Re: All-Zero-Sum Counter

Daniel Fabulich (
Sat, 20 Jun 1998 11:53:14 -0400 (EDT)

On Fri, 19 Jun 1998, Ian Goddard wrote:

> IAN: Both -C (each individual measure of
> foreshortened length) and -2C (their sum)
> are derived by relation to 0 (the 0 point
> of measure, being the observer's ship) and
> 0 - (-2) = +2
> -2 - 0 = -2
> the net difference between -2 and 0 is zero;
> and thus the inverse matrix is equally true:
> A B
> A 0 -
> B - 0
> A B
> A + 0
> B 0 +

Wait a second. This says that A relative to A is non-zero; to my reading,
A is + relative to A. This is complete absurdity: nothing is large
relative to itself.

> Because size is relative, A appears to in-
> crease in size when measuring B.

A does not increase in size relative to itself; so A relative to A must
always be 0. A DECREASES in size relative to B. There is no perspective
from which A increases in size. I might accept a second matrix if it were
valid, but that particular one isn't.

> As A looks
> at B, observer A could equally assume he had
> increased in size just as the observer in one
> train looking at another train next to him out
> the window seeing motion and may assume either
> his train or the other is in motion, and either
> assumption is equally correct. Observer B also
> assumes he may have increased in size. So net
> difference is found in the sum of the matrices.
> One size-matrix doesn't contain all difference.

This would be true if A were actually stretching relative to anything.
However, NO perspective sees A increase in size. I would grant you that
in normal Newtonian physics if you saw me shrinking relative to you, it
would be a safe presumption that you were stretching relative to me, but
not in special relativity. I am shrinking relative to you AND you are
shrinking relative to me.

> Dan, it's a great counter(!) worthy of further
> investigation, but it can only be sustained if
> it can be shown that size is not relative and
> thus that one matrix is more valid, but even
> that "more vs less" valid is another 0 sum.

Well, perhaps you meant a different second matrix; any matrix which shows
A non-zero relative to itself is fishy, I'd say. Perhpas you'd want to
say that:

A 0 +
B + 0

is true if we compare the second situation to the first, rather than the
first to the second. The only problem with this matrix is that it is
never true; B is never large relative to A. B is never large relative to
ANYTHING; at rest it is zero, and in motion it is negative relative to A.
Since you cannot find any perspective in which A or B are positive, I
sincerely doubt your ability to get back to 0 from the -.

Good luck!