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Topic: What is or is not a paradox?
Replies: 13   Last Post: Jan 2, 2013 10:17 AM

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herbert glazier

Posts: 192
Registered: 7/26/10
Re: What is or is not a paradox?
Posted: Jan 2, 2013 10:17 AM
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On Dec 31 2012, 11:58 am, kenseto <seto...@att.net> wrote:
> On Dec 31, 2:31 am, Sylvia Else <syl...@not.at.this.address> wrote:
>
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>

> > 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>

>
> There is no inertial frame exists on earth ....does that mean that SR
> is not valid
> on earth?


Well think of this. "Time in a plane flying east is less than that for
those flying west". The Earth speed of rotation sees to it. Get the
picture TreBert



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