On 06.01.2013 07:23, Sylvia Else wrote:
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.
It's a variant of the old Dingle argument,
@t1/@t2 = @t2/@t1 is a contradiction.
(@ = partial derivative)
See:
http://tinyurl.com/ah3ctmm
Koobee's response:
http://tinyurl.com/a9jkwxp
What Koobee Wublee wrote that you have quoted was an application of
the Lorentz transform in a specific scenario. You don’t understand
all that, and apparently, you don’t know what you are talking about as
usual. It is laughable that a college professor from the University
of Trondheim would attempt to swindle his way out using irrelevant,
bull**** claims. shrug
You are cornered. Why don’t you stay in the topic of discussion?
shrug
His arguments were as lethal and to the point as always. :-)
--
Paul
http://www.gethome.no/paulba/