Re: A Geometrical Proof of the Non-invariance of the Spacetime Interval

[Hugh Haskell <hhaskell@MINDSPRING.COM>]

At 18:28 -0500 1/21/03, Robert Cohen wrote:
> Help me out here.
>
> One person is going at v=0.  The other is going at v<<c.

v = 0 relative to a frame attached, say, to the earth, and v << c
relative to that frame, I assume.

> Each measures event A (breakfast) occurring at a particular point in
> space-time (x1,t1) and event B (lunch) occurring at a particular point
> in space-time (x2,t2).
>
> Are x1 and x2 measured differently by each observer, assuming v<<c?
>
> Are t1 and t2 measured differently by each observer, assuming v<<c?
>
> DR says yes to the first and no to the second, which implies that
> the four-vector is not invariant.
>
> HH says yes to both, but that they change in just the way to keep
> the four-vector invariant.
>
> It seems to me that at v<<c, the answer is no to both and so the
> four-vector is invariant.  But what do I know?

Since the scale factor gamma is only = 1 when v = 0 is the rel. vel.
between the two frames, I assert that My statement is always correct,
but since when v << c, gamma become indistinguishable from 1 to many
decimal places, the differences between my measurements and your
measurements will be essentially nil, so, to the precision you
desire, you are also correct. But you get less correct as v
increases, while I am correct at all velocities.

Hugh
--

Hugh Haskell
<mailto:haskell@ncssm.edu>
<mailto:hhaskell@mindspring.com>

(919) 467-7610

Never ask someone what computer they use. If they use a Mac, they
will tell you. If not, why embarrass them?
                       --Douglas Adams
******************************************************