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 a

** 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...


.... 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.