Conjurer 16 Posted November 5 (edited) Herman Minkowski was one of Einsteins professors at the university where he graduated from. After Einstein developed his special theory of relativity, Herman Minkowski proposed a new idea to him in order to describe spacetime, since the time dilation equation received a lot of attention in his paper On the Electrodynamics of Moving Bodies. He proposed a rather simple idea about how to create a framework of a coordinate plane which could obey the rules of relativity, and it would have to replace a normal coordinate plane in which time dilation wouldn't show up. He started out by saying that all distances in the coordinate plane should be calculated by multiplying the speed of light, times time. Where the speed of light is "c" and time is represented by "t". Then a distance (d) would be equal to the speed of light, times time, d=ct. This brought about a problem called the light clock example, where they tried to solve for time dilation in this new coordinate plane. A problem started where the only solution anyone could obtain for the problem was t'=t/sqrt(1-(v^2)/(c^2)). This started to become a problem, because this was not the same equation that Einstein developed in his paper On the Electrodynamics of Moving Bodies, which was t'=t sqrt(1-(v^2)/(c^2)). No one was able to solve this problem for over 100 years, since the theory came out. It then became commonly accepted to use both equations, since they both ended up coming out with approximately the same answer when the light clock equation was considered to be the change in time, and the equation from the original paper was considered to be the proper time. Then I have come up with the correct solution or proof that the proper time can be obtained in Minkowski Space, so a mathematical framework can exist which considers spacetime as a type of aether in a mathematical construct. The problem with the light clock proof when applying to Minkowski Spacetime is that the time variables were not assigned correctly. Then the equation doesn't need to be considered as being in Hertz of a tick of a clock. By doing this, the proof could then be calculated using Pythagorean's Theorem, even though the two objects are not connected bodies forming a right triangle. (ct')^2+(vt)^2=(ct)^2 c^2t'^2+v^2t^2=c^2t^2 c^2t'^2=c^2t^2-v^2t^2 c^2t'^2=c^2t^2(1-(v^2)/(c^2)) t'^2=t^2(1-(v^2)/(c^2)) t'=t sqrt(1-(v^2)/(c^2)) Then the value for the equation comes out to the accurate equation that has been shown to be the correct answer which has been found in repeated experiments. Then relativistic problems can be done mathematically assuming that there is an aether or fabric of spacetime which obeys the rules of the Special Theory of Relativity. Spacetime would have to be an actual physical object in this instance, because it assumes that there is a physical mechanism which allows two completely independent objects to be related to each other via Pythagorean's Theorem, which is the same as a physical limitation put on triangles that have sides which must always make physical contact with each other in way that is governed by this rule. It assumes there is also a lattice in a higher dimension, because they are related by a square. Then squaring an equation puts it into more dimensions. This dimension is not seen by the observer in the light clock example. In the same way there are a number of squares on each side of a triangle that add up to the number of squares on the hypotenuse, it is assuming that there is a lattice with a number of squares in a higher dimension that adds up to the other sides of an unconnected object through the speed of light limitation. This lattice then connects all observers, so that no other object can travel at a different relative speed to that of light. Then the speed of light remains constant to all observers. Edited November 5 by Conjurer 0 Share this post Link to post Share on other sites

studiot 1471 Posted November 5 (edited) It is indeed a little tricky to get the right variables in the right place for the moving light clock and then to transpose the results into a transformation formula between systems. But this was done a long time ago so there is no problem with the maths and no aether is required. The reciprocal relation between the square root factors comes directly from the comparison between time and distance calculation in both the stationary (call it A) and moving system (call it B). The time estimates are [math]\frac{{2l}}{{\sqrt {{c^2} - {v^2}} }}[/math] according to A and [math]\frac{{2l}}{{\sqrt {{c^2}} }}[/math] according to B When you divide one by the other you will recover the fact that A sees the time recorded by B a lengthened by the correct factor . Edited November 5 by studiot 0 Share this post Link to post Share on other sites

Strange 3257 Posted November 5 And no extra dimensions are required, either. 0 Share this post Link to post Share on other sites

Conjurer 16 Posted November 6 17 hours ago, studiot said: But this was done a long time ago so there is no problem with the maths and no aether is required. The older solutions took a lot more steps. I am not even sure how they originally got to the correct solution, because it was too rigorous to be explained anywhere. This is the quickest solution to get to it in such few of steps. Then it requires a relativistic aether like Minkowski's spacetime in order to get to the solution in this few of steps. Since all distances are measured with the speed of light in relation to their dilated time, the Minkowski spacetime would have to actually warp to maintain the correct values like physical rubber bands all connected together. It is expressing it as a physical framework, instead as only an observational difference. 18 hours ago, studiot said: When you divide one by the other you will recover the fact that A sees the time recorded by B a lengthened by the correct factor . How do you figure? The 2l would just cancel out. Where did you even get a 2 from? 18 hours ago, Strange said: And no extra dimensions are required, either. Working in Minkowski spacetime is essential to the work in most mainstream theories dealing with higher dimensions about the universe as a whole. It has proven to be just as accurate as any other type of relativistic theory. It is also the simplest way to define it. 0 Share this post Link to post Share on other sites

Strange 3257 Posted November 6 6 hours ago, Conjurer said: Working in Minkowski spacetime is essential to the work in most mainstream theories dealing with higher dimensions about the universe as a whole. It has proven to be just as accurate as any other type of relativistic theory. It is also the simplest way to define it. There are no extra dimensions. Minkowski spacetime, and even GR, just use the four dimensions of space and time. 6 hours ago, Conjurer said: Since all distances are measured with the speed of light in relation to their dilated time, the Minkowski spacetime would have to actually warp to maintain the correct values like physical rubber bands all connected together. It is expressing it as a physical framework, instead as only an observational difference. Distances can change without requiring them to be made of rubber or anything else. Space-time is purely about geometry: distances and angles. 0 Share this post Link to post Share on other sites

studiot 1471 Posted November 6 (edited) 15 hours ago, Conjurer said: The older solutions took a lot more steps. I am not even sure how they originally got to the correct solution, because it was too rigorous to be explained anywhere. This is the quickest solution to get to it in such few of steps. Then it requires a relativistic aether like Minkowski's spacetime in order to get to the solution in this few of steps. Since all distances are measured with the speed of light in relation to their dilated time, the Minkowski spacetime would have to actually warp to maintain the correct values like physical rubber bands all connected together. It is expressing it as a physical framework, instead as only an observational difference. How do you figure? The 2l would just cancel out. Where did you even get a 2 from? Working in Minkowski spacetime is essential to the work in most mainstream theories dealing with higher dimensions about the universe as a whole. It has proven to be just as accurate as any other type of relativistic theory. It is also the simplest way to define it. Minkowski is neither the simplest nor the most modern or most integrated with modern mathematics. The modern version uses vector spaces and an alternative inner product to the pythagorean one of Minkowski, which require an imaginary coordinate axis generated by incorporating the square root of minus 1 in the definition. Where did I get the 2 from? Well since you introduced light clocks I assumed you would understand that light clocks work by virtue of a double journey (to a mirror and back), hence the 2. And yes the 2l cancels. But I am still at a loss to understand where you see a difficulty. Here is what Einstein actually wrote in his paper (translated) where he develops the equation you refer to. Are you sure you are applying it to the correct quantity? Edited November 6 by studiot 0 Share this post Link to post Share on other sites

Conjurer 16 Posted November 7 3 hours ago, studiot said: But I am still at a loss to understand where you see a difficulty. I have studied relativity independently, and most of the people that write books about it seem unaware of this easier proof that could be used to explain the theory to first year students. I plan to teach math sometime after I graduate within a year. It seemed like this type of proof could be used to teach them, and I wanted to know what people thought about it. I wasn't sure what people would think of my interpretation of the math by assuming that two separate reference frames can be assumed to be forming one single light triangle. 0 Share this post Link to post Share on other sites

studiot 1471 Posted November 7 (edited) 17 hours ago, Conjurer said: I have studied relativity independently, and most of the people that write books about it seem unaware of this easier proof that could be used to explain the theory to first year students. I plan to teach math sometime after I graduate within a year. It seemed like this type of proof could be used to teach them, and I wanted to know what people thought about it. I wasn't sure what people would think of my interpretation of the math by assuming that two separate reference frames can be assumed to be forming one single light triangle. I have some more (hopefully helpful) thoughts about this post but meanwhile can you say what is in conflict with this presentation of a light clock? No bastard triangles are needed. I did ask if you are applying your analysis to the wrong thing and I ask again as you did not answer. The point is that we should be comparing the second (unit of time) as measured by each observer, as each will observe a different number of units, but it is the relative length of those units that is being transformed (dilated). So to A, B's units will appear dilated (longer) so the square root factor being less than 1 is on the bottom of the fraction in the transformation from his units to B's. This is the usual formula. However when A considers the transformation from B's units to his the square root factor will appear on the top (Einsteins formula) as he need to shorten B's apparently longer units to match his own. Edited November 7 by studiot Clarification 2 Share this post Link to post Share on other sites

Conjurer 16 Posted November 8 8 hours ago, studiot said: I have some more (hopefully helpful) thoughts about this post but meanwhile can you say what is in conflict with this presentation of a light clock? No bastard triangles are needed. I did ask if you are applying your analysis to the wrong thing and I ask again as you did not answer. In figure 2.4, they get the Lorentz Factor as a part of the solution. In the equation for the Proper Time, the Lorentz Factor would be inverted. So, it doesn't come out exactly to what has been found to be in experiments, since experiments show that the Proper Time is the more correct equation. They give slightly different answers. Then it would seem like, yes, bastard triangles are needed in order to obtain the Proper Time in this example. 0 Share this post Link to post Share on other sites

studiot 1471 Posted November 8 On 05/11/2018 at 2:05 AM, Conjurer said: He started out by saying that all distances in the coordinate plane should be calculated by multiplying the speed of light, times time. Where the speed of light is "c" and time is represented by "t". Then a distance (d) would be equal to the speed of light, times time, d=ct. This brought about a problem called the light clock example, where they tried to solve for time dilation in this new coordinate plane. A problem started where the only solution anyone could obtain for the problem was t'=t/sqrt(1-(v^2)/(c^2)). This started to become a problem, because this was not the same equation that Einstein developed in his paper On the Electrodynamics of Moving Bodies, which was t'=t sqrt(1-(v^2)/(c^2)). No one was able to solve this problem for over 100 years, since the theory came out. It then became commonly accepted to use both equations, since they both ended up coming out with approximately the same answer when the light clock equation was considered to be the change in time, and the equation from the original paper was considered to be the proper time. Then I have come up with the correct solution or proof that the proper time can be obtained in Minkowski Space, so a mathematical framework can exist which considers spacetime as a type of aether in a mathematical construct. 1) There is no problem to solve, except perhaps your understanding of what is going on. 2) I posted the part of Einstein's original paper where he states the equation you refer to above. Did you read it? Because if you had read it you would surely have the following where he explains what his equation is calculating. Quote Einstein What is the rate of this clock when viewed from the stationary system? Note also that he clearly states the clock is in the moving system. Here is a definition of proper time Quote McComb A measurement made on a body which is in the rest frame of the body is called a proper mesurement for example proper length and proper time. So very very clearly, Einstein was not talking about proper time (although this term was not introduced till later and I think not by him) This is why I keep asking if you are applying your analysis to the correct bodies. Please note that the example from Turner I quoted included a diagram and some numbered equations, as well as an explanation of the meaning of the symbols in those equations. Einstein also did this in his paper. I see that you wish to teach Mathematics and I recommend you adopt this convention. It not only helps one organise one's own thought it help communication with your students and, if you make then do it, helps you unravel what they are doing or think they are doing. In relation to this I see you mixed up the figures and equations in my last post. 17 hours ago, Conjurer said: In figure 2.4, they get the Lorentz Factor as a part of the solution 0 Share this post Link to post Share on other sites

Conjurer 16 Posted Friday at 01:06 AM On 11/6/2018 at 2:08 PM, studiot said: This is the image you linked to me of Einsteins paper, where it clearly shows the solution to the proper time after the word "therefore"... The funny looking "t" there is called tau, and it is exactly the same as t' from how I got to it using Minkowski spacetime. By definition, it is the correct time that would be read on a clock from another inertial frame. The part of the equation after the second equal sign in that same line can just be dropped for the equation, and then it is exactly the same equation as the proper time... 0 Share this post Link to post Share on other sites

studiot 1471 Posted Friday at 08:27 AM 7 hours ago, Conjurer said: This is the image you linked to me of Einsteins paper, where it clearly shows the solution to the proper time after the word "therefore"... The funny looking "t" there is called tau, and it is exactly the same as t' from how I got to it using Minkowski spacetime. By definition, it is the correct time that would be read on a clock from another inertial frame. The part of the equation after the second equal sign in that same line can just be dropped for the equation, and then it is exactly the same equation as the proper time... Indeed we are agreed that this is the equation in Einstein's paper you are referring to. This is the first time you have agreed this so this is progress. However, going back to the A(stationary) and B (moving systems) systems - or the K and k systems (Einstein) t is the time measured on a clcok in the A system tau is the time measured in the B system The transformation is the connection between t and tau; that is tau as a function of t Therefore the transformation gives the time as viewed in the stationary system. But the definition of proper time is the time as viewed in the moving system. If you weren't so determined to bulldoze this discussion, you would be in danger of learning something. 0 Share this post Link to post Share on other sites

vanholten 3 Posted Saturday at 03:03 PM (edited) On 09/11/2018 at 2:06 AM, Conjurer said: Interesting is that Einstein wrote: “What is the rate of the clock, when viewed from the stationary system?” The formula that follows does not show a parameter for the “rate” of the clock (or frequency regarding the light clock). Suppose two observers, stationary compared to their own light clocks, are in relative motion. They uniformly move parallel away from each other over an agreed fixed distance. At the end of the trajectory both observers will have counted an equal number of reflector hits for both clocks, the moving clock and the stationary clock. This outcome is supported by the formula for time dilation. The presumed relativistic lengthening of the path of a photon has no effect on the rate of the photon hitting the reflectors. Because if that were the case the factor 2L would not apply and the formula would fall short. Since the frequency of both stationary and moving light clocks is identical to both observers in relative motion, I am afraid time dilation is pointless. Edited Saturday at 03:44 PM by vanholten 0 Share this post Link to post Share on other sites

Janus 858 Posted Saturday at 08:08 PM 4 hours ago, vanholten said: Interesting is that Einstein wrote: “What is the rate of the clock, when viewed from the stationary system?” The formula that follows does not show a parameter for the “rate” of the clock (or frequency regarding the light clock). Suppose two observers, stationary compared to their own light clocks, are in relative motion. They uniformly move parallel away from each other over an agreed fixed distance. At the end of the trajectory both observers will have counted an equal number of reflector hits for both clocks, the moving clock and the stationary clock. This outcome is supported by the formula for time dilation. The presumed relativistic lengthening of the path of a photon has no effect on the rate of the photon hitting the reflectors. Because if that were the case the factor 2L would not apply and the formula would fall short. Since the frequency of both stationary and moving light clocks is identical to both observers in relative motion, I am afraid time dilation is pointless. By " agreed fixed distance" I will assume that you mean that as clock 1 and clock 2 pass each othe,r then both clocks will agree that point A( toward which clock 1 is moving) is an equal distance away as point B ( the point clock 2 is moving). The problem with your assertion, is that while both the observers will agree that both Clock 1 and 2 will have ticked off the same number of reflector hits upon reaching their respective point, they will not agree that both clocks reached those points simultaneously. Each observer will determine that his clock reached its point before the other clock did. This is result of the relativity of simultaneity. It can also be demonstrated by considering the relativistic addition of velocities. Assume that each clock is moving at 0.8 c relative to the point half way between points A and B ( the point where they pass it other). Each observer will measure the other clocks velocity relative to themselves as being (0.8c+0.8c)?(1+0.8c(0.8c)/c^{2}) = 0.9756098c And thus the difference in velocity between the midpoint and the other clock as being 0.1756098c compared to their own velocity relative to the midpoint of 0.8c. Since we also know that each observer measures the distance between midpoint and either point A or B as being the same, he also has to conclude that he will reach his point first. Thus when he reaches his point and registers X reflector hits he also knows that the other clock will have to registered fewer than X reflector hits at that moment ( since the other clock has yet to reach its point). Thus the other clock has to have ticked slower than his clock, and he measures time dilation in the other clock. 0 Share this post Link to post Share on other sites

studiot 1471 Posted Saturday at 08:33 PM 5 hours ago, vanholten said: The formula that follows does not show a parameter for the “rate” of the clock Actually it does. It is just that the procedure Einstein adopted to ensure this by correct zeroing of the clock was not part of my excerpt. Would you like to see that part or can you look it up for yourself? 0 Share this post Link to post Share on other sites

vanholten 3 Posted Sunday at 12:06 AM (edited) 4 hours ago, Janus said: By " agreed fixed distance" I will assume that you mean that as clock 1 and clock 2 pass each othe,r then both clocks will agree that point A( toward which clock 1 is moving) is an equal distance away as point B ( the point clock 2 is moving). The problem with your assertion, is that while both the observers will agree that both Clock 1 and 2 will have ticked off the same number of reflector hits upon reaching their respective point, they will not agree that both clocks reached those points simultaneously. Each observer will determine that his clock reached its point before the other clock did. This is result of the relativity of simultaneity. It can also be demonstrated by considering the relativistic addition of velocities. Assume that each clock is moving at 0.8 c relative to the point half way between points A and B ( the point where they pass it other). Each observer will measure the other clocks velocity relative to themselves as being (0.8c+0.8c)?(1+0.8c(0.8c)/c^{2}) = 0.9756098c And thus the difference in velocity between the midpoint and the other clock as being 0.1756098c compared to their own velocity relative to the midpoint of 0.8c. Since we also know that each observer measures the distance between midpoint and either point A or B as being the same, he also has to conclude that he will reach his point first. Thus when he reaches his point and registers X reflector hits he also knows that the other clock will have to registered fewer than X reflector hits at that moment ( since the other clock has yet to reach its point). Thus the other clock has to have ticked slower than his clock, and he measures time dilation in the other clock. Thanks for your explanation. But your arguments don't convince me. In my perception relativity adds an unnecessary complication to this setup. When observer 1 and clock 1 reach the end of the route at point A so do the observer 2 and clock 2 reach the opposite end of the route at point B. The light from A to reach B has to travel the same distance as the light from B to reach A. Both observers know the light has to travel this distances, because they already agreed upon the length of the route in advance. Therefore they also agreed on the amount of light seconds it takes to receive the information from the opposite end of the route. It is just a common delay in receiving information caused by the finite speed of light. I don't see the need for relativity or time dilation. You refer to the simultaneity of relativity. However reasoned from each observer the opposite clock can never finish simultaneously with their own clock. Not due to any special relativistic effects, but simply because simultaneity can never apply to merely two separated events. Simultaneity requires at least three events to manifest itself. Information coming from one event always has to bridge the distance to the opposite event. There is an inevitable distance in between an emitter and a receiver that due to the infinite speed of light results in a certain delay. From the receiving viewpoint, the actual emission took place in the past and from the emitting viewpoint the actual reception takes place in the future. In case of two separated events simultaneity is absent per definition. Simultaneity needs at least two emitters to bridge an equal distance to the same receiver. Then from the point of view of the receiver the signals from both emitters occur simultaneously, that is, if both emissions took place simultaneously. Edited Sunday at 12:21 AM by vanholten 0 Share this post Link to post Share on other sites

Janus 858 Posted Sunday at 02:16 AM 1 hour ago, vanholten said: The light from A to reach B has to travel the same distance as the light from B to reach A. Both observers know the light has to travel this distances, because they already agreed upon the length of the route in advance. Therefore they also agreed on the amount of light seconds it takes to receive the information from the opposite end of the route. It is just a common delay in receiving information caused by the finite speed of light. I don't see the need for relativity or time dilation. No. The measurements I was talking about are after we have accounted for the delay due to light propagation. It what is left over after you have factored this effect out. Thus in this example, including the light propagation delay, observer 1 will visually see, clock 2 running at a rate ~1.2% as fast as his own, but after taking into account the time delay due to the increasing distance between them will conclude that clock 2 is actually running ~21% as fast as his own. 1 hour ago, vanholten said: You refer to the simultaneity of relativity. However reasoned from each observer the opposite clock can never finish simultaneously with their own clock. Not due to any special relativistic effects, but simply because simultaneity can never apply to merely two separated events. Simultaneity requires at least three events to manifest itself. Information coming from one event always has to bridge the distance to the opposite event. There is an inevitable distance in between an emitter and a receiver that due to the infinite speed of light results in a certain delay. From the receiving viewpoint, the actual emission took place in the past and from the emitting viewpoint the actual reception takes place in the future. In case of two separated events simultaneity is absent per definition. Simultaneity needs at least two emitters to bridge an equal distance to the same receiver. Then from the point of view of the receiver the signals from both emitters occur simultaneously, that is, if both emissions took place simultaneously. You are again confusing light delay effects with Relativistic ones. If the receiver and sender are not moving relative to each other, each of them will agree as the simultaneity of events. So for example. If you you the source sends a signal to two receivers , each 1 light sec from him in opposite directions he will conclude that the light will arrive at both receivers simultaneously. Someone at either of the receivers, will also agree that the light reached both at the same time even though each of them will not actually see the signal arrive at the other receiver until 2 sec after his own signal arrived. This is not what Relativity of simultaneity means. Relativity of Simultaneity is this: As clocks 1and 2 pass the midpoint between A and B, a light is emitted from the midpoint. According to the observer that remains at the midpoint, this light travels outward at c and reaches points A and B simultaneously. However according to the observer with clock 1, this light expands outward at c from him. The distance between point A and himself is decreasing at 0.8c and the distance between point B and himself is increasing at 0.8c. Point A runs into the light before the light in the other direction catches up to point B. The light does not reach both points simultaneously. According to the observer with clock 2, the light expands outward at c relative to him, The distance to point B is decreasing at 0.8c and the distance to point A increasing at 0.8c. the light runs into point B before it catches up to point A. Again the light does not reach both points simultaneously, but the order is reversed to that which happened according to clock 1's observer. 0 Share this post Link to post Share on other sites

Conjurer 16 Posted Sunday at 04:54 AM (edited) On 11/9/2018 at 1:27 AM, studiot said: Indeed we are agreed that this is the equation in Einstein's paper you are referring to. This is the first time you have agreed this so this is progress. What is this? Some kind of government cover-up? Is this supposed to be some kind of joke or prank or something? I stated this in my original post when I made this thread in the last line of the second paragraph. "This started to become a problem, because this was not the same equation that Einstein developed in his paper On the Electrodynamics of Moving Bodies, which was \[ t'=t \sqrt{1-v^2/c^2} \] ." It seems like I was unable to present the information in a way you could understand it more clearly or you failed to actually read it. I recommend you go over it and read it again, so you can see the significance of the derivations I was talking about and the actual point I was trying to make. I am simply showing a new derivation in Minkowski spacetime which has been unknown, and that is why text don't use this to explain relativity. Then they explain it using the light clock example which comes out to an inaccurate equation, which is not the same as Einsteins original equation in his paper, but the Lorentz Factor is the inverse of itself. 14 hours ago, vanholten said: Since the frequency of both stationary and moving light clocks is identical to both observers in relative motion, I am afraid time dilation is pointless. If the object traveling were two spaceships and they launched a beacon at the starting line, they would come back to the beacon showing that both of their clocks no longer showed the same time as the beacon at the starting line or position, if the beacon just remained stationary the whole time. They would both observe each others time slow down as they are moving and this creates the Twin Paradox. Edited Sunday at 05:13 AM by Conjurer -1 Share this post Link to post Share on other sites

studiot 1471 Posted yesterday at 03:07 PM On 11/11/2018 at 4:54 AM, Conjurer said: What is this? Some kind of government cover-up? Is this supposed to be some kind of joke or prank or something? I stated this in my original post when I made this thread in the last line of the second paragraph. "This started to become a problem, because this was not the same equation that Einstein developed in his paper On the Electrodynamics of Moving Bodies, which was t′=t1−v2/c2−−−−−−−−√ ." It seems like I was unable to present the information in a way you could understand it more clearly or you failed to actually read it. I recommend you go over it and read it again, so you can see the significance of the derivations I was talking about and the actual point I was trying to make. I am simply showing a new derivation in Minkowski spacetime which has been unknown, and that is why text don't use this to explain relativity. Then they explain it using the light clock example which comes out to an inaccurate equation, which is not the same as Einsteins original equation in his paper, but the Lorentz Factor is the inverse of itself. You quote an equation. and say it is due to Einstein. I agreed that Einstein wrote the equation and found the actual part of the paper where he wrote it. And you reply by accusing me of a government cover up. ????????????????????????????????????? You also claim this equation is at variance with some other equation writen by "they" , whoever they might be. And yes I also agree that two equations are different. And I have pointed out that these equations refer to the calculation of different quantites, so I don't find it suprising that they are different. The associated personal abuse is neither warranted nor welcome. 0 Share this post Link to post Share on other sites

vanholten 3 Posted yesterday at 06:51 PM On 10-11-2018 at 9:33 PM, studiot said: Actually it does. It is just that the procedure Einstein adopted to ensure this by correct zeroing of the clock was not part of my excerpt. Would you like to see that part or can you look it up for yourself? I only found the German version. If you would be so kind? 0 Share this post Link to post Share on other sites

Strange 3257 Posted 23 hours ago 1 hour ago, vanholten said: I only found the German version. If you would be so kind? I guess that is because of your language preferences in the search engine, as various links to the English text come up first for me. http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf 0 Share this post Link to post Share on other sites

studiot 1471 Posted 23 hours ago 1 hour ago, vanholten said: I only found the German version. If you would be so kind? 0 Share this post Link to post Share on other sites

vanholten 3 Posted 9 hours ago Thanks Studiot and Strange. 0 Share this post Link to post Share on other sites