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TWIN PARADOX IN EINSTEIN 1905 PAPER
http://isites.harvard.edu/fs/docs/ic...les/chap11.pdf
pp. 14-15: "It's definitely true that when the two twins are standing next to each other (that is, when they are in the same frame), we can't have both B younger than A, and A younger than B. So what is wrong with the reasoning at the end of the previous paragraph? The error lies in the fact that there is no "one frame" that B is in. The inertial frame for the outward trip is different from the inertial frame for the return trip. The derivation of our time-dilation result applies only to one inertial frame. Said in a different way, B accelerates when she turns around, and our time-dilation result holds only from the point of view of an inertial observer. The symmetry in the problem is broken by the acceleration. If both A and B are blindfolded, they can still tell who is doing the traveling, because B will feel the acceleration at the turnaround. Constant velocity cannot be felt, but acceleration can be." http://www.fourmilab.ch/etexts/einstein/specrel/www/ ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein June 30, 1905: "...light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body.....From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by tv^2/2c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B. It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be tv^2/2c^2 second slow." Assume that, instead of the single clock at A, a large number of clocks start moving, simultaneously, from different points, with speed v, along various lines: some along a straight line (no acceleration), others along a polygonal line (multiple instantaneous accelerations), others along a curve ("continuous" acceleration). When two such clocks meet, they will show the same reading, in accordance with Einstein's text. But one of the two clocks may have been inertial all along whereas the inertiality of the other may have been disrupted many times and even absent in some intervals. And yet they show the same reading. Therefore, there can be no physical explanation of the twin paradox. It is not a paradox but, rather, a genuine contradiction following inevitably from Einstein's false light postulate. Pentcho Valev |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
"Pentcho Valev" wrote in message ps.com... [snip] Assume that, instead of the single clock at A, a large number of clocks start moving, simultaneously, from different points, with speed v, along various lines: some along a straight line (no acceleration), others along a polygonal line (multiple instantaneous accelerations), others along a curve ("continuous" acceleration). When two such clocks meet at B, in the stationary system K of A and B, they will show the same reading, in accordance with Einstein's text. But one of the two clocks may have been inertial all along whereas the inertiality of the other may have been disrupted many times and even absent in some intervals. And yet they show the same reading. Therefore, there can be no physical explanation of the twin paradox. It is not a paradox but, rather, a genuine contradiction following inevitably from Einstein's false light postulate. Or perhaps it shows what a stupid git you are. Everything is seen from the same inertial frame ("stationary system") In this frame all the clocks have a the same constant v, so the integral Dtau = integral { sqrt( 1 - v(t)^2 ) dt } is the same for all clocks. In the twin paradox the stay at home twin remains in the inertial frame. The traveler is the one with the constant velocity. Dirk Vdm |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
The twin paradox is an observation made at any distance. It is the
"observation" that is relative to the "observer" and it is correct in that context because of the time it takes light to travel from event B to observer A. And, the increased time it takes with increased distance. An increased differential of velocities translates to increased curvature of spacetime in the triangulation between B' and B and A. "B'" being actual position at the exact moment observer at "A" is observing the position -- and/or state -- to be "B." Thus "B" (including clock) is always negative (-) in time to both "B'" (0) (including clock) and "A" (0) (including clock). This is the setup at the moment of arrival of information at the speed of light to designated A (0). If one wants to back up time to the moment of the propagation of light, B and A are on an equal footing, 0 = 0. Then the projection into the future is B (0) to A' (+) rather than A (0). Per that very same projection into the future, though, one has to also project B (0) to B' (+). But Relativity doesn't care where the observer [was] in spacetime at the moment of the propagation of light at B, it only cares where the observer [is] in spacetime at the very moment he makes his observation, and what his observation will be at that very moment. In which case the setup is event B (-) (rather than object B' (0)) to object observer A (0). GLB |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
On 29 sep, 12:42, Pentcho Valev wrote:
http://isites.harvard.edu/fs/docs/ic...les/chap11.pdf pp. 14-15: "It's definitely true that when the two twins are standing next to each other (that is, when they are in the same frame), we can't have both B younger than A, and A younger than B. So what is wrong with the reasoning at the end of the previous paragraph? The error lies in the fact that there is no "one frame" that B is in. The inertial frame for the outward trip is different from the inertial frame for the return trip. The derivation of our time-dilation result applies only to one inertial frame. Said in a different way, B accelerates when she turns around, and our time-dilation result holds only from the point of view of an inertial observer. The symmetry in the problem is broken by the acceleration. If both A and B are blindfolded, they can still tell who is doing the traveling, because B will feel the acceleration at the turnaround. Constant velocity cannot be felt, but acceleration can be." http://www.fourmilab.ch/etexts/einstein/specrel/www/ ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein June 30, 1905: "...light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body.....From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by tv^2/2c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B. It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be tv^2/2c^2 second slow." Assume that, instead of the single clock at A, a large number of clocks start moving, simultaneously, from different points, with speed v, along various lines: some along a straight line (no acceleration), others along a polygonal line (multiple instantaneous accelerations), others along a curve ("continuous" acceleration). When two such clocks meet, they will show the same reading, in accordance with Einstein's text. But one of the two clocks may have been inertial all along whereas the inertiality of the other may have been disrupted many times and even absent in some intervals. And yet they show the same reading. Therefore, there can be no physical explanation of the twin paradox. It is not a paradox but, rather, a genuine contradiction following inevitably from Einstein's false light postulate. Why do you say that one of your moving clocks "may have been inertial all along"? Even for the more simple case of a single right line trajectory with constant velocity you have the "inertiality" disrupted at the start and end points. ALL moving clocks are then non-inertial ones. However, Einstein applies to all of them the Lorentz transform (LT) to calculate the time delay. See at the end of paragraph 4 how the LT is applied to a gravitational accelerated clock at the Earth equator (the unique real world example in all the paper). Evidently there exist no symmetry or reciprocity between "the stationary" and the "moving" frames, what put for me in serious doubt the meaning of the LT as derived and used by Einstein in his 1905 June 30 paper, comparing with its today general accepted role as a relation between time-space coordinates for any pair of inertial frames. Why do you blame the second postulate? Read how it is stated at the beginning of paragraph 2: {Any light ray moves in the "stationary" system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body}. And what about the "moving" frame? Is the light also moving there at velocity c? Why Einstein doesn't write here velocity c in ALL frames? My interpretation is that the "moving" frames are not necessarily the inertial ones that we considered today, but only some type of "instantaneous" ones (the clock at the equator is considered not accelerated in an instant, moving at "constant" velocity in that instant). I know that Einstein manages this type of frames in the years before the development of GR. For more details about this see my following thread in this group: Hierarchical Inertial System (HIS) in Einstein's 1905 Relativity http://groups.google.com.cu/group/sc...6b2ccdb1ef7ff6 Pentcho Valev RVHG (Rafael Valls Hidalgo-Gato |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
On 29 sep, 13:37, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote: "Pentcho Valev" wrote in glegroups.com... [snip] Assume that, instead of the single clock at A, a large number of clocks start moving, simultaneously, from different points, with speed v, along various lines: some along a straight line (no acceleration), others along a polygonal line (multiple instantaneous accelerations), others along a curve ("continuous" acceleration). When two such clocks meet at B, in the stationary system K of A and B, Why do you change what Pentcho says? You don't need it to prove him wrong. Two of Pentcho's clocks can meet at any place in the stationary system K showing the same reading, not necessarily at B. they will show the same reading, in accordance with Einstein's text. But one of the two clocks may have been inertial all along whereas the inertiality of the other may have been disrupted many times and even absent in some intervals. And yet they show the same reading. Therefore, there can be no physical explanation of the twin paradox. It is not a paradox but, rather, a genuine contradiction following inevitably from Einstein's false light postulate. Or perhaps it shows what a stupid git you are. Sure that Pentcho is wrong here (but why to insult him for it?) Everything is seen from the same inertial frame ("stationary system") In this frame all the clocks have a the same constant v, so the integral Dtau = integral { sqrt( 1 - v(t)^2 ) dt } is the same for all clocks. In the twin paradox the stay at home twin remains in the inertial frame. The traveler is the one with the constant velocity. Almost total agreement (the traveller can't have constant velocity in all his tour, he needs to accelerate at the start and at the end). Don't you think that the use of a unique inertial frame in the twin case needs a little more comment? How do you choose it? After all, relativity manages multiple inertial frames, no? Per example, can you select an inertial system where both twins have the same arbitrary velocity just before the starting of the tour by one of them? Dirk Vdm RVHG (Rafael Valls Hidalgo-Gato) |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
wrote in message ups.com... On 29 sep, 13:37, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO- SperM.hotmail.com wrote: "Pentcho Valev" wrote in glegroups.com... [snip] Assume that, instead of the single clock at A, a large number of clocks start moving, simultaneously, from different points, with speed v, along various lines: some along a straight line (no acceleration), others along a polygonal line (multiple instantaneous accelerations), others along a curve ("continuous" acceleration). When two such clocks meet at B, in the stationary system K of A and B, Why do you change what Pentcho says? Because it corresponds to the Einstein quote and Poncho thinks he is talking "in accordance with Einstein's text": | http://www.fourmilab.ch/etexts/einstein/specrel/www/ | ON THE ELECTRODYNAMICS OF MOVING BODIES | By A. Einstein June 30, 1905: | "...light is always propagated in empty space with a definite velocity | c which is independent of the state of motion of the emitting | body.....From this there ensues the following peculiar consequence. If | at the points A and B of K there are stationary clocks which, viewed | in the stationary system, are synchronous; and if the clock at A is | moved with the velocity v along the line AB to B, then on its arrival | at B the two clocks no longer synchronize, but the clock moved from A | to B lags behind the other which has remained at B by tv^2/2c^2 (up to | magnitudes of fourth and higher order), t being the time occupied in | the journey from A to B ... You don't need it to prove him wrong. Of course I don't need to prove him wrong. I just felt like rubbing his nose in his doo. Two of Pentcho's clocks can meet at any place in the stationary system K showing the same reading, not necessarily at B. That is the place labeled B. they will show the same reading, in accordance with Einstein's text. "in accordance with Einstein's text". Get the picture? But one of the two clocks may have been inertial all along whereas the inertiality of the other may have been disrupted many times and even absent in some intervals. And yet they show the same reading. Therefore, there can be no physical explanation of the twin paradox. It is not a paradox but, rather, a genuine contradiction following inevitably from Einstein's false light postulate. Or perhaps it shows what a stupid git you are. Sure that Pentcho is wrong here (but why to insult him for it?) Yes, why, does one insult an idiot who constantly begs to be insulted? Because he loves being insulted. It is called "providing a service". You, of all idiots, rusting away in your debunked Hierarchical Frame should know that. Dirk Vdm |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
On Sep 29, 11:42 am, Pentcho Valev wrote:
http://isites.harvard.edu/fs/docs/ic...les/chap11.pdf pp. 14-15: "It's definitely true that when the two twins are standing next to each other (that is, when they are in the same frame), we can't have both B younger than A, and A younger than B. So what is wrong with the reasoning at the end of the previous paragraph? The error lies in the fact that there is no "one frame" that B is in. The inertial frame for the outward trip is different from the inertial frame for the return trip. The derivation of our time-dilation result applies only to one inertial frame. Said in a different way, B accelerates when she turns around, and our time-dilation result holds only from the point of view of an inertial observer. The symmetry in the problem is broken by the acceleration. If both A and B are blindfolded, they can still tell who is doing the traveling, because B will feel the acceleration at the turnaround. Constant velocity cannot be felt, but acceleration can be." http://www.fourmilab.ch/etexts/einstein/specrel/www/ ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein June 30, 1905: "...light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body.....From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by tv^2/2c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B. It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be tv^2/2c^2 second slow." Assume that, instead of the single clock at A, a large number of clocks start moving, simultaneously, from different points, with speed v, along various lines: some along a straight line (no acceleration), others along a polygonal line (multiple instantaneous accelerations), others along a curve ("continuous" acceleration). When two such clocks meet, they will show the same reading, in accordance with Einstein's text. But one of the two clocks may have been inertial all along whereas the inertiality of the other may have been disrupted many times and even absent in some intervals. And yet they show the same reading. Therefore, there can be no physical explanation of the twin paradox. It is not a paradox but, rather, a genuine contradiction following inevitably from Einstein's false light postulate. Pentcho Valev Who is the dumb from Harvard to write this: "If both A and B are blindfolded, they can still tell who is doing the traveling, because B will feel the acceleration at the turnaround. Constant velocity cannot be felt, but acceleration can be."? According to Newton's law, constant velocity cannot be felt. According to the Principle of Equivalence, acceleration cannot be felt. We can only feel if a force is not acting evenly over our body. A local tension or compression is generated which can be 'felt' by our central nerves system. If the force is acting evenly over our body then we cannot feel the acceleration at all. If you are free falling within an enclosed container under a uniform gravitational field, you cannot tell that you are not in an inertial frame. |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
snapdragon31 a écrit :
Who is the dumb from Harvard to write this: "If both A and B are blindfolded, they can still tell who is doing the traveling, because B will feel the acceleration at the turnaround. Constant velocity cannot be felt, but acceleration can be."? 利用电解二氧化锰(EMD)和碳酸锂为原料 ,采用熔融浸渍法合成了尖晶石型锂锰 化物LiMn2O4,并用*重分析(TGA)、粉末X射 线衍射技术*究了合成条件对产物的 晶体结构、电化*性能的影响.*究结 表明,在合成的后*阶段反应时间的 * 对产物的晶体结构和电化*性能的影 很大,时间长,会使 LiMn2O4分解为 Li2MnO3和Mn2O3;LiMn2O4的初始放电比容量也 随反应时间的延长而下降.在最佳 条件下合成的LiMn2O4的首次放电比容量 达132.4mAh/g,50次循环后的放电比容 量还保持在125.6 mAh/g的水平. According to Newton's law, constant velocity cannot be felt. According to the Principle of Equivalence, acceleration cannot be felt. 和碳酸锂为原料,采用熔融浸渍法合成 尖晶石型锂锰氧化物LiMn2O4,并用*重 析. We can only feel if a force is not acting evenly over our body. A local tension or compression is generated which can be 'felt' by our central nerves system. If the force is acting evenly over our body then we cannot feel the acceleration at all. If you are free falling within an enclosed container under a uniform gravitational field, you cannot tell that you are not in an inertial frame. 在合成的后*阶段反应时间的长*对 物的晶体结构和电化在合成的后*阶 反 应时间的长*对产物的晶体结构和电 在合成的后*阶段反应时间的长*对 物 的晶体结构和电化. -- kd 应时间的长 |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
kduc, tu as raison !
Si l'on pose : ( ????? ) ( ??????? ) / ( ??????????? ) = 0 et que l'on simplifie, on obtient : 0 / 0 = 0 CQFD. -- http://www.lempel.net http://www.les-os-du-cosmos.fr B. Lempel ______________________________________ "kduc" a crit dans le message de news: ... | snapdragon31 a crit : | | Who is the dumb from Harvard to write this: "If both A and B are | blindfolded, they can still tell who is doing the traveling, because | B | will feel the acceleration at the turnaround. Constant velocity | cannot | be felt, but acceleration can be."? | | ????????(EMD)???????,????????????????? | ??LiMn2O4,??????(TGA)???X????????????????? | ?????????????.??????,??????????????? | ???????????????????,???,?? LiMn2O4??? | Li2MnO3?Mn2O3;LiMn2O4????????????????????.??? | ??????LiMn2O4??????????132.4mAh/g,50????????? | ?????125.6 mAh/g???. | | .... Kouik... Sans regrets. |
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TWIN PARADOX IN EINSTEIN 1905 PAPER
"snapdragon31" wrote in message oups.com... : On Sep 29, 11:42 am, Pentcho Valev wrote: : http://isites.harvard.edu/fs/docs/ic...les/chap11.pdf : pp. 14-15: "It's definitely true that when the two twins are standing : next to each other (that is, when they are in the same frame), we : can't have both B younger than A, and A younger than B. So what is : wrong with the reasoning at the end of the previous paragraph? The : error lies in the fact that there is no "one frame" that B is in. The : inertial frame for the outward trip is different from the inertial : frame for the return trip. The derivation of our time-dilation result : applies only to one inertial frame. Said in a different way, B : accelerates when she turns around, and our time-dilation result holds : only from the point of view of an inertial observer. The symmetry in : the problem is broken by the acceleration. If both A and B are : blindfolded, they can still tell who is doing the traveling, because B : will feel the acceleration at the turnaround. Constant velocity cannot : be felt, but acceleration can be." : : http://www.fourmilab.ch/etexts/einstein/specrel/www/ : ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein June 30, 1905: : "...light is always propagated in empty space with a definite velocity : c which is independent of the state of motion of the emitting : body.....From this there ensues the following peculiar consequence. If : at the points A and B of K there are stationary clocks which, viewed : in the stationary system, are synchronous; and if the clock at A is : moved with the velocity v along the line AB to B, then on its arrival : at B the two clocks no longer synchronize, but the clock moved from A : to B lags behind the other which has remained at B by tv^2/2c^2 (up to : magnitudes of fourth and higher order), t being the time occupied in : the journey from A to B. It is at once apparent that this result still : holds good if the clock moves from A to B in any polygonal line, and : also when the points A and B coincide. If we assume that the result : proved for a polygonal line is also valid for a continuously curved : line, we arrive at this result: If one of two synchronous clocks at A : is moved in a closed curve with constant velocity until it returns to : A, the journey lasting t seconds, then by the clock which has remained : at rest the travelled clock on its arrival at A will be tv^2/2c^2 : second slow." : : Assume that, instead of the single clock at A, a large number of : clocks start moving, simultaneously, from different points, with speed : v, along various lines: some along a straight line (no acceleration), : others along a polygonal line (multiple instantaneous accelerations), : others along a curve ("continuous" acceleration). : : When two such clocks meet, they will show the same reading, in : accordance with Einstein's text. But one of the two clocks may have : been inertial all along whereas the inertiality of the other may have : been disrupted many times and even absent in some intervals. And yet : they show the same reading. Therefore, there can be no physical : explanation of the twin paradox. It is not a paradox but, rather, a : genuine contradiction following inevitably from Einstein's false light : postulate. : : Pentcho Valev : : Who is the dumb from Harvard to write this: "If both A and B are : blindfolded, they can still tell who is doing the traveling, because : B : will feel the acceleration at the turnaround. Constant velocity : cannot : be felt, but acceleration can be."? : : According to Newton's law, constant velocity cannot be felt. : According to the Principle of Equivalence, acceleration cannot be : felt. : : We can only feel if a force is not acting evenly over our body. A : local tension or compression is generated which can be 'felt' by our : central nerves system. If the force is acting evenly over our body : then we cannot feel the acceleration at all. : : If you are free falling within an enclosed container under a uniform : gravitational field, you cannot tell that you are not in an inertial : frame. It's a tricky point, because if the travelling twin reverses direction by swinging around a star he feels no force, whereas if he fires his engine to accomplish the reversal then he does. Acceleration plays no part in the twin paradox anyway. The blunder is in the statement: -- 'we establish by definition that the "time" required by light to travel from A to B equals the "time" it requires to travel from B to A' -- Albert Einstein This gif shows that cannot be. http://www.androcles01.pwp.blueyonde...rt/tAB=tBA.gif "I know that in the past Androcles has used tAB=tBA to claim Einstein thinks tau_AB = tau_BA (the travel times in the frame in which the clocks at A and B are moving)." -- Blind "I'm not a troll" Poe. http://www.fourmilab.ch/etexts/einst...ures/img21.gif |
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