and experienced acceleration. But, to analyze the plain facts of the "twin paradox", all you really need to know is that twin B followed a crooked line. Place your ruler on a diagonal on the graph of Minkowski space, draw the line out to Alpha Centauri. Turn the ruler, draw the returning line. Ignore the impossibly sharp corners. Use special relativity to analyze the Lorentz transform for each section of B's world line.

Oh, and to try to answer some of your specific questions, the gravitational field, under either Newtonian or relativistic mechanics, is a conservative field. Its curl is zero. If it weren't, conservation of energy would be violated, and we could make a perpetual motion machine by having a planet run around in circles picking up energy. The "curl=0" aspect of gravity under general relativity is more complicated, because true vector fields have to be on Minkowski space, but it still conserves energy. When Mercury orbits the Sun, its perihelion precesses because of relativistic effects, but its energy is conserved.

I'll try to think some more about your black hole question and get back to you.

Simeon 00:00, 4 August 2010 (EDT)

- Simeon, thanks so much for your long explanation! All of that made sense to me, and cleared up the issue of field as property of space vs. path specific to person vs. curved spacetime. Also, yes, I noticed the similarity between exact differential equations and deriving potential functions from vector fields - the former I studied in Differential Equations and the latter I studied in Multivariable Calculus.

- So, if I understand you properly, gravity both (a) curves spacetime and (b) creates a conservative field. (I derived (a) from your point: "You get into complicated issues of curved coordinate systems (you're in one now; it's what you perceive as "gravity"!)", and (b) from your point: "the gravitational field, under either Newtonian or relativistic mechanics, is a conservative field. Its curl is zero.") However, you're saying that if you are in an accelerating reference frame (such as a quickly-spinning merry-go-round) only (a) occurs; there is no field. Is this correct? If so, why is there this discrepancy between the two? PhyllisS 22:54, 4 August 2010 (EDT)

Yes, gravity curves spacetime. It looks as though you're ready to go to the next level. It has to do with curved *coordinate systems* vs. curved *spacetime*. This is why GR is so complicated. But here goes.

First, we have to recognize that, at some level, we could say that energy is *not* conserved. Spacecraft use "slingshot maneuvers" around one planet to gain extra energy on their way to another planet. I assume you've heard of this. Cassini used three such maneuvers, twice around Venus and once around Earth. So what was going on? If you look at a coordinate system centered on Venus, you would see Cassini come in and go out again, with complete conservation of energy. But, in a coordinate system fixed around the solar system, Venus was moving, so Cassini came in at low speed and went out at high speed. We "stole" some energy out of Venus's orbit. So, to be *really* correct, we have to say that gravity is a conservative field in the absence of **moving** gravitating bodies.

But, in the larger sense, energy is conserved. Always. Newtonian or relativistic. (But in relativity, the mass figures into the equation. Let's not worry about that just now.)

Now we get to the really cool stuff. Your comment above suggests that you are ready for it.

You will feel a "fictitious force" whenever you are in a "curved spacetime coordinate system". A curved coordinate system would include things like polar coordinates on the plane, or spherical coordinates in 3 dimensions. But this is 4-dimensional spacetime. So I'd like you to take my word for this. On a rotating merry-go-round, your 4-dimensional spacetime coordinate system is curved. (This doesn't require relativity, special or general, to formulate this.) The 3-dimensional slice of it is flat, but, when you bring in time, and the spatial coordinates are accelerating, the overall coordinate system is curved. This creates fictitious forces. In the case of the merry-go-round, the forces are the centrifugal force and the Coriolis force. There are also the fictitious "acceleration G forces" in a rocket. But **the spacetime itself isn't curved**. It's like polar coordinates on the plane. Yes, the coordinate system is curved, but the plane isn't. There are Cartesian coordinates on the plane also. Similarly, an observer on the ground is in a flat coordinate system, and doesn't see any fictitious forces. He just sees the mechanism of the ride pushing you inward as you go around. Your recoil against that acceleration is what you perceive as the centrifugal force.

So the moral of the story would seem to be: you can choose a flat coordinate system that exposes the fictitious force for what it is. It's a perception from the curved spacetime coordinate system that the observer is operating in.

But how about gravity? In the case of gravity, **spacetime itself is curved**. If spacetime (properly called a "manifold") is curved, every coordinate system is curved, and the fictitious force which is gravity is inescapable.

So your statement "gravity curves spacetime", is exactly correct, but you have to distinguish "curving spacetime" and "curving some particular coordinate system". The actual curvature involves things called "Riemann's tensor" and "Ricci's tensor".

The number of people that understand this is way more than the 3 that Arthur Eddington claimed (it was more than 3 but probably less than 10 at the time), but it's still a pretty complicated subject.

Simeon 00:03, 5 August 2010 (EDT)

- Simeon: Ah, that does make a lot of sense now. Thanks so much for explaining this to me. Hmm, that really is an elegant concept -- wow. Cool! PhyllisS 13:50, 5 August 2010 (EDT)