Special Relativity

Ian Goddard (igoddard@netkonnect.net)
Tue, 23 Jun 1998 21:09:55 -0400

At 07:55 PM 6/23/98 -0400, Daniel Fabulich wrote:

>> IAN: Ya, and 0 is larger than -1, and
>> thus A's ruler is larger relative to B's.
>Not in this case, you silly silly boy. You don't seem to understand that
>from B's position, A is smaller, not larger. So you can't ever say that A
>is larger than 0, where the 0 is self. Nothing is larger than 0 in this
>example. However, B IS smaller than 0 according to A, and A is smaller
>than 0 according to B. Since there is no positive sign, you cannot add 0s
>and -s to get 0.

IAN: Dan, over the span of about 5 posts, you've
continually failed to accurately represent what I
am saying so blatantly as to exceed belief. I am
NOT saying A is larger as seen by B nor am I say-
ing that A is larger relative to A -- which are
the only two cases you allow me to make (that
I don't), both of which are manifestly false.

That I'm not saying what you say I say couldn't
be clearer, I have even provided illustrations
showing that each sees the other's ruler SHRINK
and each sees its ruler LARGER than the others.

Just because you choose to omit data from your
analysis does not make your analysis correct,
in fact it makes it a false net measurement.
It's your fallacy of "partial difference."

You seem to be under the impression that your
scenario has violated the symmetry of measure,
that A measures "smaller" free from reference
to something "larger" than the smaller. There
are 2 observers that can make 4 valid measure-
ments, that's 8 measurements in all, but your
chart only includes half of them, as you say:

>So, here are ALL of the measurements you have available to you:
>A measuring A: 0
>A measuring B: - (never +, uses A's ruler!)
>B measuring B: 0
>B measuring A: -

IAN: That's half, here are all measurments:

After B passed by A near the speed of light,
A filled out this chart of all differences:

A | 0 -1 |
| |
B | 1 0 |

(AA) Difference between A and A = 0
(AB) Difference between A and B = A is 1 larger
(BA) Difference between B and A = B is -1 smaller
(BB) Difference between B and B = 0

Observer B measured these differences:

A | 0 1 |
| |
B |-1 0 |

(AA) Difference between A and A = 0
(AB) Difference between A and B = A is -1 smaller
(BA) Difference between B and A = B is 1 larger
(BB) Difference between B and B = 0

Because a measure of relative size is simply a
measure of difference, we could've ascertained
a priori that the change in size would sum to
0 by observing that as a rule, the sum of
all differences between all numerical
measurements always equals zero:

0 1 2 3
0 | 0 1 2 3 |
| |
1 |-1 0 1 2 |
| |
2 |-2 -1 0 1 |
| |
3 |-3 -2 -1 0 |

VISIT IAN WILLIAMS GODDARD --------> http://Ian.Goddard.net

"A new scientific truth does not triumph by convincing its
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