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Simplified Twin Paradox Resolution.



 
 
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  #11  
Old January 6th 13, 01:44 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Lord Androcles, Zeroth Earl of Medway[_6_]
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Posts: 58
Default Simplified Twin Paradox Resolution.

"Sylvia Else" wrote in message ...

On 6/01/2013 7:33 PM, Koobee Wublee wrote:


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


They're in different places. How are you going to do the comparing? It's
not a trivial question.

Sylvia.
=============================================
Bwahahahahahahaha!
Of course its a trivial question, the rod has always been in a different
place to the barn. just find a coordinate transformation between them. You
are supposed to know all about coordinate transformations. Show us the
Phuckwit Duck Transform.

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When I get my O.B.E. I'll be an earlobe.

Ads
  #12  
Old January 6th 13, 02:21 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Vilas Tamhane
external usenet poster
 
Posts: 52
Default 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, 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.
  #13  
Old January 6th 13, 02:42 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Sylvia Else
external usenet poster
 
Posts: 1,063
Default Simplified Twin Paradox Resolution.

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.


  #14  
Old January 6th 13, 02:47 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Paul B. Andersen[_7_]
external usenet poster
 
Posts: 70
Default Simplified Twin Paradox Resolution.

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/
  #15  
Old January 6th 13, 03:25 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Vilas Tamhane
external usenet poster
 
Posts: 52
Default Simplified Twin Paradox Resolution.

On Jan 6, 11:23*am, 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.


Nobody is observing anybody. Everybody has right to write down
equations.
  #16  
Old January 7th 13, 12:43 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Sylvia Else
external usenet poster
 
Posts: 1,063
Default Simplified Twin Paradox Resolution.

On 7/01/2013 1:25 AM, Vilas Tamhane wrote:
On Jan 6, 11:23 am, 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.


Nobody is observing anybody. Everybody has right to write down
equations.


And then? Koobee is trying to show that the equations contain an
inconsistency, which is to say, that they cannot be a description of any
reality.

Sylvia.
  #17  
Old January 7th 13, 07:28 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 6, 5:47 am, "Paul B. Andersen" 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


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


Excellent documentations, paul. When you are plagued with these
embarrassing blunders, at least, you have a skill in good
documentations. Koobee Wublee is indeed very grateful that you are
able to document His great posts. Seriously, paul. All that good
documentations still did not save you from that job in the private
industry, did it? When someone charging in claiming a Doppler shift
in 10^-8 should be seriously considered in adjusting the carrier
frequencies for compensations, the management just have to do the best
for either parties. :-)

By the way, Koobee Wublee never uses the partial derivative like what
you have done. dt still basically the rate of time flow when
comparing two observers. Thus, total derivative has to be
considered. shrug

Or better yet, if you are still confused with the Lorentz transform,
why don’t you look into the equations describing Minkowski spacetime
which Koobee Wublee has included in this post? That should leave no
confusion about what Dingle had to say was actually visionary. Well,
not quite. The stuff is so simple that it is a big surprise when all
the so-called bright minds in the scientific communities have so much
trouble understanding. What a shame, no? shrug

His arguments were as lethal and to the point as always. :-)


You bet, paul. Glad you are finding amusement amid these gross
blunders of yours. In doing so, you started personal attacks. Not
until Koobee Wublee pointed out to you, you have now calmed down. By
the way, have you finished the JAVA applet yet with the twins
traveling using the exact same acceleration profile? Please also
leave an adjustable coasting time with no acceleration in the
program. please and thanks in advance
  #18  
Old January 7th 13, 02:32 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Paul B. Andersen[_7_]
external usenet poster
 
Posts: 70
Default Simplified Twin Paradox Resolution.

On 07.01.2013 07:28, Koobee Wublee wrote:

Seriously, paul. All that good
documentations still did not save you from that job in the private
industry, did it? When someone charging in claiming a Doppler shift
in 10^-8 should be seriously considered in adjusting the carrier
frequencies for compensations, the management just have to do the best
for either parties. :-)


This is about the GPS again.

What "someone charging in claiming":
http://tinyurl.com/bdzm4k

The reason, and only reason, why the frequency standard is
corrected for relativistic effects is to make the SV clock
run synchronously with the ground clocks.

That the carrier and shipping frequencies also are adjusted is
just a side effect because all frequencies are derived from
the same frequency standard.

The carrier frequencies, like all other frequencies,
are at the receiver Doppler shifted between ± 3E-7.
The satellites are moving!
This is equivalent to a frequency shift of the carriers
in the order of kHz. Since the bandwidth of the channels
is ca. 20 MHz, this is of no consequence for the receiver.

The Doppler shift may be almost a thousand times more than
the minute GR-correction, so of bloody course the -4.4647E-10
offset is of no consequence whatsoever for the receiver!
AND NOBODY EVER SAID OTHERWISE!


How can
"the Doppler shift is of no consequence for the receiver"
become
"a Doppler shift in 10^-8 should be seriously considered
in adjusting the carrier frequencies for compensations"


Either you have a serious reading comprehension problem,
or you are deliberately lying about what I have said.

I suspect the latter.

--
Paul

http://www.gethome.no/paulba/
  #19  
Old January 7th 13, 03:41 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Vilas Tamhane
external usenet poster
 
Posts: 52
Default Simplified Twin Paradox Resolution.

On Jan 7, 11:28*am, Koobee Wublee wrote:
On Jan 6, 5:47 am, "Paul B. Andersen" 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.

  #20  
Old January 7th 13, 05:40 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Dono.
external usenet poster
 
Posts: 83
Default Simplified Twin Paradox Resolution.

On Jan 7, 6:41*am, old cretin Vilas Tamhane
wrote:
This proposed reciprocity is exactly the foolishness that SR is based on..


Imbecile, you still don't get it, do you?

 




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