
Re: Simplified Twin Paradox Resolution.
Posted:
Jan 5, 2013 11:59 PM


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>

