
Re: Simplified Twin Paradox Resolution.
Posted:
Jan 6, 2013 1:23 AM


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, let?s 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 are: >> >>> ** 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 n?th 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.

