On 7/01/2013 12:21 AM, Vilas Tamhane wrote:
On Jan 6, 9:59 am, 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 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.


The formulation of special relativity looks at uniform relative motion,
but it doesn't exclude the exchange of information at the speed of light
(or indeed, at less than speed of light), including exchanges between
observers who are in relative motion.

The experiment I proposed in the orignal posting involves only uniform
relative motion and exchanges of information at the speed of light.
Futher the analysis shows an absence of inconsistencies in the rate at
which clocks measure time, and are seen to measure time.

You're attempting to assert that a twin paradox exists, but in scenarios
that special relativity is incapable of addressing. Leaving aside
whether that says anything about twin paradoxes in special relativity,
it is any case just wrong.

Sylvia.