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#11




Simplified Twin Paradox Resolution.
"Sylvia Else" wrote in message ...
On 6/01/2013 7:33 PM, Koobee Wublee wrote: With all these parameters derived, all we have to do is to compare the time elapses of the observers own clock versus whoever it is observing. So, what is the problem, PD? shrug They're in different places. How are you going to do the comparing? It's not a trivial question. Sylvia. ============================================= Bwahahahahahahaha! Of course its a trivial question, the rod has always been in a different place to the barn. just find a coordinate transformation between them. You are supposed to know all about coordinate transformations. Show us the Phuckwit Duck Transform.  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. 
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#12




Simplified Twin Paradox Resolution.
On Jan 6, 9:59*am, Koobee Wublee wrote:
On Jan 5, 5:57 pm, Sylvia Else wrote: On 5/01/2013 5:59 AM, Koobee Wublee wrote: Instead of v, lets say (B = v / c) for simplicity. *The earth is Point #0, outbound spacecraft is Point #1, and inbound spacecraft is Point #2. According to the Lorentz transform, relative speeds a ** *B_00^2 = 0, speed of #0 as observed by #0 ** *B_01^2 = B^2, speed of #1 as observed by #0 ** *B_02^2 = B^2, speed of #2 as observed by #0 ** *B_10^2 = B^2, speed of #0 as observed by #1 ** *B_11^2 = 0, speed of #1 as observed by #1 ** *B_12^2 = 4 B^2 / (1 B^2), speed of #2 as observed by #1 ** *B_20^2 = B^2, speed of #0 as observed by #2 ** *B_21^2 = 4 B^2 / (1 B^2), speed of #1 as observed by #2 ** *B_22^2 = 0, speed of #2 as observed by #2 When Point #0 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_00^2 (1 B_00^2) = dt_10^2 (1 B_10^2) = dt_20^2 (1 B_20^2) When Point #1 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_01^2 (1 B_01^2) = dt_11^2 (1 B_11^2) = dt_21^2 (1 B_21^2) When Point #2 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_02^2 (1 B_02^2) = dt_12^2 (1 B_12^2) = dt_22^2 (1 B_22^2) Where ** *dt_00 = Local rate of time flow at Point #0 ** *dt_01 = Rate of time flow at #1 as observed by #0 ** *dt_02 = Rate of time flow at #2 as observed by #0 ** *dt_10 = Rate of time flow at #0 as observed by #1 ** *dt_11 = Local rate of time flow at Point #1 ** *dt_12 = Rate of time flow at #2 as observed by #1 ** *dt_20 = Rate of time flow at #0 as observed by #2 ** *dt_21 = Rate of time flow at #1 as observed by #2 ** *dt_22 = Local rate of time flow at Point #2 So, with all the pertinent variables identified, the contradiction of the twins paradox is glaring right at anyone with a thinking brain.. shrug You assert that there are a paradox. I take it you mean in the sense that the theory gives two results for one situation, such that they are impossible to reconcile. I challenge you to show that mathematically, rather than just asserting it. Do not just point at the maths above and claim that it's obvious. PD, are you turning into a troll now? *For the nth time, the following is one such presentation of mathematics that show the contradiction in the twins paradox.    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) ** *dt2^2 = dt1^2 (1 B^2) . . . (4) Where ** *B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can coexist is when B^2 = 0. Thus, the twins paradox is very real under the Lorentz transform. shrug They do not agree with the symmetry in the above equations. Because, it is due to symmetry twin paradox exists. They say that time dilation is one way effect. They can achieve this only by making the situation asymmetrical. Situation IS asymmetrical if we tag a frame that undergoes actual acceleration. But this is against the basic principles of SR which deals with uniform relative motion and nothing else. 
#13




Simplified Twin Paradox Resolution.
On 7/01/2013 12:21 AM, Vilas Tamhane wrote:
On Jan 6, 9:59 am, Koobee Wublee wrote: On Jan 5, 5:57 pm, Sylvia Else wrote: On 5/01/2013 5:59 AM, Koobee Wublee wrote: Instead of v, lets say (B = v / c) for simplicity. The earth is Point #0, outbound spacecraft is Point #1, and inbound spacecraft is Point #2. According to the Lorentz transform, relative speeds a ** B_00^2 = 0, speed of #0 as observed by #0 ** B_01^2 = B^2, speed of #1 as observed by #0 ** B_02^2 = B^2, speed of #2 as observed by #0 ** B_10^2 = B^2, speed of #0 as observed by #1 ** B_11^2 = 0, speed of #1 as observed by #1 ** B_12^2 = 4 B^2 / (1 B^2), speed of #2 as observed by #1 ** B_20^2 = B^2, speed of #0 as observed by #2 ** B_21^2 = 4 B^2 / (1 B^2), speed of #1 as observed by #2 ** B_22^2 = 0, speed of #2 as observed by #2 When Point #0 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_00^2 (1 B_00^2) = dt_10^2 (1 B_10^2) = dt_20^2 (1 B_20^2) When Point #1 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_01^2 (1 B_01^2) = dt_11^2 (1 B_11^2) = dt_21^2 (1 B_21^2) When Point #2 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_02^2 (1 B_02^2) = dt_12^2 (1 B_12^2) = dt_22^2 (1 B_22^2) Where ** dt_00 = Local rate of time flow at Point #0 ** dt_01 = Rate of time flow at #1 as observed by #0 ** dt_02 = Rate of time flow at #2 as observed by #0 ** dt_10 = Rate of time flow at #0 as observed by #1 ** dt_11 = Local rate of time flow at Point #1 ** dt_12 = Rate of time flow at #2 as observed by #1 ** dt_20 = Rate of time flow at #0 as observed by #2 ** dt_21 = Rate of time flow at #1 as observed by #2 ** dt_22 = Local rate of time flow at Point #2 So, with all the pertinent variables identified, the contradiction of the twins paradox is glaring right at anyone with a thinking brain. shrug You assert that there are a paradox. I take it you mean in the sense that the theory gives two results for one situation, such that they are impossible to reconcile. I challenge you to show that mathematically, rather than just asserting it. Do not just point at the maths above and claim that it's obvious. PD, are you turning into a troll now? For the nth time, the following is one such presentation of mathematics that show the contradiction in the twins paradox.    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) ** dt2^2 = dt1^2 (1 B^2) . . . (4) Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can coexist is when B^2 = 0. Thus, the twins paradox is very real under the Lorentz transform. shrug They do not agree with the symmetry in the above equations. Because, it is due to symmetry twin paradox exists. They say that time dilation is one way effect. They can achieve this only by making the situation asymmetrical. Situation IS asymmetrical if we tag a frame that undergoes actual acceleration. But this is against the basic principles of SR which deals with uniform relative motion and nothing else. The formulation of special relativity looks at uniform relative motion, but it doesn't exclude the exchange of information at the speed of light (or indeed, at less than speed of light), including exchanges between observers who are in relative motion. The experiment I proposed in the orignal posting involves only uniform relative motion and exchanges of information at the speed of light. Futher the analysis shows an absence of inconsistencies in the rate at which clocks measure time, and are seen to measure time. You're attempting to assert that a twin paradox exists, but in scenarios that special relativity is incapable of addressing. Leaving aside whether that says anything about twin paradoxes in special relativity, it is any case just wrong. Sylvia. 
#14




Simplified Twin Paradox Resolution.
On 06.01.2013 07:23, Sylvia Else wrote:
On 6/01/2013 3:59 PM, Koobee Wublee wrote: On Jan 5, 5:57 pm, Sylvia Else wrote: On 5/01/2013 5:59 AM, Koobee Wublee wrote: Instead of v, lets say (B = v / c) for simplicity. The earth is Point #0, outbound spacecraft is Point #1, and inbound spacecraft is Point #2. According to the Lorentz transform, relative speeds a ** B_00^2 = 0, speed of #0 as observed by #0 ** B_01^2 = B^2, speed of #1 as observed by #0 ** B_02^2 = B^2, speed of #2 as observed by #0 ** B_10^2 = B^2, speed of #0 as observed by #1 ** B_11^2 = 0, speed of #1 as observed by #1 ** B_12^2 = 4 B^2 / (1 B^2), speed of #2 as observed by #1 ** B_20^2 = B^2, speed of #0 as observed by #2 ** B_21^2 = 4 B^2 / (1 B^2), speed of #1 as observed by #2 ** B_22^2 = 0, speed of #2 as observed by #2 When Point #0 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_00^2 (1 B_00^2) = dt_10^2 (1 B_10^2) = dt_20^2 (1 B_20^2) When Point #1 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_01^2 (1 B_01^2) = dt_11^2 (1 B_11^2) = dt_21^2 (1 B_21^2) When Point #2 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_02^2 (1 B_02^2) = dt_12^2 (1 B_12^2) = dt_22^2 (1 B_22^2) Where ** dt_00 = Local rate of time flow at Point #0 ** dt_01 = Rate of time flow at #1 as observed by #0 ** dt_02 = Rate of time flow at #2 as observed by #0 ** dt_10 = Rate of time flow at #0 as observed by #1 ** dt_11 = Local rate of time flow at Point #1 ** dt_12 = Rate of time flow at #2 as observed by #1 ** dt_20 = Rate of time flow at #0 as observed by #2 ** dt_21 = Rate of time flow at #1 as observed by #2 ** dt_22 = Local rate of time flow at Point #2 So, with all the pertinent variables identified, the contradiction of the twins paradox is glaring right at anyone with a thinking brain. shrug You assert that there are a paradox. I take it you mean in the sense that the theory gives two results for one situation, such that they are impossible to reconcile. I challenge you to show that mathematically, rather than just asserting it. Do not just point at the maths above and claim that it's obvious. PD, are you turning into a troll now? For the nth time, the following is one such presentation of mathematics that show the contradiction in the twins paradox.    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) ** dt2^2 = dt1^2 (1 B^2) . . . (4) Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can coexist is... ... never In deriving [1] and [2] you prefaced them with caveats about who is observing whom. So they relate to different measurement situations. You cannot combine them in any meaningful way. Sylvia. It's a variant of the old Dingle argument, @t1/@t2 = @t2/@t1 is a contradiction. (@ = partial derivative) See: http://tinyurl.com/ah3ctmm Koobee's response: http://tinyurl.com/a9jkwxp What Koobee Wublee wrote that you have quoted was an application of the Lorentz transform in a specific scenario. You dont understand all that, and apparently, you dont know what you are talking about as usual. It is laughable that a college professor from the University of Trondheim would attempt to swindle his way out using irrelevant, bull**** claims. shrug You are cornered. Why dont you stay in the topic of discussion? shrug His arguments were as lethal and to the point as always. :)  Paul http://www.gethome.no/paulba/ 
#15




Simplified Twin Paradox Resolution.
On Jan 6, 11:23*am, Sylvia Else wrote:
On 6/01/2013 3:59 PM, Koobee Wublee wrote: On Jan 5, 5:57 pm, Sylvia Else wrote: On 5/01/2013 5:59 AM, Koobee Wublee wrote: Instead of v, lets say (B = v / c) for simplicity. *The earth is Point #0, outbound spacecraft is Point #1, and inbound spacecraft is Point #2. According to the Lorentz transform, relative speeds a ** *B_00^2 = 0, speed of #0 as observed by #0 ** *B_01^2 = B^2, speed of #1 as observed by #0 ** *B_02^2 = B^2, speed of #2 as observed by #0 ** *B_10^2 = B^2, speed of #0 as observed by #1 ** *B_11^2 = 0, speed of #1 as observed by #1 ** *B_12^2 = 4 B^2 / (1 B^2), speed of #2 as observed by #1 ** *B_20^2 = B^2, speed of #0 as observed by #2 ** *B_21^2 = 4 B^2 / (1 B^2), speed of #1 as observed by #2 ** *B_22^2 = 0, speed of #2 as observed by #2 When Point #0 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_00^2 (1 B_00^2) = dt_10^2 (1 B_10^2) = dt_20^2 (1 B_20^2) When Point #1 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_01^2 (1 B_01^2) = dt_11^2 (1 B_11^2) = dt_21^2 (1 B_21^2) When Point #2 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_02^2 (1 B_02^2) = dt_12^2 (1 B_12^2) = dt_22^2 (1 B_22^2) Where ** *dt_00 = Local rate of time flow at Point #0 ** *dt_01 = Rate of time flow at #1 as observed by #0 ** *dt_02 = Rate of time flow at #2 as observed by #0 ** *dt_10 = Rate of time flow at #0 as observed by #1 ** *dt_11 = Local rate of time flow at Point #1 ** *dt_12 = Rate of time flow at #2 as observed by #1 ** *dt_20 = Rate of time flow at #0 as observed by #2 ** *dt_21 = Rate of time flow at #1 as observed by #2 ** *dt_22 = Local rate of time flow at Point #2 So, with all the pertinent variables identified, the contradiction of the twins paradox is glaring right at anyone with a thinking brain.. shrug You assert that there are a paradox. I take it you mean in the sense that the theory gives two results for one situation, such that they are impossible to reconcile. I challenge you to show that mathematically, rather than just asserting it. Do not just point at the maths above and claim that it's obvious. PD, are you turning into a troll now? *For the nth time, the following is one such presentation of mathematics that show the contradiction in the twins paradox.    *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) ** *dt2^2 = dt1^2 (1 B^2) . . . (4) Where ** *B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can coexist is... ... never In deriving [1] and [2] you prefaced them with caveats about who is observing whom. So they relate to different measurement situations. You cannot combine them in any meaningful way. Sylvia. Nobody is observing anybody. Everybody has right to write down equations. 
#16




Simplified Twin Paradox Resolution.
On 7/01/2013 1:25 AM, Vilas Tamhane wrote:
On Jan 6, 11:23 am, Sylvia Else wrote: On 6/01/2013 3:59 PM, Koobee Wublee wrote: On Jan 5, 5:57 pm, Sylvia Else wrote: On 5/01/2013 5:59 AM, Koobee Wublee wrote: Instead of v, lets say (B = v / c) for simplicity. The earth is Point #0, outbound spacecraft is Point #1, and inbound spacecraft is Point #2. According to the Lorentz transform, relative speeds a ** B_00^2 = 0, speed of #0 as observed by #0 ** B_01^2 = B^2, speed of #1 as observed by #0 ** B_02^2 = B^2, speed of #2 as observed by #0 ** B_10^2 = B^2, speed of #0 as observed by #1 ** B_11^2 = 0, speed of #1 as observed by #1 ** B_12^2 = 4 B^2 / (1 B^2), speed of #2 as observed by #1 ** B_20^2 = B^2, speed of #0 as observed by #2 ** B_21^2 = 4 B^2 / (1 B^2), speed of #1 as observed by #2 ** B_22^2 = 0, speed of #2 as observed by #2 When Point #0 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_00^2 (1 B_00^2) = dt_10^2 (1 B_10^2) = dt_20^2 (1 B_20^2) When Point #1 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_01^2 (1 B_01^2) = dt_11^2 (1 B_11^2) = dt_21^2 (1 B_21^2) When Point #2 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_02^2 (1 B_02^2) = dt_12^2 (1 B_12^2) = dt_22^2 (1 B_22^2) Where ** dt_00 = Local rate of time flow at Point #0 ** dt_01 = Rate of time flow at #1 as observed by #0 ** dt_02 = Rate of time flow at #2 as observed by #0 ** dt_10 = Rate of time flow at #0 as observed by #1 ** dt_11 = Local rate of time flow at Point #1 ** dt_12 = Rate of time flow at #2 as observed by #1 ** dt_20 = Rate of time flow at #0 as observed by #2 ** dt_21 = Rate of time flow at #1 as observed by #2 ** dt_22 = Local rate of time flow at Point #2 So, with all the pertinent variables identified, the contradiction of the twins paradox is glaring right at anyone with a thinking brain. shrug You assert that there are a paradox. I take it you mean in the sense that the theory gives two results for one situation, such that they are impossible to reconcile. I challenge you to show that mathematically, rather than just asserting it. Do not just point at the maths above and claim that it's obvious. PD, are you turning into a troll now? For the nth time, the following is one such presentation of mathematics that show the contradiction in the twins paradox.    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) ** dt2^2 = dt1^2 (1 B^2) . . . (4) Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can coexist is... ... never In deriving [1] and [2] you prefaced them with caveats about who is observing whom. So they relate to different measurement situations. You cannot combine them in any meaningful way. Sylvia. Nobody is observing anybody. Everybody has right to write down equations. And then? Koobee is trying to show that the equations contain an inconsistency, which is to say, that they cannot be a description of any reality. Sylvia. 
#17




Simplified Twin Paradox Resolution.
On Jan 6, 5:47 am, "Paul B. Andersen" wrote:
On 6/01/2013 3:59 PM, Koobee Wublee wrote: Instead of v, lets say (B = v / c) for simplicity. The earth is Point #0, outbound spacecraft is Point #1, and inbound spacecraft is Point #2. According to the Lorentz transform, relative speeds a ** B_00^2 = 0, speed of #0 as observed by #0 ** B_01^2 = B^2, speed of #1 as observed by #0 ** B_02^2 = B^2, speed of #2 as observed by #0 ** B_10^2 = B^2, speed of #0 as observed by #1 ** B_11^2 = 0, speed of #1 as observed by #1 ** B_12^2 = 4 B^2 / (1 B^2), speed of #2 as observed by #1 ** B_20^2 = B^2, speed of #0 as observed by #2 ** B_21^2 = 4 B^2 / (1 B^2), speed of #1 as observed by #2 ** B_22^2 = 0, speed of #2 as observed by #2 When Point #0 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_00^2 (1 B_00^2) = dt_10^2 (1 B_10^2) = dt_20^2 (1 B_20^2) When Point #1 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_01^2 (1 B_01^2) = dt_11^2 (1 B_11^2) = dt_21^2 (1 B_21^2) When Point #2 is observed by all, the Minkowski spacetime (divided by c^2) is: ** dt_02^2 (1 B_02^2) = dt_12^2 (1 B_12^2) = dt_22^2 (1 B_22^2) Where ** dt_00 = Local rate of time flow at Point #0 ** dt_01 = Rate of time flow at #1 as observed by #0 ** dt_02 = Rate of time flow at #2 as observed by #0 ** dt_10 = Rate of time flow at #0 as observed by #1 ** dt_11 = Local rate of time flow at Point #1 ** dt_12 = Rate of time flow at #2 as observed by #1 ** dt_20 = Rate of time flow at #0 as observed by #2 ** dt_21 = Rate of time flow at #1 as observed by #2 ** dt_22 = Local rate of time flow at Point #2 So, with all the pertinent variables identified, the contradiction of the twins paradox is glaring right at anyone with a thinking brain. shrug    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) ** dt2^2 = dt1^2 (1 B^2) . . . (4) Where ** B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can coexist is when B^2 = 0. Thus, the twins paradox is very real under the Lorentz transform. shrug It's a variant of the old Dingle argument, @t1/@t2 = @t2/@t1 is a contradiction. (@ = partial derivative) See: http://tinyurl.com/ah3ctmm Koobee's response: http://tinyurl.com/a9jkwxp What Koobee Wublee wrote that you have quoted was an application of the Lorentz transform in a specific scenario. You dont understand all that, and apparently, you dont know what you are talking about as usual. It is laughable that a college professor from the University of Trondheim would attempt to swindle his way out using irrelevant, bull**** claims. shrug You are cornered. Why dont you stay in the topic of discussion? shrug Excellent documentations, paul. When you are plagued with these embarrassing blunders, at least, you have a skill in good documentations. Koobee Wublee is indeed very grateful that you are able to document His great posts. Seriously, paul. All that good documentations still did not save you from that job in the private industry, did it? When someone charging in claiming a Doppler shift in 10^8 should be seriously considered in adjusting the carrier frequencies for compensations, the management just have to do the best for either parties. :) By the way, Koobee Wublee never uses the partial derivative like what you have done. dt still basically the rate of time flow when comparing two observers. Thus, total derivative has to be considered. shrug Or better yet, if you are still confused with the Lorentz transform, why dont you look into the equations describing Minkowski spacetime which Koobee Wublee has included in this post? That should leave no confusion about what Dingle had to say was actually visionary. Well, not quite. The stuff is so simple that it is a big surprise when all the socalled bright minds in the scientific communities have so much trouble understanding. What a shame, no? shrug His arguments were as lethal and to the point as always. :) You bet, paul. Glad you are finding amusement amid these gross blunders of yours. In doing so, you started personal attacks. Not until Koobee Wublee pointed out to you, you have now calmed down. By the way, have you finished the JAVA applet yet with the twins traveling using the exact same acceleration profile? Please also leave an adjustable coasting time with no acceleration in the program. please and thanks in advance 
#18




Simplified Twin Paradox Resolution.
On 07.01.2013 07:28, Koobee Wublee wrote:
Seriously, paul. All that good documentations still did not save you from that job in the private industry, did it? When someone charging in claiming a Doppler shift in 10^8 should be seriously considered in adjusting the carrier frequencies for compensations, the management just have to do the best for either parties. :) This is about the GPS again. What "someone charging in claiming": http://tinyurl.com/bdzm4k The reason, and only reason, why the frequency standard is corrected for relativistic effects is to make the SV clock run synchronously with the ground clocks. That the carrier and shipping frequencies also are adjusted is just a side effect because all frequencies are derived from the same frequency standard. The carrier frequencies, like all other frequencies, are at the receiver Doppler shifted between ± 3E7. The satellites are moving! This is equivalent to a frequency shift of the carriers in the order of kHz. Since the bandwidth of the channels is ca. 20 MHz, this is of no consequence for the receiver. The Doppler shift may be almost a thousand times more than the minute GRcorrection, so of bloody course the 4.4647E10 offset is of no consequence whatsoever for the receiver! AND NOBODY EVER SAID OTHERWISE! How can "the Doppler shift is of no consequence for the receiver" become "a Doppler shift in 10^8 should be seriously considered in adjusting the carrier frequencies for compensations" Either you have a serious reading comprehension problem, or you are deliberately lying about what I have said. I suspect the latter.  Paul http://www.gethome.no/paulba/ 
#19




Simplified Twin Paradox Resolution.
On Jan 7, 11:28*am, Koobee Wublee wrote:
On Jan 6, 5:47 am, "Paul B. Andersen" wrote: On 6/01/2013 3:59 PM, Koobee Wublee wrote: Instead of v, lets say (B = v / c) for simplicity. *The earth is Point #0, outbound spacecraft is Point #1, and inbound spacecraft is Point #2. According to the Lorentz transform, relative speeds a ** *B_00^2 = 0, speed of #0 as observed by #0 ** *B_01^2 = B^2, speed of #1 as observed by #0 ** *B_02^2 = B^2, speed of #2 as observed by #0 ** *B_10^2 = B^2, speed of #0 as observed by #1 ** *B_11^2 = 0, speed of #1 as observed by #1 ** *B_12^2 = 4 B^2 / (1 B^2), speed of #2 as observed by #1 ** *B_20^2 = B^2, speed of #0 as observed by #2 ** *B_21^2 = 4 B^2 / (1 B^2), speed of #1 as observed by #2 ** *B_22^2 = 0, speed of #2 as observed by #2 When Point #0 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_00^2 (1 B_00^2) = dt_10^2 (1 B_10^2) = dt_20^2 (1 B_20^2) When Point #1 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_01^2 (1 B_01^2) = dt_11^2 (1 B_11^2) = dt_21^2 (1 B_21^2) When Point #2 is observed by all, the Minkowski spacetime (divided by c^2) is: ** *dt_02^2 (1 B_02^2) = dt_12^2 (1 B_12^2) = dt_22^2 (1 B_22^2) Where ** *dt_00 = Local rate of time flow at Point #0 ** *dt_01 = Rate of time flow at #1 as observed by #0 ** *dt_02 = Rate of time flow at #2 as observed by #0 ** *dt_10 = Rate of time flow at #0 as observed by #1 ** *dt_11 = Local rate of time flow at Point #1 ** *dt_12 = Rate of time flow at #2 as observed by #1 ** *dt_20 = Rate of time flow at #0 as observed by #2 ** *dt_21 = Rate of time flow at #1 as observed by #2 ** *dt_22 = Local rate of time flow at Point #2 So, with all the pertinent variables identified, the contradiction of the twins paradox is glaring right at anyone with a thinking brain.. shrug    *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) ** *dt2^2 = dt1^2 (1 B^2) . . . (4) Where ** *B^2 = B1^2 = B2^2 The only time the equations (3) and (4) can coexist is when B^2 = 0. 
#20




Simplified Twin Paradox Resolution.
On Jan 7, 6:41*am, old cretin Vilas Tamhane
wrote: This proposed reciprocity is exactly the foolishness that SR is based on.. Imbecile, you still don't get it, do you? 
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