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#1




Simplified Twin Paradox Resolution.
On Jan 3, 5:52 pm, Sylvia Else wrote:
Instead, on Earth there is a clock and a camcorder with a very powerful telescopic lens and a tranceiver There is also a spacecraft travelling at velocity v towards Earth, similarly equipped. For simplicity, we treat Earth + clock + camcorder + transceivers as a single point. Similarly for the spacecraft. After some time T in its own frame, the spacecraft encounters a most similar spacecraft headed towards Earth at velocity v relative to Earth. As they pass, the first spacecraft transmits its entire camcorder recording to the second spacecraft, and the second spacecraft's clock is set to the value shown by the first spacecraft's clock. The camcorder on the second spacecraft then starts recording, continuing the recording it just received. 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 Did PD get it right? [snipped garbage] Of course not, what do you expect from an Einstein Dingleberry? shrug After abandoning the projection crap, is Tom very busy trying to find an excuse to resolve this through projection again? Koobee Wublee is not surprised since Tom has already decided SR as a valid hypothesis despite lack of any professional validations. shrug POINT BLANK IN YOUR FACES, PD, PAUL (ANDERSEN), AND TOM (ROBERTS)! shrug 
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#2




Simplified Twin Paradox Resolution.
On 4/01/2013 5:03 PM, Koobee Wublee wrote:
On Jan 3, 5:52 pm, Sylvia Else wrote: Instead, on Earth there is a clock and a camcorder with a very powerful telescopic lens and a tranceiver There is also a spacecraft travelling at velocity v towards Earth, similarly equipped. For simplicity, we treat Earth + clock + camcorder + transceivers as a single point. Similarly for the spacecraft. After some time T in its own frame, the spacecraft encounters a most similar spacecraft headed towards Earth at velocity v relative to Earth. As they pass, the first spacecraft transmits its entire camcorder recording to the second spacecraft, and the second spacecraft's clock is set to the value shown by the first spacecraft's clock. The camcorder on the second spacecraft then starts recording, continuing the recording it just received. 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 Prove it. Show that you can properly derive an equation that is manifestly false. Sylvia. 
#3




Simplified Twin Paradox Resolution.
stop hiding behind teh spacetimey orthodoxy;
get Lanscoz' book and use the quaternions (vector mechanics). Instead, on Earth there is a clock and a camcorder with a very powerful Show that you can properly derive an equation that is manifestly false. 
#4




Simplified Twin Paradox Resolution.
On Jan 3, 11:10 pm, PD as Sylvia Else wrote:
On 4/01/2013 5:03 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 Prove it. Show that you can properly derive an equation that is manifestly false. Prove what, PD? The equation describing spacetime? If you have to ask, you are indeed a moron. It is a good thing that you are an ex collegeprofessor of physics. May God have mercy on your ex students. shrug 
#5




Simplified Twin Paradox Resolution.
On 5/01/2013 5:59 AM, Koobee Wublee wrote:
On Jan 3, 11:10 pm, PD as Sylvia Else wrote: On 4/01/2013 5:03 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 Prove it. Show that you can properly derive an equation that is manifestly false. Prove what, PD? The equation describing spacetime? If you have to ask, you are indeed a moron. It is a good thing that you are an ex collegeprofessor of physics. May God have mercy on your ex students. shrug It was I who wrote "Prove it". 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. Sylvia. 
#6




Simplified Twin Paradox Resolution.
"Sylvia Else" wrote in message ...
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. Sylvia. ====================================== As has been proven already, if the twin travels a distance x in the frame of Earth then the distance xi that Earth must travel in the frame of the traveller is greater then x. The time for the traveller to move a distance x is t as measured by the Earth clock, hence v = x/t. The time for the Earth to move a distance xi is tau, and tau is less than t as all relativists agree, moving clocks run slow, the traveller does not age as rapidly as the twin on Earth. The velocity of Earth in the frame of the traveller is therefore upsilon = xi/tau and I leave it to the competent schoolchild to compute the value of upsilon when v = 0.866c, which if done correctly shows a clear paradox, upsilon c. Do not just point at v = upsilon and claim it is obvious.  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. 
#7




Simplified Twin Paradox Resolution.
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 
#8




Simplified Twin Paradox Resolution.
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. 
#9




Simplified Twin Paradox Resolution.
On Jan 5, 10:23 pm, Sylvia Else 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 ... never What? When (B^2 = 0), equations (3) and (4) become the following. ** dt1^2 = dt2^2 Why do you say never, PD? shrug 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. You are very correct, and they are not combined. Each equation, (1) or (2), has its own Lorentz transformation via these spacetime equations. shrug 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 
#10




Simplified Twin Paradox Resolution.
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. 
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