Simplified Twin Paradox Resolution.
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, lets 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 nth 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.
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