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Re: Simplified Twin Paradox Resolution.
Posted:
Jan 5, 2013 11:59 PM
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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 co-exist is when B^2 = 0.
Thus, the twins? paradox is very real under the Lorentz transform.
<shrug>
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