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Old December 31st 12, 08:31 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 5:04 PM, Koobee Wublee wrote:
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)


I assume you meant to write

dt1^2 = 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.


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).

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


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

Sylvia.
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