
Re: Simplified Twin Paradox Resolution.
Posted:
Jan 6, 2013 3:33 AM


On Jan 5, 10:23 pm, Sylvia Else wrote: > On 6/01/2013 3:59 PM, 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> > > >    > > > 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 observer?s own clock versus whoever it is observing. So, what is the problem, PD? <shrug>

