# All-Zero-Sum Counter... Not!

Ian Goddard (igoddard@netkonnect.net)
Sat, 20 Jun 1998 19:01:35 -0400

At 11:53 AM 6/20/98 -0400, Daniel Fabulich wrote:

>> 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 B
>A 0 +
>B + 0

IAN: My placing the + under A was an error.
What you write is the proper "got larger" matrix,
and as I shall show, it's as valid as the first.

>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 -.

IAN: A reduction (-) of B in size by 1
unit relative to A (0) is expressed as
-1. -1 is LESS than 0, 0 is 1 more (+)
then -1, so in fact A has gotten larger!

Size is 100% relative. There is no absolute
size, and that's what your case requires.
That the second matrix is valid is clearly
proven by the fact that -1 is less than
0 and 0 is 1 more than -1. If we have 2
things in the universe, A and B, and B
gets smaller, it's equally true that A
gets larger, because size is relative.

So the second matrix that defines the
zero sum of all valid data = valid data.

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