On 4/16/2015 11:45 AM, Odd Bodkin wrote:
> On 4/16/2015 1:33 AM, Koobee Wublee wrote:
>> On Wednesday, April 15, 2015 at 1:23:26 PM UTC-7, Odd Bodkin wrote:
>>> On 4/15/2015 3:15 PM, Koobee Wublee wrote:
>>
>>>> The Twin paradox has been shown to be a fatal contradiction by Koobee
>>>> Wublee and many others based on mathematical axioms.
>>>
>>> Oh, that's really interesting.
>>
>> You are a terrible liar! <shrug>
>>
>>> Which mathematical axioms have you used, and can you show the proof?
>>
>> For instance, the following are good starts for you. <shrug>
>>
>>
https://groups.google.com/d/msg/sci.physics.relativity/1eWIUe5gZMM/Yf-2nu9pllUJ
>>
>
Let's actually extract your language from this post.
=========================================================
**** Background: To decide who is right or wrong, we must first look
at the bigger picture where a system involves 4 parties --- 2
observers (#1 and #2) and 2 observed (#3 and #4). Writing down the
Lorentz transform for time transformation only, we have the following
where equation 2) and 4) are the inverse transforms of 1) and 3)
respectively. Otherwise, 1) and 2) still belong to the same
transformation, and 3) and 4) belong to another transformation.
1) dt13 = (dt23 – [B21] * d[s23] / c) / sqrt(1 – B21^2)
2) dt23 = (dt13 – [B12] * d[s13] / c) / sqrt(1 – B12^2)
And
3) dt14 = (dt24 – [B21] * d[s24] / c) / sqrt(1 – B21^2)
4) dt24 = (dt14 – [B12] * d[s14] / c) / sqrt(1 – B12^2)
Where
** [s23] = Position vector of #3 as observed by #2
** [B21] c = Velocity vector of #1 as observed by #2
** [] * [] = Dot product of two vectors
==========================================================
Without any loss of generality we can revert back to the form of the
Lorentz transforms without the differentials.
**** Background: To decide who is right or wrong, we must first look
at the bigger picture where a system involves 4 parties --- 2
observers (#1 and #2) and 2 observed (#3 and #4). Writing down the
Lorentz transform for time transformation only, we have the following
where equation 2) and 4) are the inverse transforms of 1) and 3)
respectively. Otherwise, 1) and 2) still belong to the same
transformation, and 3) and 4) belong to another transformation.
1) t13 = (t23 – [B21] * [s23] / c) / sqrt(1 – B21^2)
2) t23 = (t13 – [B12] * [s13] / c) / sqrt(1 – B12^2)
And
3) t14 = (dt24 – [B21] * [s24] / c) / sqrt(1 – B21^2)
4) t24 = (dt14 – [B12] * [s14] / c) / sqrt(1 – B12^2)
Where
** [s23] = Position vector of #3 as observed by #2
** [B21] c = Velocity vector of #1 as observed by #2
** [] * [] = Dot product of two vectors
Now please notice that B21 has one sign in the event that 1 and 2 are
receding from each other and the opposite sign in the event that 1 and 2
are approaching each other. Clearly, it would be a silly and KW-like
mistake to use the same expression to discuss a trip involving both
cases, without tracking the sign change.
But let's go back to what you were saying in your post:
================================================================
Of course, in the twins’ paradox, there each of the two observers is
observed by the other as the observed. Thus, it is a matter of
writing the above 2 transformations into the equations equating how
the time flow rate at #1 differs from #2.....
Step 1: Discard equations 3) and 4).
Step 2: Replace #3 in equations 1) and 2) with #1 or #2 (#2 in the
following example):
1) dt12 = (dt22 – [B21] * d[s22] / c) / sqrt(1 – B21^2)
2) dt22 = (dt12 – [B12] * d[s12] / c) / sqrt(1 – B12^2)
===================================================================
And again, without loss of generality, one can always revert to the
non-differential form of these equations:
1) t12 = (t22 – [B21] * [s22] / c) / sqrt(1 – B21^2)
2) t22 = (t12 – [B12] * [s12] / c) / sqrt(1 – B12^2)
Now, let's take a look at this equation [1] carefully here. Let's let
observer 2 be the traveling twin, and let's suppose for the sake of
argument that the turn-around of observer 2 is instantaneous. Let's make
that moment of turnaround to be at location s22 and time t22 according
to observer 2. Now, we'll consider a moment just BEFORE that turnaround,
t22-e, where e is small, and a moment just AFTER that turnaround t22+e,
where e is small. When this happens, observer 2 is not at s22 but at
s22-f, where f is also small and related to e; please notice that the
location is the SAME value both before and after the turnaround because,
well, it's turned around.
The other thing that has changed is that B21 has completely changed
sign. Now, we said earlier that it would be a completely stupid and
KW-like mistake to use the same variable both before and after the
turnaround when it has changed sign, so we should keep track of it by
calling it, say, B before the turnaround and -B after the collision, and
removing the dot-product notation.
Thus at time t22-e, we have
1) t12 = (t22-e - (B)(s22-f)/c) / sqrt(1-B^2)
and at time t22+e, we have
1') t12' = (t22+e + (B)(s22-f)/c) / sqrt(1-B^2)
So if we wanted to find out what happens in t12 over the small
differential in time t22, which we'll call dt22 = 2e, then we can
subtract these two equations to find that
t12' - t12 = (dt22 + 2(B)(s22-f)/c) / sqrt(1-B^2)
Zoot alors! t12'-t12 is not small at all, even though dt22 is small.
In fact if we neglect the size of e and f, then we find out that just in
turning around, the duration of the turnaround, as seen by the earth has
taken a huge amount of time!
t12' - t12 = (2(B)(s22)/c) / sqrt(1-B^2)
This is of course dramatically different than the dt12 expression that
KW produced in his post, which was the consequence of using the same
variable B21 for both legs of the journey when in fact the value of B21
changes sign, a completely silly mistake that KW would rightfully feel
like an idiot for committing.