ALEXANDER AND ANDREW ABIAN WATCHES
Alexander Abian
Let A and B be two distinct points in ANY SPACE (finite or infinite
dimensional where the notion of PATH is meaningful) and let P be any
given path connecting A to B
/\/-----\
A./ \
\-------/\/--.B
Alexander and Andrew are at rest at the point A and have
their watches synchronized. Andrew moves on the path P toward
the point B and Alexander follows Andrew on P but moving
ALWAYS BEHIND Andrew (except at the points A and B, of course)
Andrew reaches point B and waits until his father Alexander gets
to the point B. They meet and stay at the point B.
After a while they check their watches:
WHICH OF THE FOLLOWINGS WILL BE THE CASE
1. Andrew's and Alexanders watches show the same time
2. Andrew's watch is slower than Alexander's
3. Andrew's watch is faster than Alexander's
THE POWER AND THE SIGNIFICANCE OF THE ABOVE EXAMPLE consists in
the fact that no formula is used, no velocity of light enters
into consideration no coordinate systems enter into discussion,
no extra specific conditions are imposed on the motions.
THIS EXAMPLE CAN BE EXPLAINED to a grammar school student to point
out the possible discrepancy between Newton's and Einstein's answers
to the above question.
To satisfy the physicists the ABOVE IS AN EXPERIMENT, EXPERIMENT,
EXPERIMENT!! a thing that they always consider the ULTIMATE SUPREME
TEST OF TRUTH!!!
--
-------------------------------------------------------------------------
ABIAN TIME-MASS EQUIVALENCE FORMULA T = A m^2 in Abian units.
ALTER EARTH'S ORBIT AND TILT TO STOP GLOBAL DISASTERS AND EPIDEMICS.
JOLT THE MOON TO JOLT THE EARTH INTO A SANER ORBIT.ALTER THE SOLAR SYSTEM.
REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT TO CREATE A BORN AGAIN EARTH(1990)
THERE WAS A BIG SUCK AND DILUTION OF PRIMEVAL MASS INTO THE VOID OF SPACE
Wayne Throop
: Alexander and Andrew are at rest at the point A and have
: their watches synchronized. Andrew moves on the path P toward
: the point B and Alexander follows Andrew on P but moving
: ALWAYS BEHIND Andrew (except at the points A and B, of course)
: Andrew reaches point B and waits until his father Alexander gets
: to the point B. They meet and stay at the point B.
: After a while they check their watches:
:
: WHICH OF THE FOLLOWINGS WILL BE THE CASE
:
: 1. Andrew's and Alexanders watches show the same time
: 2. Andrew's watch is slower than Alexander's
: 3. Andrew's watch is faster than Alexander's
Why the switch from "show the same time" to "watch is [faster/slower]"?
Do you mean "show a greater time" and "show a lesser time"?
Presuming that's so, any of the three are possible; not enough
information is given to eliminate any of the three cases. If, however,
both Alexander and Andrew traverse the arbitrary path with the same
velocity (wrt the end points A and B), then (1) is the correct answer.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Gerald L. O'Barr
In <7ml31e$lb6$1...@news.iastate.edu>
ab...@iastate.edu (Alexander Abian) wrote:
----------
Let A and B be two distinct points in ANY SPACE (finite
or infinite dimensional where the notion of PATH is
meaningful) and let P be any given path connecting A to B
/\/-----\
A./ \
\-------/\/--.B
Alexander and Andrew are at rest at the point A and have
their watches synchronized. Andrew moves on the path P
toward the point B and Alexander follows Andrew on P but
moving ALWAYS BEHIND Andrew (except at the points A and B,
of course)
Andrew reaches point B and waits until his father
Alexander gets to the point B. They meet and stay at the
point B. After a while they check their watches:
WHICH OF THE FOLLOWINGS WILL BE THE CASE
1. Andrew's and Alexanders watches show the same time
2. Andrew's watch is slower than Alexander's
3. Andrew's watch is faster than Alexander's
O'Barr comments:
Hi! Alexander Abian! It has been a while since I have
answered one of your posts! I believe this is a good
problem. I saw Wayne's answer. Wayne is smarter than I
am. I did not catch the inference where you ask `is' the
watch slower or faster. I just assume that you were
asking if the watch showed more or less time. I am sure
that this is what you were doing in as much as the intent
appears to set out the only three possible results, and
the three possible results must then include the
assumption of just showing more of less time.
Wayne was correct: There is not enough here to give a
responsible answer. Then he went on to give an answer
that, IMO, was not complete. I would state the following.
If this is just SR (no gravity present) and all points
along each path for each person were `experienced' at the
same velocity, and the entire path was inertial and static
for the entire problem, then Throop's answer is
reasonable. But if this is a GR problem, or the path is
experiencing any rotation effects, etc., then their times
could differ even if their travel times or specific path
rates were identical. For one specific example, if the
two end points were at two different gravity levels, the
clocks would differ.
Thanks!!!!
--
Gerald L. O'Barr fl...@access1.net
Read: http://www.access1.net/flaco
Read Pete Brown's Aether FAQ at:
http://magna.com.au/~prfbrown/aeth_faq.htm
Read Jan 99 issue of Physics Today about the ether!
Alexander Abian
YES,in (2) and (3) I was asking "less time" or "more time"
You may first consider SR case and for your answewr(s)
give an explicit example.
Let A and B be two distinct points in ANY SPACE (finite
or infinite dimensional where the notion of PATH is
meaningful) and let P be any given path connecting A to B
/\/-----\
A./ \
\-------/\/--.B
Alexander and Andrew are at rest at the point A and have
their watches synchronized. Andrew moves on the path P
toward the point B and Alexander follows Andrew on P but
moving ALWAYS BEHIND Andrew (except at the points A and B,
of course)
Andrew reaches point B and waits until his father
Alexander gets to the point B. They meet and stay at the
point B. After a while they check their watches:
WHICH OF THE FOLLOWINGS WILL BE THE CASE
1. Andrew's and Alexander's watches show the same time
2. Andrew's watch shows less time than Alexander's
3. Andrew's watch shows more time than Alexander's
Bernard HP Gilroy, maximum proconsul
<fl...@access1.net> wrote:
>
> In <7ml31e$lb6$1...@news.iastate.edu>
>ab...@iastate.edu (Alexander Abian) wrote:
>
>----------
> Let A and B be two distinct points in ANY SPACE (finite
>or infinite dimensional where the notion of PATH is
>meaningful) and let P be any given path connecting A to B
>
> /\/-----\
> A./ \
> \-------/\/--.B
>
>
> Alexander and Andrew are at rest at the point A and have
>their watches synchronized. Andrew moves on the path P
>toward the point B and Alexander follows Andrew on P but
>moving ALWAYS BEHIND Andrew (except at the points A and B,
>of course)
> Andrew reaches point B and waits until his father
>Alexander gets to the point B. They meet and stay at the
>point B. After a while they check their watches:
>
> WHICH OF THE FOLLOWINGS WILL BE THE CASE
>
> 1. Andrew's and Alexanders watches show the same time
> 2. Andrew's watch is slower than Alexander's
> 3. Andrew's watch is faster than Alexander's
>
>
>
> Wayne was correct: There is not enough here to give a
>responsible answer. Then he went on to give an answer
>that, IMO, was not complete. I would state the following.
>If this is just SR (no gravity present) and all points
Um, this problem _can't_ be SR. The two begin at rest
relative to each other. Then, one is "ahead" -- meaning he has some
velocity relative to the other. This means there _must_ have been
acceleration, and therefore SR is simply not applicable.
Yeah, it's a weasel way out. I'd have to dig out my old grad
GR books and actually _think_ to give a real answer, and that's too
much to ask for July. :)
Alexander Abian
Let A and B be two distinct points in ANY SPACE (finite or infinite
dimensional where the notion of PATH is meaningful) and let P be any
given path connecting A to B
/\/-----\
A./ \
\-------/\/--.B
Alexander and Andrew are at rest at the point A and have
heir watches synchronized. Andrew moves on the path P toward
he point B and Alexander follows Andrew on P but moving
ALWAYS BEHIND Andrew (except at the points A and B, of course)
Andrew reaches point B and waits until his father Alexander gets
to the point B. They meet and stay at the point B.
After a while they check their watches:
WHICH OF THE FOLLOWINGS WILL BE THE CASE
1. Andrew's and Alexander's watches show the same time
2. Andrew's watch shows less time than Alexander's
3. Andrew's watch shows more time than Alexander's
THE POWER AND THE SIGNIFICANCE OF THE ABOVE EXAMPLE consist in
the fact that no formula is used, no velocity of light enters
into consideration no coordinate systems enter into discussion,
no extra specific conditions are imposed on the motions.
THIS EXAMPLE CAN BE EXPLAINED to a grammar school student to point
out the possible discrepancy between Newton's and Einstein's answers
to the above question by considering some typical cases.
To satisfy the physicists: the ABOVE IS AN EXPERIMENT, EXPERIMENT,
EXPERIMENT!! a thing that they always consider the PENULTIMATE SUPREME
TEST OF THE TRUTH!!!
mad...@myremarq.com
100 mph.
And another train leaves Miami, at 50 mph. When they cross,
Witch train will be closer to NY?? After alex and Adrian
meet there Watches will again be synchronized. If they where
not both in the same time frame How could they meet?
Sent from RemarQ http://www.remarq.com/?z The Internet's Discussion Network
The fastest and easiest way to search and participate in Usenet - Free!
Gerald L. O'Barr
"mad...@myremarq.com" wrote
> 100 mph.
> And another train leaves Miami, at 50 mph. When they cross,
> Witch train will be closer to NY??
O'Barr comments:
Assuming that the tracks are straight between these two
cities, etc., etc., etc.
If the trains have finite length, and the point that they
'meet' is considered to be the first location on dual tracks
where any one part of one train is directly accross from the
other, then this point, for each train is an equal distance
from NY, and after they each have totally crossed this point,
then the train from Miami is the closest. If there is only
one track, then all bets are off. The train from NY will
probably remain the closest!
This sounds like the problem where a plane crashed exactly
on the border between the U.S. and Canada. And people did not
know where to bury the survivors! Do you bury them in Canada
or in the U.S.??????
--
Gerald L. O'Barr fl...@access1.net
Ramji Venkateswaran
> This sounds like the problem where a plane crashed exactly
> on the border between the U.S. and Canada. And people did not
> know where to bury the survivors! Do you bury them in Canada
> or in the U.S.??????
>
Pardon me,
But how exactly do you bury survivors? Don't they generally tend
to object. I did this to a $user of mine one day, and the
Operations manager wasn't impressed.
Ramji
--
Ramji Venkateswaran Internet Technology Group PLC
113-123 Upper Richmond Road, Putney, London, SW15 2TL
Tel: 0181 957 1119
Wayne Throop
: If they where not both in the same time frame How could they meet?
I don't know what you are talking about, but it isn't relativity.
There's no such concept as "being in a time frame" in relativity.
What do YOU mean by that phrase? Please be specific.
Jackie
>
> Pardon me,
>
> But how exactly do you bury survivors? Don't they generally tend
> to object. I did this to a $user of mine one day, and the
> Operations manager wasn't impressed.
>
> Ramji
Hi Ramji, maybe in the UK you don't bury the survivors, but over here in
North America we do.
It saves an incredible amount of paperwork and reduces costs immensely.
Barry
mad...@myremarq.com
distant from all destinations. regardless of the length,
position, orientation, or any other variable that has to do
with the tracks. as long as they cross.
At that point all variables are set aside, and both trains
are equally distant from all destinations. The same would be
true of adrian and alex. After they arrive at point B having
traveled down path P there watches would be synchronized, If
not they would not meet. If I ask you to meet me at the
corner of 5th and maple, at 3:00 pm est but you arrive at
6:00pm and I get there around 2:00 pm how would we see each
other?
So to answer the question: What do I mean buy a time frame,
I would have to describe it as a Common time shared between
two independent clocks, within this frame the two clock will
always be synchronized. Even If they are not so outside the
frame. For us to meet at the corner of 5th and Maple we must
share the same frame in time and space.
Wayne Throop
: Like the two trains, When they cross they are equally distant from all
: destinations. regardless of the length, position, orientation, or any
: other variable that has to do with the tracks. as long as they cross.
: At that point all variables are set aside, and both trains are equally
: distant from all destinations. The same would be true of adrian and
: alex. After they arrive at point B having traveled down path P there
: watches would be synchronized, If not they would not meet.
So people can't meet unless their watches are synchronized.
Hmmmm. You know, I could have SWORN my watch had a different
reading than this fellow I met yesterday.
The whole notion that "people can't meet" unless they are "at the same
time" (meaning, having had equal elapsed times from some common event)
presumes a bizarre notion of meta-time; that somebody arriving at an
event will somehow repeal or revoke their presense at that event "after"
they have "gone forward in time".
Both bizarre and unnecessary.
In SR, the two watches could be different, and the watch owners still meet.
Because in SR, there's no "meta-time" which would cause them to be
able to miss each other despite arriving at a common event in space and time.
In lorentzian ether theory, the two watches could also be different, and
the watch owners still meet. Because in LET, there's no guarantee that
ideal pocketwatches keep correct time.
mad...@myremarq.com
different readings how did you determine that they where not
synchronized? ( please be specific )
Have you ever seen anyone someplace they where not? ( pbs )
There is no need for watches to keep correct time. once they
enter a common time frame, they will move to synchronisity.
(The Doppler effect) like the train tracks, they could be as
different as you like, they just have to have some frame of
reference in common. the time frame itself is completely
arbitrary and depends on the position of the clocks wrt each
other.
Sent from RemarQ http://www.remarq.com/?z The Internet's Discussion Network
The fastest and easiest way to search and participate in Usenet - Free!
Wayne Throop
: There is no need for watches to keep correct time. once they enter a
: common time frame, they will move to synchronisity.
I asked you before to define this odd notion of "entering a common time
frame". I've never heard of it before. It has nothing to do with
relativity in particular or physics in general. Or rather, if it does,
it's some invention of yours, and not in the literature of the field,
as near as I can tell. Has anybody else heard of this term in this context?
: When you met your friend yesterday and your watches had different
: readings how did you determine that they where not synchronized? (
: please be specific )
His watch read 17:08. Mine read 17:12. Therefore, they were not synchronized.
Perhaps you can explain what YOU mean by "synchronized"?
I think it's just greek for "same time". As in, "we synchronized
our watches by setting them to 11pm at my mark".
In this context, "syntonized" is used to imply the same *rate*.
And finally, like it or not, SR implies that two clocks, each having
the same proper rate can become unsynchronized. Of course, they
do not have the same rate in some inertial frame, but then that's
not the proper rate.
To borrow a bit from Daryl McCullough, in both ether theory and SR, the
reading on a clock depends on the amount of time between since it was
reset times the rate at which the clock reading changes. In SR, the
amount of time depends on the path, and the rate is invariant. In LET,
the amount of time is invariant, and the rate depends upon the path.
In either case, two people can meet with clocks that have different readings.
Note specifically: the question was in terms of WHAT THE CLOCKS READ.
Since LET and SR are prefectly logical physical theories which show how
the clocks could aquire different readings, it seems silly to claim
that the clocks must not differ at the reunion. At the very least,
please describe the physical theory under which this assertion is made.
It is, by the way, known to be wrong; specifically, it disagrees
with the H&K experiment, where clocks were separated and reunited.
LET and SR match reality in thise matter.
The notion that time and clock rate are both invariant does not.
It's just that simple.
mad...@myremarq.com
times or rates?
I said the time in the clock would move to a synchronized
common time, or something like that. moving time is not
invariant. I just said that there was a kind of area, that I
call a time frame, in witch clocks share a common time. And
I think I said that the frame itself was arbitrary and
depended on the ( position) of the clock wrt each other, or
something. Within this time frame clocks can share
information, with very little delay. Like gravity in large
galaxy clusters. Or like the Idea of an instant reaction to
an action, but at a distance. In both SR and LET the clocks
are responding to the path, space time, or either, I just
think that the (you claim) well documented observation of
variant clocks and rates, is interactive to a degree.
So in my time frame distant fast moving large objects will
not appear in a past position, they will shrink and slow
down till I see the object in my present time at it's
present location. to a degree.
How else can strings of clusters billions of light years in
length form coherent structures? if information about the
position of a galaxy is delayed to the other galaxies in
time, how do they choreograph their positions in the
cluster?
I wonder just how simple it really is?
Not that I would assume to have all the answers tucked under
my hat.
Im just not convince that we know what time is?
* Sent from RemarQ http://www.remarq.com The Internet's Discussion Network *