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- - ** What is or is not a paradox?**
(*http://www.spacebanter.com/showthread.php?t=192049*)

What is or is not a paradox?On Dec 30, 4:17 pm, Sylvia Else wrote:
I started writing a post about this yesterday, then scrubbed it - too What is a paradox in special relativity (hereinafter SR)? I've expressed the view that to contain a paradox, SR has to predict, from different frames, outcomes that are mutually incompatible. An example that comes to mind (though not directly arising) from a recent discussion is that in one frame, there is massive destruction on a citywide scale, and in another other frame, nothing much happens. Clearly, if SR were to make such predictions for two frames, it would have to be regarded as seriously wanting. Of course, it does no such thing. But people seem to want to regard measurements in two frames as mutually incompatible if they give different results. I am at a loss to understand why people would seek to regard those different results as constituting a paradox that invalidates SR (well, leaving intellectual dishonesty aside). From the Lorentz transformations, you can write down the following equation per Minkowski spacetime. Points #1, #2, and #3 are observers. They are observing the same target. ** c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2 Where ** dt1 = Time flow at Point #1 ** dt2 = Time flow at Point #2 ** dt3 = Time flow at Point #3 ** ds1 = Observed target displacement segment by #1 ** ds2 = Observed target displacement segment by #2 ** ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2) Where ** B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** dt1^2 (1 B1^2) = dt2^2 . . . (1) Where ** B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** dt1^2 = dt2^2 (1 B2^2) . . . (2) Where ** B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** B1^2 = B2^2 Thus, equations (1) and (2) become the following equations. ** dt1^2 (1 B^2) = dt2^2 . . . (3) ** dt2^2 = dt2^2 (1 B^2) . . . (4) Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can co-exist is when B^2 = 0. Thus, the twins paradox is very real under the Lorentz transform. shrug Within the Lorentz transform, the little professor from Norway, paul andersen, was able to play a mathemagic trick, and he is not alone. In doing so, the Minkowski spacetime was not recognized in his little applet. He is out in the very left field chasing chickens again. shrug Tom and other self-styled physicists have recognized that fault and moved on to claim a mythical proper time flow where all local time flow is a projection of this absolute time flow. Oh, excuse Koobee Wublee. Not absolute time but proper time whatever $hit it is. However, these guys cannot explain why the projection did not cancel out on the traveling twins return trip. So, equations (3) and (4) are still indicating the paradox regardless if projection or not. shrug Well, sooner or later, these bozos are going to wake up someday and ask themselves what the fvck was I thinking?. Guess what? The time projection crap is the last piece of float the self-styled physicists are clinging on to. Take that away. They will sink. That is why the self-styled physicists are very reluctant to give the time projection crap a serious thought. shrug |

What is or is not a paradox?On 31/12/2012 5:04 PM, Koobee Wublee wrote:
On Dec 30, 4:17 pm, Sylvia Else wrote: I started writing a post about this yesterday, then scrubbed it - too What is a paradox in special relativity (hereinafter SR)? I've expressed the view that to contain a paradox, SR has to predict, from different frames, outcomes that are mutually incompatible. An example that comes to mind (though not directly arising) from a recent discussion is that in one frame, there is massive destruction on a citywide scale, and in another other frame, nothing much happens. Clearly, if SR were to make such predictions for two frames, it would have to be regarded as seriously wanting. Of course, it does no such thing. But people seem to want to regard measurements in two frames as mutually incompatible if they give different results. I am at a loss to understand why people would seek to regard those different results as constituting a paradox that invalidates SR (well, leaving intellectual dishonesty aside). From the Lorentz transformations, you can write down the following equation per Minkowski spacetime. Points #1, #2, and #3 are observers. They are observing the same target. ** c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2 Where ** dt1 = Time flow at Point #1 ** dt2 = Time flow at Point #2 ** dt3 = Time flow at Point #3 ** ds1 = Observed target displacement segment by #1 ** ds2 = Observed target displacement segment by #2 ** ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2) Where ** B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** dt1^2 (1 B1^2) = dt2^2 . . . (1) Where ** B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** dt1^2 = dt2^2 (1 B2^2) . . . (2) Where ** B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** B1^2 = B2^2 Thus, equations (1) and (2) become the following equations. ** dt1^2 (1 B^2) = dt2^2 . . . (3) ** dt2^2 = dt2^2 (1 B^2) . . . (4) I assume you meant to write dt1^2 = dt2^2 (1 - B^2) . . . (4) Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can co-exist is when B^2 = 0. Which tells us nothing more than that when two observers observe each other, the situation is symmetrical. Each will measure the same time for equivalent displacements of the other. Or more simply, they share a common relative velocity (save for sign). Thus, the twins paradox is very real under the Lorentz transform. shrug blink Where did that come from? The twin "paradox" involves bringing the two twins back together, which necessitates accelerating at least one of them, making their frame non-inertial./blink Sylvia. |

What is or is not a paradox?On Dec 30, 11:31 pm, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote: From the Lorentz transformations, you can write down the following equation per Minkowski spacetime. Points #1, #2, and #3 are observers. They are observing the same target. ** c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2 Where ** dt1 = Time flow at Point #1 ** dt2 = Time flow at Point #2 ** dt3 = Time flow at Point #3 ** ds1 = Observed target displacement segment by #1 ** ds2 = Observed target displacement segment by #2 ** ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2) Where ** B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** dt1^2 (1 B1^2) = dt2^2 . . . (1) Where ** B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** dt1^2 = dt2^2 (1 B2^2) . . . (2) Where ** B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** B1^2 = B2^2 Thus, equations (1) and (2) become the following equations. ** dt1^2 (1 B^2) = dt2^2 . . . (3) ** dt2^2 = dt2^2 (1 B^2) . . . (4) I assume you meant to write dt1^2 = dt2^2 (1 - B^2) . . . (4) No, Koobee Wublee meant every letter in the equations (3) and (4). shrug Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can co-exist is when B^2 = 0. Which tells us nothing more than that when two observers observe each other, the situation is symmetrical. Each will measure the same time for equivalent displacements of the other. Or more simply, they share a common relative velocity (save for sign). The symmetry is everything about the twins paradox. shrug Thus, the twins paradox is very real under the Lorentz transform. shrug blink Where did that come from? Have you not been reading Koobee Wublee? Did Koobee Wublee not say the Lorentz transform? shrug The twin "paradox" involves bringing the two twins back together, which necessitates accelerating at least one of them, making their frame non-inertial./blink So, you believe in the nonsense of Born? He was the first one to propose acceleration thing breaking the symmetry. Can you show any mathematics that support your/Borns claim? No self-styled physicists have now believed in such nonsense. shrug |

What is or is not a paradox?"Sylvia Else" wrote in message ...
blink Where did that come from? The twin "paradox" involves bringing the two twins back together, which necessitates accelerating at least one of them, making their frame non-inertial./blink Sylvia. ================================================== "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 1/2 tv^2/c^2 second slow." -- Einstein. blink/ Non-inertial? Where did that come from? The twin "paradox" involves bringing the two twins back together, which necessitates keeping one at absolute rest, but the phenomena of electrodynamics as well as of mechanics possess no properties corresponding to the idea of absolute rest. Oh wait, I get it. You are discussing Phuckwit Duck's special relativity, not Einstein's special relativity. /blink -- This message is brought to you from the keyboard of Lord Androcles, Zeroth Earl of Medway. When I get my O.B.E. I'll be an earlobe. |

What is or is not a paradox?On 31/12/2012 7:49 PM, Koobee Wublee wrote:
On Dec 30, 11:31 pm, Sylvia Else wrote: On 31/12/2012 5:04 PM, Koobee Wublee wrote: From the Lorentz transformations, you can write down the following equation per Minkowski spacetime. Points #1, #2, and #3 are observers. They are observing the same target. ** c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2 Where ** dt1 = Time flow at Point #1 ** dt2 = Time flow at Point #2 ** dt3 = Time flow at Point #3 ** ds1 = Observed target displacement segment by #1 ** ds2 = Observed target displacement segment by #2 ** ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2) Where ** B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** dt1^2 (1 B1^2) = dt2^2 . . . (1) Where ** B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** dt1^2 = dt2^2 (1 B2^2) . . . (2) Where ** B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** B1^2 = B2^2 Thus, equations (1) and (2) become the following equations. ** dt1^2 (1 B^2) = dt2^2 . . . (3) ** dt2^2 = dt2^2 (1 B^2) . . . (4) I assume you meant to write dt1^2 = dt2^2 (1 - B^2) . . . (4) No, Koobee Wublee meant every letter in the equations (3) and (4). (2) doesn't become (4) just be writing B for B2. shrug Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can co-exist is when B^2 = 0. Which tells us nothing more than that when two observers observe each other, the situation is symmetrical. Each will measure the same time for equivalent displacements of the other. Or more simply, they share a common relative velocity (save for sign). The symmetry is everything about the twins paradox. shrug In the classical twins paradox, there is no symmetry. The travelling twin has to change velocities in order to be able to get back to the stay at home twin. To get symmetry, both twins have to travel, and if the travel is really symmetrical, their ages will match when they return. Thus, the twins paradox is very real under the Lorentz transform. shrug blink Where did that come from? Have you not been reading Koobee Wublee? Did Koobee Wublee not say the Lorentz transform? shrug The twin "paradox" involves bringing the two twins back together, which necessitates accelerating at least one of them, making their frame non-inertial./blink So, you believe in the nonsense of Born? He was the first one to propose acceleration thing breaking the symmetry. Can you show any mathematics that support your/Borns claim? No self-styled physicists have now believed in such nonsense. shrug The symmetry can be broken without acceleration though to bring an actual person back then involves cloning. It's simpler to forget the twin, and just take a clock whose time is copied onto another clock going in the opposite direction halfway through the travel. But the symmetry is still broken, and once that happens, you have no paradox. Sylvia. |

What is or is not a paradox?On Dec 31, 2:31*am, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote: On Dec 30, 4:17 pm, Sylvia Else wrote: I started writing a post about this yesterday, then scrubbed it - too What is a paradox in special relativity (hereinafter SR)? I've expressed the view that to contain a paradox, SR has to predict, from different frames, outcomes that are mutually incompatible. An example that comes to mind (though not directly arising) from a recent discussion is that in one frame, there is massive destruction on a citywide scale, and in another other frame, nothing much happens. Clearly, if SR were to make such predictions for two frames, it would have to be regarded as seriously wanting. Of course, it does no such thing. But people seem to want to regard measurements in two frames as mutually incompatible if they give different results. I am at a loss to understand why people would seek to regard those different results as constituting a paradox that invalidates SR (well, leaving intellectual dishonesty aside). *From the Lorentz transformations, you can write down the following equation per Minkowski spacetime. *Points #1, #2, and #3 are observers. *They are observing the same target. ** *c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2 Where ** *dt1 = Time flow at Point #1 ** *dt2 = Time flow at Point #2 ** *dt3 = Time flow at Point #3 ** *ds1 = Observed target displacement segment by #1 ** *ds2 = Observed target displacement segment by #2 ** *ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** *dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2) Where ** *B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** *dt1^2 (1 B1^2) = dt2^2 . . . (1) Where ** *B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** *dt1^2 = dt2^2 (1 B2^2) . . . (2) Where ** *B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** *B1^2 = B2^2 Thus, equations (1) and (2) become the following equations. ** *dt1^2 (1 B^2) = dt2^2 . . . (3) ** *dt2^2 = dt2^2 (1 B^2) . . . (4) I assume you meant to write dt1^2 = dt2^2 (1 - B^2) . . . (4) Where ** *B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can co-exist is when B^2 = 0. Which tells us nothing more than that when two observers observe each other, the situation is symmetrical. Each will measure the same time for equivalent displacements of the other. Or more simply, they share a common relative velocity (save for sign). Thus, the twins paradox is very real under the Lorentz transform. shrug blink Where did that come from? The twin "paradox" involves bringing the two twins back together, which necessitates accelerating at least one of them, making their frame non-inertial./blink There is no inertial frame exists on earth ....does that mean that SR is not valid on earth? |

What is or is not a paradox?On Dec 31 2012, 1:48 am, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote: From the Lorentz transformations, you can write down the following equation per Minkowski spacetime. Points #1, #2, and #3 are observers. They are observing the same target. ** c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2 Where ** dt1 = Time flow at Point #1 ** dt2 = Time flow at Point #2 ** dt3 = Time flow at Point #3 ** ds1 = Observed target displacement segment by #1 ** ds2 = Observed target displacement segment by #2 ** ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2) Where ** B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** dt1^2 (1 B1^2) = dt2^2 . . . (1) Where ** B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** dt1^2 = dt2^2 (1 B2^2) . . . (2) Where ** B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** B1^2 = B2^2 Thus, equations (1) and (2) become the following equations [respectively]. ** dt1^2 (1 B^2) = dt2^2 . . . (3) ** dt1^2 = dt2^2 (1 B^2) . . . (4) Where ** B^2 = B1^2 = B2^2 (2) doesn't become (4) just be writing B for B2. Are you complaining about the typo? It is corrected above. shrug The only time the equations (3) and (4) can co-exist is when B^2 = 0. The symmetry is everything about the twins paradox. shrug In the classical twins paradox, there is no symmetry. The travelling twin has to change velocities in order to be able to get back to the stay at home twin. [snip more nonsense] This is the second time, you are asked to show the math that shows this acceleration breaking the symmetry nonsense. There is no way you can, and that is because you are totally wrong just like Born. shrug So, you believe in the nonsense of Born? He was the first one to propose acceleration thing breaking the symmetry. Can you show any mathematics that support your/Borns claim? No self-styled physicists have now believed in such nonsense. shrug The symmetry can be broken without acceleration though to bring an actual person back then involves cloning. It's simpler to forget the twin, and just take a clock whose time is copied onto another clock going in the opposite direction halfway through the travel. But the symmetry is still broken, and once that happens, you have no paradox. You have no idea what you are talking about, and there is no need to discuss any further. shrug |

What is or is not a paradox?On Jan 1, 11:46 am, Tom Roberts wrote:
OK, consider the case in which "home" is at rest in an inertial frame; one twin stays at home in a centrifuge, spinning at a few tens of meters per second with a proper acceleration of 1g; the other twin blasts off in a rocket accelerating at 1g for a year on his clock, turns around with a maneuver that maintains the 1g acceleration, and returns home executing another maneuver that beings him to rest at home (the trip takes 4 years on the rocket twin's clock: 1y speed up headed out, 1y slow down, 1y speed up headed back, 1y slow down, stop). The rocket twin is unequivocally younger when they meet again, because his average speed relative to the inertial frame of home is much greater than that of the centrifuge twin, even though their proper accelerations are equal. Just as Koobee Wublee has predicted: http://groups.google.com/group/sci.p...c3288cbc93ca08 ** Desperation facing with paradox leads to ** Divine visions of showing no paradox to ** Euphoria of rejoicing no paradox to ** Careful examination of these divine visions to ** what the fvck was I thinking? rude awakening to ** Desperation once again and the cycle repeats. shrug The divine visions a 1) Acceleration breaking the symmetry by Born 2) Spacetime time diagram of a few lines 3) Mathemagic trick by paul andersen and others 4) Coordinate time as projection of proper time The last time, Tom was in 4) so happy with the projection thing. Apparently, he had gone through the rude awakening stage and more desperation. Now, he is back to the first one. That is the life of a self-styled physicist. Koobee Wublee thinks that is a fvcked up life style, but what does Koobee Wublee know? Perhaps, Tom enjoys stuck in these cycles chasing his own tail. shrug Anyway, one can simply redesign the scenario to have both twins traveling where each twin will experience the same acceleration profile and thus nullifying the effect of acceleration. Given an arbitrary amount of time to allow both twins coasting with null acceleration but with a significant speed between them, the mutual time dilation building up will obviously never be rectified. Thus, showing the paradox is very real. shrug Bottom line: here it is speed relative to the inertial frame of home that matters, not acceleration. The actual computation involves integrating the metric over the path followed by each twin between meetings, and comparing the results; when written in terms of home's inertial coordinates, this integral involves only speed, not acceleration, not position, and not direction. Remember that in SR (no gravity) it simply is not possible for two twins that both move inertially to separate and rejoin. Tom, care to show some math to back up your babbling? shrug |

What is or is not a paradox?On Dec 31 2012, 11:58*am, kenseto wrote:
On Dec 31, 2:31*am, Sylvia Else wrote: On 31/12/2012 5:04 PM, Koobee Wublee wrote: On Dec 30, 4:17 pm, Sylvia Else wrote: I started writing a post about this yesterday, then scrubbed it - too What is a paradox in special relativity (hereinafter SR)? I've expressed the view that to contain a paradox, SR has to predict, from different frames, outcomes that are mutually incompatible. An example that comes to mind (though not directly arising) from a recent discussion is that in one frame, there is massive destruction on a citywide scale, and in another other frame, nothing much happens. Clearly, if SR were to make such predictions for two frames, it would have to be regarded as seriously wanting. Of course, it does no such thing. But people seem to want to regard measurements in two frames as mutually incompatible if they give different results. I am at a loss to understand why people would seek to regard those different results as constituting a paradox that invalidates SR (well, leaving intellectual dishonesty aside). *From the Lorentz transformations, you can write down the following equation per Minkowski spacetime. *Points #1, #2, and #3 are observers. *They are observing the same target. ** *c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2 Where ** *dt1 = Time flow at Point #1 ** *dt2 = Time flow at Point #2 ** *dt3 = Time flow at Point #3 ** *ds1 = Observed target displacement segment by #1 ** *ds2 = Observed target displacement segment by #2 ** *ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** *dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2) Where ** *B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** *dt1^2 (1 B1^2) = dt2^2 . . . (1) Where ** *B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** *dt1^2 = dt2^2 (1 B2^2) . . . (2) Where ** *B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** *B1^2 = B2^2 Thus, equations (1) and (2) become the following equations. ** *dt1^2 (1 B^2) = dt2^2 . . . (3) ** *dt2^2 = dt2^2 (1 B^2) . . . (4) I assume you meant to write dt1^2 = dt2^2 (1 - B^2) . . . (4) Where ** *B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can co-exist is when B^2 = 0. |

What is or is not a paradox?On Wed, 02 Jan 2013 07:17:18 -0800, G=EMC^2 wrote:
On Dec 31 2012, 11:58Β*am, kenseto wrote: On Dec 31, 2:31Β*am, Sylvia Else wrote: On 31/12/2012 5:04 PM, Koobee Wublee wrote: On Dec 30, 4:17 pm, Sylvia Else wrote: I started writing a post about this yesterday, then scrubbed it - too What is a paradox in special relativity (hereinafter SR)? I've expressed the view that to contain a paradox, SR has to predict, from different frames, outcomes that are mutually incompatible. An example that comes to mind (though not directly arising) from a recent discussion is that in one frame, there is massive destruction on a citywide scale, and in another other frame, nothing much happens. Clearly, if SR were to make such predictions for two frames, it would have to be regarded as seriously wanting. Of course, it does no such thing. But people seem to want to regard measurements in two frames as mutually incompatible if they give different results. I am at a loss to understand why people would seek to regard those different results as constituting a paradox that invalidates SR (well, leaving intellectual dishonesty aside). Β*From the Lorentz transformations, you can write down the Β*following equation per Minkowski spacetime. Β*Points #1, #2, and #3 are observers. Β*They are observing the same target. ** Β*c^2 dt1^2 β ds1^2 = c^2 dt2^2 β ds2^2 = c^2 dt3^2 β ds3^2 Where ** Β*dt1 = Time flow at Point #1 ** Β*dt2 = Time flow at Point #2 ** Β*dt3 = Time flow at Point #3 ** Β*ds1 = Observed target displacement segment by #1 ** Β*ds2 = Observed target displacement segment by #2 ** Β*ds3 = Observed target displacement segment by #3 The above spacetime equation can also be written as follows. ** Β*dt1^2 (1 β B1^2) = dt2^2 (1 β B2^2) = dt3^2 (1 β B3^2) Where ** Β*B^2 = (ds/dt)^2 / c^2 When #1 is observing #2, the following equation can be deduced from the equation above. ** Β*dt1^2 (1 β B1^2) = dt2^2 . . . (1) Where ** Β*B2^2 = 0, #2 is observing itself Similarly, when #2 is observing #1, the following equation can be deduced. ** Β*dt1^2 = dt2^2 (1 β B2^2) . . . (2) Where ** Β*B1^2 = 0, #1 is observing itself According to relativity, the following must be true. ** Β*B1^2 = B2^2 Thus, equations (1) and (2) become the following equations. ** Β*dt1^2 (1 β B^2) = dt2^2 . . . (3) ** Β*dt2^2 = dt2^2 (1 β B^2) . . . (4) I assume you meant to write dt1^2 = dt2^2 (1 - B^2) . . . (4) Where ** Β*B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can co-exist is when B^2 = 0. Which tells us nothing more than that when two observers observe each other, the situation is symmetrical. Each will measure the same time for equivalent displacements of the other. Or more simply, they share a common relative velocity (save for sign). Thus, the twinsβ paradox is very real under the Lorentz transform. shrug blink Where did that come from? The twin "paradox" involves bringing the two twins back together, which necessitates accelerating at least one of them, making their frame non-inertial./blink There is no inertial frame exists on earth ....does that mean that SR is not valid on earth? Well think of this. "Time in a plane flying east is less than that for those flying west". The Earth speed of rotation sees to it. Get the picture TreBert Treeb is right. Every schoolkid knows that if you fly east, it's a time machine. Every time you go around the earth you go back in time a day! Get the Picture? |

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