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

[Hugh Haskell <hhaskell@MINDSPRING.COM>]

At 23:04 -0800 1/20/03, David Rutherford wrote:
> I'm not denying that you _can_ say that you had breakfast and lunch in
> the same place. I'm arguing that it's not valid, geometrically to do so.
> When you (the moving observer) determine the spatial distance between
> breakfast and lunch at lunchtime, you put the breakfast end of the tape
> measure at San Diego, not Los Angeles, and the lunch end of the tape
> measure at San Diego. The Los Angeles observers put the breakfast end of
> the tape measure at Los Angeles and the lunch end of the tape measure at
> San Diego. The locations, in space, of breakfast at the time of lunch,
> are what both frames consider to have been the locations of breakfast,
> at the time of breakfast, since both frames say that the locations have
> not changed, in their frame, since the time of breakfast. These
> locations don't coincide in space, thus, they don't coincide in
> spacetime.
No.

All I know for certain is what goes on in *my* reference frame. If a
location in my reference frame happens to coincide with one point in
another reference frame at one time and a different point in that
reference frame at another time, so be it. The person in the Los
Angeles/San Diego reference frame can measure the distance between
those two points with a tape measure. I cannot. All I can do is
calculate the distance by multiplying the relative speed between the
two frames by the time *I* measure on the clock I carry with me. The
times and the distances measured by observers in the two reference
frames will not be the same. But since the Lorentz transformation
allows us to find the relationships between those times and
distances, it is also true that both observers will find the
magnitude of the spacetime interval between the two to be the same.
That is, assuming that the x-axis for both frames lies along the
LA-San Diego vector, that both will get the same value for x^2 -
c^2*t^2, as measured by instruments in their own reference frames. It
doesn't mean that either x or t will be seen to be the same in both
frames, because they won't.

Hugh
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Hugh Haskell
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<mailto:hhaskell@mindspring.com>

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