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



I hesitate to get into this fray because I know very little about
relativity but here goes.

David Rutherford wrote:

I'm just saying that they lead to a paradox when you compare the
assigned locations of breakfast, in the two frames, at the time of
lunch, which is when you "draw the lines" between breakfast and lunch,
in the two frames.
[snip]
Therefore, both
lines (from breakfast) start at time zero, but at different spatial
locations. That means that they start at different points in
spacetime,
thus, they don't coincide in spacetime and cannot describe a
geometically invariant quantity (the spacetime four-vector between the
events).

to which Hugh Haskell wrote:

No.

[snip]
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.

Help me out here.

One person is going at v=0. The other is going at v<<c.

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?

Obviously, I need some help here.
____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301