
Re: Simplified Twin Paradox Resolution.
Posted:
Jan 6, 2013 8:21 AM


On Jan 6, 9:59 am, Koobee Wublee <koobee.wub...@gmail.com> 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 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.

