View Single Post
  #9  
Old January 6th 13, 09:33 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Koobee Wublee
external usenet poster
 
Posts: 815
Default Simplified Twin Paradox Resolution.

On Jan 5, 10:23 pm, Sylvia Else wrote:
On 6/01/2013 3:59 PM, 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


- - -


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


... never


What? When (B^2 = 0), equations (3) and (4) become the following.

** dt1^2 = dt2^2

Why do you say never, PD? shrug

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.


You are very correct, and they are not combined. Each equation, (1)
or (2), has its own Lorentz transformation via these spacetime
equations. shrug

With all these parameters derived, all we have to do is to compare the
time elapses of the observer’s own clock versus whoever it is
observing. So, what is the problem, PD? shrug