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#1 January 4th 13, 07:03 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Koobee Wublee external usenet poster Posts: 815 Simplified Twin Paradox Resolution.

On Jan 3, 5:52 pm, Sylvia Else wrote:

Instead, on Earth there is a clock and a camcorder with a very powerful
telescopic lens and a tranceiver There is also a spacecraft travelling
at velocity v towards Earth, similarly equipped. For simplicity, we
treat Earth + clock + camcorder + transceivers as a single point.
Similarly for the spacecraft.

After some time T in its own frame, the spacecraft encounters a most
similar spacecraft headed towards Earth at velocity v relative to Earth.
As they pass, the first spacecraft transmits its entire camcorder
recording to the second spacecraft, and the second spacecraft's clock is
set to the value shown by the first spacecraft's clock. The camcorder on
the second spacecraft then starts recording, continuing the recording it

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

Did PD get it right?

[snipped garbage]

Of course not, what do you expect from an Einstein Dingleberry?
shrug

After abandoning the projection crap, is Tom very busy trying to find
an excuse to resolve this through projection again? Koobee Wublee is
not surprised since Tom has already decided SR as a valid hypothesis
despite lack of any professional validations. shrug

POINT BLANK IN YOUR FACES, PD, PAUL (ANDERSEN), AND TOM (ROBERTS)!
shrug
#2 January 4th 13, 08:10 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Sylvia Else external usenet poster Posts: 1,063 Simplified Twin Paradox Resolution.

On 4/01/2013 5:03 PM, Koobee Wublee wrote:
On Jan 3, 5:52 pm, Sylvia Else wrote:

Instead, on Earth there is a clock and a camcorder with a very powerful
telescopic lens and a tranceiver There is also a spacecraft travelling
at velocity v towards Earth, similarly equipped. For simplicity, we
treat Earth + clock + camcorder + transceivers as a single point.
Similarly for the spacecraft.

After some time T in its own frame, the spacecraft encounters a most
similar spacecraft headed towards Earth at velocity v relative to Earth.
As they pass, the first spacecraft transmits its entire camcorder
recording to the second spacecraft, and the second spacecraft's clock is
set to the value shown by the first spacecraft's clock. The camcorder on
the second spacecraft then starts recording, continuing the recording it

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

Prove it. Show that you can properly derive an equation that is
manifestly false.

Sylvia.

#3 January 4th 13, 07:10 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 1treePetrifiedForestLane external usenet poster Posts: 974 Simplified Twin Paradox Resolution.

stop hiding behind teh spacetimey orthodoxy;
get Lanscoz' book and use the quaternions (vector mechanics).

Instead, on Earth there is a clock and a camcorder with a very powerful

Show that you can properly derive an equation that is manifestly false.

#4 January 4th 13, 07:59 PM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Koobee Wublee external usenet poster Posts: 815 Simplified Twin Paradox Resolution.

On Jan 3, 11:10 pm, PD as Sylvia Else wrote:
On 4/01/2013 5:03 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

Prove it. Show that you can properly derive an equation that is
manifestly false.

Prove what, PD? The equation describing spacetime? If you have to
ask, you are indeed a moron. It is a good thing that you are an ex-
college-professor of physics. May God have mercy on your ex-
students. shrug
#5 January 6th 13, 02:57 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Sylvia Else external usenet poster Posts: 1,063 Simplified Twin Paradox Resolution.

On 5/01/2013 5:59 AM, Koobee Wublee wrote:
On Jan 3, 11:10 pm, PD as Sylvia Else wrote:
On 4/01/2013 5:03 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

Prove it. Show that you can properly derive an equation that is
manifestly false.

Prove what, PD? The equation describing spacetime? If you have to
ask, you are indeed a moron. It is a good thing that you are an ex-
college-professor of physics. May God have mercy on your ex-
students. shrug

It was I who wrote "Prove it".

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.

Sylvia.
#6 January 6th 13, 04:00 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Lord Androcles, Zeroth Earl of Medway[_6_] external usenet poster Posts: 58 Simplified Twin Paradox Resolution.

"Sylvia Else" wrote in message ...

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.

Sylvia.

======================================
As has been proven already, if the twin travels a distance x in the frame of
Earth then the distance xi that Earth must travel in the frame of the
traveller is greater then x. The time for the traveller to move a distance x
is t as measured by the Earth clock, hence v = x/t.
The time for the Earth to move a distance xi is tau, and tau is less than t
as all relativists agree, moving clocks run slow, the traveller does not age
as rapidly as the twin on Earth.
The velocity of Earth in the frame of the traveller is therefore upsilon =
xi/tau and I leave it to the competent school-child to compute the value of
upsilon when v = 0.866c, which
if done correctly shows a clear paradox, upsilon c.
Do not just point at v = upsilon and claim it is obvious.

-- 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.

#7 January 6th 13, 05:59 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Koobee Wublee external usenet poster Posts: 815 Simplified Twin Paradox Resolution.

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

- - -

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
#8 January 6th 13, 07:23 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Sylvia Else external usenet poster Posts: 1,063 Simplified Twin Paradox Resolution.

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

- - -

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  and  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.

#9 January 6th 13, 09:33 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Koobee Wublee external usenet poster Posts: 815 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  and  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
#10 January 6th 13, 11:57 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
 Sylvia Else external usenet poster Posts: 1,063 Simplified Twin Paradox Resolution.

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.

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