geordief 108 Posted November 19, 2016 Share Posted November 19, 2016 (edited) This news just in from a Feynman lecture http://www.feynmanlectures.caltech.edu/II_42.html "Suppose we take the earth as an example and forget that the density varies from point to point—so we won’t have to do any integrals. Suppose we were to measure the surface of the earth very carefully, and then dig a hole to the center and measure the radius. From the surface area we could calculate the predicted radius we would get from setting the area equal to 4πr24πr2. When we compared the predicted radius with the actual radius, we would find that the actual radius exceeded the predicted radius by the amount given in Eq. (42.3). The constant G/3c2G/3c2 is about 2.5×10−292.5×10−29 cm per gram, so for each gram of material the measured radius is off by 2.5×10−292.5×10−29 cm. Putting in the mass of the earth, which is about 6×10276×1027 grams, it turns out that the earth has 1.51.5 millimeters more radius than it should have for its surface area.4 Doing the same calculation for the sun, you find that the sun’s radius is one-half a kilometer too long." Can things really be as simple as this? Is this the famous curvature that launched a thousand "is spacetime "real" " threads? Sitting there below our feet all along... All our "spheres" are squashed** by the tiniest fraction ,but measurably so as a result of the mass of the Earth? Our space is intrinsically curved? ** ie not Euclidean Edited November 19, 2016 by geordief 4 Link to post Share on other sites

imatfaal 2481 Posted November 19, 2016 Share Posted November 19, 2016 A beautiful passage - nicely demonstrating the physics and incidentally the genius of Feynman as a teacher. Good find and thanks for posting Link to post Share on other sites

Mike-from-the-Bronx 1 Posted November 19, 2016 Share Posted November 19, 2016 I don't get it. We are all slightly shorter standing than when laying down. Gravity compresses us when we stand and to a lesser extent a meter stick too. Hence measuring the radius of Earth gives a slightly larger value than one would expect from measuring the circumference and applying a Euclidean formula. Even Newtonian physics tells us that. Is Feynman referring to something additional? Link to post Share on other sites

imatfaal 2481 Posted November 19, 2016 Share Posted November 19, 2016 I don't get it. We are all slightly shorter standing than when laying down. Gravity compresses us when we stand and to a lesser extent a meter stick too. Hence measuring the radius of Earth gives a slightly larger value than one would expect from measuring the circumference and applying a Euclidean formula. Even Newtonian physics tells us that. Is Feynman referring to something additional? It is nothing to do with compression. The surface area of a sphere is 4 pi r^{2 }where r is the radius. That is a Euclidean measurement - ie when there is no intrinsic curvature to the geometry of space; the mass of the earth causes an intrinsic positive curvature and the euclidean ratios no longer quite apply. The easiest way to think of an analogy is in an extrinsic curvature situation - a curved 2d surface in our 3d world; that is to say the surface of a sphere. Draw a large circle on the ground, go to the centre of the circle and measure the distance to the circumference by walking a tape measure from the centre to the line, then measure around the circumference. We would expect the ratio to be Circumference = 2 pi r; but it is not. In this example it is clear that it cannot be because the "radius" we have walked from the centre to the circumference is curved; we can easily see this as the curvature is extrinsic (ie it is a curved object embedded in our 3d space). In Feynman's example the background geometry itself is curved - there is an intrinsic curvature but it still screws up ratios and expectations which are made on the basis of flat euclidean geometry. The amount that the radius is excessive compared to flat geometry (if the sphere is constant density) is simple and directly proportional to the mass [latex] r_{excess} = \frac{G}{3c^2} \cdot M [/latex] 5 Link to post Share on other sites

koti 484 Posted November 20, 2016 Share Posted November 20, 2016 This news just in from a Feynman lecture http://www.feynmanlectures.caltech.edu/II_42.html "Suppose we take the earth as an example and forget that the density varies from point to pointso we wont have to do any integrals. Suppose we were to measure the surface of the earth very carefully, and then dig a hole to the center and measure the radius. From the surface area we could calculate the predicted radius we would get from setting the area equal to 4πr24πr2. When we compared the predicted radius with the actual radius, we would find that the actual radius exceeded the predicted radius by the amount given in Eq. (42.3). The constant G/3c2G/3c2 is about 2.5×10−292.5×10−29 cm per gram, so for each gram of material the measured radius is off by 2.5×10−292.5×10−29 cm. Putting in the mass of the earth, which is about 6×10276×1027 grams, it turns out that the earth has 1.51.5 millimeters more radius than it should have for its surface area.4 Doing the same calculation for the sun, you find that the suns radius is one-half a kilometer too long." Can things really be as simple as this? Is this the famous curvature that launched a thousand "is spacetime "real" " threads? Sitting there below our feet all along... This is a great, simple example of spacetime curvature, thanks for posting it. Feynman was a lover of simplifying concepts thats why he is so popular. I think we were both involved in that "is spacetime real" thread a couple of months ago where I had the same notion as you are having right now - if it curves, it has to be real. We went through Gravity probe B, frame dragging, etc. Ultimately I was rightly smacked by swansont and studiot and came to the conclusion that its not right to deal with spacetime curvature and spacetime itself in terms of "realness", instead I decided to treat spacetime in my mind as an entity which has no structure but still plays on the same team with mass, velocity and gravity. Im still struggling with this in my mind but got nothing better so far. I hope that there is an accurate way of describing space and spacetime in terms of it being "real" but in order to do so we need to know more about quantum gravity. Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 This is a great, simple example of spacetime curvature, thanks for posting it. Feynman was a lover of simplifying concepts thats why he is so popular. I think we were both involved in that "is spacetime real" thread a couple of months ago where I had the same notion as you are having right now - if it curves, it has to be real. We went through Gravity probe B, frame dragging, etc. Ultimately I was rightly smacked by swansont and studiot and came to the conclusion that its not right to deal with spacetime curvature and spacetime itself in terms of "realness", instead I decided to treat spacetime in my mind as an entity which has no structure but still plays on the same team with mass, velocity and gravity. Im still struggling with this in my mind but got nothing better so far. I hope that there is an accurate way of describing space and spacetime in terms of it being "real" but in order to do so we need to know more about quantum gravity. Interesting the degree of disparity between the expected radius (of the Earth) and what is measured. based on the surface area. I millimetre. That is seriously tiny but does it begin to give us a sense of the scale of this phenomenon ? I would also be interested in how these measurements were arrived at. It must be a long process and I can't see how you can easily measure the radius of the Earth and its surface area . How do they account for the different densities as you go through the Earth's interior? How do they know where the centre is? It reminds me a bit of that thread where they were measuring the curvature of the Lake Balaton http://www.scienceforums.net/topic/98386-laser-curvature-test-on-lake-balaton/?hl=%2Blake+%2Bcurvature That seems simple in comparison. Link to post Share on other sites

imatfaal 2481 Posted November 20, 2016 Share Posted November 20, 2016 I would also be interested in how these measurements were arrived at. It must be a long process and I can't see how you can easily measure the radius of the Earth and its surface area . How do they account for the different densities as you go through the Earth's interior? How do they know where the centre is? This is, unfortunately, entirely a thought experiment; one could not begin to do in reality. It relies on a uniform density sphere to do the calculation of what the radius would be in euclidean flat space etc. If you look at the calculation r_excess = G/c^2 M ie r_excess = 10^{-11} / 10^{17 }M = 10^{-28 }M The real imperfections of the earth would overwhelm such a tiny fraction - the difference is around one part in ten million million million Link to post Share on other sites

studiot 2205 Posted November 20, 2016 Share Posted November 20, 2016 Hi geordie, This is a very good question, however I am not convinced by the Feynman quote. Although, like Einstein and Newton he was a tremendous genius, he, like them, was not always right in hindsight. That is not to say he did not have a point, but, as ever, life is a bit more complicated. His 'experiment' was a thought experiment and I will return to it at the end of this post. Here is a real world example of curvature and its effects, that has real implications in the real world. If we set up a theodolite at each of three points, A, B and C on the surface of the Earth and measured (the sum of) the angles of the triangle ABC we would not get 180^{o}. We would get some larger number. This has practical significance in surveying. We do not need theoretically super accurate equipment this affects ordinary theodolites and surveying and is called 'spherical excess' Nor is it to do with the fact that the Earth is not a perfect sphere. The reason is due to the fact that the surface of the Earth is a curved manifold that cannot contain a plane triangle exactly. Both a plane and the Earth's surface are 2D manifolds 'embedded' in the same 3D space. The plane triangle, whose angles do add up to 180^{o}. 'cuts through' the underlying 3D space, in which the 'curvature' radii and centres lie. In order to calculate the correct values we have to use the more complicated equations relevant to the 2D manifold that is the surface of the Earth. The choice of sphere or more complicated figure is the realm of Geodesy. As we are 3D (or 4D in spacttime) beings we have direct access to all the dimensions concerned and can make measurements in them. So we get different measurements depending upon whether we measure in the Earth's surface manifold or the plane manifold. [sidenote] This situation is the difference between 'local' and 'global' for many phenomena in Physics and of great importance in computer graphics where the approximation to any curved surface by a plane is known as a 'Coon's patch' [/sidenote] Back to Feynman. As we have discussed before, the 'curvature' in 3D space or 4D spacetime is different in that we do not have access to another dimension where the curvature might take place. All we have are the observed equations between points in our 3 or 4 D space which are not linear or Euclidian or flat - the appropriate terms sicne we have ramped up one dimension from a plane. So Feynman was not 'wrong', but his simplification was not totally 'right' either. Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 This is, unfortunately, entirely a thought experiment; one could not begin to do in reality. It relies on a uniform density sphere to do the calculation of what the radius would be in euclidean flat space etc. If you look at the calculation r_excess = G/c^2 M ie r_excess = 10^{-11} / 10^{17 }M = 10^{-28 }M The real imperfections of the earth would overwhelm such a tiny fraction - the difference is around one part in ten million million million Ha ha . Do I get points for naivete ? I realize that Newtonian mechanics falls/has fallen at one or more hurdles but is it out of the question to run a computer simulation where we model the Earth -or any test body as a group of plastic massive bodies that settle into a ball shape under the action of Newtonian gravity? Is it possible to compare the results obtained by such a simulation with that obtained by another that ran a GR algorithm? It that as daunting a project as physical measurement? What about quantum computers ? Would they be up to the task? Link to post Share on other sites

imatfaal 2481 Posted November 20, 2016 Share Posted November 20, 2016 Ha ha . Do I get points for naivete ? I realize that Newtonian mechanics falls/has fallen at one or more hurdles but is it out of the question to run a computer simulation where we model the Earth -or any test body as a group of plastic massive bodies that settle into a ball shape under the action of Newtonian gravity? Is it possible to compare the results obtained by such a simulation with that obtained by another that ran a GR algorithm? It that as daunting a project as physical measurement? What about quantum computers ? Would they be up to the task? The settling of the ball into a sphere has little to do with the problem - the ball is a given, a uniform density globe. It does not matter how you form or simulate the ball. If you assume flat Euclidean spacetime then the usual formula will apply with perfect precision - the radius is a straight line, 4 pi times the radius squared is the surface area, and 4/3 pi times the radius cubed is the volume. In a GR setting the geometry no longer works as it did in Euclidean space. Our measurements of space are no longer simple - the shortest distances are spacetime geodesics on a curved background. It is a simple thought experiment to highlight that difference. Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 (edited) The settling of the ball into a sphere has little to do with the problem - the ball is a given, a uniform density globe. It does not matter how you form or simulate the ball. If you assume flat Euclidean spacetime then the usual formula will apply with perfect precision - the radius is a straight line, 4 pi times the radius squared is the surface area, and 4/3 pi times the radius cubed is the volume. In a GR setting the geometry no longer works as it did in Euclidean space. Our measurements of space are no longer simple - the shortest distances are spacetime geodesics on a curved background. It is a simple thought experiment to highlight that difference. In Feynman's thought experiment how is he calculating distances (ideally, if the Earth or equivalent body were amenable to precise radius measurement)? Would it be by a laser (neutrino?) beam ? Would these be 3D or 4D measurements ? Edited November 20, 2016 by geordief Link to post Share on other sites

imatfaal 2481 Posted November 20, 2016 Share Posted November 20, 2016 Hi geordie, This is a very good question, however I am not convinced by the Feynman quote. Although, like Einstein and Newton he was a tremendous genius, he, like them, was not always right in hindsight. That is not to say he did not have a point, but, as ever, life is a bit more complicated. His 'experiment' was a thought experiment and I will return to it at the end of this post. Here is a real world example of curvature and its effects, that has real implications in the real world. If we set up a theodolite at each of three points, A, B and C on the surface of the Earth and measured (the sum of) the angles of the triangle ABC we would not get 180^{o}. We would get some larger number. This has practical significance in surveying. We do not need theoretically super accurate equipment this affects ordinary theodolites and surveying and is called 'spherical excess' Nor is it to do with the fact that the Earth is not a perfect sphere. The reason is due to the fact that the surface of the Earth is a curved manifold that cannot contain a plane triangle exactly. Both a plane and the Earth's surface are 2D manifolds 'embedded' in the same 3D space. The plane triangle, whose angles do add up to 180^{o}. 'cuts through' the underlying 3D space, in which the 'curvature' radii and centres lie. In order to calculate the correct values we have to use the more complicated equations relevant to the 2D manifold that is the surface of the Earth. The choice of sphere or more complicated figure is the realm of Geodesy. As we are 3D (or 4D in spacttime) beings we have direct access to all the dimensions concerned and can make measurements in them. So we get different measurements depending upon whether we measure in the Earth's surface manifold or the plane manifold. [sidenote] This situation is the difference between 'local' and 'global' for many phenomena in Physics and of great importance in computer graphics where the approximation to any curved surface by a plane is known as a 'Coon's patch' [/sidenote] Back to Feynman. As we have discussed before, the 'curvature' in 3D space or 4D spacetime is different in that we do not have access to another dimension where the curvature might take place. All we have are the observed equations between points in our 3 or 4 D space which are not linear or Euclidian or flat - the appropriate terms sicne we have ramped up one dimension from a plane. So Feynman was not 'wrong', but his simplification was not totally 'right' either. Your example - like mine - was embedded curvature; Feynman's was an avenue to understanding the intrinsic curvature which is not so immediately apparent I am not sure where you say Feynman is not totally right - unless we are taking his idea outside the strictures of a thought-experiment. The ratio of the radius of a circle to its area is not pi in curved space, nor is the ratio of a sphere to its surface area 4 pi - this does not require the test object to be embedded in a higher dimension, it also applies to intrinsic curvature Link to post Share on other sites

studiot 2205 Posted November 20, 2016 Share Posted November 20, 2016 Your example - like mine - was embedded curvature; Feynman's was an avenue to understanding the intrinsic curvature that we cannot use theodolites etc to measure. I am not sure where you say Feynman is not totally right - unless we are taking his idea outside the strictures of a thought-experiment. The ratio of the radius of a circle to its area is not pi in curved space, nor is the ratio of a sphere to its surface area 4 pi - this does not require the test object to be embedded in a higher dimension, it also applies to intrinsic curvature With respect, this sounds like a rather rote response. I did not say we could use theodolites etc to measure the curvature, in fact I said the opposite. You can always regard a mathematical structure from different viewpoints, so you can always mathematically establish a dimension to set the radius in if you wish. Alternatively you can use the point to point properties in the working dimensions, as Feynman is proposing in his measurement of the Earth's radius. The plain fact is that space, in that direction, is non linear. Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 The settling of the ball into a sphere has little to do with the problem - the ball is a given, a uniform density globe. It does not matter how you form or simulate the ball. If you assume flat Euclidean spacetime then the usual formula will apply with perfect precision - the radius is a straight line, 4 pi times the radius squared is the surface area, and 4/3 pi times the radius cubed is the volume. In a GR setting the geometry no longer works as it did in Euclidean space. Our measurements of space are no longer simple - the shortest distances are spacetime geodesics on a curved background. It is a simple thought experiment to highlight that difference. Is it not possible to run a simulation of GR and NM (=Newtonian Mechanics?) side by side bottom the "bottom up" ? Each simulation would start with n identical particles and n would increase indefinitely. Would it serve any purpose to compare "results" at various stages or would the curvature be "baked into" GR from the start and NM would just produce a ball of increasing density towards to centre? Since we already know that NM is a less complete theory ,then would this be a waste of (quantum?) computer power? Link to post Share on other sites

imatfaal 2481 Posted November 20, 2016 Share Posted November 20, 2016 In Feynman's thought experiment how is he calculating distances (ideally, if the Earth or equivalent body were amenable to precise radius measurement)? Would it be by a laser (neutrino?) beam ? Would these be 3D or 4D measurements ? It is a thought experiment - you just measure. This is not a practical proposition. You are overthinking; what is a 3d measurement or 4d measurement anyway? With respect, this sounds like a rather rote response. Really ? That's a pretty poor way to discuss things. You said Feynman was not right - how? Your final word - which I have to presume is demonstrating your assertion is also exactly the point of Feynman's gedankan I have noticed that you responded whilst I was editing my previous post - not intentional on my side Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 It is a thought experiment - you just measure. This is not a practical proposition. You are overthinking; what is a 3d measurement or 4d measurement anyway? All measurements are 4D perforce except when the observer is equidistant from either end? Link to post Share on other sites

imatfaal 2481 Posted November 20, 2016 Share Posted November 20, 2016 Is it not possible to run a simulation of GR and NM (=Newtonian Mechanics?) side by side bottom the "bottom up" ? Each simulation would start with n identical particles and n would increase indefinitely. Would it serve any purpose to compare "results" at various stages or would the curvature be "baked into" GR from the start and NM would just produce a ball of increasing density towards to centre? Since we already know that NM is a less complete theory ,then would this be a waste of (quantum?) computer power? It is not density related - get that out of your head. Any impact of intrinsic curvature would be existent as soon as you started - but as it is in a ratio of 10^-23 to the masss then you need a huge mass before you can discern anything Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 (edited) It is not density related - get that out of your head. Any impact of intrinsic curvature would be existent as soon as you started - but as it is in a ratio of 10^-23 to the masss then you need a huge mass before you can discern anything No ,but in the NM, model that would have to be accounted for. I am just interested if the two final results could be somehow compared , The physical shape of the two balls ,I guess but the result would only be of interest if they were identical (or could be tweaked somehow to be so). I think I appreciate that the physical difference in the 2 results would be small but I don't know what order of smallness it would be. Edit: I am not sure that the connection between NM and GR is a mystery anyway Isn't NM supposed to be something of a "special case" of GR ?(I think I was told that before) http://www.scienceforums.net/topic/94702-why-is-newtonian-gravity-so-accurate/ Edited November 20, 2016 by geordief Link to post Share on other sites

Strange 4273 Posted November 20, 2016 Share Posted November 20, 2016 I am just interested if the two final results could be somehow compared , The physical shape of the two balls ,I guess but the result would only be of interest if they were identical (or could be tweaked somehow to be so). That is what Feynman is comparing: the "expected" value in Newtonian gravity (based entirely on Euclidean geometry) and the fractionally different value in GR's curved spacetime. Isn't NM supposed to be something of a "special case" of GR ?(I think I was told that before) It is an approximation to GR, in "normal" (low energy) cases. The Feynman example shows how good an approximation it is! Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 That is what Feynman is comparing: the "expected" value in Newtonian gravity (based entirely on Euclidean geometry) and the fractionally different value in GR's curved spacetime. But ,if you actually ran a simulation along NM (OK?) lines where the constituent bodies would start out identical but would be eventually shaped according to their nearness to the centre of overall mass could we not see an overall shape of the group arise that could be consistent with a non Euclidean geometry? The individual bodies would be squashed (maybe I am wrong on this point) by Newtonian gravity increasingly as they were closer to the centre of mass and this would also alter the distribution of the individual bodies . Could not this distribution (produced entirely by the mm/r^2 formula ) lead to a group distribution that was in conformity (if viewed in an appropriate way) with a non Euclidean geometry? I am fishing because I don't have the skills, so I will drop this idea if if seems implausible. Link to post Share on other sites

Strange 4273 Posted November 20, 2016 Share Posted November 20, 2016 I think you have missed the point. No simulation is necessary, because this is just calculating the geometrical effects. The individual bodies would be squashed (maybe I am wrong on this point) by Newtonian gravity increasingly as they were closer to the centre of mass and this would also alter the distribution of the individual bodies . But however much they were squashed, the relationship of area to radius would still be 4 pi r^{2} In GR, that is no longer true. Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 (edited) But however much they were squashed, the relationship of area to radius would still be 4 pi r^{2} In GR, that is no longer true I was imagining that the radius distance in my NM simulation might be defined differently. If one measured the radius by the amount of individual objects would that change the geometry appropriately. ie 1 object counts as 1 unit of distance along the radius. Any good? Edited November 20, 2016 by geordief Link to post Share on other sites

Strange 4273 Posted November 20, 2016 Share Posted November 20, 2016 Any good? Nope! Radius is radius: the distance from the centre to the surface. It doesn't matter what the sphere is made of or how it formed or how you measure it. Link to post Share on other sites

geordief 108 Posted November 20, 2016 Author Share Posted November 20, 2016 Nope! Radius is radius: the distance from the centre to the surface. It doesn't matter what the sphere is made of or how it formed or how you measure it. Can't distance be measured according to different criteria? I mean you could specify that you have to count 1 for the first object 2 for the second encountered and so on . Absurd , but just an example of an alternative method. . Would that be equivalent to a different geometry? If there were 1000001 objects instead of 1000000 could that correspond to a curvature. if the objects making up the surface area were unchanged (or less changed) in number? Link to post Share on other sites

Strange 4273 Posted November 20, 2016 Share Posted November 20, 2016 Can't distance be measured according to different criteria? Not really. It doesn't matter if you measure in furlongs or millimetres, or base 10 or base 17. As long as you are consistent between measuring radius and area. If there were 1000001 objects instead of 1000000 could that correspond to a curvature. if the objects making up the surface area were unchanged (or less changed) in number? Yes (if you mean what I think you mean). If the relationship is no longer A = 4 pi r^{2} then that represents curved geometry. Link to post Share on other sites

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