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Re: What is or is not a paradox?
Posted:
Jan 2, 2013 11:41 AM


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: >> >> >> >> >> >> >> >> >> >> > 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 coexist 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 noninertial.</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?



