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What is or is not a paradox?



 
 
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Old December 31st 12, 07:04 AM posted to sci.physics.relativity,sci.physics,sci.math,sci.astro
Koobee Wublee
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Default What is or is not a paradox?

On Dec 30, 4:17 pm, Sylvia Else wrote:
I started writing a post about this yesterday, then scrubbed it - too


What is a paradox in special relativity (hereinafter SR)?

I've expressed the view that to contain a paradox, SR has to predict,
from different frames, outcomes that are mutually incompatible. An
example that comes to mind (though not directly arising) from a recent
discussion is that in one frame, there is massive destruction on a
citywide scale, and in another other frame, nothing much happens.

Clearly, if SR were to make such predictions for two frames, it would
have to be regarded as seriously wanting. Of course, it does no such thing.

But people seem to want to regard measurements in two frames as mutually
incompatible if they give different results. I am at a loss to
understand why people would seek to regard those different results as
constituting a paradox that invalidates SR (well, leaving intellectual
dishonesty aside).


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)

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

Within the Lorentz transform, the little professor from Norway, paul
andersen, was able to play a mathemagic trick, and he is not alone.
In doing so, the Minkowski spacetime was not recognized in his little
applet. He is out in the very left field chasing chickens again.
shrug

Tom and other self-styled physicists have recognized that fault and
moved on to claim a mythical proper time flow where all local time
flow is a projection of this absolute time flow. Oh, excuse Koobee
Wublee. Not absolute time but proper time whatever $hit it is.
However, these guys cannot explain why the projection did not cancel
out on the traveling twin’s return trip. So, equations (3) and (4)
are still indicating the paradox regardless if projection or not.
shrug

Well, sooner or later, these bozos are going to wake up someday and
ask themselves “what the fvck was I thinking?”. Guess what? The time
projection crap is the last piece of float the self-styled physicists
are clinging on to. Take that away. They will sink. That is why the
self-styled physicists are very reluctant to give the time projection
crap a serious thought. shrug


 




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