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Re: What is or is not a paradox?
Posted:
Jan 2, 2013 10:17 AM


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



