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Old December 31st 12, 10:48 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Sylvia Else
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Default What is or is not a paradox?

On 31/12/2012 7:49 PM, Koobee Wublee wrote:
On Dec 30, 11:31 pm, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote:


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.


** dt1^2 (1 – B^2) = dt2^2 . . . (3)
** dt2^2 = dt2^2 (1 – B^2) . . . (4)


I assume you meant to write

dt1^2 = dt2^2 (1 - B^2) . . . (4)


No, Koobee Wublee meant every letter in the equations (3) and (4).


(2) doesn't become (4) just be writing B for B2.

shrug

Where


** B^2 = B1^2 = B2^2


The only time the equations (3) and (4) can co-exist is when B^2 = 0.


Which tells us nothing more than that when two observers observe each
other, the situation is symmetrical. Each will measure the same time for
equivalent displacements of the other. Or more simply, they share a
common relative velocity (save for sign).


The symmetry is everything about the twins’ paradox. shrug


In the classical twins paradox, there is no symmetry. The travelling
twin has to change velocities in order to be able to get back to the
stay at home twin.

To get symmetry, both twins have to travel, and if the travel is really
symmetrical, their ages will match when they return.


Thus, the twins’ paradox is very real under the Lorentz transform.
shrug


blink Where did that come from?


Have you not been reading Koobee Wublee? Did Koobee Wublee not say
the Lorentz transform? shrug

The twin "paradox" involves bringing
the two twins back together, which necessitates accelerating at least
one of them, making their frame non-inertial./blink


So, you believe in the nonsense of Born? He was the first one to
propose acceleration thing breaking the symmetry. Can you show any
mathematics that support your/Born’s claim? No self-styled physicists
have now believed in such nonsense. shrug


The symmetry can be broken without acceleration though to bring an
actual person back then involves cloning. It's simpler to forget the
twin, and just take a clock whose time is copied onto another clock
going in the opposite direction halfway through the travel.

But the symmetry is still broken, and once that happens, you have no
paradox.

Sylvia.