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Re: What is or is not a paradox?
Posted:
Jan 2, 2013 11:41 AM
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On Wed, 02 Jan 2013 07:17:18 -0800, G=EMC^2 wrote:
> 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
Treeb is right. Every schoolkid knows that if you fly east, it's a time
machine. Every time you go around the earth you go back in time a day!
Get the Picture?
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